Chemistry Form 4 • MODULE
EXERCISE / LATIHAN
1 Complete the table below / Lengkapkan jadual di bawah:
Electrolyte Electrode Factor that affects Ions present Half equation at the Half equation at the
Elektrolit Elektrod electrolysis Ion yang hadir anode and observation cathode and observation
Faktor yang Persamaan setengah di anod dan Persamaan setengah di katod
Dilute Carbon mempengaruhi
sulphuric acid Karbon elektrolisis pemerhatian dan pemerhatian
Asid sulfurik cair Carbon
Karbon Position of ion in H+, SO42–, OH– 4OH– 2H2O + O2 + 4e 2H+ + 2e H2
Concentrated the electrochemical Gas bubbles are released. Gas bubbles are released.
hydrochloric Carbon series
acid Karbon
Asid hidroklorik Concentration of H+, Cl–, OH– 2Cl– Cl2 + 2e 2H+ + 2e H2
pekat Silver electrolyte Gas bubbles are released.
Argentum Greenish yellow gas is
Silver nitrate released.
solution Carbon
Larutan Karbon Position of ion in OAgH+–, NO3–, H+, 4OH– 2H2O + O2 + 4e Ag+ + e Ag
argentum nitrat the electrochemical Gas bubbles are released. Grey shiny solid deposited.
Carbon series
Silver nitrate Karbon
solution Type of electrode OAgH+–, NO3–, H+, Ag Ag+ + e Ag+ + e Ag
Larutan Carbon Anode becomes thinner. Grey shiny solid deposited.
argentum nitrat Karbon
Position of ion in K+, I–, H+, OH– 4OH– 2H2O + O2 + 4e 2H+ + 2e H2
Dilute the electrochemical Gas bubbles are released. Gas bubbles are released.
potassium series
iodide solution
Larutan kalium Concentration of K+, I–, H+, OH– 2I– I2 + 2e 2H+ + 2e H2
iodida cair electrolyte Brown solution formed. Gas bubbles are released.
Concentrated Position of ion in OK+H,– SO42–, H+, 4OH– 2H2O + O2 +4e 2H+ + 2e H2
potassium the electrochemical Gas bubbles are released. Gas bubbles are released.
iodide solution series
Larutan kalium
iodida pekat
Dilute
potassium
sulphate
solution
Larutan kalium
sulfat cair
2 Electrolysis is carried out on a dilute potassium chloride solution using carbon electrodes. Explain how this electrolysis
occurs. Use a labelled diagram to explain your answer.
Proses elektrolisis dijalankan ke atas larutan kalium klorida cair menggunakan elektrod karbon. Jelaskan bagaimana proses elektrolisis ini
berlaku. Gunakan gambar rajah berlabel untuk menerangkan jawapan anda.
Set-up of apparatus / Susunan radas:
Carbon electrode Dilute potassium
chloride solution
Carbon electrode
m Publication Sdn. Bhd.
99
Nila
MODULE • Chemistry Form 4
Explanation / Penerangan:
– Potassium chloride solution consist of K+, H+, Cl– and OH– ions that move freely.
Larutan kalium klorida mengandungi ion K +, H +, Cl – dan OH – yang bergerak bebas.
– Cl– ion and OH– ions move to the anode.
Ion Cl – dan ion OH – bergerak ke anod.
– OH– ion is lower than Cl– ion in the electrochemical series.
Ion OH – terletak di bawah ion Cl – dalam siri elektrokimia.
– OH– ion is selectively discharged by releasing electrons to form oxygen and water molecule.
Ion OH – dipilih untuk dinyahcaskan dengan melepaskan elektron membentuk molekul oksigen dan air .
– Half equation / Persamaan setengah: 4OH– 2H2O + O2 + 4e .
– K+ ion and H+ ion move to the cathode / Ion K+ dan ion H+ bergerak ke katod.
– H+ ion is lower than K+ ion in the electrochemical series.
Ion H + terletak di bawah ion K + dalam siri elektrokimia
– H+ ion is selectively discharged by receiving electrons to form hydrogen molecules.
Ion H + dipilih untuk dinyahcaskan dengan menerima elektron membentuk molekul hidrogen .
– Half equation / Persamaan setengah: 2H + + 2e H2 .
3 Describe an experiment to determine the product of electrolysis copper(II) sulphate solution with carbon electrode.
Your answer should include the observation, confirmatory test for the product at the anode and half equation at the
electrode.
Huraikan satu eksperimen untuk menentukan hasil elektrolisis larutan kuprum(II) sulfat menggunakan elektrod karbon. Dalam jawapan
anda perlu disertakan pemerhatian, ujian pengesahan untuk hasil yang terbentuk di anod dan persamaan setengah bagi tindak balas yang
berlaku di elektrod.
Answer / Jawapan:
Apparatus / Radas : Battery / power supply, carbon electrodes, wire, electrolytic cell, test tube, Ammeter [from a
labelled diagram]
Material / Bahan : 1 mol dm–3 copper(II) sulphate solution
Carbon electrodes Copper(II) sulphate
solution
Procedure / Langkah:
(a) Pour 1 mol dm–3 copper(II) sulphate solution in the electrolytic cell until it is half full .
Masukkan larutan kuprum(II) sulfat 1 mol dm–3 ke dalam sel elektrolitik sehingga separuh penuh .
(b) The apparatus is set up as shown in the diagram. Fill the test tube with copper(II) sulphate solution and
invert the test tube on the anode .
Radas disusunkan seperti dalam gambar rajah. Isi tabung uji dengan larutan kuprum(II) sulfat dan terbalikkan tabung
uji itu pada anod .
(c) Turn on the switch / Hidupkan suis.
(d) Collect the gas produced at the anode / Kumpulkan gas yang terhasil di anod .
(e) Gas produced at the anode is tested with a glowing wooden splinter .
Gas yang terhasil di anod diuji dengan kayu uji berbara .
Nila m Publication Sdn.
100
Bhd.
Chemistry Form 4 • MODULE
Observation and half equation / Pemerhatian dan persamaan setengah:
Electrodes Observation Confirmatory test Half equation
Elektrod Pemerhatian Ujian pengesahan Persamaan setengah
Cu2+ + 2e Cu
Cathode Brown solid deposited –
Anode Gas bubbles are released – Insert the glowing wooden splinter into the test 4OH– 2H2O + O2 + 4e
tube.
– The glowing wooden splinter is lighted up.
4 Copper(II) sulphate solution is electrolysed using copper electrodes.
Larutan kuprum(II) sulfat dielektrolisis dengan menggunakan elektrod kuprum.
(a) Write the formula of all the anions present in the solution / Tuliskan formula semua anion yang terdapat dalam larutan itu.
SO 2–, OH–
4
(b) Write the half equation for the reaction at the / Tuliskan persamaan setengah untuk tindak balas di
(i) anode / anod : Cu Cu2+ + 2e
(ii) cathode / katod : Cu2+ + 2e Cu
(c) (i) From your observations, what happen to the intensity of the blue colour of the copper(II) sulphate solution
during electrolysis?
Daripada pemerhatian anda, nyatakan apakah yang berlaku ke atas keamatan warna biru larutan kuprum(II) sulfat semasa
proses elektrolisis?
The intensity of the blue colour of copper(II) sulphate remains unchanged.
(ii) Explain your answer / Jelaskan jawapan anda.
The number of copper(II) ions become copper atoms at the cathode is equal to the number of copper atoms
become copper(II) ions at the anode.
(d) If the experiment is repeated with the copper electrodes being replaced by carbon electrodes, state the name of the
products formed at the
Jika eksperimen diulangi dengan menggantikan elektrod kuprum dengan elektrod karbon, namakan hasil yang terbentuk di
(i) anode / anod: Oxygen (ii) cathode / katod: Copper
5 The diagram below shows the set-up of apparatus of an electrolytic cell.
Rajah di bawah menunjukkan susunan radas bagi sel elektrolisis.
Carbon electrode P Carbon electrode Q
Elektrod karbon P Elektrod karbon Q
Copper(II) nitrate solution
Larutan kuprum(II) nitrat
(a) Write the formula of all ions present in copper(II) nitrate solution.
Tuliskan formula semua ion yang hadir dalam larutan kuprum(II) nitrat.
Cu2+, NO3–, H+ and OH– .
(b) Write half equation for the reaction at / Tuliskan persamaan setengah di:
electrode P / elektrod P : Cu2+ + 2e Cu
electrode Q / elektrod Q : 4OH – 2H2O + O2 + 4e
(c) (i) What is the colour of copper(II) nitrate / Apakah warna larutan kuprum(II) nitrat?
Blue
(ii) What happens to the intensity of the colour of copper(II) nitrate solution? Explain your answer.
Apakah yang berlaku kepada keamatan warna larutan kuprum(II) nitrat? Jelaskan jawapan anda.
The intensity of the blue colour of copper(II) nitrate decreases. The concentration of Cu2+ decreases because
copper(II) ions receive electrons to form copper atom at the cathode.
m Publication Sdn. Bhd.
101
Nila
MODULE • Chemistry Form 4
ELECTROLYSIS IN INDUSTRY / ELEKTROLISIS DALAM INDUSTRI
1 Three uses of electrolysis in industries are / Tiga kegunaan elektrolisis dalam industri ialah:
Application Example Electrolyte Anode / Half equation Cathode / Half equation
Aplikasi Contoh Elektrolit Anod / Persamaan setengah Katod / Persamaan setengah
(a) Electroplating Silver Silver nitrate Anode / Anod: Cathode / Katod:
Silver metal Metal to be electroplated
Penyaduran logam electroplating solution
Half equation / Persamaan setengah: Half equation / Persamaan setengah:
Penyaduran perak Ag Ag+ + e Ag+ + e Ag
(b) Purification of Purification of Copper(II) Anode / Anod: Cathode / Katod:
metal copper sulphate solution Impure copper Pure copper
Penulenan
Penulenan logam Half equation / Persamaan setengah: Half equation / Persamaan setengah:
kuprum Cu Cu2+ + 2e Cu2+ + 2e Cu
(c) Metal extraction Extraction of Molten Anode / Anod: Cathode / Katod:
Pengekstrakan aluminium aluminium oxide Carbon Carbon
Pengekstrakan Half equation / Persamaan setengah:
logam Half equation / Persamaan setengah:
aluminium 2O2– O2 + 4e Al3+ + 3e Al
2 The following diagram shows the aluminium extraction process.
Rajah di bawah menunjukkan proses pengekstrakan aluminium.
Substance Z / Bahan Z
Substance Y
Bahan Y
Substance W Substance X + cryolite
Bahan W Bahan X + kriolit
(a) State the name of the following substances / Nyatakan nama bahan-bahan berikut:
W : Liquid aluminium X : Molten aluminium oxide
Y : Carbon Z : Carbon
(b) Which substance acts as anode and cathode / Bahan yang manakah bertindak sebagai anod dan katod?
Anode / Anod : Z Cathode / Katod : Y
(c) State the name of the product at anode and cathode / Namakan hasil yang diperoleh di anod dan katod.
Anode / Anod : Oxygen Cathode / Katod : Aluminium
(d) Write the ionic equation for the reactions at / Tuliskan persamaan ion bagi tindak balas yang berlaku di
anode / anod : 2O2– O2 + 4e cathode / katod : Al3+ + 3e Al
(e) Why is cryolite added to X / Mengapakan kriolit ditambah ke dalam X?
To lower down the melting point of aluminium oxide (from 2 045°C to 900°C ).
3 The diagram below shows the set-up of apparatus used in the purification of copper.
Rajah di bawah menunjukkan susunan radas yang digunakan untuk proses penulenan kuprum.
Electrode X Electrode Y
Elektrod X Elektrod Y
Electrode Z
Elektrod Z
Nila m Publication Sdn.
102
Bhd.
Chemistry Form 4 • MODULE
(a) State the name of the substance used as / Nyatakan nama bahan yang dijadikan sebagai:
electrode X / elektrod X : Impure copper
electrode Y / elektrod Y : Pure copper
electrolyte Z / elektrolit Z : Copper(II) sulphate solution
(b) Write the half equation that occur at the / Tuliskan persamaan setengah yang berlaku di
electrode X / elektrod X : Cu Cu2+ + 2e electrode Y / elektrod Y : Cu2+ + 2e Cu
(c) What are the observations at the / Apakah pemerhatian di
electrode X / elektrod X : Electrode becomes thinner
electrode Y / elektrod Y : Brown solid deposited
4 To purify metal an impure metal / Untuk menulenkan logam tak tulen:
(a) The impure metal is used as the anode / Logam tak tulen dijadikan sebagai anod.
(b) The pure metal is used as the cathode / Logam tulen dijadikan sebagai katod.
(c) The electrolyte used is an salt solution containing the ions of the purifying metal.
Elektrolit adalah larutan garam yang mengandungi ion logam yang hendak ditulenkan.
5 A student intends to electroplate an iron spoon with copper. Describe a laboratory experiment to electroplate the iron
ring. Your answer should involve the following:
Seorang pelajar bercadang untuk menyadurkan sebatang sudu besi dengan kuprum. Huraikan satu eksperimen di dalam makmal untuk
menyadur sebatang sudu besi. Jawapan anda perlu mengandungi:
– A labelled diagram showing the set-up of apparatus / Rajah berlabel menunjukkan susunan radas.
– Procedure / Kaedah.
– Half equation for the reactions at both electrodes / Persamaan setengah untuk tindak balas di kedua-dua elektrod.
– Observation at both electrodes / Pemerhatian di kedua-dua elektrod.
Answer / Jawapan:
Iron spoon Copper
Copper(II) nitrate solution
Procedure / Kaedah:
(a) Copper plate and iron spoon are cleaned with sand paper .
Kepingan kuprum dan sudu besi dibersihkan dengan kertas pasir .
(b) Copper(II) nitrate solution is poured into a beaker until half full .
Larutan kuprum(II) nitrat dituangkan ke dalam bikar sehingga separuh penuh .
(c) Iron spoon is then connected to the negative terminal of battery while the copper plate is connected to the
positive terminal of the battery// Iron spoon is made as cathode while copper plate is made as anode.
Sudu besi disambungkan kepada terminal negatif bateri dan kepingan kuprum disambungkan kepada terminal positif
bateri// Sudu besi dijadikan katod dan kepingan kuprum dijadikan anod.
(d) The iron spoon and the copper plate are dipped in the copper(II) nitrate solution as shown in the diagram.
Sudu besi dan plat kuprum dicelup ke dalam larutan kuprum(II) nitrat seperti ditunjukkan dalam rajah.
(e) The circuit is completed / Litar dilengkapkan .
(f) Half equation at the cathode / Persamaan setengah di katod : Cu2+ + 2e Cu .
(g) Observation of the cathode: Brown solid is deposited / Pemerhatian di katod: pepejal perang terenap.
(h) Half equation at the anode / Persamaan setengah di anod : Cu Cu2+ + 2e .
(i) Observation of the anode / Pemerhatian di anod : Copper plate becomes thinner .
m Publicat ion Sdn. Bhd.
103
Nila
MODULE • Chemistry Form 4
6 To electroplate an object with metal / Untuk menyadur sesuatu objek dengan logam: katod ..
(a) The metal object to be electroplated is made to be cathode / Objek yang hendak disadur dijadikan
(b) The electroplating metal is made to be anode / Logam penyadur dijadikan anod ..
(c) The electrolyte used is an aqueous salt solution containing the ions of the electroplating metal.
Elektrolit yang digunakan adalah larutan akueus garam yang mengandungi ion logam penyadur.
ELECTROCHEMICAL SERIES / SIRI ELEKTROKIMIA
1 Electrochemical Series is an arrangement of metals according to their tendency to release electrons to form a
positive ion.
Siri Elektrokimia ialah susunan logam mengikut kecenderungan melepaskan elektron membentuk ion bercas positif .
2 The position of metal atoms in Electrochemical Series / Kedudukan atom logam dalam Siri Elektrokimia:
K, Na, Ca, Mg, Al, Zn , Fe, Sn ,Pb, Cu, Ag
Tendency of metal atom to release/donate electrons increases (electropositivity increases)
Kecenderungan untuk atom logam melepaskan/menderma elektron bertambah (keelektropositifan bertambah)
3 The position of metal ions (cation) in the Electrochemical Series / Kedudukan ion logam (kation) dalam Siri Elektrokimia:
K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, *H+, Cu2+
Tendency of metal ion (cation) to receive/gain electrons increases
Kecenderungan untuk ion logam (kation) untuk menerima elektron bertambah
*H+ is also in the series of ion because it is present in aqueous solution of any electrolyte (salt solution/acid/alkali)
* H+ juga terdapat dalam siri ion kerana kehadiran ion H+ dalam elektrolit larutan akueus (larutan garam/asid/alkali)
METAL DISPLACEMENT REACTION / TINDAK BALAS PENYESARAN LOGAM
1 The metal which is situated at a higher position (higher tendency to release electron) in the Electrochemical Series is
able to displace metals below it from its salt solution .
Logam yang berada di kedudukan atas (kecenderungan melepaskan elektron yang tinggi) dalam Siri Elektrokimia dapat menyesarkan
logam yang di bawahnya daripada larutan garam logam tersebut.
2 Example / Contoh:
Experiment / Eksperimen Observation / Pemerhatian Remark / Catatan
Silver nitrate solution
Larutan argentum nitrat – Copper strip becomes Inference / Inferens: .
thinner . – The grey solid is silver .
Copper
Kuprum Kepingan kuprum Pepejal kelabu adalah argentum .
menipis .
– The blue solution is copper(II) nitrate
– A grey solid Larutan biru adalah kuprum(II) nitrat .
deposited.
Pepejal kelabu terenap. Explanation / Penerangan:
– Silver ion receives electrons to form silver atom.
– The colourless
solution turns blue. Ion argentum menerima elektron membentuk atom argentum .
Larutan tidak berwarna – Copper atom releases electrons to form copper(II) ion .
bertukar menjadi biru.
Atom kuprum melepaskan elektron membentuk ion kuprum(II) .
– Copper has displaced silver from silver nitrate solution.
Kuprum telah menyesarkan argentum dari larutan argentum nitrat.
Cu + 2AgNO3 Cu(NO3)2 + 2Ag .
– Copper is more electropositive than silver// Copper is
above silver in the Electrochemical Series of metal.
Kuprum adalah lebih elektropositif daripada argentum //Kuprum
terletak di atas argentum dalam Siri Elektrokimia logam.
Nila m Publication Sdn.
104
Bhd.
Chemistry Form 4 • MODULE
Copper(II) sulphate – Magnesium strip Inference / Inferens:
solution becomes thinner . – The brown solid is copper .
Kepingan magnesium
Larutan kuprum (II) sulfat menipis . Pepejal perang adalah kuprum .
Magnesium – The brown solid – The colourless solution is magnesium sulphate .
Magnesium deposited. .
Pepejal perang terenap. Larutan tidak berwarna adalah magnesium sulfat
Zinc nitrate solution
Larutan zink sulfat – The blue solution turn Explanation / Penerangan: .
colourless. – Copper(II) ion receives electrons to form copper atom.
Copper / Kuprum Larutan biru bertukar
menjadi tidak berwarna. Ion kuprum(II) menerima elektron membentuk atom kuprum.
– Magnesium atom releases electrons to form magnesium ion
No observable changes.
Tiada perubahan yang dapat Atom magnesium melepaskan elektron membentuk ion magnesium .
diperhatikan.
– Magnesium has displaced copper from copper(II) sulphate
solution.
Magnesium telah menyesarkan kuprum dari larutan kuprum(II) sulfat.
Mg + Cu SO4 MgSO4 + Cu .
– Magnesium is more electropositive than copper// Magnesium
is above copper in the Electrochemical Series of metal.
Magnesium adalah lebih elektropositif daripada kuprum// magnesium
terletak di atas kuprum dalam Siri Elektrokimia logam.
Inference / Inferens:
– No reaction occur.
Tiada tindak balas berlaku.
Explanation / Penerangan:
– Copper cannot displace zinc from zinc sulphate solution.
Kuprum tidak boleh menyesarkan zink daripada larutan zink sulfat.
– Copper is more electropositive than zinc// Copper is below
zinc in the Electrochemical Series of metal.
Kuprum adalah kurang elektropositif daripada zink // kuprum terletak
di bawah zink dalam Siri Elektrokimia logam.
VOLTAIC CELL (CHEMICAL CELL) / SEL RINGKAS (SEL KIMIA)
1 A cell that produces electrical energy when chemical reactions occur in it.
Sel yang menghasilkan tenaga elektrik apabila berlaku tindak balas kimia di dalamnya.
2 Energy change in voltaic cell is chemical energy to electrical energy .
Perubahan tenaga dalam sel ringkas ialah dari tenaga kimia kepada tenaga elektrik .
3 Produced when two different metals are dipped in an electrolyte and are connected by an external circuit .
Terhasil apabila dua logam berlainan dicelup dalam elektrolit dan disambung dengan litar luar .
4 The voltage of chemical cell depends on the distance between the two metals in the Electrochemical Series, where
the further the distance between them, the higher is the voltage.
Voltan sel kimia bergantung pada jarak antara dua logam dalam Siri Elektrokimia di mana semakin jauh dua logam dalam Siri
Elektrokimia, semakin tinggi voltannya.
5 A more electropositive metal becomes the negative terminal of the cell. A less electropositive metal becomes the
positive terminal:
Logam yang lebih elektropositif akan menjadi terminal negatif sel. Logam yang kurang elektropositif akan menjadi terminal
positif sel:
m Publication Sdn. Bhd.
105
Nila
MODULE • Chemistry Form 4
Electrical current produced is detected by the galvanometer G
(Chemical energy Electrical energy)
Arus elektrik terhasil dikesan oleh galvanometer
(Tenaga kimia Tenaga elektrik)
Negative terminal / Terminal negatif : _ + Positive terminal / Terminal positif :
• More electropositive metal. • Less electropositive metal.
Logam lebih elektropositif. Logam kurang elektropositif.
• Metal atom will release electrons that will • The electrons that flow from the external
flow through the external circuit. Metal atom circuit are received by the positive ion in
becomes metal ion (becomes thinner). the electrolyte through this terminal.
Atom logam akan melepaskan elektron yang akan Elektron yang akan mengalir dari litar luar diterima
oleh ion positif dalam elektrolit melalui terminal ini.
mengalir di litar luar. Atom logam menjadi ion logam
(semakin nipis).
6 Example of simple voltaic cell / Contoh voltan sel ringkas:
V
Copper Magnesium
Kuprum Magnesium
Copper(II) sulphate solution
Larutan kuprum(II) sulfat
(a) Magnesium electrode is a negative terminal because magnesium is more electropositive than copper :
Elektrod magnesium adalah terminal negatif kerana magnesium lebih elektropositif daripada kuprum :
– Magnesium atom releases electrons to form magnesium ion, Mg2+.
Atom magnesium melepaskan elektron untuk membentuk ion magnesium, Mg2+.
– Half equation / Persamaan setengah : Mg Mg2+ + 2e .
– Magnesium electrode becomes thinner / Elektrod magnesium menjadi nipis .
– Electron flows through external circuit to the copper electrode.
Elektron mengalir melalui litar luar ke elektrod kuprum .
(b) Copper electrode is a positive terminal because copper is less electropositive than magnesium :
magnesium :
Elektrod kuprum adalah terminal positif kerana kuprum kurang elektropositif daripada
– Electrons from magnesium flow through external circuit to copper electrode.
Elektron dari magnesium mengalir melalui litar luar ke elektrod kuprum.
– Copper(II) ion in the electrolyte receives electron to form copper atom.
Ion kuprum(II) dalam elektrolit menerima elektron untuk membentuk atom kuprum.
– Half equation / Persamaan setengah : Cu+ + 2e Cu .
– Brown solid is deposited on the surface of copper electrode.
Pepejal perang terenap di permukaan elektrod kuprum.
(c) The concentration of copper(II) sulphate decreases because copper(II) ions discharged to copper atom at the
positive terminal. The intensity of blue colour of copper(II) sulphate decreases.
Kepekatan larutan kuprum(II) sulfat berkurang kerana ion kuprum(II) dinyahcaskan kepada atom kuprum. Keamatan
warna biru larutan kuprum(II) sulfat berkurang.
(d) If the magnesium metal is replaced with a zinc metal, the voltage reading decreases because zinc is nearer to
copper in the electrochemical series.
Jika logam magnesium digantikan dengan logam zink, bacaan voltan akan berkurang kerana zink lebih dekat dengan kuprum
dalam siri elektrokimia.
Nila m Publication Sdn.
106
Bhd.
Chemistry Form 4 • MODULE
7 Daniell cell / Sel Daniell
(a) It is an example of voltaic cell which consists of zinc electrode dipped in zinc sulphate solution, copper electrode
dipped in copper(II) sulphate solution and connected by a salt bridge or porous pot.
Merupakan satu contoh sel kimia yang terdiri daripada elektrod zink yang dicelup ke dalam larutan zink sulfat, elektrod kuprum
dicelupkan ke dalam larutan kuprum(II) sulfat dan dihubungkan dengan titian garam atau pasu berliang.
Zn / ZnSO4 // CuSO4 / Cu
(b) The function of porous pot or salt bridge is to allow the flow of ions through it so that the electric circuit is
completed.
Fungsi pasu berliang atau titian garam adalah untuk membenarkan ion-ion mengalir melaluinya dan melengkapkan litar.
(c) The diagram below shows the set-up of apparatus of Daniell cell.
Rajah di bawah menunjukkan susunan radas bagi sel Daniell.
Copper Sulphuric acid Copper Zinc / Zink
Kuprum Asid sulfurik Kuprum Zinc sulphate
Zink sulfat
Copper(II) sulphate Zinc Copper(II) sulphate
solution Zink solution Porous pot
Zinc sulphate Pasu berliang
Larutan kuprum(II) sulfat Zink sulfat Larutan kuprum(II) sulfat
(d) Zinc electrode is a negative terminal because zinc is more electropositive than copper :
daripada kuprum :
Elektrod zink adalah terminal negatif kerana zink adalah lebih elektropositif
– Zinc atom releases electron to form zinc ion, Zn2+.
Atom zink melepaskan elektron untuk membentuk ion zink, Zn2+.
– Half equation / Persamaan setengah : Zn Zn2+ + 2e .
– Zinc electrode becomes thinner / Elektrod zink menjadi nipis ..
– Electrons flow through external circuit to the copper electrode.
Elektron mengalir melalui litar luar ke elektrod kuprum .
(e) Copper electrode is a positive terminal because copper is less electropositive than zinc :
Elektrod kuprum adalah terminal positif kerana kuprum kurang elektropositif daripada zink :
– Electrons from zinc electrode flow through external circuit to copper electrode.
Elektron dari zink mengalir melalui litar luar ke elektrod kuprum.
– Copper(II) ion in the electrolyte receives electron to form copper atom.
Ion kuprum(II) dalam elektrolit menerima elektron untuk membentuk atom kuprum.
– Half equation / Persamaan setengah : Cu2+ + 2e Cu .
– Brown solid is deposited on the surface of copper electrode.
Pepejal perang terenap di permukaan elektrod kuprum.
(f) The concentration of copper(II) sulphate decreases because copper(II) ions are discharged to copper atoms. The
intensity of blue colour of copper(II) sulphate decreases.
Kepekatan larutan kuprum(II) sulfat berkurang kerana ion kuprum(II) telah dinyahcaskan kepada atom kuprum. Keamatan
warna biru kuprum(II) sulfat berkurang.
(g) If zinc metal is replaced with a magnesium metal, the voltage reading increases because magnesium is
further from copper in the Electrochemical Series.
Jika logam zink digantikan dengan logam magnesium, bacaan voltan bertambah kerana jarak antara magnesium dengan kuprum
lebih jauh daripada jarak antara zink dengan kuprum dalam Siri Elektrokimia.
m Publication Sdn. B hd.
107
Nila
MODULE • Chemistry Form 4
8 Four main uses of the Electrochemical Series / Kegunaan utama Siri Elektrokimia:
(a) To predict the terminal of chemical cell / Untuk meramalkan terminal sel kimia
– The more electropositive metal is the negative terminal of the cell.
Logam yang lebih elektropositif ialah terminal negatif sel.
– The less electropositive metal is the positive terminal of the cell.
Logam yang kurang elektropositif ialah terminal positif sel.
(b) To predict the voltage of chemical cell / Untuk meramalkan voltan sel kimia
– The further the distance between two metals in the Electrochemical Series, the higher is the voltage of the
chemical cell.
Semakin jauh jarak antara dua logam dalam Siri Elektrokimia, semakin tinggi bacaan voltan sel kimia.
(c) To predict the metal displacement reactions / Untuk meramalkan tindak balas penyesaran logam
– The more electropositive metal can displace a less electropositive metal from its salt solution.
Logam yang lebih elektropositif dapat menyesarkan logam yang kurang elektropositif daripada larutan garamnya.
(d) To predict the selected ion to be discharged at the electrode in an electrolysis
Untuk meramalkan pemilihan ion untuk dinyahcas di elektrod dalam proses elektrolisis
EXERCISE / LATIHAN
1 The table below shows the results of an experiment to construct the Electrochemical Series through the ability of metals
to displace other metals from their salt solution.
Jadual di bawah menunjukkan keputusan eksperimen untuk membina Siri Elektrokimia berdasarkan keupayaan suatu logam untuk
menyesarkan logam lain dari larutan garamnya.
Experiment I / Eksperimen I Experiment II / Eksperimen II
P nitrate solution R nitrate solution
Larutan P nitrat Larutan R nitrat
Zinc / Zink Zinc / Zink
Metal P is displaced, blue colour solution turn colourless. No reaction.
Logam P disesarkan, larutan biru bertukar menjadi tanpa warna. Tiada tindak balas.
(a) Based on the results in the table, arrange metal P, zinc and R in descending order of electropositivity.
Berdasarkan keputusan dalam jadual, susunkan logam P, zink dan R dalam tertib menurun keelektropositifan.
R, Zn, P
(b) Based on the observation in Experiment I / Berdasarkan pemerhatian dalam Eksperimen I,
(i) state the name the suitable metal P / namakan logam yang sesuai bagi P.
Copper .
(ii) zinc can displace metal P from P nitrate solution. Explain.
zink boleh menyesarkan logam P daripada larutan P nitrat. Terangkan.
Zinc is more electropositive than P. .
(iii) write the chemical equation for the reaction / tuliskan persamaan kimia untuk tindak balas.
Zn + Cu(NO3 )2 Zn(NO3 )2 + Cu
Nila m Publication Sdn.
108
Bhd.
Chemistry Form 4 • MODULE
2 The diagram below shows the set-up of the apparatus to arrange metals W, X, Y and Z based on the potential difference
of the metals.
Rajah di bawah menunjukkan susunan radas bagi eksperimen untuk menentukan kedudukan logam W, X, Y dan Z berdasarkan beza upaya
logam.
V
Metal electrode Metal electrode
Elektrod logam Elektrod logam
Electrolyte / Elektrolit
The table below shows the results of the experiment.
Jadual di bawah menunjukkan keputusan eksperimen.
Pair of metals Potential difference (V) Negative terminal
Pasangan logam Beza keupayaan (V) Terminal negatif
0.50 X
W and X 0.30 Y
W dan X 1.10 Z
X and Y
X dan Y
W and Z
W dan Z
(a) Arrange metals W, X, Y and Z in descending order of the electropositivity of metal.
Susunkan logam W, X, Y dan Z dalam tertib menurun keelektropositifan logam.
Z, Y, X, W .
(b) (i) Metals X and Z are used as electrodes in the diagram. State which metal acts as positive terminal.
Logam X dan Z digunakan sebagai terminal dalam rajah. Nyatakan logam yang manakah akan bertindak sebagai terminal
positif.
Metal X
(ii) Give reason for your answer in (b)(i) / Berikan sebab untuk jawapan anda di (b)(i).
Metal X is less electropositive than metal Z. .
(c) Predict the voltage of the cell in (b)(i) / Ramalkan nilai voltan dalam sel di (b)(i).
0.6 V
3 The diagram below shows the set-up of apparatus for two types of cell.
Rajah di bawah menunjukkan susunan radas untuk dua jenis sel.
Copper Copper Zinc
Kuprum Kuprum Zink
Copper(II) sulphate solution Cell Y / Sel Y
Larutan kuprum(II) sulfat
Cell X / Sel X
m Publication Sdn. B hd.
109
Nila
MODULE • Chemistry Form 4
Complete the following table to compare cell X and cell Y :
Lengkapkan jadual berikut untuk membandingkan sel X dan sel Y :
Description Cell X Cell Y
Perkara Sel X Sel Y
Type of cell Electrolytic cell Chemical cell
Jenis sel
Chemical energy Electrical energy
The energy change Electrical energy Chemical energy
Perubahan tenaga Cu2+, H+, SO42–, OH–
Ion presence in the Cu2+, H+, SO42–, OH– Negative terminal / Terminal negatif : Zinc
electrolyte Positive terminal / Terminal positif : Copper
Ion hadir dalam Negative terminal / Terminal negatif : Zn Zn2+ + 2e
Positive terminal / Terminal positif : Cu2+ + 2e Cu
elektrolit Negative terminal / Terminal negatif :
Zinc electrode becomes thinner
Electrode Anode / Anod: Copper Positive terminal / Terminal positif :
Elektrod Cathode / Katod: Copper Brown solid deposited
Electrolyte / Elektrolit:
Half equation Anode / Anod: Cu Cu2+ + 2e Intensity blue colour of copper(II) sulphate decreases
Persamaan setengah Cathode / Katod: Cu2+ + 2e Cu
Observation Anode / Anod:
Pemerhatian Copper electrode becomes thinner
Cathode / Katod:
Brown solid deposited
Electrolyte / Elektrolit:
Intensity blue colour of copper(II) sulphate
remains unchanged
4 The diagram below shows the set-up of apparatus for an experiment.
Rajah di bawah menunjukkan susunan radas bagi suatu eksperimen.
V
Zinc / Zink –+ Anode Cathode
Cell A / Set A
Zinc sulphate solution Copper Copper
Larutan zink sulfat Kuprum Kuprum
Copper(II) sulphate solution
Larutan kuprum(II) sulfat Porous pot Copper(II) sulphate solution
Pasu berliang Larutan kuprum(II) sulfat
Cell B / Set B
Nila tion Sdn.(a) In the above diagram, label
Dalam gambar rajah di atas, label
(i) the positive terminal and negative terminal Cell A,
terminal positif dan terminal negatif bagi Sel A,
(ii) anode and cathode in Cell B.
anod dan katod bagi Sel B.
(b) What is the energy change in Cell A and Cell B?
Apakah perubahan tenaga dalam Sel A dan Sel B?
Cell A / Sel A : Chemical energy to electrical energy
Cell B / Sel B : Electrical energy to chemical energy
m Publica
110
Bhd.
m Publicat Chemistry Form 4 • MODULE hd.
(c) What is the function of the porous pot in cell A?
Apakah fungsi pasu berliang dalam Sel A?
To allow the movement of ions through it.
(d) Referring to Cell A.
Merujuk kepada Sel A.
(i) What is the observation at zinc electrode?
Apakah pemerhatian di elektrod zink?
Zinc electrode becomes thinner.
(ii) Write the half equation for the reaction at zinc electrode.
Tuliskan persamaan setengah untuk tindak balas di elektrod zink.
Zn Zn2+ + 2e
(iii) What is the observation at copper electrode / Apakah pemerhatian di elektrod kuprum?
Brown solid deposited.
(iv) Write the half equation for the reaction at copper electrode.
Tuliskan persamaan setengah untuk tindak balas di elektrod kuprum.
Cu2+ + 2e Cu .
(v) After 30 minutes, what is the colour change of the copper(II) sulphate solution? Explain why.
Selepas 30 minit, apakah perubahan warna larutan kuprum(II) sulfat? Jelaskan mengapa.
– The intensity of blue colour decreases.
– Copper(II) ions are discharged to form copper atoms.
– Concentration of copper(II) ions in copper(II) sulphate decreases.
(e) Referring to Cell B.
Merujuk kepada Sel B.
(i) What is the observation at the anode?
Apakah pemerhatian di anod?
Copper electrode becomes thinner.
(ii) Write the half equation for the reaction at the anode.
Tuliskan persamaan setengah untuk tindak balas di anod.
Cu Cu2+ + 2e
(iii) What is the observation at the cathode?
Apakah pemerhatian di katod?
Brown solid deposited.
(iv) Write the half equation for the reaction at copper electrode.
Tuliskan persamaan setengah untuk tindak balas di katod.
Cu2+ + 2e Cu
(f) The intensity of blue colour of copper(II) sulphate solution in the Cell B remains unchanged during the experiment.
Explain why.
Keamatan warna biru larutan kuprum(II) sulfat dalam Sel B tidak berubah semasa eksperimen. Jelaskan mengapa.
– The concentration of copper(II) sulphate remain unchanged.
– The rate of copper(II) ions discharged to copper atom at the cathode equals to the rate of copper atoms form
copper(II) ions at the anode.
ion Sdn. B
111
Nila
MODULE • Chemistry Form 4
Objective Questions / Soalan Objektif
1 Which of the following is an electrolyte? C Copper electrode becomes Copper electrode becomes
Antara berikut, yang manakah adalah elektrolit? thicker thinner
A Glacial ethanoic acid
Asid etanoik glasial Elektrod kuprum semakin tebal Elektrod kuprum semakin nipis
B Molten naphthalene
Naftalena lebur D Gas bubbles are released Copper electrode becomes
C Aqueous solution of zinc chloride Gelembung gas dibebaskan thicker
Larutan akueus zink klorida
D Hydrogen chloride in methylbenzene Elektrod kuprum semakin tebal
Hidrogen klorida dalam metilbenzena
4 The diagram below shows the set-up of apparatus of an
electrolysis process.
Rajah di bawah menunjukkan sususnan radas untuk proses elektrolisis.
2 The diagram below shows the set-up of apparatus used Electrolyte
to electrolyse substance X. Elektrolit
Rajah di bawah menunjukkan susunan radas untuk elektrolisis
bahan X.
Carbon electrode P Q Carbon electrode
Elektrod karbon Elektrod karbon
Carbon electrodes Which of the following electrolytes produce oxygen gas at
Elektrod karbon electrode Q?
Antara elektrolit berikut, yang manakah membebaskan gas oksigen pada
Substance X
Bahan X elektrod Q?
Heat I 1.0 mol dm–3 hydrochloric acid
Panaskan Asid hidroklorik 1.0 mol dm–3
Which of the following compounds can light up the bulb II 1.0 mol dm–3 sulphuric acid
when used as substance X? Asid sulfurik 1.0 mol dm–3
Antara berikut, yang manakah boleh menyalakan mentol apabila
digunakan sebagai bahan X? III 1.0 mol dm–3 potassium iodide solution
A Copper(II) nitrate / Kuprum(II) nitrat Larutan kalium iodida 1.0 mol dm–3
B Lead(II) iodide / Plumbum(II) iodida
C Zinc carbonate / Zink karbonat IV 1.0 mol dm–3 nitric acid
D Sodium carbonate / Natrium karbonat Asid nitrik 1.0 mol dm–3
A I and II only / I dan II sahaja
B II and III only / II dan III sahaja
C II and IV only / II dan IV sahaja
D II, III and IV only / II, III dan IV sahaja
3 The diagram below shows the set-up of apparatus for 5 The table below shows the observation of electrolysis of a
electrolysis of copper(II) sulphate solution. substance Q using carbon electrode.
Rajah di bawah menunjukkan susunan radas untuk elektrolisis larutan Jadual di bawah menunjukkan pemerhatian bagi elektrolisis bahan Q
kuprum(II) sulfat. menggunakan elektrod karbon.
Electrode Observation
Elektrod Pemerhatian
Copper Copper electrode Y Anode A greenish-yellow gas released
electrode X Elektrod kuprum Y Anod Gas kuning kehijauan terbebas
Elektrod Copper(II) Cathode A colorless gas which burns with a ‘pop’
kuprum X sulphate solution Katod sound is released
Larutan kuprum(II) Gas tanpa warna terbakar dengan bunyi ‘pop’
sulfat dibebaskan
What can be observed at the electrodes X and Y after What is substance Q?
30 minutes? Apakah bahan Q?
Apakah yang dapat diperhatikan pada elektrod X dan Y selepas 30 A 1.0 mol dm–3 of hydrochloric acid.
minit?
Asid hidroklorik 1.0 mol dm–3.
XY B 1.0 mol dm–3 of potassium nitrate solution.
A Copper electrode becomes Copper electrode becomes Larutan natrium nitrat 1.0 mol dm–3.
thinner thicker C 1.0 mol dm–3 of copper(II) chloride solution.
Elektrod kuprum semakin nipis Elektrod kuprum semakin tebal Larutan kuprum(II) klorida 1.0 mol dm–3.
D 1.0 mol dm–3 of magnesium bromide solution.
B Copper electrode becomes Gas bubbles are released
Larutan magnesium bromida 1.0 mol dm–3.
m Publica
thinner Gelembung gas dibebaskan
112
Elektrod kuprum semakin nipis
Nila tion Sdn.
Bhd.
Chemistry Form 4 • MODULE
6 The diagram below shows the set-up of apparatus of a 8 The table below shows the information about three voltaic
chemical cell that shows the direction of electron flow from cells.
zinc to metal Q. Jadual di bawah menunjukkan maklumat tentang tiga sel kimia.
Rajah di bawah menunjukkan susunan radas sel kimia yang menunjukkan Pair of metals Positive Potential difference (V)
arah pengaliran elektron ke logam Q. Pasangan logam terminal Beza upaya (V)
Terminal positif
W, Z Z 3.1
Q Zinc X, Y Y 0.3
Zink W, X X 1.8
Dilute sodium What is the potential difference of the voltaic cell when metal
chloride solution Y is paired with metal Z?
Larutan natrium
klorida cair Apakah beza upaya sel kimia apabila logam Y dipasangkan dengan logam
Z?
What is metal Q? A 1.0 V C 2.1 V
B 1.3 V D 2.8 V
Apakah logam Q?
A Copper 9 The diagram below shows the set-up of apparatus in a chemical
cell and electrolytic cell.
Kuprum Rajah di bawah menunjukkan susunan radas bagi sel kimia dan sel
B Iron
elekrolisis.
Besi
C Aluminium
Aluminium
D Magnesium
Magnesium
7 The diagram below shows the set-up of apparatus used to Zinc PQ Copper RS Copper
purify impure copper by using electrolysis method. Zink Kuprum Kuprum
Rajah di bawah menunjukkan susunan radas yang digunakan untuk
Zinc
menulenkan kuprum tak tulen dengan menggunakan kaedah elektrolisis. sulphate
solution
Larutan
zink sulfat Copper(II) sulphate solution
Larutan kuprum(II) sulfat
XY Which of the following is the observation at electrode R?
Antara berikut, yang manakah merupakan pemerhatian pada elektrod
Electrolyte Z
Elektrolit Z R?
A Electrode R becomes thinner
Which of the following shows the correct position of pure
copper and impure copper? Elektrod R semakin nipis
Antara berikut, yang manakah adalah kedudukan yang betul untuk B Electrode R becomes thicker
kuprum tulen dan kuprum tak tulen? Elektrod R semakin tebal
C A colourless gas is released
Gas tanpa warna terbebas
D A brown solid is deposited
Pepejal perang terenap
Electrode X Electrode Y Electrolyte Z 10 The table below shows the results of an experiment to study the
Elektrod X Elektrod Y Elektrolit Z displacement of metal from its solution using other metals.
Jadual di bawah menunjukkan keputusan eksperimen untuk mengkaji
A Impure copper Pure copper Copper(II) sulphate
Kuprum tak tulen Kuprum tulen solution penyesaran logam daripada larutan garamnya menggunakan logam
Larutan kuprum(II)
lain.
sulfat
Metal Nitrate of Q Nitrate of S
B Pure copper Impure copper Copper(II) sulphate Logam Nitrat bagi Q Nitrat bagi S
Kuprum tulen Kuprum tak tulen solution
Larutan kuprum(II) P
sulfat Q–
C Impure copper Pure copper Sulphuric acid – reaction occur / tindak balas berlaku
Kuprum tak tulen Kuprum tulen Asid sulfurik
– no reaction / tiada tindak balas
D Pure copper Impure copper Sulphuric acid Which of the following is the arrangement of metals P, Q and
Kuprum tulen Kuprum tak tulen Asid sulfurik
R in ascending order of the tendency of the metals to form
ions?
Antara berikut, yang manakah adalah susunan logam P, Q dan R dalam
susunan menaik kecenderungan logam membentuk ion?
A P, S, Q C S, P, Q
B Q, S, P D S, Q, P
m Publication Sdn. Bhd.
113
Nila
MODULE • Chemistry Form 4 ACID AND BASES
6 ASID DAN BES
ACID AND BASES / ACID DAN BES
• ACID / ASID
– To state the meaning of acid, give examples and write chemical equations and observations for the reaction of acids:
Menyatakan maksud asid, memberi contoh dan menulis persamaan tindak balas kimia dan pemerhatian bagi tindak balas asid:
(i) with carbonates / dengan karbonat (ii) with metals / dengan logam (iii) with bases / dengan bes
• BASICITY OF AN ACID / KEBESAN ASID
– To state the meaning of basicity of an acid and to write equations for the ionisation of monoprotic and diprotic acids.
Menyatakan maksud kebesan asid dan menulis persamaan pengionan asid monoprotik dan diprotik.
– To relate the basicity of acid/alkali with pH values / Mengaitkan kebesan asid /alkali dengan nilai pH.
• BASE / ALKALI / BES / ALKALI
– To state the meaning of base and to correlate base with alkali / Menyatakan maksud bes dan mengaitkan bes dengan alkali.
– To write chemical equations involving alkalis with acids and ammonium salts.
Menulis persamaan tindak balas kimia alkali dengan asid dan dengan garam ammonium.
ROLE OF WATER IN ACIDS AND ALKALI / PERANAN AIR DALAM ASID DAN ALKALI
– To explain why the acid and alkali properties are shown in the presence of water.
Menerangkan mengapa sifat asid dan alkali ditunjukkan dengan kehadiran air.
– To explain why the acid and alkali properties do not show in the absence of water or in non-water solvent.
Menerangkan mengapa sifat asid dan alkali tidak ditunjukkan tanpa kehadiran air atau dalam pelarut bukan air.
pH SCALE / SKALA pH
– To state the meaning of pH / Menyatakan maksud pH.
– To relate the pH value with the concentration of H+ ion for the acids and OH– ions for alkalis.
Mengaitkan nilai pH dengan kepekatan ion H+ bagi asid dan ion OH– bagi alkali.
• STRONG / WEAK ACID AND STRONG / WEAK ALKALI / ASID KUAT / LEMAH DAN ALKALI KUAT / LEMAH
– To list examples and equations for the ionisation of strong / weak acid and strong / weak alkali.
Menyenaraikan contoh dan menulis persamaan pengionan bagi asid kuat / lemah dan alkali kuat / lemah.
– To relate the pH value with the strength of acid / alkali / Mengaitkan nilai pH dengan kekuatan asid / alkali.
ACID AND ALKALI CONCENTRATION / KEPEKATAN ASID DAN ALKALI
– To state the meaning of concentration in g dm–3 and mol dm–3 / Menyatakan maksud kepekatan dalam unit g dm–3 dan mol dm–3.
– To state the meaning of standard solution and to describe the preparation of standard solution.
Menyatakan maksud larutan piawai dan menghuraikan eksperimen penyediaan larutan piawai. MV
1 000
– To solve various problems with calculations related to the preparation of standard solution using n = .
Menyelesaikan pelbagai masalah pengiraan berkaitan penyediaan larutan piawai menggunakan formula n = MV .
1 000
• NEUTRALISATION OF ACID AND ALKALI / TINDAK BALAS PENEUTRALAN ASID DAN ALKALI
– To describe the titration of acid with alkali and to calculate acid / alkali concentrations if a standard solution are given.
Menghuraikan titratan asid dengan alkali dan menghitung kepekatan asid / alkali jika satu larutan piawai diberikan.
– To describe the type of indicators used and the colour changes at the end-point.
Menyatakan jenis penunjuk yang digunakan dan perubahan warna penunjuk pada takat akhir.
– To solve numerical problems involving neutralisation / Menyelesaikan masalah pengiraan berkaitan peneutralan.
Nila m Publication Sdn.
114
Bhd.
Chemistry Form 4 • MODULE
ACID / ASID
1 Acid is a chemical substance which ionises in water to produce hydrogen ion.
Asid ialah bahan kimia yang mengion dalam air menghasilkan ion hidrogen.
2 Acid tastes sour and turns moist blue litmus to red.
Asid mempunyai rasa yang masam dan menukar kertas litmus biru lembap menjadi merah.
3 Example of acid is hydrochloric acid / Contoh asid ialah asid hidroklorik :
(a) Hydrogen chloride gas is a *covalent compound exist in the form of molecule.
Gas hidrogen klorida ialah *sebatian kovalen wujud dalam bentuk molekul.
(b) As hydrogen chloride dissolves in water, hydrogen chloride molecule ionises to hydrogen ion and chloride ion in
aqueous solution. This aqueous solution is called hydrochloric acid.
Apabila hidrogen klorida melarut dalam air, molekul hidrogen klorida mengion kepada ion hidrogen dan ion klorida dalam larutan
akueus. Larutan akueus itu dipanggil asid hidroklorik.
HCl (aq / ak ) H+ (aq / ak ) + Cl– (aq / ak )
Hydrochloric acid / Asid hidroklorik
Hydrogen ion / Ion hidrogen Chloride ion / Ion klorida
(c) An aqueous hydrogen ion, H+(aq) is actually the hydrogen ion combined with water molecule to form
hydroxonium ion, H3O+. However this ion can be written as H+.
Ion hidrogen akueus, H+(ak) ialah ion hidrogen yang bergabung dengan molekul air membentuk ion hidroksonium, H3O+. Walau
bagaimanapun, ion ini boleh ditulis sebagai H+.
HCl (g) + H 2O(l/ce) HIo3nO h+ y(adqro/axko)n ium + Cl– (aq/ak) The ionisation of hydrochloric acid is
Hydrogen chloride Ion klorida
Hidrogen klorida Ion hidroksonium Ion klorida represented as:
IHo3nO h+ y d r o x o n i u m H+(aq/ak) + H2O Pengionan asid hidroklorik diwakili oleh:
Ion hidrogen HCl (aq/ak) H+ (aq/ak) + Cl– (aq/ak)
Ion hidroksonium Ion hidrogen
4 Basicity of an acid is the number of ionisable of hydrogen atom per molecule of an acid molecule in an aqueous
solution / Kebesan asid ialah bilangan atom hidrogen yang boleh mengion bagi setiap molekul asid dalam larutan akueus.
– Monoprotic: One acid molecule ionises to one hydrogen ion.
Monoprotik: Satu molekul asid mengion kepada satu ion hidrogen.
– Diprotic: One acid molecule ionises to two hydrogen ion.
Diprotik: Satu molekul asid mengion kepada dua ion hidrogen.
– Triprotic: One acid molecule ionises to three hydrogen ion.
Triprotik: Satu molekul asid mengion kepada tiga ion hidrogen.
Hydrochloric is monoprotic acid because one molecule of hydrochloric acid ionises to one hydrogen ion.
Asid hidroklorik ialah sejenis asid monoprotik kerana satu molekul asid hidroklorik mengion kepada satu ion hidrogen.
5 Examples of acid and their basicity / Contoh-contoh asid dan kebesannya:
Ionisation of acid Number of hydrogens ion produce Basicity of acid
Pengionan asid per molecule of acid Kebesan asid
Monoprotic
HNO3 (aq/ak ) H+(aq) + NO3–(aq) Bilangan ion hidrogen dihasilkan bagi setiap Diprotic
Nitric acid Hydrogen ion Nitrate ion molekul asid Triprotic
Asid nitrik Ion hidrogen One
Ion nitrat
H2SO4 (aq/ak ) 2H+(aq) Two
Sulphuric acid Hydrogen ion + SO42–(aq)
Asid sulfurik Ion hidrogen Sulphate ion Three
H3PO4 (aq/ak ) 3H+(aq) Ion sulfat
Phosphoric acid Hydrogen ion
Asid fosforik Ion hidrogen + PO43–(aq)
Phosphate ion
Ion fosfat
*ECthHa3CnOoiOcH a c(iadq /ak ) CH3COO–(aq) + H+(aq) One Monoprotic
Ethanoate ion Hydrogen ion
Asid etanoik Ion etanoat Ion hidrogen
*Not all hydrogen atoms in ethanoic acid are ionisable / *Bukan semua ion hidrogen dalam asid etanoik boleh mengion
m Publication Sdn. Bhd.
115
Nila
MODULE • Chemistry Form 4
BASES / BES
1 Bases is a chemical substance that reacts with acid to produce salt and water only. For example,
Bes ialah sejenis bahan kimia yang bertindak balas dengan asid menghasilkan garam dan air sahaja. Contohnya,
(a) Copper(II) oxide (a base) reacts with sulphuric acid to produce copper(II) sulphate (a salt) and water.
Kuprum(II) oksida (bes) bertindak balas dengan asid sulfurik menghasilkan kuprum(II) sulfat (garam) dan air.
CuO + H2SO4 CuSO4 + H2O
(b) Zinc hydroxide (a base) reacts with hydrochloric acid to produce zinc chloride (a salt) and water.
Zink hidroksida (bes) bertindak balas dengan asid hidroklorik menghasilkan zink klorida (garam) dan air.
Zn(OH)2 + 2HCl ZnCl2 + H2O
2 Most bases are metal oxide or metal hydroxide which are ionic compound. Example of bases are magnesium oxide, zinc
oxide, sodium hydroxide and potassium hydroxide.
Kebanyakan bes ialah oksida logam atau hidroksida logam yang merupakan sebatian ion. Contoh-contoh bes ialah magnesium oksida, zink
oksida, natrium hidroksida dan kalium hidroksida.
3 The bases that can dissolve in water (soluble bases) are known as alkali.
Bes yang boleh melarut dalam air (bes larut) dikenali sebagai alkali.
4 Sodium hydroxide and potassium hydroxide are soluble in water and they are called alkali whereas magnesium oxide
and zinc oxide are called bases as they are insoluble in water.
Natrium hidroksida dan kalium hidroksida larut dalam air dan dipanggil sebagai alkali manakala magnesium oksida dan zink oksida
dipanggil sebagai bes kerana tidak terlarut dalam air.
5 Alkali is a base that is soluble in water and ionises to hydroxide ion. For example,
Alkali ialah bes yang larut dalam air dan mengion kepada ion hidroksida. Contohnya,
(a) Sodium hydroxide dissolves in water and ionises to hydroxide ion.
Natrium hidroksida terlarut dalam air dan mengion kepada ion hidroksida.
NaOH (aq/ak) Na+ (aq/ak) + OH– (aq/ak)
(b) Ammonia solution is obtained by dissolving ammonia molecule in water, ionisation occur to produce a hydroxide
ion, OH–.
Larutan ammonia diperoleh dengan melarutkan molekul ammonia dalam air, pengionan berlaku menghasilkan ion hidroksida, OH–.
NH3 (g) + H2O (l/ce) NH4+ (aq/ak) + OH– (aq/ak)
(c) Other examples of alkalis are barium hydroxide and calcium hydroxide.
Contoh alkali lain adalah barium hidroksida dan kalsium hidroksida.
6 Alkali tastes bitter, slippery and turns moist red litmus to blue.
Alkali mempunyai rasa yang pahit, licin dan menukarkan kertas litmus merah lembap kepada biru.
EXERCISE / LATIHAN
Complete the following table / Lengkapkan jadual berikut :
Soluble base (alkali) / Bes terlarut (alkali) Insoluble base / Bes tak terlarut
Name / Nama Formula / Formula Ionisation equation / Persamaan pengionan Name / Nama Formula / Formula
Sodium oxide Na2O NaNOaH2O(a(sq)) + H 2NOa + (a q2)N +aO OHH(–a (qa)q) Copper(II) oxide CuO
Natrium oksida Kuprum(II) oksida
Potassium oxide K2O KOKH2O(a(sq)) + H 2KO+ (aq) 2 +KO OHH(–a (qa)q) Copper(II) hydroxide Cu(OH)2
Kalium oksida Kuprum(II) hidroksida
Ammonia NH3 NH3(g)+ H2O NH4+(aq) + OH–(aq) Zinc hydroxide Zn(OH)2
Ammonia Zink hidroksida
Sodium hydroxide NaOH NaOH(aq) Na+ (aq) + OH– (aq) Aluminium oxide Al2O3
Natrium hidroksida Aluminium oksida
Potassium hydroxide KOH KOH(aq) K+ (aq) + OH– (aq) Lead(II) hydroxide Pb(OH)2
Kalium hidroksida Plumbum(II) hidroksida
Barium hydroxide Ba(OH)2 Ba(OH)2(aq) Ba2+(aq) + 2OH– (aq) Magnesium hydroxide Mg(OH)2
Barium hidroksida Magnesium hidroksida
Bases that can dissolve in water (soluble bases) are known as alkali / Bes yang larut dalam air (bes larut) dipanggil alkali
Nila m Publication Sdn.
116
Bhd.
Chemistry Form 4 • MODULE
CHEMICAL PROPERTIES OF ACID / SIFAT-SIFAT KIMIA ASID
1 Acid react with metal, base / alkali and metal carbonate / Asid bertindak balas dengan logam, bes/alkali dan karbonat logam:
Chemical properties Example of experiment Observation Remark
Sifat-sifat kimia Contoh eksperimen Pemerhatian Catatan
1 Acid + Metal Salt + Hydrogen Zinc + Hydrochloric acid – The grey solid Chemical equation:
Persamaan kimia:
Asid + Logam Garam + Hidrogen Zink + Asid hidroklorik dissolves.
Mg + 2HCl MgCl2 + H2
* Acid react with the metals that are Pepejal kelabu
Inference / Inferens :
more electropositive than hydrogen Lighted wooden terlarut. – Magnesium reacts with
in electrochemical series, acids do not splinter
react with copper and silver (type of Kayu uji menyala – Gas bubbles hydrochloric acid.
reaction is displacement, the metals are released. Magnesium bertindak balas
that are placed above hydrogen in Hydrochloric acid When a dengan asid hidroklorik.
Electrochemical Series can displace Asid hidroklorik burning – Hydrogen gas is
hydrogen from acid) wooden released.
Magnesium powder splinter is Gas hidrogen terbebas.
* Asid bertindak balas dengan logam-logam Serbuk magnesium placed at the
mouth of the
yang lebih elektropositif daripada hidrogen (a) About 5 cm3 of dilute
test tube,
dalam Siri Elektrokimia, asid tidak bertindak hydrochloric acid is poured
‘pop sound’ is
balas dengan kuprum dan argentum (jenis into a test tube.
tindak balas ialah penyesaran, logam-logam Sebanyak 5 cm3 asid hidroklorik produced.
di atas hidrogen dalam Siri Elektrokimia cair dimasukkan ke dalam tabung Gelembung gas
boleh menyesarkan hidrogen daripada asid) uji. dibebaskan.
* Application of the reaction: (b) One spatula of magnesium Apabila kayu
* Aplikasi tindak balas: powder is added to the acid. uji menyala
– Preparation of soluble salt (Topic Satu spatula serbuk magnesium didekatkan pada
Salt) ditambah kepada asid. mulut tabung
Penyediaan garam terlarut (Tajuk (c) A burning wooden splinter is uji, bunyi ‘pop’
placed at the mouth of the test dihasilkan.
Garam) tube.
– Preparation of hydrogen gas in
Kayu uji menyala diletakkan pada
determination of the empirical
formula of copper(II) oxide (Topic mulut tabung uji.
Chemical Formula and Equation) (d) The observations are recorded.
Penyediaan gas hidrogen dalam Semua pemerhatian direkodkan.
menentukan formula empirik kuprum(II)
oksida (Tajuk Formula dan Persamaan
Kimia)
2 Acid + Metal carbonate Salt + Calcium carbonate + Nitric acid – The white Chemical equation:
Water + Carbon dioxide Kalsium karbonat + Asid nitrik solid Persamaan kimia:
Asid + Karbonat logam Garam + Air + dissolves.
Karbon dioksida Hydrochloric acid Lime water Pepejal putih CaCO3 + 2HCl
Asid hidroklorik Air kapur
*Application of the reaction: terlarut. CaCl2 + H2O + CO2
*Aplikasi tindak balas: Calcium carbonate / Kalsium karbonat
– Preparation of soluble salt (Topic – Gas bubbles Inference / Inferens :
Salt) (a) About 5 cm3 of dilute are released. – Calcium carbonate
hydrochloric acid is poured When the
Penyediaan garam terlarut (Tajuk into a test tube. gas passed reacts with nitric acid.
Sebanyak 5 cm3 asid hidroklorik through lime Kalsium karbonat
Garam) water, the lime bertindak balas dengan
– Confirmatory test for anion cair dimasukkan ke dalam tabung water turns asid hidroklorik.
chalky. – Carbon dioxide gas
carbonate ion in qualitative uji. Gelembung gas is released.
analysis of salt (Topic Salt) (b) One spatula of calcium Gas karbon dioksida
terbebas. Apabila terbebas.
Ujian pengesahan bagi ion karbonat carbonate powder is added
dalam analisis kualitatif garam (Tajuk into the test tube. gas tersebut
Garam) Satu spatula serbuk kalsium
dilalukan melalui
karbonat dimasukkan ke dalam asid.
(c) The gas released is passed air kapur, air
through lime water as shown kapur menjadi
in the diagram.
Gas yang dibebaskan dilalukan keruh.
melalui air kapur seperti
ditunjukkan dalam rajah.
(d) The observations are recorded.
Semua pemerhatian direkodkan.
m Publicat ion Sdn. Bhd.
117
Nila
MODULE • Chemistry Form 4
3 Acid + Base / Alkali Salt + Water Copper(II) oxide + Sulphuric acid – The black Chemical equation:
Asid + Bes / Alkali Garam + Air Kuprum(II) oksida + Asid sulfurik solid Persamaan kimia:
dissolves.
*Acid neutralises base/alkali Sulphuric acid / Asid sulfurik Pepejal hitam CuO + H2SO4
* Asid meneutralkan bes/alkali
*Application of the reaction: Copper(II) oxide / Kuprum(II) oksida terlarut. CuSO4 + H2O
(a) Dilute hydrochloric acid is
*Aplikasi tindak balas: – The colourless Inference / Inferens :
– Preparation of soluble salt (Topic poured into a beaker until half solution turns – Copper(II) oxide reacts
full. blue.
Salt) Asid hidroklorik cair dimasukkan Larutan tanpa with sulphuric acid.
dalam bikar hingga separuh penuh. Kuprum(II) oksida
Penyediaan garam terlarut (Tajuk (b) The acid is warmed gently. warna bertukar bertindak balas dengan
Garam) Asid dihangatkan. asid sulfurik.
(c) One spatula of copper(II) oxide menjadi biru.
powders added to the acid. – The blue solution is
Satu spatula serbuk kuprum(II) copper(II) sulphate .
oksida ditambahkan kepda asid
tersebut. Larutan biru tersebut ialah
(d) The mixture is stirred with a kuprum(II) sulfat .
glass rod.
Campuran dikacau dengan rod kaca.
(e) The observations are recorded.
Semua pemerhatian direkodkan.
2 Write the chemical formulae for the following compounds / Tuliskan formula kimia bagi sebatian berikut:
Compound / Sebatian Chemical formulae / Formula kimia Compound / Sebatian Chemical formulae / Formula kimia
Hydrochloric acid HCl Magnesium oxide MgO
Asid hidroklorik Magnesium oksida
Nitric acid HNO3 Calcium oxide CaO
Asid nitrik Kalsium oksida
Sulphuric acid H2SO4 Copper(II) oxide CuO
Asid sulfurik Kuprum(II) oksida
Ethanoic acid CH3COOH Lead(II) oxide PbO
Asid etanoik Plumbum(II) oksida
Sodium hydroxide NaOH Sodium nitrate NaNO3
Natrium hidroksida Natrium nitrat
Potassium hydroxide KOH Potassium sulphate K2SO4
Kalium hidroksida Kalium sulfat
Calcium hydroxide Ca(OH)2 Barium hydroxide Ba(OH)2
Kalsium hidroksida Barium hidroksida
Sodium carbonate Na2CO3 Sodium chloride NaCl
Natrium karbonat Natrium klorida
Magnesium hydroxide Mg(OH)2 Magnesium Mg
Magnesium hidroksida Magnesium
Ammonium sulphate (NH4 )2SO4 Zinc Zn
Ammonium sulfat Zink
Hydroxide ion OH– Sodium Na
Ion hidroksida Natrium
Sodium sulphate Na2SO4 Calcium carbonate CaCO3
Natrium sulfat Kalsium karbonat
Carbon dioxide CO2 Hydrogen gas H2
Karbon dioksida Gas hidrogen
Copper(II) carbonate CuCO3 Sodium oxide Na2O
Kuprum(II) karbonat Natrium oksida
Water H2O Magnesium nitrate Mg(NO3 )2
Air Magnesium nitrat
Nila m Publication Sdn.
118
Bhd.
Chemistry Form 4 • MODULE
3 Ionic equation / Persamaan ion :
Ionic equation shows particles that change during chemical reaction.
Persamaan ion menunjukkan zarah yang berubah semasa tindak balas kimia.
Example / Contoh :
(i) Reaction between sulphuric acid and sodium hydroxide solution:
Tindak balas antara asid sulfurik dengan larutan natrium hidroksida:
Write balanced equation / Tulis persamaan seimbang :
H2SO4 + 2NaOH Na2SO4 + 2H2O
Write the formula of all the particles in the reactants and products:
Tulis formula bagi semua zarah dalam bahan tindak balas dan hasil tindak balas:
2H+ + SO42– + 2Na+ + 2OH– 2Na+ + SO42– + 2H2O
Remove all the particles in the reactants and products which remain unchanged:
Keluarkan semua zarah dalam bahan dan hasil tindak balas yang tidak berubah:
2H+ + SO42– + 2Na+ + 2OH– 2Na+ + SO42– + 2H2O
Ionic equation / Persamaan ion :
2H+ + 2OH– 2H2O ⇒ H+ + OH– H2O
(ii) Reaction between zinc oxide and hydrochloric acid / Tindak balas antara zink dengan asid hidroklorik :
Write balanced equation / Tulis persamaan seimbang :
2HCl + Zn ZnCl2 + H2
Write the formula of all the particles in the reactants and products:
Tulis formula bagi semua zarah dalam bahan tindak balas dan hasil tindak balas:
2H+ + 2Cl– + Zn Zn2+ + 2Cl– + H2
Remove all the particles in the reactants and products which remain unchanged:
Keluarkan semua zarah dalam bahan dan hasil tindak balas yang tidak berubah:
2H+ + 2Cl– + Zn Zn2+ + 2Cl– + H2
Ionic equation / Persamaan ion :
2H+ + Zn Zn2+ + H2
4 Write the chemical equations and ionic equation for the following reactions:
Tulis persamaan kimia dan persamaan ion untuk tindak balas berikut:
Reactant / Bahan tindak balas Chemical equations / Persamaan kimia Ionic equation / Persamaan ion
Hydrochloric acid and #magnesium oxide MgO + 2HCl MgCl2 + H2O 2H+ + MgO Mg2+ + H2O
Asid hidroklorik dan #magnesium oksida
Hydrochloric acid and sodium hydroxide HCl + NaOH NaCl + H2O H+ + OH– H2O
Asid hidroklorik dan natrium hidroksida
Hydrochloric acid and magnesium 2HCl + Mg MgCl2 + H2 2H+ + Mg Mg2+ + H2
Asid hidroklorik dan magnesium
Hydrochloric acid and #calcium carbonate 2HCl + CaCO3 CaCl2 + CO2 + H2O 2H+ + CaCO3 Ca2+ + CO2 + H2O
Asid hidroklorik dan #kalsium karbonat
Sulphuric acid and zinc H2SO4 + Zn ZnSO4 + H2 2H+ + Zn Zn2+ + H2
Asid sulfurik dan zink
Sulphuric acid and #zinc oxide H2SO4 + ZnO ZnSO4 +H2O 2H+ + ZnO Zn2+ + H2O
Asid sulfurik dan #zink oksida
Sulphuric acid and #zinc carbonate H2SO4 +ZnCO3 ZnSO4 + CO2 + H2O 2H+ + ZnCO3 Zn2+ + CO2 + H2O
Asid sulfurik dan #zink karbonat
Nitric acid and #copper(II) oxide 2HNO3 + CuO Cu(NO3)2 + H2O 2H+ + CuO Cu2+ + H2O
Asid nitrik dan #kuprum(II) oksida
Nitric acid and sodium hydroxide HNO3 + NaOH NaNO3 + H2O H+ + OH– H2O
Asid nitrik dan natrium hidroksida
# Ions in magnesium oxide, calcium carbonate, zinc oxide, zinc carbonate and copper(II) oxide cannot be separated because
the compounds are insoluble in water and the ions do not ionise.
# Ion dalam magnesium oksida, kasium karbonat, zink oksida, zink karbonat dan kuprum(II) oksida tidak boleh diasingkan kerana sebatian tersebut
tidak larut dalam air dan ion-ionnya tidak mengion.
m Publication Sdn. B hd.
119
Nila
MODULE • Chemistry Form 4
CHEMICAL PROPERTIES OF ALKALIS / SIFAT KIMIA ALKALI
Chemical properties Write the balance chemical equation for the reaction
Sifat-sifat kimia Tuliskan persamaan kimia seimbang bagi tindak balas
1 Alkali + Acid Salt + Water (a) Potassium hydroxide and sulphuric acid
Alkali + Asid Garam + Air Kalium hidroksida dan asid sulfurik :
*Alkali neutralises acid / Alkali meneutralkan asid. H2SO4 + 2KOH K2SO4 + 2H2O
*Application of the reaction / Aplikasi tindak balas :
(b) Barium hydroxide and hydrochloric acid:
– Preparation of soluble salt (Topic Salt) Barium hidroksida dan asid hidroklorik:
Penyediaan garam terlarut (Tajuk Garam)
2HCl + Ba(OH)2 BaCl2 + H2O
2 Alkali + Ammonium salt Salt + Water + Ammonia gas (c) Ammonium chloride and potassium hydroxide:
Alkali + Garam Ammonium Garam + Air + Gas ammonia Ammonium klorida dan kalium hidroksida:
*Ammonia gas is released when alkali is heated with KOH + NH4Cl KCl + H2O + NH3
ammonium salt. Ammonia gas has pungent smell and turn
moist red litmus paper to blue. (d) Ammonium sulphate and sodium hydroxide:
*Gas ammonia dibebaskan apabila alkali dipanaskan dengan garam Ammonium sulfat dan natrium hidroksida:
ammonium. Gas ammonia mempunyai bau yang sengit dan menukar
kertas litmus merah lembap kepada biru. 2NaOH + (NH4)2SO4 Na2SO4 + 2H2O + 2NH3
*Application of the reaction / Aplikasi tindak balas : (e) 2OH–(aq/ak) + Mg2+(aq/ak) Mg(OH)2(p)
– Confirmatory test for cations ammonium in qualitative Magnesium hydroxide
analysis of salt (Topic Salt) (white precipitate)
Ujian pengesahan kation ammonium dalam analisis kualitatif garam Magnesium hidroksida
(Tajuk Garam)
(mendakan putih)
3 Alkali + Metal ion Insoluble metal hydroxide (f) 2OH–(aq/ak) + Cu2+(aq/ak)
Alkali + Ion logam Logam hidroksida tak larut Cu(OH)2(p)
Copper(II) hydroxide
*Most of the metal hydroxides are insoluble. (blue precipitate)
*Kebanyakan logam hidroksida tak terlarut. Kuprum(II) hidroksida
*Hydroxides of transition element metals are coloured.
(mendakan biru)
*Hidroksida bagi logam peralihan adalah berwarna.
*Application of the reaction / Aplikasi tindak balas :
– Confirmatory test for cations in qualitative analysis of salt
(Topic Salt)
Ujian pengesahan bagi kation dalam analisis kualitatif garam (Tajuk
Garam)
Nila tion Sdn.ROLE OF WATER AND THE PROPERTIES OF ACID / PERANAN AIR DAN SIFAT ASID
1 An acid shows its acidic properties when it is dissolved in water.
Asid menunjukkan sifat keasidannya apabila terlarut dalam air.
2 Acid molecules ionise in aqueous solution to form hydrogen ions. The presence of hydrogen ions is needed for the acid
to show its acidic properties.
Molekul asid mengion dalam larutan akueus membentuk ion hidrogen. Kehadiran ion hidrogen diperlukan oleh asid untuk menunjukkan
sifat keasidannya.
3 Acid will remain in the form of molecules in two conditions / Asid akan kekal dalam bentuk molekul dengan dua keadaan:
(a) Without the presence of water for example dry hydrogen chloride gas and *glacial ethanoic acid.
Tanpa kehadiran air seperti gas hidrogen klorida kering dan *asid etanoik glasial
(b) Acid is dissolved in *organic solvent for example solution of hydrogen chloride in methylbenzene and ethanoic
acid in propanone.
Asid dilarutkan dalam *pelarut organik seperti larutan hidrogen klorida dalam metilbenzena dan asid etanoik dalam propanon.
* Glacial ethanoic acid is pure ethanoic acid / Asid etanoik glasial ialah asid etanoik tulen.
* Organic solvent is covalent compound that exist as liquid at room temperature such as propanone,
methylbenzene and trichloromethane.
*Pelarut organik ialah sebatian kovalen yang wujud dalam bentuk cecair pada suhu bilik seperti propanon, metilbenzena dan
triklorometana.
m Publica
120
Bhd.
Chemistry Form 4 • MODULE
4 Example / Contoh : Solution of hydrogen chloride in Solution of hydrogen chloride in
Glacial ethanoic acid methylbenzene water (hydrochloric acid)
Asid etanoik glasial Larutan hidrogen klorida dalam metilbenzena Larutan hidrogen klorida dalam air
(asid hidroklorik)
CH3COOH HCl HCl HCl HCl H+ Cl- Cl- Cl-
CH3COOH CH3COOH HCl HCl H+ H+
CH3COOH
CH3COOH CH3COOH HCl HCl HCl H+ Cl- H+
HCl Cl-
CH3COOH
Methylbenzene / Metilbenzena Water / Air
• Glacial ethanoic acid molecules do not • Hydrogen chloride molecules in • Hydrogen chloride molecule in water
ionise . methylbenzene do not ionise . ionises :
Molekul asid etanoik glasial tidak mengion . Molekul hidrogen klorida dalam metilbenzena Molekul hidrogen klorida dalam air
• Glacial ethanoic exist as molecule tidak mengion . mengion :
only, no hydrogen ions present.
• Hydrogen chloride exist as molecule HCl (aq/ak) H+ (aq/ak) + Cl– (aq/ak)
Etanoik glasial hanya terdiri daripada only, there are no hydrogen ions • Hydrogen ions and chloride ions
molekul CH3COOH sahaja, tiada ion present. present.
Hidrogen klorida wujud sebagai molekul
hidrogen hadir. Ion hidrogen dan ion klorida hadir.
sahaja, tiada ion hidrogen hadir.
• Hydrogen chloride in water
• Glacial ethanoic acid and hydrogen chloride in methylbenzene does not show acidic (hydrochloric acid) shows acidic
properties: properties:
Asid etanoik glasial dan hidrogen klorida dalam metilbenzena tidak menunjukkan sifat asid: Hidrogen klorida dalam air (asid hidroklorik)
(i) They do not react with metal, base or metal carbonate. menunjukkan sifat asid:
Sebatian tersebut tidak bertindak balas dengan logam, bes dan karbonat logam. (i) Hydrochloric acid react with
(ii) They do not turn blue litmus paper to red . metal, base or metal carbonate.
Sebatian tersebut tidak menukarkan warna kertas litmus biru kepada merah . Asid hidroklorik bertindak balas
• There are no free moving ions, hydrogen chloride in methylbenzene and glacial dengan logam, bes dan karbonat logam.
ethanoic acid cannot conduct electricity (non-electrolyte).
Tidak wujud ion bebas bergerak , asid etanoik glasial dan hidrogen klorida dalam metilbenzena (ii) Hydrogen ions turn blue
litmus paper to red .
tidak dapat mengkonduksikan elektrik (bukan elektrolit). Ion hidrogen menukarkan warna kertas
litmus biru kepada merah .
• There are free moving ions,
hydrochloric acid can conduct
electricity (electrolyte).
Terdapat ion yang bebas bergerak , asid
hidroklorik dapat mengkonduksikan
elektrik (elektrolit).
ROLE OF WATER AND THE PROPERTIES OF ALKALI / PERANAN AIR DAN SIFAT ALKALI
1 In the presence of water, an alkali dissolves and ionises to produce hydroxide ions. For example potassium hydroxide
solution and ammonia solution.
Dengan kehadiran air, alkali melarut dan mengion menghasilkan ion hidroksida. Contohnya larutan kalium hidroksida dan larutan
ammonia:
KOH(aq/ak) K+(aq/ak) + OH–(aq/ak)
NH3(g) + H2O(l/ce) NH4+(aq/ak) + OH–(aq/ak)
2 Without water or in organic solvents, no hydroxide ions are produced, so the alkaline properties are not shown.
Tanpa air atau dalam pelarut organik, tiada ion hidroksida yang dihasilkan, maka sifat-sifat alkali tidak ditunjukkan.
m Publication Sdn. Bhd.
121
Nila
MODULE • Chemistry Form 4
EXERCISE / LATIHAN
1 The diagram below shows the apparatus set-up to investigate the role of water and other solvent in showing the
properties of acid and the observations made from the investigation.
Rajah di bawah menunjukkan susunan radas untuk mengkaji peranan air atau pelarut lain dalam menunjukkan sifat asid serta pemerhatian
yang dibuat.
Experiment / Eksperimen I II
Set-up of apparatus
Susunan radas
Hydrochloric acid in water Hydrochloric acid in
Asid hidroklorik dalam air tetrachloromethane
Magnesium ribbon Asid hidroklorik dalam tetraklorometana
Pita magnesium Magnesium ribbon
Pita magnesium
Observation • Bubbles of gas are released • No bubble of gas
Pemerhatian Gelembung gas dibebaskan Tiada gelembung gas
• Magnesium ribbon dissolves
Pita magnesium larut
(a) What is meant by acid / Apakah yang dimaksudkan dengan asid ?
Acid is a chemical substance which ionises in water to produce hydrogen ion.
(b) (i) Name the bubble of gas released in Experiment I / Namakan gas yang terbebas dalam Eksperimen I.
Hydrogen gas
(ii) Write the chemical equation for the formation of the bubbles in Experiment I.
Tulis persamaan kimia untuk pembentukan gelembung gas dalam Eksperimen I.
Mg + 2HCl MgCl2 + H2
(iii) Write the ionic equation for the chemical equation in (b)(ii).
Tulis persamaan ion untuk persamaan kimia dalam (b)(ii).
Mg + 2H+ Mg2+ + H2
(c) Compare observation in Experiment I and Experiment II. Explain your answer.
Bandingkan pemerhatian dalam Eksperimen I dan Eksperimen II. Terangkan jawapan anda.
– Hydrochloric acid in water in Experiment I reacts with magnesium.
Asid hidroklorik dalam air dalam Eksperimen I bertindak balas dengan magnesium.
– Hydrochloric acid in tetrachloromethane in Experiment I do not react with magnesium.
Asid hidroklorik dalam tetraklorometana dalam Eksperimen II tidak bertindak balas dengan magnesium.
– Hydrochloric acid in water ionises to H+ / Asid hidroklorik dalam air mengion kepada ion H+:
HCl H+ + Cl–
– H+ ions react with magnesium atom to produce hydrogen molecule:
Ion H+ bertindak balas dengan atom magnesium untuk menghasilkan molekul hidrogen:
Mg + 2H+ Mg2+ + H2
– Hydrochloric acid in tetrachloromethane remains in the form of molecule . No hydrogen ion present.
Asid hidroklorik dalam tetraklorometana kekal dalam bentuk molekul . Tiada ion hidrogen hadir.
Nila m Publication Sdn.
122
Bhd.
Chemistry Form 4 • MODULE
2 The diagram below shows the set-up of apparatus to prepare two solutions of ammonia in solvent X and solvent Y. A
piece of red litmus paper is dropped into each beaker.
Gambar rajah di bawah menunjukkan susunan radas bagi menyediakan dua larutan ammonia dalam pelarut X dan pelarut Y. Sekeping
kertas litmus merah dimasukkan ke dalam setiap bikar.
Ammonia Ammonia
Ammonia Ammonia
Solvent X Solvent Y
Pelarut X Pelarut Y
Beaker A / Bikar A Beaker B / Bikar B
The table below shows the observation on the red litmus paper in solvent X and solvent Y.
Jadual di bawah menunjukkan pemerhatian ke atas kertas litmus merah dalam pelarut X dan pelarut Y.
Solution / Larutan Observation / Pemerhatian
Ammonia in solvent X in beaker A The red litmus paper turns blue.
Ammonia dalam pelarut X dalam bikar A Kertas litmus merah bertukar menjadi biru.
Ammonia in solvent Y in beaker B No visible change in the colour of red litmus paper.
Ammonia dalam pelarut Y dalam bikar B Tiada perubahan yang nyata pada warna kertas litmus merah.
(a) Name possible substances that can be solvent X and solvent Y.
Namakan bahan-bahan yang mungkin bagi pelarut X dan pelarut Y.
Solvent X / Pelarut X : Water
Solvent Y / Pelarut Y : Propanone / methylbenzene / trichloromethane
(b) Explain the difference in the observation on the beakers A and B.
Terangkan perbezaan antara pemerhatian dalam bikar A dengan bikar B.
– Ammonia gas in beaker A is dissolved in water, ammonia molecules ionise to ammonium ion and
hydroxide ions:
Gas ammonia dalam bikar A larut dalam air, molekul ammonia mengion kepada ion ammonium dan
ion hidroksida : NH3 (g) + H2O (l/ce) NH 4+ (ak) + OH– (ak)
– The presence of hydroxide ions change the red litmus paper to blue.
Kehadiran ion-ion hidroksida menukar kertas litmus merah kepada biru.
– Ammonia gas in beaker B is dissolved in propanone / methylbenzene / trichloromethane , ammonia
molecules do not ionise .
Gas ammonia dalam bikar B larut dalam propanon / metilbenzena / triklorometana , molekul ammonia tidak
mengion .
– No hydroxide ions present, the red litmus paper remains unchanged.
Tiada ion hidroksida , warna merah kertas litmus tidak berubah.
(c) (i) Between solution in beakers A and B, which one is an electrolyte and non-electrolyte? Explain your answer.
Antara larutan dalam bikar A dangan bikar B, yang manakah elektrolit dan bukan elektrolit? Terangkan jawapan anda.
– Solution in beaker A is an electrolyte , it contains free moving ions from the
ionisation of ammonia molecules in water.
Larutan dalam bikar A ialah elektrolit , ia mengandungi ion-ion yang bebas bergerak daripada
pengionan molekul ammonia dalam air.
– Solution in beaker B is a non-electrolyte , ammonia molecules do not ionise in
propanone / methylbenzene / trichloromethane .
Larutan dalam bikar B bukan elektrolit , molekul ammonia tidak mengion dalam
propanon / metilbenzena / triklorometana .
m Publication Sdn. Bhd.
123
Nila
MODULE • Chemistry Form 4
(ii) Draw a labelled diagram to show the set-up of apparatus used to show the electrical conductivity of an
electrolyte.
Lukiskan gambar rajah berlabel yang menunjukkan susunan radas yang digunakan untuk menunjukkan kekonduksian arus
elektrik bagi sesuatu elektrolit.
CEalerbkotrno delkeacrtrboodne EClaerkbtroond eklaecrtbroonde
EElekctroliytte
THE pH SCALE / SKALA pH
1 The pH is a scale of numbers to measure the degree of acidity and alkalinity of an aqueous solution based on the
concentration of hydrogen ions, H+ or hydroxide ions, OH–.
Skala pH ialah skala bernombor untuk mengukur darjah keasidan dan kealkalian suatu larutan akueus berdasarkan kepekatan ion
hidrogen, H+ atau ion hidroksida, OH–.
2 The pH scale has the range of number from 0 to 14 / Skala pH bernombor dari 0 hingga 14 :
pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
pH < 7: pH = 7 pH > 7:
• Acidic solution / Larutan berasid. Neutral • Alkaline solution / Larutan beralkali.
• The lower the pH value, the higher is the Neutral • The higher the pH value, the higher is the
concentration of hydrogen ion, H+. concentration of hydroxide ion, OH–.
Semakin rendah nilai pH, semakin tinggi Semakin tinggi nilai pH, semakin tinggi
kepekatan ion hidrogen, H+. kepekatan ion hidroksida, OH–.
3 The pH of an aqueous solution can be measured by / Nilai pH bagi sesuatu larutan akueus boleh diukur dengan menggunakan:
(a) pH meter / Meter pH
(b) Acid-base indicator / Penunjuk asid-bes
Complete the following table / Lengkapkan jadual berikut :
Indicator Acid / Asid Colour / Warna Alkali / Alkali
Penunjuk Neutral / Neutral
Litmus solution / Larutan litmus Red Purple Blue
Methyl orange / Metil jingga Red Orange Yellow
Phenolphthalein / Fenolftalein Colourless Colourless Pink
Universal indicator / Penunjuk universal Red Green Purple
Nila tion Sdn.THE STRENGTH OF ACID AND ALKALI / KEKUATAN ASID DAN ALKALI
1 The strength of acid and alkali depend on the degree of ionisation or dissociation of the acid and alkali in water.
Kekuatan asid dan alkali bergantung pada darjah pengionan asid dan alkali dalam air.
(a) A strong acid is an acid that ionises completely in water to produce high concentration of hydrogen ion, H+.
Asid kuat ialah asid yang mengion sepenuhnya dalam air menghasilkan kepekatan ion hidrogen, H+ yang tinggi.
(b) A weak acid is an acid that partially ionises in water to produce low concentration of hydrogen ion, H+.
Asid lemah ialah asid yang mengion separa dalam air menghasilkan kepekatan ion hidrogen, H+ yang rendah.
(c) A strong alkali is an alkali that ionises completely in water to produce high concentration of hydroxide ion,
OH–.
Alkali kuat ialah alkali yang mengion sepenuhnya dalam air menghasilkan kepekatan ion hidroksida, OH– yang tinggi.
(d) A weak alkali is an alkali that partially ionises completely in water to produce low concentration of hydroxide
ion, OH–.
Alkali lemah ialah alkali yang mengion separa dalam air menghasilkan kepekatan ion hidroksida, OH– yang rendah.
m Publica
124
Bhd.
Chemistry Form 4 • MODULE
2 Example of different strength of acid and alkali / Contoh asid dan alkali dengan kekuatan yang berbeza.
Acid / Alkali Example Ionisation equation Explanation Particles
Asid / Alkali Contoh Persamaan ion Penerangan present
Zarah-zarah
yang hadir
Strong acid Hydrochloric HCl (aq/ak) All hydrogen chloride molecules that H+ and Cl–
Asid kuat
acid, HCl H+ (aq/ak ) + Cl– (aq/ak ) dissolve in water ionises completely into H+ dan Cl–
Asid
hydrogen ions and chloride ions.
hidroklorik,
Semua molekul hidrogen klorida melarut dalam
HCl
air dan mengion sepenuhnya kepada ion
hidrogen dan ion klorida .
Nitric acid, HNO3(aq/ak ) All nitric acid ionises completely in H+ and NO3–
HNO3 H+ (aq/ak ) + NO3– (aq/ak ) water into hydrogen ions and nitrate H+ dan NO3–
ions.
Asid nitrik, Semua asid nitrik mengion sepenuhnya dalam
HNO3
air kepada ion hidrogen dan ion nitrat .
Sulphuric H2SO4 (aq/ak ) All sulphuric acid ionises completely H+ and SO42–
acid, H2SO4 2H+ (aq/ak ) + SO42– (aq/ak ) into hydrogen ions and sulphate ions. H+ dan SO42–
Semua asid sulfurik mengion sepenuhnya dalam
Asid sulfurik,
H2SO4 air kepada ion hidrogen dan ion sulfat .
Weak acid Ethanoic CH3COOH (aq/ak ) Ethanoic acid partially ionises in water CCHHH+33CCOOOO–H a, nd
Asid lemah acid, CH3COO– (aq/ak ) + H+ (aq/ak ) into etahnoate ions and hydrogen CH3COOH,
CH3COOH ions. Some remain in the form of CH3COOH CH3COO– dan H+
Strong alkali Asid etanoik, H2CO3 (aq/ak )
Alkali kuat CH3COOH 2H+ (aq/ak ) + CO32– (aq/ak ) molecules . H2CO3, H+ and
Asid etanoik mengion separa kepada ion CO32–
Carbonic NaOH (aq/ak ) H2CO3, H+ dan
acid, H2CO3 Na+(aq) + OH– (aq) etanoat dan ion hidrogen . Sebahagian lagi CO32–
Asid karbonik, kekal dalam bentuk molekul CH3COOH.
H2CO3 Na+ and OH–
Carbonic acid partially ionises in water Na+ dan OH–
Sodium into carbonate ions and hydrogen ion. Some
hydroxide, remain in the form of H2CO3 molecules .
NaOH Sebahagian asid karbonik mengion dalam air
Natrium kepada ion karbonat dan ion hidrogen. Sebahagian
hidroksida, lagi kekal dalam bentuk molekul H2CO3.
NaOH
Sodium hydroxide ionises completely in
water into sodium ions and hydroxide
ions.
Natrium hidroksida mengion sepenuhnya dalam
air kepada ion natrium dan ion hidroksida .
Potassium KOH (aq/ak ) Potassium hydroxide ionises completely K+ and OH–
hydroxide, K+ (aq) + OH– (aq) in water into potassium ions and K+ dan OH–
KOH
Kalium Ba(OH)2 (aq/ak ) hydroxide ions. Ba2+ and OH–
hidroksida, Ba2+ (aq) + 2OH– (aq) Kalium hidroksida mengion sepenuhnya dalam air Ba2+ dan OH–
KOH kepada ion kalium dan ion hidroksida .
NH3 (g)+ H2 O(l/ce)
Barium NH4+(aq/ak ) + OH–(aq/ak ) Barium hydroxide ionises completely in
hydroxide, water into barium ions and hydroxide
Ba(OH)2 ions.
Barium Barium hidroksida mengion sepenuhnya dalam
hidroksida, air kepada ion barium dan ion hidroksida .
Ba(OH)2
Weak alkali Ammonia partially ionises in water into NH3, NH4+ and
Alkali lemah Ammonia ammonium ions and hydroxide ions, OH–
solution,
NH3(aq) some remain in the form of NH3 molecules . NH3, NH4+ dan
Larutan Ammonia mengion separa dalam air kepada OH–
ammonia,
NH3(ak) ion ammonium dan ion hidroksida ,
sebahagian lagi kekal dalam bentuk molekul NH3.
m Publication Sdn. Bhd.
125
Nila
MODULE • Chemistry Form 4
CONCENTRATION OF ACID AND ALKALI / KEPEKATAN ASID DAN ALKALI
1 A solution is a homogeneous mixture formed when a solute is dissolved in a solvent. For example copper(II) sulphate
solution is prepared by dissolving copper(II) sulphate powder (solute) in water (solvent).
Larutan adalah campuran homogen yang terbentuk apabila bahan larut dilarutkan dalam pelarut. Contohnya larutan kuprum(II) sulfat
disediakan dengan melarutkan serbuk kuprum(II) sulfat (bahan larut) di dalam air (pelarut).
2 Concentration of a solution the quantity of solute in a given volume of solution which is usually 1 dm3 of solution.
Kepekatan sesuatu larutan ialah kuantiti bahan terlarut dalam isi padu larutan yang tertentu, biasanya isi padu 1 dm3 larutan.
3 Concentration can be expressed in two ways / Kepekatan boleh diwakili dengan dua cara :
(a) Mass of solute in gram per 1 dm3 solution, g dm–3/ Jisim bahan larut dalam gram bagi setiap 1 dm3 larutan, g dm–3.
Concentration of solution (g dm–3) = Mass of solute in gram (g) / Jisim bahan larut dalam gram (g)
Kepekatan larutan (g dm–3) Volume of solution (dm3) / Isi padu larutan (dm3)
(b) Number of moles of solute in 1 dm3 solution, mol dm–3 / Bilangan mol bahan larut dalam 1 dm3 larutan, mol dm–3.
Concentration of solution (mol dm–3) = Number of mole of solute (mol) / Bilangan mol bahan larut (mol)
Kepekatan larutan (mol dm–3) Volume of solution (dm3) / Isi padu larutan (dm3)
4 The concentration in mol dm–3 is called molarity or molar concentration. The unit mol dm–3 can be represented by ‘M’.
Kepekatan dalam mol dm–3 dipanggil sebagai kemolaran atau kepekatan molar. Unit mol dm–3 boleh diwakili dengan‘M’.
Molarity = Number of mole of solute (mol) / Bilangan mol bahan larut (mol) n = Number of moles of solute
Volume of solution (dm3) / Isi padu larutan (dm3) Bilangan mol bahan terlarut
Kemolaran
M = Concentration in mol dm–3
Number of mole of solute (mol) = Molarity × Volume (dm3) (molarity)
Bilangan mol bahan larut (mol) Kemolaran × Isi padu (dm3) Kepekatan dalam mol dm–3
n = MV (kemolaran)
V = Volume of solution in dm3
n = Mv
1 000 Isi padu larutan dalam dm3
v = Volume of solution in cm3
Isi padu larutan dalam cm3
5 The concentration of a solution can be converted from mol dm–3 to g dm–3 and vice versa.
Kepekatan larutan boleh ditukar daripada mol dm–3 kepada g dm–3 dan sebaliknya.
× molar mass of the solute / jisim molar bahan terlarut
mol dm–3 g dm–3
÷ molar mass of the solute / jisim molar bahan terlarut
6 The pH value of an acid or an alkali depends on the concentration of hydrogen ions or hydroxide ions:
Nilai pH bagi asid atau alkali bergantung pada kepekatan ion hidrogen atau ion hidroksida:
The higher the concentration of hydrogen ions in acidic solution, the lower the pH value.
Semakin tinggi kepekatan ion hidrogen dalam larutan berasid, semakin rendah nilai pH.
The higher the concentration of hydroxide ions in alkaline solution, the higher the pH value.
Semakin tinggi kepekatan ion hidroksida dalam larutan beralkali, semakin tinggi nilai pH.
7 The pH value of an acid or an alkali is depends on / Nilai pH bagi asid atau alkali bergantung pada:
(a) The strength of acid or alkali / Kekuatan asid atau alkali
– the degree of ionisation or dissociation of the acid and alkali in water / darjah pengionan asid atau alkali dalam air.
(b) Molarity of acid or alkali / Kemolaran asid atau alkali
– the concentration of acid or alkali in mol dm–3 / kepekatan bahan terlarut dalam mol dm–3.
(c) Basicity of an acid / Kebesan asid
– the number ionisable of hydrogen atom per molecule of an acid molecule in an aqueous solution.
bilangan atom hidrogen per molekul asid yang terion dalam larutan akueus.
Nila m Publication Sdn.
126
Bhd.
Example / Contoh:
The diagram below shows the reading of pH meter for different types and concentration of acids. The aim of the experiment is to investigate the relationship between
concentration of hydrogen ions with the pH value. Compare the concentration of hydrogen ions and the pH value of the following acids. Explain your answer.
Rajah di bawah menunjukkan bacaan pH meter untuk pelbagai jenis dan kepekatan asid. Tujuan eksperimen adalah untuk mengkaji hubungan antara kepekatan ion hidrogen dengan nilai pH.
Bandingkan kepekatan ion hidrogen dan nilai pH untuk asid-asid yang berikut. Terangkan jawapan anda.
Experiment I II III
Eksperimen
pH meter 1.21 2.98 1.15 2.25 1.21 3.45
reading
Bacaan pH
meter
0.1 mol dm–3 HCl 0.01 mol dm–3 HCl 0.05 mol dm–3 H2SO4 0.05 mol dm–3 HCl 0.1 mol dm–3 HCl 0.1 mol dm–3 CH3COOH
Compare – Hydrochloric acid is a strong acid ionises – Sulphuric acid is a strong diprotic acid. – Hydrochloric acid is a strong acid ionises completely
concentration completely in water to hydrogen ion. Asid sulfurik adalah asid kuat diprotik . in water to hydrogen ion.
of H+ and pH Asid hidroklorik adalah asid kuat yang Asid hidroklorik adalah asid kuat yang mengion
value – 0.05 mol dm-3 of sulphuric acid ionises air kepada ion hidrogen. lengkap dalam
Bandingkan mengion lengkap dalam air kepada ion hidrogen. completely in water to form 0.1 mol dm–3
hydrogen ion: – 0.1 mol dm–3 of hydrochloric acid ionises to form
kepekatan ion – 0.1 mol dm–3 of hydrochloric acid ionises to Asid sulfurik 0.05 mol dm–3 mengion lengkap kepada 0.1 mol dm–3 hydrogen ion:
form 0.1 mol dm–3 hydrogen ion: Asid hidroklorik 0.1 mol dm–3 mengion lengkap kepada 0.1 mol dm–3
hidrogen dan Asid hidroklorik 0.1 mol dm–3 mengion kepada 0.1 mol dm–3 ion hidrogen:
0.1 mol dm–3 ion hidrogen: ion hidrogen:
nilai pH
HCl H+ + Cl–
0.1 mol dm–3 0.1 mol dm–3 H2SO4 2H+ + SO42– HCl H+ + Cl–
0.05 mol dm–3 0.1 mol dm–3 0.1 mol dm–3 0.1 mol dm–3
– 0.01 mol dm–3 of hydrochloric acid ionises to
form 0.01 mol dm–3 hydrogen ion: – Hydrochloric acid is a strong monoprotic – Ethanoic acid is a weak acid ionises partially in
Asid hidroklorik 0.01 mol dm–3 mengion kepada acid. water to produce lower concentration hydrogen ion.
0.01 mol dm–3 ion hidrogen: Asid hidroklorik adalah asid kuat monoprotik . Asid etanoik adalah asid lemah mengion separa dalam air
HCl H+ + Cl– – 0.05 mol dm–3 of ionises completely in water menghasilkan kepekatan ion hidrogen yang lebih rendah .
0.01 mol dm–3 0.01 mol dm–3 to form 0.05 mol dm–3 hydrogen ion: – 0.1 mol dm–3 of ethanoic acid ionises to less than
Asid hidroklorik 0.05 mol dm–3 mengion lengkap dalam
– Concentration hydrogen ion in 0.1 mol dm–3 of 0.1 mol dm–3 hydrogen ion:
hydrochloric acid is higher than air menghasilkan 0.05 mol dm–3 ion hidrogen: Asid etanoik 0.1 mol dm–3 mengion kurang daripada 0.1 mol dm–3
0.01 mol dm–3 of hydrochloric acid.
Kepekatan ion hidrogen dalam asid hidroklorik HCl H+ + Cl– ion hidrogen:
0.1 mol dm–3 lebih tinggi daripada asid
hidroklorik 0.01 mol dm–3. 0.05 mol dm–3 0.05 mol dm–3 CH3COOH(aq/ak ) l e s s 0 t.h1a mnH/ok+lu rd a m n g – 3 d a r i+ CH3COO–(aq/ak )
0.1 mol dm–3
– The pH value of 0.1 mol dm–3 of hydrochloric – Concentration hydrogen ion in 0.05 mol dm–3
acid is lower than 0.01 mol dm–3 of of sulphuric acid is double of (higher than) – Concentration hydrogen ion in 0.1 mol dm–3 of
0.05 mol dm–3 of hydrochloric acid.
Kepekatan ion hidrogen dalam asid sulfurik hydrochloric acid is higher than of 0.1 mol dm–3 of
ethanoic acid.
0.05 mol dm–3 adalah dua kali ganda (lebih tinggi)
Kepekatan ion hidrogen dalam asid hidroklorik 0.1 mol dm–3 lebih
daripada asid hidroklorik 0.05 mol dm–3.
Chemistry Form 4 • MODULE
ion Sdn. B
127
hydrochloric acid. – The pH value of 0.05 mol dm–3 of sulphuric tinggi daripada asid etanoik 0.1 mol dm–3.
Nilai pH bagi asid hidroklorik 0.1 mol dm–3 lebih
acid is lower than 0.05 mol dm–3 of – The pH value of 0.1 mol dm–3 of hydrochloric acid lower
rendah daripada asid hidroklorik 0.01 mol dm–3. hydrochloric acid. than of 0.1 mol dm–3 of ethanoic acid.
Nilai pH bagi asid sulfurik 0.05 mol dm–3 lebih rendah Nilai pH bagi asid hidroklorik 0.1 mol dm–3 lebih rendah daripada
m Publicat
Nila daripada asid hidroklorik 0.05 mol dm–3. asid etanoik 0.1 mol dm–3.
hd.
MODULE • Chemistry Form 4
PREPARATION OF STANDARD SOLUTION / PENYEDIAAN LARUTAN PIAWAI
1 Standard solution is a solution in which its concentration is accurately known.
Larutan piawai ialah larutan yang kepekatannya diketahui dengan tepat.
2 The steps taken in preparing a standard solution are:
Langkah-langkah yang diambil dalam menyediakan larutan piawai adalah:
(a) Calculate the mass of solute needed to give the required volume and molarity.
Hitung jisim bahan larut yang diperlukan untuk menghasilkan isi padu dan kemolaran yang dikehendaki.
(b) The solute is weighed / Bahan larut ditimbang.
(c) The solute is completely dissolved in distilled water and then transferred to a volumetric flask partially filled with
distilled water.
Bahan larut dilarutkan sepenuhnya dalam air suling dan dipindahkan kepada kelalang volumetrik yang sebahagiannya sudah diisi
dengan air suling.
(d) Distilled water is added to the calibration mark of the volumetric flask and the flask is inverted to make sure
thorough mixing.
Air suling ditambah ke dalam kelalang volumetrik hingga tanda senggatan dan kelalang volumetrik ditelangkupkan beberapa kali
untuk memastikan campuran sekata.
PREPARATION OF A SOLUTION BY DILUTION / PENYEDIAAN LARUTAN SECARA PENCAIRAN
Adding water to the standard solution lowered the concentration of the solution. Since no solute is added, the amount of
solute in the solution before and after dilution remains unchanged:
Penambahan air kepada larutan piawai merendahkan kepekatan larutan tersebut. Oleh kerana tiada bahan terlarut yang ditambah, kandungan
bahan terlarut dalam larutan sebelum dan selepas pencairan tidak berubah:
Number of mol of solute before dilution = Number of mole of solute after dilution
Bilangan mol bahan terlarut sebelum pencairan Bilangan mol bahan terlarut selepas pencairan
M1V1 = M2V2
Therefore / Oleh itu, 1 000 1 000
M1V1 = M2V2
M1 = Initial concentration of the solute / Kemolaran larutan awal
V1 = Initial volume of the solution in cm3 / Isipadu larutan awal dalam cm3
M2 = Final concentration of the solute / Kemolaran larutan akhir
V2 = Final volume of the solution in cm3 / Isipadu larutan akhir dalam cm3
Example / Contoh :
1 (a) What is meant by a standard solution / Apakah yang dimaksudkan dengan larutan piawai ?
Standard solution is a solution in which its concentration is accurately known.
(b) (i) You are given solid sodium hydroxide. Describe the procedure to prepare 500 cm3 of 1.0 mol dm–3 sodium
hydroxide solution. [Relative atomic mass: H = 1; O = 16; Na = 23]
Anda diberi pepejal natrium hidroksida. Huraikan kaedah untuk menyediakan 500 cm3 larutan natrium hidroksida
1.0 mol dm–3. [Jisim atom relatif: H = 1, O = 16, Na = 23]
Calculate the mass of sodium hydroxide / Hitung jisim natrium hidroksida :
– Molar mass of NaOH / Jisim molar NaOH = 23 + 16 + 1 = 40 g mol–1
– Mol NaOH / Bilangan mol NaOH = 500 × 1.0/1 000 = 0.5 mol
– Mass of NaOH / Jisim NaOH = mol / Bilangan mol × molar mass / Jisim molar
= 0.5 mol × 40 g mol–1 = 20.0 g
Preparation of 500 cm3 1.0 mol dm–3 sodium hydroxide: .
Penyediaan 500 cm3 larutan natrium hidroksida 1.0 mol dm–3:
m Publica
– Weigh exactly 20.0 g of NaOH accurately using a weighing bottle
128
Timbang 20.0 g NaOH dengan tepat menggunakan botol penimbang .
Nila tion Sdn.
Bhd.
Chemistry Form 4 • MODULE
– Dissolve 20.0 g of NaOH in distilled water in a beaker.
Larutkan 20.0 g NaOH dalam air suling di dalam bikar.
– Transfer the content into a 500 cm3 volumetric flask .
Pindahkan kandungan ke dalam kelalang volumetrik 500 cm3.
– Rinse the beaker with distilled water and transfer all the contents into the volumetric flask .
Bilas bikar dengan air suling dan pindahkan semua kandungan ke dalam kelalang volumetrik .
– Distilled water is added to the volumetric flask until the calibration mark .
Air suling ditambah kepada kelalang volumetrik hingga tanda senggatan .
– The volumetric flask is closed tightly with stopper and inverted a few times to get
homogeneous solution.
Kelalang volumetrik ditutup dengan penutup dan ditelangkupkan beberapa kali untuk mendapatkan larutan
yang homogen.
(ii) Describe how you would prepare 250 cm3 of 0.1 mol dm–3 sodium hydroxide from the above solution.
Huraikan bagaimana anda menyediakan 250 cm3 larutan natrium hidroksida 0.1 mol dm–3 daripada larutan di atas.
Calculate the volume of 1 mol dm–3 sodium hydroxide used:
Hitung isi padu natrium hidroksida 1 mol dm–3 yang digunakan:
– M1 × V1 = M2 × V2
– V1 = M2 × V2 = 0.1 ×1 250 = 25 cm3
M1
Preparation of 250 cm3 1.0 mol dm–3 sodium hydroxide solution:
Penyediaan 250 cm3 larutan natrium hidroksida 1.0 mol dm–3:
– A pipette is filled with 25 cm3 of 1.0 mol dm–3 sodium hydroxide solution.
Sebuah pipet diisi dengan 25 cm3 larutan natrium hidroksida 1.0 mol dm–3.
– 25 cm3 of 1.0 mol dm–3 NaOH is transferred into 250 cm3 volumetric flask .
25 cm3 NaOH 1.0 mol dm–3 dipindahkan kepada kelalang volumetrik 250 cm3.
– Distilled water is added to the volumetric flask until the calibration mark .
Air suling ditambah kepada kelalang volumetrik hingga tanda senggatan .
– The volumetric flask is closed tightly with stopper and inverted a few times to get
homogeneous solution.
Kelalang volumetrik ditutup dengan penutup dan ditelangkupkan beberapa kali untuk mendapatkan larutan
yang homogen.
EXERCISE / LATIHAN
1 The table below shows the pH value of a few substances / Jadual di bawah menunjukkan nilai pH bagi beberapa bahan.
Substance / Bahan pH value / Nilai pH
Ethanoic acid 0.1 mol dm–3 / Asid etanoik 0.1 mol dm–3 3
Hydrochloric acid 0.1 mol dm–3 / Asid hidroklorik 0.1 mol dm–3 1
Glacial ethanoic acid / Asid etanoik glasial 7
(a) (i) What is meant by weak acid and strong acid / Apakah yang dimaksudkan dengan asid lemah dan asid kuat ?
Weak acid / Asid lemah : An acid that partially ionises in water to produce low concentration of hydrogen ion,
H+.
Strong acid / Asid kuat : An acid that completely ionises in water to produce high concentration of hydrogen
ion, H+.
m Publication Sdn. Bhd.
129
Nila
MODULE • Chemistry Form 4
(ii) Between ethanoic acid and hydrochloric acid, which acid has the higher concentration of H+ ion? Explain
your answer.
Antara asid etanoik dengan asid hidroklorik, asid manakah mempunyai kepekatan ion H+ yang lebih tinggi? Terangkan
jawapan anda.
– Hydrochloric acid has higher concentration of H+ than ethanoic acid.
Asid hidroklorik mempunyai kepekatan ion H+ yang lebih tinggi berbanding dengan asid etanoik.
– Hydrochloric acid is a strong acid which ionises completely in water to produce higher
concentration of H+:
Asid hidroklorik ialah asid kuat yang mengion sepenuhnya dalam air untuk menghasilkan kepekatan ion H+
yang lebih tinggi : H+(aq/ak ) + Cl–(aq/ak )
HCl (aq/ak)
– Ethanoic acid is a weak acid which ionises partially in water to produce lower concentration of
H+:
Asid etanoik ialah asid lemah yang mengion separa dalam air untuk menghasilkan kepekatan ion H+
yang lebih rendah : CH3COO– (aq/ak ) + H+ (aq/ak )
CH3COOH (aq/ak)
(iii) Why do ethanoic acid and hydrochloric acid have different pH value?
Mengapakah asid etanoik dan asid hidroklorik mempunyai nilai pH yang berbeza?
– The concentration H+ in hydrochloric acid is higher , the pH value is lower .
Kepekatan H+ dalam asid hidroklorik tinggi , nilai pH lebih rendah .
– The concentration H+ in ehanoic acid is lower , the pH value is higher .
Kepekatan H+ dalam asid etanoik rendah , nilai pH lebih tinggi .
(b) Glacial ethanoic acid has a pH value of 7 but a solution of ethanoic acid has a pH value less than 7. Explain the
observation.
Asid etanoik glasial mempunyai nilai pH 7 tetapi asid etanoik mempunyai nilai pH yang kurang daripada 7. Terangkan pemerhatian
tersebut.
– Glacial ethanoic acid molecules do not ionise . Glacial ethanoic acid consists of only CH3COOH
glmacoialel ceutlheasno.i Tc haec iCdH is3C 7O.OH molecules are neutral . No hydrogen ions present. The pH value of
Molekul asid etanoik glasial tidak mengion . Asid etanoik glasial hanya terdiri daripada molekul CH3COOH
sahaja. Molekul CH3COOH adalah neutral. Tiada ion hidrogen hadir. Nilai pH asid etanoik glasial adalah 7.
– Ethanoic acid ionises partially in water to produce ethanoate ions and hydrogen ions causes the
solution to have acidic property. The pH value of the solution is less than 7.
Asid etanoik mengion separa dalam air untuk menghasilkan ion etanoat dan ion hidrogen yang menyebabkan
larutan mempunyai sifat asid . Nilai pH bagi larutan tersebut adalah kurang daripada 7.
2 The table shows the pH value of a few solution / Jadual berikut menunjukkan nilai pH bagi beberapa jenis larutan berbeza.
Solution / Larutan P Q R S T U
pH 1 3 5 7 11 14
(a) (i) Which solution has the highest concentration of hydrogen ion?
Larutan manakah yang mempunyai kepekatan ion hidrogen yang paling tinggi?
Solution P
(ii) Which solution has the highest concentration of hydroxide ion?
Larutan yang manakah mempunyai kepekatan ion hidroksida yang paling tinggi?
Solution U
(b) Which is the following solution could be / Larutan manakah yang mungkin
(i) 0.01 mol dm–3 of hydrochloric acid / 0.01 mol dm–3 asid hidroklorik ? Q
R
(ii) 0.01 mol dm–3 of ethanoic acid / 0.01 mol dm–3 asid etanoik ? T
(iii) 0.1 mol dm–3 ammonia aqueous / 0.1 mol dm–3 larutan ammonia ?
Nila m Publication Sdn.
130
Bhd.
Chemistry Form 4 • MODULE
(iv) 1 mol dm–3 of hydrochloric acid / 1 mol dm–3 asid hidroklorik ? P
(v) 1 mol dm–3 sodium hydroxide solution / 1 mol dm–3 larutan natrium hidroksida ? U
(vi) 1 mol dm–3 potassium sulphate solution / 1 mol dm–3 larutan kalium sulfat ? S
(c) (i) State two solutions which react to form neutral solution.
Nyatakan dua larutan yang bertindak balas untuk membentuk larutan neutral.
P/Q/R and T/U // Hydrochloric acid/ethanoic acid with ammonia aqueous/sodium hydroxide solution.
(ii) Which solutions will produce carbon dioxide gas when calcium carbonate powder is added?
Larutan manakah menghasilkan gas karbon dioksida apabila ditambah serbuk kalsium karbonat?
P/Q // Hydrochloric acid/ethanoic acid
3 The molarity of sodium hydroxide solution is 2 mol dm–3. What is the concentration of the solution in g dm–3?
[RAM: Na = 23, 0 = 16, H = 1]
Kemolaran larutan natrium hidroksida ialah 2 mol dm–3. Apakah kepekatan larutan tersebut dalam g dm–3?
[JAR: Na = 23, O = 16, H = 1]
Answer / Jawapan : 80 g dm–3
4 Calculate the molarity of the solution obtained when 14 g potassium hydroxide is dissolved in distilled water to make
up 500 cm3 solution. [RAM: K = 39, H = 1, O = 16]
Hitung kemolaran larutan yang diperoleh apabila 14 g kalium hidroksida dilarutkan dalam air suling untuk menyediakan larutan yang
berisi padu 500 cm3. [JAR: K = 39, H = 1, O = 16]
Answer / Jawapan : 0.5 mol dm–3
5 Calculate the molarity of a solution which is prepared by dissolving 0.5 mol of hydrogen chloride, HCl in distilled water
to make up 250 cm3 solution.
Hitung kemolaran larutan yang disediakan dengan melarutkan 0.5 mol hidrogen klorida, HCl dalam air suling untuk menyediakan larutan
yang berisi padu 250 cm3.
Answer / Jawapan : 2 mol dm–3
6 How much of sodium hydroxide in gram should be dissolved in water to prepare 500 cm3 of 0.5 mol dm–3 sodium
hydroxide solution? [RAM: Na = 23, O = 16, H = 1]
Berapakah jisim natrium hidroksida dalam gram yang patut dilarutkan dalam air untuk menyediakan 500 cm3 larutan natrium hidroksida
0.5 mol dm–3? [JAR: Na = 23, O = 16, H = 1]
Answer / Jawapan : 10 g
7 300 cm3 water is added to 200 cm3 hydrochloric acid, 1 mol dm–3. What is the resulting molarity of the solution?
Jika 300 cm3 air ditambah kepada 200 cm3 asid hidroklorik 1 mol dm–3, apakah kemolaran bagi larutan yang dihasilkan?
Answer / Jawapan : 0.4 mol dm–3
m Publication Sdn. Bhd.
131
Nila
MODULE • Chemistry Form 4
8 Calculate the volume of nitric acid, 1 mol dm–3 needed to be diluted by distilled water to obtain 500 cm3 of nitric acid,
0.1 mol dm–3.
Hitung isi padu asid nitrik 1 mol dm–3 yang diperlukan untuk dilarutkan oleh air suling bagi menghasilkan 500 cm3 asid nitrik 0.1 mol dm–3.
Answer / Jawapan : 50 cm3
9 (a) Compare the number of mol of H+ ions which are present in 50 cm3 of 1 mol dm–3 of sulphuric acid and 50 cm3 of
1 mol dm–3 of hydrochloric acid. Explain your answer.
Bandingkan bilangan mol ion H+ yang hadir dalam 50 cm3 asid sulfurik 1 mol dm–3 dengan 50 cm3 asid hidroklorik 1 mol dm–3.
Terangkan jawapan anda.
Acid 50 cm3 of 1 mol dm–3 of sulphuric acid 50 cm3 of 1 mol dm–3 of hydrochloric acid
Asid 50 cm3 asid sulfurik 1 mol dm–3 50 cm3 asid hidroklorik 1 mol dm–3
Calculate number of Number of mol of sulphuric acid = 50 × 1 Number of mol of hydrochloric acid = 50 × 1
hydrogen ion, H+ 1 000 1 000
Hitung bilangan mol
= 0.05 mol = 0.05 mol
ion hidrogen, H+
H2SO4 2H+ + SO42– HCl H+ + Cl–
From the equation, From the equation,
1 mol of H2SO4 : 2 mol of H+ 1 mol of HCl : 1 mol of H+
0.05 mol of H2SO4 : 0.1 mol of H+ 0.05 mol of HCl : 0.05 mol of H+
Compare the number The number of H+ in 50 cm3 of 1 mol dm–3 of sulphuric acid is twice of the number of H+ in
of hydrogen ions 50 cm3 of 1 mol dm–3 of hydrochloric acid.
Bandingkan bilangan
ion hidrogen
Explanation Sulphuric acid is diprotic acid whereas hydrochloric acid is monoprotic acid.
Penerangan 1 mol of sulphuric ionises to 2 mol of H+ whereas 1 mol of hydrochloric acid ionises to 1 mol of H+.
The number of H+ in the same volume and concentration of both acids is doubled in sulphuric acid
compared to hydrochloric acid.
(b) Suggest the volume of 1 mol dm–3 of hydrochloric acid that has the same number of H+ with 50 cm3 of 1 mol dm–3
of sulphuric acid.
Cadangkan isi padu asid hidroklorik 1 mol dm–3 yang mempunyai bilangan ion H+ yang sama dengan 50 cm3 asid sulfurik 1 mol dm–3.
100 cm3
NEUTRALISATION / PENEUTRALAN
1 Neutralisation is the reaction between an acid and a base to form only salt and water:
Peneutralan ialah tindak balas antara asid dan bes untuk membentuk garam dan air sahaja:
Acid / Asid + Base / Bes Salt / Garam + Water / Air
Example / Contoh :
HCl (aq/ak ) + NaOH (aq/ak ) MNga(CNl O(a3)q2/ a(akq) /+ak )H +2O H(l2/Oce )(l/ce)
2HNO3 (aq/ak ) + MgO (s/p)
2 In neutralisation, the acidity of an acid is neutralised by an alkali. At the same time the alkalinity of an alkali is
neutralised by an acid. The hydrogen ions in the acid react with hydroxide ions in the alkali to produce water:
Dalam peneutralan, keasidan asid dineutralkan oleh alkali. Pada masa yang sama, kealkalian alkali dineutralkan oleh asid. Ion hidrogen
dalam asid bertindak balas dengan ion hidroksida dalam alkali:
H+ (aq/ak) + OH– (aq/ak) H2O (l/ce)
Nila m Publication Sdn.
132
Bhd.
Chemistry Form 4 • MODULE
3 Application of neutralisation in daily life / Aplikasi peneutralan dalam kehidupan seharian :
Application Example
Aplikasi Contoh
Agriculture 1 Acidic soil is treated with powdered soda lime (calcium oxide, CaO), limestone (calcium carbonate,
Agrikultur CaCO3)or ashes of burnt wood.
Tanah berasid dirawat dengan serbuk kapur tohor (kalsium oksida, CaO), batu kapur (kalsium karbonat, CaCO3) atau abu
daripada kayu api.
2 Basic soil is treated with compost. The acidic gas from the decomposition of compost neutralises the
alkalis in basic soil.
Tanah berbes dirawat dengan kompos. Gas berasid yang terbebas daripada penguraian kompos meneutralkan alkali
dalam tanah berbes.
3 The acidity of water farming is controlled by adding soda lime , (calcium oxide, CaO).
Keasidan air dalam pertanian dikawal dengan menambah kapur tohor (kalsium oksida, CaO).
Industries 1 Acidic gases emitted by industries are neutralised with soda lime , (calcium oxide, CaO) before the gases
Industri are released into air.
Gas-gas berasid yang dibebaskan oleh kilang dineutralkan dengan kapur tohor (kalsium oksida, CaO), sebelum gas-gas tersebut
dibebaskan ke udara.
2 Organic acid produced by bacteria in latex neutralises by ammonia and prevents coagulation.
Ammonia meneutralkan asid organik yang dihasilkan oleh bakteria dalam lateks dan mencegah penggumpalan.
Health 1 Excess acid in the stomach is neutralised with its anti-acids that contain bases such as aluminium hydroxide ,
Kesihatan calcium carbonate and magnesium hydroxide .
Anti-asid mengandungi bes seperti aluminium hidroksida , kalsium karbonat dan magnesium hidroksida untuk
meneutralkan asid berlebihan dalam perut.
2 Toothpastes contain bases (such as magnesium hydroxide) to neutralise the acid produced by bacteria in
mouth.
Ubat gigi mengandungi bes (seperti magnesium hidroksida) untuk meneutralkan asid yang dihasilkan oleh bakteria dalam
mulut.
3 Baking powder (sodium hydrogen carbonate) is used to cure alkaline bee stings.
Serbuk penaik (natrium hidrogen karbonat) digunakan untuk merawat sengatan lebah yang beralkali.
4 Vinegar (Ethanoic acid) is used to cure acidic wasp sting.
Cuka (asid etanoik) digunakan untuk merawat sengatan tebuan yang berasid.
4 An acid-base titration / Pentitratan asid-bes :
(a) It is a technique used to determine the volume of an acid required to neutralise a fixed volume of an alkali with
the help of acid-base indicator.
Ianya adalah teknik yang digunakan untuk menentukan isi padu asid yang diperlukan untuk meneutralkan isi padu tertentu alkali
dengan bantuan penunjuk asid-bes. Penunjuk yang biasa digunakan adalah fenolftalein dan metil jingga.
(b) Steps taken are / Langkah-langkah yang diambil adalah :
(i) An exact volume of alkali is measured with a pipette and poured into a conical flask.
Isi padu alkali yang tepat diukur dengan pipet dan dituang ke dalam kelalang kon.
(ii) A few drops of indicator are added to the alkali / Beberapa titik penunjuk ditambah kepada alkali.
(iii) A burette is filled with an acid. An acid is added drop by drop into the alkali in the conical flask until the
indicator changes colour, indicating the pH of neutral solution produced.
Buret diisi dengan asid. Asid ditambah setitik demi setitik kepada alkali dalam kelalang kon sehingga warna penunjuk bertukar,
menunjukkan pH larutan neutral telah dihasilkan.
(c) When the acid has completely neutralised the given volume of an alkali, the titration has reached the end point.
Apabila asid sudah lengkap meneutralkan isi padu alkali yang diberi, pentitratan telah mencapai takat akhir.
(d) The end point is the point in the titration at which the indicator changes colour.
Takat akhir ialah takat dalam pentitratan di mana penunjuk bertukar warna.
(e) The commonly used indicators are phenolphthalein and methyl orange.
Penunjuk yang biasa digunakan adalah fenolftalein dan metil jingga.
m Publicat ion Sdn. Bhd.
133
Nila
MODULE • Chemistry Form 4
5 The general steps used in any calculation involving neutralisation:
Langkah umum dalam penghitungan yang melibatkan peneutralan:
Step / Langkah 1 : Write the balanced equation / Tulis persamaan yang seimbang.
Step / Langkah 2 : Write the information from the question above the equation.
Tulis maklumat daripada soalan di atas persamaan.
Step / Langkah 3 : Write the information from the chemical equation below the equation (number of moles of substance
involved).
Tulis maklumat daripada persamaan kimia di bawah persamaan (bilangan mol bahan yang terlibat).
Step / Langkah 4 : Change the information to mole / Tukar maklumat kepada mol.
Step / Langkah 5 : Use the relationship between the number of moles of the substances in Step 3.
Guna hubungan di antara bilangan mol bahan-bahan dalam Langkah 3.
Step / Langkah 6 : Convert the number of mol to the required unit with the formula:
Tukar bilangan mol kepada unit yang diperlukan dengan menggunakan formula:
n = Mv atau n = MV
1 000
n = Number of moles of solute / Bilangan mol bahan terlarut
M = Concentration in mol dm–3 (molarity) / Kepekatan dalam mol dm–3 (kemolaran)
V = Volume of solution in dm3 / Isi padu larutan dalam dm3
v = Volume of solution in cm3 / Isi padu larutan dalam cm3
EXERCISE / LATIHAN
1 50 cm3 of 1 mol dm–3 sodium hydroxide solution is neutralised by 25 cm3 of sulphuric acid. Calculate the concentration
of sulphuric acid in mol dm–3 and g dm–3. [RAM: H = 1, S = 32, O = 16]
50 cm3 larutan natrium hidroksida 1 mol dm–3 dineutralkan oleh 25 cm3 asid sulfurik. Hitung kepekatan asid sulfurik dalam mol dm–3 dan
g dm–3. [JAR: H = 1, S = 32, O = 16]
M = 1 mol dm–3 M = ? Concentration of H2SO4 = n mol
V = 50 cm3 V = 25 cm3 V dm3
2NaOH + H2SO4 Na2SO4 + 2H2O = 0.025 mol = 1 mol dm–3
25
50 1 000 dm3
1 000
Number of mol of NaOH = 1 × = 0.05 mol =Co n1c menotlr adtmio–n3 ×o f( 2H 2×S 1O 4+ 32 + 16 × 4) g mol–1
= 98 g dm–3
F rom the equation, 0.025 mmooll ooff NNaaOOHH :: 10 .m02o5l omf oHl 2oSfO H42SO4
2 Calculate the volume of 2 mol dm–3 sodium hydroxide needed to neutralise 100 cm3 of 1 mol dm–3 hydrochloric acid.
Hitung isi padu larutan natrium hidroksida 2 mol dm–3 yang diperlukan untuk meneutralkan 100 cm3 asid hidroklorik 1 mol dm–3.
M = 2 mol dm–3 M = 1 mol dm–3
V = ? cm3 V = 100 cm3
NaOH + HCl NaCl + H2O
Number of mol of HCl = 1 × 100 = 0.1 mol
1 000
From the equation, 1 mol of HCl : 1 mol of mol NaOH
0.1 mol of HCl : 0.1 mol of mol NaOH
Volume of NaOH = n mol
M mol dm–3
= 0.1 mol
2 mol dm–3
= 0.05 dm–3
= 50 cm3
Nila m Publication Sdn.
134
Bhd.
Chemistry Form 4 • MODULE
3 Experiment I / Eksperimen I
1 mol dm–3 of nitric acid is used to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution.
Asid nitrik 1 mol dm–3 digunakan untuk meneutralkan 100 cm3 larutan natrium hidroksida 1 mol dm–3.
Experiment II / Eksperimen II
1 mol dm–3 of sulphuric acid is used to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution.
Asid sulfurik 1 mol dm–3 digunakan untuk meneutralkan 100 cm3 larutan natrium hidroksida 1 mol dm–3.
Compare the volume of acids needed to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution in Experiment I
and Experiment II. Explain your answer.
Bandingkan isi padu asid yang diperlukan untuk meneutralkan 100 cm3 larutan natrium hidroksida 1 mol dm–3 dalam Eksperimen I dan
Eksperimen II. Terangkan jawapan anda.
Answer / Jawapan:
Experiment Experiment I Experiment II
Eksperimen Eksperimen I Eksperimen II
Balanced equation
Persamaan kimia NaOH + HNO3 NaNO3 + H2O 2NaOH + H2SO4 Na2SO4 + 2H2O
Calculation
Pengiraan Mol of KOH / Bilangan mol NaOH = 1 × 100 Mol of KOH / Bilangan mol NaOH = 1 × 100
1 000 1 000
Comparison and
explanation = 0.1 mol = 0.1 mol
Perbandingan dan
penerangan From the equation / Daripada persamaan : From the equation / Daripada persamaan :
1 mol NaOH : 1 mol HNO3 2 mol NaOH / NaOH : 1 mol H2SO4
0.1 mol NaOH : 0.1 mol HNO3 0.1 mol NaOH / NaOH : 0.05 mol H2SO4
Mv Mv
Mol of HCl / Bilangan mol HNO3 = 1 000 Mol of H2SO4 / Bilangan mol H2SO4 = 1 000
M = Concentration of HNO3 / Kepekatan HNO3 M = Concentration of H2SO4 / Kepekatan H2SO4
v = Volume of HNO3 in cm3 / Isi padu HNO3 dalam cm3 v = Volume of H2SO4 in cm3 / Isi padu H2SO4 dalam cm3
1 mol dm–3 × v 1 mol dm–3 × v
1 000 = 0.1 mol 1 000 = 0.1 mol
v = 100 cm3 v = 50 cm3
– The volume of acid needed in Experiment II is doubled of Experiment I.
Isi padu asid nitrik yang diperlukan adalah dua kali ganda dalam Eksperimen I dibandingkan dengan Eksperimen II.
– Sulphuric acid is diprotic acid while nitric acid is monoprotic .
Asid sulfurik adalah asid diprotik manakala asid nitrik adalah asid monoprotik .
– One mol of sulphuric ionises two mol of H+, one mol nitric acid ionises to one mol of H+.
Satu mol asid sulfurik mengion kepada dua mol ion H+ manakala satu mol asid nitrik mengion kepada satu mol
ion H+.
– The number of H+ in the same volume and concentration of both acids is doubled in sulphuric acid
compared to hydrochloric acid.
Bilangan ion H+ dalam isi padu dan kepekatan yang sama bagi kedua-dua asid adalah dua kali ganda dalam asid
sulfurik dibandingkan dengan asid nitrik.
4 Diagram below shows the apparatus set-up for the titration of potassium hydroxide solution with sulphuric acid.
Gambar rajah di bawah menunjukkan susunan radas bagi pentitratan larutan kalium hidroksida dengan asid sulfurik.
0.5 mol dm–3 sulphuric acid
Asid sulfurik 0.5 mol dm–3
50 cm3 of 1 mol dm3 potassium hydroxide solution + methyl orange m Publicat hd.
50 cm3 larutan kalium hidroksida 1 mol dm3 + metil jingga
0.5 mol dm–3 sulphuric acid is titrated to 50 cm3 of 1 mol dm3 potassium hydroxide solution and methyl orange is used
as indicator.
Asid sulfurik 0.5 mol dm–3 ditambahkan kepada 50 cm3 larutan kalium hidroksida 1 mol dm3 dan metil jingga digunakan sebagai penunjuk.
ion Sdn. B
135
Nila
MODULE • Chemistry Form 4
(a) (i) Name the reaction between sulphuric acid and potassium hydroxide.
Namakan tindak balas antara asid sulfurik dengan kalium hidroksida.
Neutralisation
(ii) Name the salt formed in the reaction / Namakan garam yang terbentuk dalam tindak balas tersebut.
Potassium sulphate
(b) Suggest an apparatus that can be used to measure 25.0 cm3 of potassium hydroxide solution accurately.
Cadangkan radas yang boleh digunakan untuk mengukur 25.0 cm3 larutan kalum hidroksida dengan tepat.
Pipette
(c) What is the colour of methyl orange / Apakah warna metil jingga dalam
(i) in potassium hydroxide solution / larutan kalium hidroksida?
Red
(ii) in sulphuric acid / asid sulfurik?
Yellow
(iii) at the end point of the titration / pada titik akhir pentitratan?
Orange
(d) (i) Write a balanced equation for the reaction that occurs / Tuliskan persamaan seimbang bagi tindak balas yang berlaku.
2KOH + H2SO4 K2SO4 + 2H2O
(ii) Calculate the volume of the 0.1 mol dm–3 sulphuric acid needed to completely react with 50 cm3 of 1 mol dm3
potassium hydroxide.
Hitung isi padu asid sulfurik 0.1 mol dm–3 yang diperlukan untuk bertindak balas dengan lengkap dengan 50 cm3 larutan
kalium hidroksida 1 mol dm–3. 50
1 000
Number of mol KOH = 1 × = 0.05 mol
From the equation, 0.025 mmooll ooff KKOOHH :: 10 .m02o5l omf oHl 2oSfO H4 2SO4
Volume of H2SO4 = n mol
M mol dm–3
= 0.025 mol
1 mol dm–3
= 0.025 dm3
(e) (i) = 25 cm3
The experiment is repeated with 0.1 mol dm–3 hydrochloric acid to replace sulphuric acid. Predict the volume
of hydrochloric acid needed to neutralise 50.0 cm3 potassium hydroxide solution.
Eksperimen diulang dengan menggunakan asid hidroklorik 0.1 mol dm–3 untuk menggantikan asid sulfurik. Ramalkan isipadu asid
hidroklorik yang diperlukan untuk meneutralkan 50.0 cm3 larutan kalium hidroksida.
50 cm3 // double the volume of sulphuric acid
(ii) Explain your answer in (e)(i) / Terangkan jawapan anda di (e)(i).
– Hydrochloric acid is a monoprotic acid whereas sulphuric acid is a diprotic acid.
Asid hidroklorik ialah asid monoprotik manakala asid sulfurik ialah asid diprotik .
– The same volume and concentration of both acids, hydrochloric acid contains half the number
of mole of H+ ions as in sulphuric acid.
Pada isi padu dan kepekatan yang sama untuk kedua-dua asid, asid hidroklorik mengandungi separuh bilangan
mol ion H+ daripada asid sulfurik.
Nila m Publication Sdn.
136
Bhd.
Chemistry Form 4 • MODULE
Objective Questions / Soalan objektif
1 Which of the following substances changes blue litmus 5 The table below shows the concentration of hydrochloric
paper to red when dissolved in water? acid and ethanoic acid.
Antara bahan berikut, yang manakah menukarkan warna kertas Jadual di bawah menunjukkan kepekatan asid hidroklorik dan
litmus merah kepada biru apabila dilarutkan dalam air? asid etanoik.
A Sulphur dioxide Acid Concentration / mol dm–3
Asid Kepekatan / mol dm–3
Sulfur dioksida
B Carbon dioxide
Karbon dioksida Hydrochloric acid 0.1
Asid hidroklorik
C Lithium oxide
Litium oksida Ethanoic acid 0.1
Asid etanoik
D Sodium carbonate
Natrium karbonat Which of the following statements is true about both
acids?
Antara berikut, yang manakah adalah betul tentang kedua-dua
2 The table below shows the pH value of four acids which
asid?
have the same concentration. A Both are strong acids
Jadual di bawah menunjukkan nilai pH empat larutan yang
Kedua-duanya adalah asid kuat
mempunyai kepekatan yang sama.
Solution / Larutan pH value / Nilai pH B Both acids are strong electrolyte
Kedua-duanya adalah elektrolit yang kuat
P 2 C The pH value of both acid are equal
Q 7 Nilai pH kedua-dua asid adalah sama
R 12 D 50 cm3 of each acid need 50 cm3 of 0.1 mol dm–3 of
sodium hydroxide to be neutralised
S 13 50 cm3 setiap asid memerlukan 50 cm3 larutan natrium
Which of the following solutions has the highest hidroksida 0.1 mol dm–3 untuk dineutralkan
concentration of hydroxide ion? 6 The molarity of sodium hydroxide solution 0.5 mol dm–3.
What is the concentration of the solution in g dm–3?
Antara larutan berikut, yang manakah mempunyai kepekatan ion [Relative atomic mass: H = 1, O =16, Na = 23]
Kemolaran larutan natrium hidroksida adalah 0.5 mol dm–3.
hidroksida paling tinggi?
Apakah kepekatan larutan itu dalam g dm–3?
A P CR
B Q DS
3 Which of the following pairs of reactants would result [Jisim atom relatif: H = 1, O =16, Na = 23]
in a reaction?
Antara pasangan bahan tindak balas berikut, yang manakah akan A 20 C 80
B 40 D 120
menghasilkan tindak balas? 7 What is the volume of 2.0 mol dm–3 potassium hydroxide
solution needed to prepare 500 cm3 of 0.1 mol dm–3
A Sulphuric acid and copper(II) sulphate solution
Asid sulfurik dan larutan kuprum(II) sulfat potassium hydroxide solution?
B Nitric acid and magnesium oxide Berapakah isi padu larutan kalium hidroksida 2.0 mol dm–3
Asid nitrik dan magnesium oksida
diperlukan untuk menyediakan 500 cm3 larutan kalium hidroksida
C Hydrochloric acid and sodium nitrate solution
Asid hidroklorik dan larutan natrium nitrat 1 mol dm–3?
D Ethanoic acid and sodium sulphate solution A 100 cm3 C 200 cm3
Asid etanoik dan larutan natrium sulfat B 150 cm3 D 250 cm3
4 Which of the following reactions will not produce any 8 Which of the following solutions have the same
gas? concentration of hydrogen ions, H+, as in 0.1 mol dm–3
Antara tindak balas berikut, yang manakah tidak akan sulphuric acid, H2SO4?
Antara asid berikut, yang manakah mempunyai kepekatan ion
membebaskan sebarang gas?
A Copper metal with sulphuric acid hidrogen, H+ yang sama dengan asid sulfurik 0.1 mol dm–3?
A 0.1 mol dm–3 hydrochloric acid
Logam kuprum dengan asid sulfurik
B Zinc metal with hydrochloride acid Asid hidroklorik 0.1 mol dm–3
B 0.1 mol dm–3 carbonic acid
Logam zink dengan asid hidroklorik
Asid karbonik 0.1 mol dm–3
C Ammonium chloride with calcium hydroxide
Ammonium klorida dengan kalsium hidroksida C 0.2 mol dm–3 ethanoic acid
Asid etanoik 0.2 mol dm–3
D Sodium carbonate hydrochloric acid
Natrium karbonat dengan asid hidroklorik D 0.2 mol dm–3 nitric acid
Asid nitrik 0.2 mol dm–3
m Publication Sdn. Bhd.
137
Nila
MODULE • Chemistry Form 4
9 Which of the following sodium hydroxide solutions have 10 The diagram below shows 25.0 cm3 of 1.0 mol dm–3 of
concentration of 1.0 mol dm–3? sulphuric acid and 50.0 cm3 of 1.0 mol dm–3 of sodium
[Relative atomic mass: H=1, O=16, Na =23] are added hydroxide solution to form solution A.
Antara larutan natrium hidroksida berikut, yang manakah Rajah di bawah menunjukkan 25.0 cm3 asid sulfurik 1.0 mol dm–3 dan
mempunyai kepekatan 1.0 mol dm–3? 50.0 cm3 larutan natrium hidroksida 1.0 mol dm–3 ditambah
[JAR: H = 1, O = 16, Na = 23] bersama untuk menghasilkan larutan A.
I 5 g of sodium hydroxide dissolved in distilled water 50 cm3 of 1 mol dm–3 of hydroxide solution
to make 250 cm3 of solution. 50 cm3 larutan natrium hidroksida 1 mol dm–3
5 g natrium hidroksida dilarutkan dalam air suling
menjadikan 250 cm3 larutan.
II 20 g of sodium hydroxide dissolved in distilled water
to make 500 cm3 of solution.
20 g natrium hidroksida dilarutkan dalam air suling
menjadikan 500 cm3 larutan. 25 cm3 of 2.0 mol dm–3
III 250 cm3 of 2 mol dm–3 sodium hydroxide solution is sulphuric acid Solution A / Larutan A
added to distilled water to make 1 dm3 of solution. 25 cm3 asid sulfrik 2.0 mol dm–3
250 cm3 larutan natrium hidroksida 2 mol dm–3 ditambah air
suling menjadikan 1 dm3 larutan. Which of the following is true about the solution A?
IV 500 cm3 of 2 mol dm–3 sodium hydroxide solution is Antara berikut, yang manakah adalah benar tentang larutan A?
added to distilled water to make 1 dm3 of solution. A The solution has a pH value of 7
500 cm3 larutan natrium hidroksida 2 mol dm–3 ditambah air
Larutan itu menpunyai nilai pH 7
suling menjadikan 1 dm3 larutan.
A I and III only B The solution will react with any acid
I dan III sahaja Larutan itu boleh bertindak balas dengan sebarang asid
B II and III only
C The solution turns a red litmus paper blue
II dan III sahaja
C II and IV only Larutan itu menukarkan warna kertas litmus merah kepada biru
II dan IV sahaja D The solution will react with zinc to produce hydrogen
gas
Larutan itu bertindak balas dengan zink untuk menghasilkan gas
hidrogen
D I, II, III and IV
I, II, III dan IV
Nila m Publication Sdn.
138
Bhd.
Chemistry Form 4 • MODULE
7 SALT
GARAM
PREPARATION OF SALTS / PENYEDIAAN GARAM
• THE MEANING OF SALTS / MAKSUD GARAM
– To write the meaning of salts and the formulae for all types of salt that are commonly found in this topic.
Menyatakan maksud garam dan menulis formula semua jenis garam yang biasa ditemui dalam tajuk ini.
• THE SOLUBILITY OF SALTS / KETERLARUTAN GARAM
– To determine the solubility of nitrate, sulphate, carbonate and chloride salts.
Menentukan keterlarutan semua garam nitrat, sulfat, karbonat dan klorida.
• EXPERIMENTS FOR THE PREPARATION OF SALTS BASED ON SOLUBILITY
EKSPERIMEN PENYEDIAAN GARAM BERDASARKAN KETERLARUTAN
– To determine the suitable methods for the preparation of salts based on solubility:
Menentukan kaedah yang sesuai bagi penyediaan garam berdasarkan keterlarutan:
i. Acid + metal / Asid + logam ii. Acid + metal oxides / Asid + oksida logam
iii. Acid + alkali / Asid + alkali iv. Acid + metal carbonate / Asid + karbonat logam
v. Double decomposition reaction / Tindak balas penguraian ganda dua
– To describe the experiments for each method of preparation and explain the rationale for each step.
Menghuraikan eksperimen bagi setiap jenis kaedah penyediaan serta menerangkan rasional setiap langkah.
CALCULATION ON QUANTITY OF REACTANTS/PRODUCTS [ QUANTITATIVE ANALYSIS ]
PENGHITUNGAN KUANTITI BAHAN/HASIL [ ANALISIS KUANTITATIF]
• CONTINUOUS VARIATIONS METHODS / KAEDAH PERUBAHAN BERTERUSAN
– To describe the methods of experiment to determine the formulae of insoluble salts.
Menghuraikan eksperimen bagi kaedah penentuan formula garam tak larut.
• SOLVING VARIOUS PROBLEMS RELATING TO QUANTITY OF REACTANTS/PRODUCTS IN SOLID, LIQUID AND GAS
FORMS
MENYELESAIKAN PELBAGAI MASALAH BERKAITAN KUANTITI BAHAN DALAM BENTUK PEPEJAL, LARUTAN DAN GAS
– Using the formula / Menggunakan formula:
i. n = MV ii. Mole / Mol = Mass / Jisim
1 000 RAM/RMM/RFM / JAR/JMR/JFR
iii. The molar volume of gas at room temperature and s.t.p / Isi padu molar gas pada suhu bilik dan s.t.p
IDENTIFICATION OF IONS [ QUALITATIVE ANALYSIS ] / MENGENAL ION [ ANALISIS KUALITATIF ]
• ACTION OF HEAT ON SALTS / KESAN HABA KE ATAS GARAM
– To state the colour of the residue of lead(II) oxide, zinc oxide and copper(II) oxide.
Menyatakan warna baki bagi plumbum(II) oksida, zink oksida dan kuprum(II) oksida.
– To state the confirmatory tests for carbon dioxide and nitrogen dioxide.
Menyatakan ujian pengesahan bagi gas karbon dioksida dan nitrogen dioksida.
– To write the equations of the decomposition of carbonate and nitrate salts.
Menulis persamaan penguraian semua garam karbonat dan nitrat.
• CONFIRMATORY TEST CATIONS AND ANIONS / UJIAN PENGESAHAN KATION DAN ANION
– To state the confirmatory tests for all cations using sodium hydroxide and ammonia solution.
Menghuraikan ujian pengesahan semua kation menggunakan natrium hidroksida dan larutan ammonia.
– To state the confirmatory tests to differentiate Al3+ and Pb2+.
Menghuraikan ujian untuk membezakan Al3+ dan Pb2+.
– To state the confirmatory tests for anions of sulphate, nitrate, carbonate and chloride.
Menghuraikan ujian pengesahan anion sulfat, nitrat, karbonat dan klorida.
m Publication Sdn. Bhd.
139
Nila
MODULE • Chemistry Form 4
PREPARATION OF SALT / PENYEDIAAN GARAM
1 A salt is a compound formed when the hydrogen ion in an acid is replaced with metal ion or ammonium ion. Example:
Sodium chloride, copper(II) sulphate, potassium nitrate and ammonium sulphate.
Garam ialah sebatian ion yang terhasil apabila ion hidrogen daripada asid diganti oleh ion logam termasuk ion ammonium.
Contoh: natrium klorida, kuprum(II) sulfat, kalium nitrat dan ammonium sulfat.
2 Write the formulae of the salts in the table below by replacing hydrogen ion in sulphuric acid, hydrochloric acid, nitric
acid and carbonic acid with metal ions or ammonium ion.
Tuliskan formula kimia garam berikut dengan menggantikan ion hidrogen dalam asid sulfurik, asid hidroklorik, asid nitrik dan asid
karbonik dengan ion logam atau ion ammonium:
Metal ion Sulphate salt (from H2SO4) Chloride salt (from HCl) Nitrate salt (from HNO3) Carbonate salt (from H2CO3)
Ion logam Garam sulfat (dari H2SO4 ) Garam klorida (dari HCl) Garam nitrat (dari HNO3 ) Garam karbonat (dari H2CO3 )
Na2SO4 Na2CO3
Na+ NaCl NaNO3
K+ K2SO4 KCl KNO3 K2CO3
Mg2+ MgSO4 MgCl2 Mg(NO3 )2 MgCO3
Ca2+ CaSO4 CaCl2 Ca(NO3 )2 CaCO3
Al3+ Al2(SO4 )3 AlCl3 Al(NO3 )3 Al2(CO3 )3
Zn2+ ZnSO4 ZnCl2 Zn(NO3 )2 ZnCO3
Fe2+ FeSO4 FeCl2 Fe(NO3 )2 FeCO3
Pb2+ PbSO4 PbCl2 Pb(NO3 )2 PbCO3
Cu2+ CuSO4 CuCl2 Cu(NO3 )2 CuCO3
Ag+ Ag2SO4 AgCl AgNO3 Ag2CO3
NH4+ (NH4 )2SO4 NH4Cl NH4NO3 (NH4 )2CO3
Ba2+ BaSO4 BaCl2 Ba(NO3 )2 BaCO3
3 Solubility of salts in water: / Keterlarutan garam dalam air:
(a) All K+, Na+ and NH4+ salts are soluble. / Semua garam K+, Na+ dan NH4+ larut.
(b) All nitrate salts are soluble. / Semua garam nitrat larut.
(c) All carbonate salts are insoluble except K2CO3, Na2CO3 and (NH4)2CO3.
Semua garam karbonat tak larut kecuali K2CO3, Na2CO3 dan (NH4 )2CO3.
(d) All sulphate salts are soluble except CaSO4, PbSO4 and BaSO4.
Semua garam sulfat larut kecuali CaSO4, PbSO4 dan BaSO4.
(e) All chloride salts are soluble except PbCl2 and AgCl. / Semua garam klorida larut kecuali PbCl2 dan AgCl.
* Based on the solubility of the salts in water, shade the insoluble salts in the above table.
* Berdasarkan keterlarutan garam dalam air, lorekkan garam yang tak larut dalam jadual di atas.
4 Method used to prepare salt depends on the solubility of the salt.
Kaedah penyediaan garam bergantung pada keterlarutan garam tersebut.
Soluble salts are prepared from the reactions between an acid with a metal/ base/ metal carbonate:
Garam terlarut disediakan melalui tindak balas antara asid dengan logam/bes/karbonat logam:
i. Acid + metal / Asid + logam salt + hydrogen / garam + hidrogen Acid + *base salt + water
ii. Acid + metal oxide / Asid + oksida logam salt + water / garam + air
iii. Acid + alkali / Asid + alkali salt + water / garam + air Asid + *bes garam + air
iv. Acid + metal carbonate / Asid + karbonat logam salt + water + carbon dioxide / garam + air + karbon dioksida
* Most bases are metal oxide or metal hydroxide. / Kebanyakan bes adalah oksida logam atau hidroksida logam.
* All metal oxides and hydroxides are insoluble in water except Na2O, K2O, NaOH and KOH.
Semua oksida logam dan hidroksida logam tidak larut dalam air kecuali Na2O, K2O, NaOH dan KOH.
* Alkali is a base that soluble in water and ionises to hydroxide ion.
Alkali ialah bes yang larut dalam air dan mengion menjadi ion hidroksida.
Nila m Publication Sdn.
140
Bhd.
PREPARATION OF SOLUBLE AND INSOLUBLE SALT / PENYEDIAAN GARAM LARUT DAN GARAM TAK LARUT
1 Salts are prepared based on their solubility as shown in the flow chart below:
Garam disediakan berdasarkan keterlarutannya sebagaimana yang ditunjukkan pada carta aliran di bawah:
Preparation of salt / Penyediaan garam
Soluble salt Insoluble salt
Garam larut Garam tak larut
Method III / Kaedah III
Salts / Garam K+, Na+, NH4+ Other than / Garam selain K+, Na+, NH4+ The salt is prepared by precipitation
Method I / Kaedah I Method II / Kaedah II method. (Double decomposition
reaction)
The salt is prepared by titration method of acid The salt is prepared by reacting acid with insoluble metal/metal oxide/ Garam ini disediakan melalui kaedah
and alkali using an indicator.
metal carbonate: pemendakan. (Tindak balas penguraian ganda
Garam ini disediakan melalui kaedah pentitratan di antara
Garam ini disediakan melalui tindak balas antara asid dengan logam/oksida logam/ dua).
asid dan alkali dengan menggunakan penunjuk. – Mix two solutions containing cations
karbonat logam yang tak larut:
– Acid + Alkali Salt + Water and anions of insoluble salts.
(Neutralisation Reaction) – Acid + Metal Salt + Hydrogen (Displacement reaction) Campur dua larutan yang mengandungi kation
Asid + Alkali Garam + Air Asid + Logam Garam + Hidrogen (Tindak balas penyesaran) dan anion garam tak larut.
– Stir with glass rod.
(Tindak balas Peneutralan) – Acid + Metal oxide Salt + Water (Neutralisation Reaction)
Asid + Oksida bes Garam + Air (Tindak balas Peneutralan) Kacau dengan rod kaca.
– Filter using filter funnel.
– Acid + Metal carbonate Salt + Water + Carbon Dioxide
Turas dengan corong turas.
Asid + Karbonat logam Garam + Air + Karbon dioksida – Rinse the residue with distilled water.
– A titration is conducted to determine the volume – Add metal/metal oxide/metal carbonate powder until excess into a fixed Bilas baki dengan air suling.
of acid needed to neutralise a fixed volume of an – Dry the residue by pressing it between
alkali with the aid of an indicator. volume of the heated acid
Pentitratan dijalankan dengan menentukan isi padu asid Tambah serbuk logam/oksida logam/karbonat logam ke dalam isi padu tetap asid filter papers.
Keringkan baki dengan menekankan antara
yang diperlukan untuk meneutralkan alkali yang isi yang dihangatkan sehingga berlebihan.
kertas turas.
padunya sudah ditetapkan dengan menggunakan penunjuk. – Filter the mixture to remove excess metal/metal oxide/metal carbonate
– The same volume of acid is then added to the Turas campuran tersebut untuk mengeluarkan pepejal logam/oksida logam/karbonat
same volume of alkali without any indicator to
obtain pure and neutral salt solution. logam yang berlebihan.
Isi padu asid yang sama juga ditambah kepada isi padu
alkali yang sama tanpa penunjuk untuk mendapatkan
garam yang tulen dan neutral.
Chemistry Form 4 • MODULE
ion Sdn. B
141
Nila m Publicat – Evaporate the filtrate until it becomes a saturated solution/ Sejatkan hasil turasan hingga larutan tepu.
– Dip in a glass rod, if crystals are formed, the solution is saturated.
Celupkan dengan rod kaca, jika hablur terbentuk dengan serta merta, larutan adalah tepu.
– Cooled at room temperature / Biarkan sejuk pada suhu bilik.
– Filter and dry the salt crystals by pressing them between filter papers.
Turas dan keringkan hablur garam dengan menekan antara kertas turas.
hd.
Nila
2 Steps to Prepare Soluble Salt/Langkah Penyediaan Garam Larut
Method II:/Kaedah II:
Soluble salt except K+, Na+ and NH4+ / Garam larut selain K+, Na+ dan NH4+
Bhd.tion Sdn. MODULE • Chemistry Form 4
m Publica
142
Method I:/Kaedah I:
Soluble salt of K+, Na+ and NH4+
Garam larut K+, Na+ dan NH4+
• Stir the mixture with a glass rod . • Filter the mixture to separate • The salt solution is poured into • Measure and pour 50 cm3
excess metal /metal oxide evaporating dish of 1 mol dm–3 any
Kacau campuran dengan /metal carbonate with the . alkali into a conical
rod kaca . salt solution . flask. Add a few drops of
menggunakan Turas campuran tersebut untuk Larutan garam dituangkan dalam phenolphthalein.
mengasingkan bahan berlebihan iaitu mangkuk penyejat . Sukat dan tuangkan 50 cm3
• Add metal/metal oxide / metal logam/oksida logam/karbonat logam sebarang alkali berkepekatan
carbonate powder to the acid • Evaporate the salt solution until 1 mol dm–3 ke dalam
until excess . dengan larutan garam . saturated solution is formed. kelalang. Tambah beberapa
titis fenolftalein.
Tambah serbuk logam / oksida logam The residue is Sejatkan larutan sehingga larutan tepu
/ karbonat logam kepada asid sehingga metal /metal oxide terbentuk. • 1 mol dm–3 of any acid
/metal carbonate . is titrated to the alkali
berlebihan . Saturated until neutral by using an
Baki adalah logam/ salt solutions indicator. The volume of
Excess of metal/ logam oksida/ acid used is recorded.
metal oxide/ logam karbonat . Larutan garam 1 mol dm–3 sebarang
tepu asid dititratkan kepada alkali
metal carbonate sehingga neutral menggunakan
The filtrate is Heat penunjuk. Isi padu asid yang
Logam/oksida logam/ salt solution Panaskan digunakan dicatat.
karbonat logam
Heat Hasil turasan ialah . • Repeat the titration
Panaskan yang berlebihan. larutan garam . without the indicator
to get pure and neutral
• Measure and pour 50 – 100 cm3 of 0.5 – 2 mol dm–3 of any acid • Cool it at room temperature until crystals salts are salt solution.
and pour into a beaker. formed. Ulang titratan tanpa penunjuk
Sejukkan pada suhu bilik sehingga hablur garam terbentuk. untuk mendapatkan larutan
garam yang
Sukat dan tuangkan 50 – 100 cm3 sebarang asid berkepekatan Salt crystals tulen dan neutral .
Hablur garam
0.5 – 2 mol dm–3 dan tuangkan ke dalam bikar. Acid
Asid
• Add metal/metal oxide/ metal carbonate powder into the acid and heat
gently Alkali
. Alkali
Tambahkan serbuk logam/ oksida logam/ karbonat logam pada isi padu asid yang • Filter the mixture to separate
tetap sambil dihangatkan perlahan-lahan . the salt crystals .
• Dry the salt crystals by Turaskan campuran tersebut untuk mengasingkan
pressing them between filter hablur garam .
Residue is
Acid papers. salt crystals
Asid Keringkan hablur garam dengan
menekan antara kertas turas.
Baki adalah
hablur garam
Heat Salt crystals
Panaskan
Chemistry Form 4 • MODULE
3 Steps to Prepare Insoluble Salt / Penyediaan Garam Tak Larut
Insoluble salts are prepared by the precipitation method through double decomposition reactions.
Garam tak larut disediakan dengan cara pemendakan melalui tindak balas penguraian ganda dua.
(i) In this reaction, the precipitate of insoluble salt is formed when two different solutions that contain the cation and
anion of the insoluble salt are mixed.
Dalam tindak balas ini, mendakan garam tak larut terbentuk apabila dua larutan berbeza yang mengandungi kation dan anion
garam tak terlarut dicampurkan.
(ii) The insoluble salt is obtained as a residue of a filtration.
Garam tak terlarut tersebut diperoleh daripada baki penurasan.
Method III: Preparation of Insoluble XnYm Salt by Double Decomposition Reaction
Kaedah III: Penyediaan Garam Tak Larut XnYm Melalui Tindak balas Penguraian Ganda Dua
1) Measure and pour 50 – 100 cm3 2) Measure and pour 50 – 100 cm3
of 0.5 – 2 mol dm–3 of aqueous of 0.5 – 2 mol dm–3 of
solution contains X m+ cation. aqueous solution contains Yn– anion
Sukat dan tuangkan 50 – 100 cm3 into another beaker.
larutan berkepekatan 0.5 – 2 mol dm–3 Sukat dan tuangkan 50 – 100 cm3
mengandungi kation Xm+ ke dalam bikar. larutan berkepekatan 0.5 – 2
mol dm–3 mengandungi anion Yn– ke dalam
bikar yang lain.
Precipitate of XnYm 3) Mix both solutions and stir the mixture with
salt is formed. XnYm glass rod .
Mendakan garam
Campur dan kacaukan campuran menggunakan
terbentuk.
rod kaca .
The residue is XnYm
salt.
4) Filter the mixture and rinse the precipitate
Mendakan adalah garam
XnYm . with distilled water . The residue is XnYm
salt.
Turas campuran dan bilas mendakan itu menggunakan air suling.
Baki ialah garam XnYm.
Salt XnYm 5) Press the precipitate between filter papers to dry it.
Garam XnYm Tekankan mendakan antara kertas turas untuk mengeringkannya.
ion Sdn. Bm Publicathd.
143
Nila
MODULE • Chemistry Form 4
Complete the following table: Yn– XnYm Ion equation/Persamaan ion
Lengkapkan jadual berikut: I– [KI] PbI2
BaSO4 Pb2+ + I– PbI2
Xm+ SO42–[ Na2SO4 ] AgCl
Pb2+ [Pb(NO3)2] Cl– [NaCl] Ba2+ + SO42– BaSO4
Ag+ + Cl– AgCl
Ba2+[ BaCl2 ]
Ag+ [AgNO3]
Ca2+ [Ca(NO3)2] CO32– [Na2CO3] CaCO3 Ca2+ + CO32– CaCO3
4 Complete the following table by writing “S” for soluble salts and “IS” for insoluble salts. Write all the possible chemical
equations to prepare soluble salts and two chemical equations for insoluble salts.
Lengkapkan jadual berikut dengan menulis “L” bagi garam larut dan “TL” bagi garam tak larut. Tuliskan semua persamaan kimia dalam
penyediaan garam larut dan dua persamaan kimia bagi garam tak larut.
Salt “S” / “IS” Chemical equations
Garam “L” / “TL” Persamaan kimia
Zinc chloride S Zn + 2HCl Z nZC nZlC2n l+2C l+ 2H + H2 2COO2 + H2O
Zink klorida
ZnCO3 + 2HCl
ZnO + 2HCl
Sodium nitrate S NaOH + HNO3 NaNO3 + H2O
Natrium nitrat
Silver chloride IS AgNO3 + HCl A AggCCl l+ + H NNaON3O3
Argentum klorida S AgNO3 + NaCl
IS
Copper(II) sulphate S CuO + H2SO4 Cu CSuOS4O +4 + H 2COO2 + H2O
Kuprum(II) sulfat IS CuCO3 + H2SO4
S
Lead(II) sulphate Pb(NO3)2 + H2SO4 P PbbSSOO4 4+ + 2 2HNNaON3O3
Plumbum(II) sulfat Pb(NO3)2 + Na2SO4
Aluminium nitrate A2AAll22l(O C+3O + 36) 3H6 +NHO N63OH 3N O 32 A2lA(Nl (2ONA3Ol)(33N )+3O + 33) 3H3 +2H 23OCO2 + 3H2O
Aluminium nitrat
Pb(NO3)2 + 2HCl P PbbCCl2l 2+ + 2 2HNNaON3O3
Lead(II) chloride Pb(NO3)2 + 2NaCl
Plumbum(II) klorida
Mg + 2HNO3 M Mg(gN(NO3O)23 )+2 + H H2 2O
Magnesium nitrate MgO + 2HNO3
Magnesium nitrat
MgCO3 + 2HNO3 Mg(NO3)2 + CO2 + H2O
Potassium chloride S KOH + HCl KCl + H2O
Kalium klorida
Lead(II) nitrate S PbO + 2HNO3 Pb P(bN(ON3O)23 )+2 + H 2COO2 + H2O
Plumbum(II) nitrat PbCO3 + 2HNO3
Barium sulphate IS BaCl2 + H2SO4 B BaaSSOO4 4+ + 2 2HNCal Cl
Barium sulfat BaCl2 + Na2SO4
Nila m Publication Sdn.
144
Bhd.
Chemistry Form 4 • MODULE
EXERCISE / LATIHAN
1 The diagram below shows the set-up of apparatus to prepare soluble salt Y.
Rajah di bawah menunjukkan susunan radas bagi menyediakan garam larut Y.
Nitric acid
Asid nitrik
25 cm3 of 1 mol dm–3 potassium hydroxide solution
+ phenolphthalein
25 cm3 larutan kalium hidroksida 1 mol dm-3 + fenolftalein
Phenolphthalein is used as an indicator in a titration between nitric acid and sodium hydroxide solution. 25 cm3 of nitric
acid completely neutralises 25 cm3 of 1 mol dm–3 potassium hydroxide solution. The experiment is repeated by reacting
25 cm3 of 1 mol dm–3 potassium hydroxide solution with 25 cm3 nitric acid without phenolphthalein. Salt Y is formed
from the reaction.
Fenolftalein digunakan sebagai penunjuk dalam pentitratan antara asid nitrik dengan larutan kalium hidroksida. 25 cm3 asid nitrik
meneutralkan 25 cm3 larutan kalium hidroksida 1 mol dm–3. Eksperimen ini diulang dengan menindakbalaskan 25 cm3 larutan kalium
hidroksida 1 mol dm–3 dengan 25 cm3 asid nitrik tanpa fenolftalein. Garam Y terbentuk daripada tindak balas ini.
(a) Name salt Y.
Nyatakan nama garam Y.
Potassium nitrate
(b) Write a balanced equation for the reaction that occurs.
Tuliskan persamaan seimbang bagi tindak balas yang berlaku.
HNO3 + KOH KNO3 + H2O
(c) Calculate the concentration of nitric acid.
Hitungkan kepekatan asid nitrik tersebut.
Mol of NaOH = 1 × 1 = 0.025 mol
1 000
From the equation, 1 mol of KOH :: 10 .m02o5l omf o lH oNf OH3NO3
0.025 mol of KOH
C oncentration of HNO3, M
25
0.025 = M × 1 000
M = 1 mol dm–3
(d) Why is the experiment is repeated without phenolphthalein?
Mengapakah eksperimen ini diulang tanpa menggunakan fenolftalein?
To get pure and neutral salt solution Y.
(e) Describe briefly how a crystal of salt Y is obtained from the salt solution.
Huraikan secara ringkas bagaimana hablur garam Y diperoleh daripada larutan garamnya.
– The salt solution is poured into an evaporating dish.
– The solution is heated to evaporate the solution until one third its original volume// a saturated solution formed.
– The saturated solution is allowed to cool until salt crystals Y are formed.
– The crystals are filtered and dried by pressing them between filter papers.
(f) Name two other salts that can be prepared with the same method.
Namakan dua garam lain yang boleh disediakan dengan kaedah yang sama.
Potassium/sodium/ammonium salt. Example: potassium nitrate, sodium sulphate.
(g) State the type of reaction in the preparation of the salts.
Nyatakan jenis tindak balas dalam penyediaan garam ini.
Neutralisation
m Publication Sdn. B hd.
145
Nila
MODULE • Chemistry Form 4
2 The following is the steps in the preparation of dry copper(II) sulphate crystals.
Berikut adalah langkah-langkah dalam penyediaan hablur garam kuprum(II) sulfat kering.
Step I: Copper(II) oxide powder is added a little at a time with constant stirring to the heated 50 cm3 of
1 mol dm–3 sulphuric acid until some of it no longer dissolve.
Langkah I: Serbuk kuprum(II) oksida ditambahkan, sedikit demi sedikit sambil dikacau ke dalam 50 cm3 asid sulfurik
1 mol dm-3 yang dipanaskan sehingga serbuk itu tidak boleh larut lagi.
Step II: The mixture is filtered.
Langkah II: Campuran dituras.
Step III: The filtrate is poured into an evaporating dish and heated to evaporate the solution until one
third of its original.
Langkah III: Hasil turasan dipanaskan di dalam mangkuk penyejat sehingga isi padunya menjadi satu pertiga
daripada isi padu asal.
Step IV: The salt solution is allowed to cool at room temperature for the crystallisation to take place.
Langkah IV: Hasil turasan itu dibiarkan sejuk ke suhu bilik sehingga penghabluran berlaku.
Step V: The crystals formed are filtered and dried by pressing them between filter papers.
Langkah V: Hablur yang terbentuk dituraskan dan dikeringkan dengan menekan antara kertas turas.
(a) (i) State two observations during Step I.
Nyatakan dua pemerhatian pada Langkah I.
– Black solid dissolve
– Colourless solution turns black
(ii) Write a balance chemical equation for the reaction that occur in Step I.
Tuliskan persamaan kimia seimbang bagi tindak balas yang berlaku dalam Langkah I.
CuO + H2SO4 CuSO4 +H2O
(iii) State the type of reaction in the preparation of the salts.
Nyatakan jenis tindak balas yang berlaku dalam penyediaan garam.
Neutralisation
(b) Why is copper(II) oxide powder added until some of it no longer dissolve in Step I?
Mengapakah serbuk kuprum(II) oksida ditambah pada larutan tersebut sehingga ia tidak boleh melarut lagi dalam Langkah I?
To make sure that all sulphuric acid has reacted.
(c) What is the purpose of heating in Step III?
Apakah tujuan pemanasan dalam Langkah III?
To evaporate the water and copper(II) sulphate solution becomes saturated
(d) What is the colour of copper(II) sulphate?
Apakah warna kuprum(II) sulfat?
Blue
Nila tion Sdn.(e) What is the purpose of filtration in
Apakah tujuan penurasan dalam
(i) Step II? / Langkah II?
– To remove the excess copper(II) oxide.
– To obtain copper(II) sulphate solution as a filtrate
(ii) Step V? / Langkah V?
To obtain copper(II) sulphate crystals as a residue.
m Publica
146
Bhd.
Chemistry Form 4 • MODULE
(f) Draw the a labelled diagram to show the set-up of apparatus used Step II and Step III.
Lukiskan gambar rajah berlabel untuk menunjukkan susunan alat radas yang digunakan dalam Langkah II dan Langkah III.
Excess of Filter paper Copper(II) sulphate solution
copper(II) oxide
Heat
Copper(II) sulphate
solution
(g) Can copper powder replace copper(II) oxide in the experiment? Explain your answer.
Bolehkah serbuk kuprum digunakan untuk menggantikan kuprum(II) oksida dalam eksperimen ini? Terangkan jawapan anda.
Cannot. Copper is less electropositive than hydrogen in the electrochemical series, copper cannot displace hydrogen
from the acid.
(h) Name other substance that can replace copper(II) oxide to prepare the same salt. Write a balance chemical equation
for the reaction that occur.
Namakan sebatian lain yang dapat menggantikan kuprum(II) oksida dalam penyediaan garam yang sama. Tuliskan persamaan kimia
yang seimbang bagi tindak balas yang berlaku.
Substance / Garam larut : Copper(II) carbonate
Balance equation / Persamaan seimbang : CuCO3 + H2SO4 CuSO4 + H2O + CO2
3 The diagram below shows the flow chart for the preparation of lead(II) nitrate and lead(II) sulphate through reaction I
and II.
Rajah di bawah menunjukkan carta aliran bagi penyediaan plumbum(II) nitrat dan plumbum(II) sulfat melalui tindak balas I dan II.
Lead(II) carbonate Reaction I Lead(II) nitrate Reaction II Lead(II) sulphate
Plumbum(II) karbonat Tindak balas I Plumbum(II) nitrat Tindak balas II Plumbum(II) sulfat
(a) (i) What is meant by salt?
Apakah maksud garam?
Salts are ionic compounds produced when hydrogen ion from acid is replaced with metal ion including
ammonium ion.
(ii) Based on the flow chart above, classify the above salt to soluble salt and insoluble salt.
Berdasarkan carta aliran di atas, kelaskan garam-garam tersebut kepada garam larut dan garam tak larut.
Soluble salt / Garam larut : Lead(II) nitrate
Insoluble salt / Garam tak larut : Lead(II) carbonate, Lead(II) sulphate
(b) (i) Describe how lead(II) nitrate solution is obtained in reaction I.
Terangkan bagaimana larutan plumbum(II) nitrat diperoleh daripada tindak balas I.
– Measure and pour 50 cm3 of 1 mol dm–3 nitric acid in a beaker.
Sukat sebanyak 50 cm3 asid nitrik 1 mol dm-3 dan tuangkan ke dalam bikar.
– Lead(II) carbonate powder is added to the acid in the beaker until excess .
Serbuk plumbum(II) karbonat ditambahkan kepada asid di dalam bikar sehingga berlebihan .
– Stir the mixture with a glass rod.
Campuran tersebut dikacau dengan rod kaca.
– The mixture in the beaker is filtered.
Campuran dituraskan.
– The filtrate is lead(II) nitrate solution .
larutan plumbum(II) nitrat.
Hasil turasan ialah
m Publication Sdn. B hd.
147
Nila
MODULE • Chemistry Form 4
(ii) Write a balanced chemical equation for the reaction that occur.
Tuliskan persamaan kimia yang seimbang bagi tindak balas yang berlaku.
PbCO3 + HNO3 Pb(NO3)2 + H2O + CO2
(c) (i) Describe how to prepare pure and dry lead(II) sulphate in reaction II.
Huraikan bagaimana cara menyediakan plumbum(II) sulfat yang tulen dan kering dalam tindak balas II.
– 50 cm3 of 1 mol dm–3 lead(II) nitrate solution is added to 50 cm3 of 1 mol dm–3
sodium sulphate solution in a beaker.
50 cm3 larutan plumbum(II) nitrat 1 mol dm–3 ditambahkan kepada 50 cm3 larutan natrium sulfat
1 mol dm–3 ke dalam bikar.
– The mixture is stirred with glass rod.
Campuran tersebut dikacau dengan rod kaca.
– The mixture is filtered. The white precipitate of lead(II) sulphate is collected as the residue.
Campuran dituraskan. Mendakan putih plumbum(II) sulfat dikumpulkan sebagai baki.
– The precipitate is rinsed with distilled water .
Mendakan tersebut dibilas dengan air suling .
– The precipitate is pressed between sheets of filter papers to dry it.
Mendakan tersebut ditekan antara kertas turas .
(ii) Write an ionic equation the reaction that occur.
Tuliskan persamaan ion bagi tindak balas yang berlaku.
Pb2+ + SO42– PbSO4
(iii) Name the type of reaction that occur in reaction II.
Namakan jenis tindak balas yang berlaku dalam tindak balas II.
Double decomposition reaction
(iv) What is the step taken to make sure that pure lead(II) sulphate in reaction II is pure?
Apakah langkah yang diambil untuk memastikan plumbum(II) sulfat dalam tindak balas II tulen?
The precipitate is rinsed with distilled water.
(d) (i) Can lead(II) sulphate be prepared by adding excess of lead(II) nitrate to calcium(II) sulphate followed by
filtration. Explain your answer.
Bolehkah plumbum(II) sulfat disediakan dengan menambahkan plumbum(II) nitrat berlebihan kepada kalsium(II) sulfat dan
diikuti dengan penurasan? Terangkan jawapan anda.
– Cannot.
– Calcium sulphate is insoluble salt, it cannot form a solution and there are no free moving ions.
– Double decomposition reaction cannot occur.
(ii) Can lead(II) sulphate be prepared by adding excess of lead(II) oxide to sulphuric acid. Explain your answer.
Bolehkah plumbum(II) sulfat disediakan dengan menambahkan plumbum(II) oksida berlebihan kepada asid sulfurik? Terangkan
jawapan anda.
– Cannot.
– Lead(II) sulphate and lead(II) oxide are insoluble, both cannot be separated by filtration.
– The insoluble lead(II) sulphate will prevent lead(II) oxide to undergo further reaction with sulphuric acid.
Nila m Publication Sdn.
148
Bhd.