Modul Kimia Onsoi (TeamTenom)

Modul Kimia Onsoi! 250

What is concentration? Meaning :
Apakah kepekatan?
Maksud

Quantity of solute dissolved in a unit volume of solution

Kuantiti bahan terlarut yang larut dalam unit isipadu larutan

Formula:
Concentration in g dm-3 = Mass of solute (g)

Volume of solution (dm3)
Concentration in mol dm-3 = Number of moles in solute (mole)

Volume of solution (dm3)

What is molarity? Number of moles of solute found in 1 dm3
Apakah kemolaran? Bilangan mol bahan larut dalam 1 dm3

What is the formula n = MV
relate number of mole,n 1000
with molarity, M and
volume of solution, V? n = number of moles // bilangan mole
Apakah formula yang M = Molarity in mol dm-3 // kemolaran dalam mol dm-3
menghubungkaitkan V = Volume of solute // Isipadu pelarut
bilangan mol, n dengan
kemolaran, M dan
isipadu larutan, V?

Formula to change molar mass (jisim molar relatif)

concentration in g dm _3 ⇄Concentration (g dm-3) Molarity (mol dm-3)
to mol dm_3 and vice

versa

Formula untuk molar mass (jisim molar relatif)

menukurkan kepekatan

dalam g dm-3 kepada
mol dm-3 dan sebaliknya

1. Calculate the = 18 g
concentration in g dm-3 0.75 dm3

for 18 g of sodium = 24.0 g dm-3

hydroxide, NaOH

dissolved in water to

produce 750 cm3

Hitungkan kepekatan

dalam g dm-3 bagi 18 g

natrium hidroksida,

NaOH yang larut dalam

air 750 cm3

Modul Kimia Onsoi! 251

2. Calculate molarity = 0.1 mol
when 0.1 mol solid 0.5 dm3
calcium chloride, CaCl2
dissolve in 500 cm3 of = 0.2 mol dm_3
distilled water.
Hitungkan kemolaran
apabila 0.1 mol pepejal
kalsium klorida larut
dalam 500 cm3 air
suling.

3. What is the = 3.6 24
dm3
concentration of 3.6 g
dm-3 lithium hydroxide = 0.15 mol dm-3

solution, LiOH to
molarity ,mol dm-3.

(Relative atomic mass :

H=1, Li= 7, O=16)

Apakah kepekatan 3.6 g
dm-3 litium hidroksida
dalam unit mol dm-3

(Jisim atom relatif : H=1,

Li = 7, O =16)

4. Calculate num of n = MV 1000
= (0.2)(2.5) 1000
moles of sodium = 0.0005 mole

hydroxide, NaOH in 2.5
dm3 of 0.2 mol dm-3

sodium hydroxide

solution

Hitungkan bilangan mol

natrium hidroksida,

NaOH dalam 2.5 dm3

larutan natrium

hidroksida 0.2 mol dm-3

Modul Kimia Onsoi! 252

STANDARD SOLUTIONS
LARUTAN PIAWAI

What is a standard Solution with known its concentration

solution? Larutan yang diketahui kepekatannya

Apakah itu larutan

piawai?

How to prepare 1. Calculate the mass of solid X needed using formula n= MV
standard solution 1000, then mass = mol x RFM
Bagaimana Kira jisim pepejal X yang diperlukan menggunakan formula
menyediakan larutan n=MV 1000, kemudian jisim = mol x JFR
piawai 2. Weight the mass solid using electronic balance
Timbang berat pepejal dengan penimbang elektronik

3. Add 100 cm3 water to the solid X in a beaker
Tambahkan 100 cm3 air suling kepada pepejal X di dalam
bikar

4. Stir the mixture until solid X completely dissolved in distilled
water
Kacau campuran sehingga pepejal X larut sepenuhnya dalam
air suling

5. Transfer the solution X into volumetric flask via filter funnel
Pindahkan larutan X ke dalam kelalang volumetrik melalui
corong turas
6. Rinse the beaker and filter funnel with distilled water and
transfer its remaning solution into volumetric flask
Bilas bikar dan corong turas dengan air suling dan pindahkan
larutan ke dalam kelalang volumetrik

Modul Kimia Onsoi! 253

7. Remove filter funnel and add distiled water until it almost
reach calibration mark.
Alihkan corong turas dan tambahkan air suling sehingga ia
hampir mencecah tanda kalibrasi.
8. Drop slowly distilled water using dropper until meniscus
level is aligned on calibration mark
Titiskan dengan perlahan air suling menggunakan penitis
sehingga aras meniskus berada pada garis kalibrasi

9. Close volumetric flask with stopper. Shake well by inverting
the flask several time.
Tutup kelalang volumetrik dengan penutup. Goncangkan
secara terbalik kelalang berulang kali

What is dilution Preparing solution of known concentration by adding water into
method? kaedah concentrated standard solution/ stock solution to produce dilute
Apakah solution
pencairan? Menyediakan larutan diketahui kepekatan dengan menambahkan air
suling kepada larutan piawai pekat / larutan stok untuk meghasilkan
lauratn lebih cair

What is being added to Water
Air
solution during dilution?

Apakah yang ditambah

ke dalam larutan

semasa proses

pencairan?

What happen to Concentration : Reduce / Berkurang
Kepekatan
concentration and Number of mole solute : Same / Sama
Bilangan mol bahan larut
number of mole of

solute after dilution?

Apakah yang berlaku

kepada kepekatan dan

bilangan mol pelarut

selepas pencairan?

Modul Kimia Onsoi! 254

How to prepare 1. Calculate the volume of solution need to take out, V1 using
standard solution by formula M1V1 = M2V2
dillution? Kirakan isipadu larutan yang perlu diambil , V1 dengan formula
Bagaimanakah M1V1 = M2V2
menyediakan larutan 2. Pour stock solution into a beaker
piawai melalui kaedah Tuang larutan stok ke dalam bikar
pencairan?

3. Pipette out calculated volume of solution, V1.
Pipetkan isipadu V1 keluar

4. Transfer V1 larutan ke dalam kelalang volumetrik
Pindahkan V1 larutan ke dalam kelalang volumetrik

5. Add distilled water until solution approaches calibration mark.
Tambahkan air suling sehingga menghampiri tanda kalibrasi
6. Add distilled water slowly with dropper until reach calibration
mark.
Tambahkan air suling perlahan-lahan dengan penitis sehingga
mencecah garis kalibrasi.

7. Close the volumetric flask and shake well by inverting it
several times to get a homogenous solution
Tutupkan kelalang volumetrik dan goncang secara terbalik
beberapa kali untuk mendapat larutan yang homogen

Modul Kimia Onsoi! 255

1. Calculate the new M1 =1.5 mol dm-3

molarity of hydrochloric V1 =25 cm3

acid, HCl produced if 25 M =?
2

cm3 of 1.5 mol dm-3 V2 = 150 cm3

hydrochloric caid is M1 V1 = M2 V2

diluted to produce 150 (1.5)( 25) = M2( 150)
M2 = 0.25 mol dm-3
cm3 of hydrochloric

acid.

Hitungkan kemolaran

baru bagi asid

hidroklorik, HCl terhasil

jika 25 cm3 asid
hidroklorik 1.5 mol dm-3

dicairkan untuk

menghasilkan asid

hidroklorik 150 cm3.

2. Calculate mass of N = MV 1000= Mass
= (1.0)(1000) 1000= Mass
sodium hydroxide pellet
Mass= 40.0 g
needed to produce

1000 cm3 sodium

hydroxide solution with

molarity 1.0 mol dm-3

(Relative atomic mass

H=1, O=16, Na=23)

Hitungkan jisim pelet

natrium hidroksida yang

diperlukan untuk

menghasilkan larutan
1000 cm3 larutan

natrium hidroksida

dengan kemolaran 1.0

mol dm-3.

(Jisim atom relatif H=1,

O=16, Na=23)

NEUTRALISATION Modul Kimia Onsoi! 256
PENEUTRALAN
Reaction between acid and alkali to
What is neutralisation? produce salt and water only
Apakah itu peneutralan? Tindakbalas antara asid dan alkali
menghasilkan garam dan air sahaja

Modul Kimia Onsoi! 257

Write chemical equation and ionic equation for 1. Hydrochloric acid reacts with sodium
following reactions hydroxide
Tulis persamaan kimia dan persamaan ion bagi
tindak balas berikut Asid hidroklorik bertindakbalas dengan
natrium hidroksida

Chemical equation : HCl + NaOH →NaCl +

H2O

Persamaan kimia
Ionic equation : H+ + OH-→H2O

Persamaan ion
2. Nitric acid reacts with potassium
hydroxide

Asid nitrik bertindakbalas dengan kalium
hidroksida

Chemical equation : HNO3 + KOH →

KNO3 + H2O

Persamaann kimia
Ionic equation : H+ + OH-→H2O

Persamaan ion
3. Sulphuric acid reacts with sodium
hydroxide

Asid sulfurik bertindakbalas dengan
natrium hidroksida

Chemical equation : H2SO4 + 2 NaOH →

Na2SO4 + 2 H2O

Persamaann kimia
Ionic equation : 2 H+ +2 OH-→2 H2O

Persamaan ion
4. Ethanoic acid with potassium hydroxide

Asid etanoik bertindabalas dengan
kalium hidroksida

Chemical equation : CH3COOH + KOH →

CH3COOK + H2O
Persamaann kimia
Ionic equation : H+ + OH-→H2O
Persamaan ion

Modul Kimia Onsoi! 258

Write application of neutralisation in daily life. Agriculture / Pertanian
Tuliskan aplikasi peneutralan dalam kehidupan 1. Adding soda lime (calcium oxide) or
seharian limestone(calcium carbonate) or ashes of
burnt wood to acidic soil
Menambahkan batu kapur kepada tanah
berasid untuk mengurangkan keasidan
tanah.
2.Neutralisation between salt and alkali
produce salt as a fertilisers
Tindakbalas peneutralan antara asid dan
alkali menghasilkan garam iaitu baja
Industries / Perindustrian
1. Alkali neutralised acid in latex produced
by bacteria and prevent it from
coagulation.
Alkali meneutralkan asid di dalam lateks
yang dihasilkan bakteria dan menghalang
latks membeku
2. Acidic gases emitted by industries
neutralised with soda lime
Gas berasid dari industri dineutralkan
dengan batu kapur.
Health / Kesihatan :
1. Baking soda is used to cure acidic bee
sting
Serbuk penaik untuk merawat bisa sengat
lebah yang berasid
2. Vinegar is used to cure alkaline wasp
sting
Cuka untuk merawat bisa tebuan yang
beralkali

What is acid-based titration? Quantitative analysis method to determine
Apakah maksud pentitritan asid-bes? the volume of acid needed to completely
neutralise a given volume of alkali and
vice versa.
Analisis kuantitatif untuk menentukan
isipadu asid yang diperlukan untuk
meneutralkan dengan lengkap isipadu
alkali diketahui atau sebaliknya.

What is end-point? Modul Kimia Onsoi! 259
Apakah maksud takat akhir?
Point in titration at which acid-base
indicator changes it colour.
Titik dalam pentitratan apabila penunjuk
asid-bes berubah warna

Why white tile is used in titration? To observe colour change of acid-base
Kenapa jubin putih digunakan dalam pentitritan? indicator easily
Untuk melihat perubahan warna penunjuk
asid-bes dengan jelas

Why we should not rinse inside the conical flask Prevent droplets of alkali stick on the wall
with alkali before beginning the titration? of conical flask which will cause its volume
Kenapa kita tidak boleh membilas bahagian dalam more than actual volume
kelalang kon dengan larutan alkali sebelum Halang titisan larutan alkali melekat pada
pentitratan? dinding kelalang kon yang menyebabkan
isipadu alkali melebihi isipadu sepatutnya

Write the formula used to solve neutralisation Ma = Molarity of acid// Kemolaran asid
problem. Va = Volume of acid // Isipadu asid
Tuliskan formula digunakan untuk menyelesaikan Mb = Molarity of alkali // Kemolaran alkali
masalah penutralan. Vb = Volume of alkali // Isipadu alkali
a = mole of acid
b = mole of alkali
MaVa = a
MbVb b

1. 20 cm3 of 0.25 mol dm-3 sodium hydroxide HCl + NaOH →NaCl + H2O
solution, NaOH is neutralised with 0.2 mol dm-3 (0.2)Va = 1
hydrochloric acid, HCl. Calculate the volume of the (0.25)(20) 1

hydrochloric acid needed for this neutralisation Va = 25 cm3
reaction.
20 cm3 larutan natrium hidroksida, NaOH 0.25 mol
dm-3 dineutralkan dengan 0.2 mol dm-3 asid

hidroklorok, HCl. Kirakan isipadu adi hidroklorik =,
Hl yang diperlukan untuk tindak balas peneutralan
ini.

Modul Kimia Onsoi! 260

2. 4.05 g of zinc oxide, ZnO is needed to complete 2 HNO3 + ZnO → Zn(NO3)2 + H2O
the neutralisation of 50 cm3 of nitric acid, HNO3 .
Calculate the concentration of acid in mol dm-3 Num of moles ZnO , n = 0.05 mol
1 mol of ZnO → 2 mol HNO3
(Relative atomic mass H=1, N=14,O=16, Zn=65) 0.05 mol ZnO → 0.1 mol HNO3
Concentration of HNO3, M = = 2.0 mol dm-
4.05 g zink oksida bertindakbalas dengan lengkap
dalam peneutralan dengan 50 cm3 asid nitrik, 3
HNO3. Hitungkan kepekatan asid dalam mol dm-3

(Jisim atom relatif H=1, N=14,O=16, Zn=65)

SALTS
SALTS

What is salt? An ionic compound formed when hydrogen ion from acid is
Apakah garam? replaced with the metal ion or ammonium ion.
Sebation ion terbentuk apabila ion hidrogen daripada asid diganti
dengan ion logam atau ion ammonium.

What is type of compound Ionic compoud
in salt? Sebatian ion
Apakah jenis sebation
garam?

How are salt crystals When saturated salt solution is cooled down
formed? hablur garam Apabila larutan garam tepu disejukkan
Bagaimana
diperolehi?

What are characeristics of 1. Has flat surface, straight sides and sharp verticles
salt crystals? Mempunyai permukan rata, lurus dan bucu tajam
Apakah ciri-ciri hablur 2. Has a fixed angle between two adjacent surfaces
garam? Mempunyai sudut antara dua permukaan bersebelahan
3. Has a specific geometrical shape such as cube, cuboid,
rhombus and prism
Mempunyai bentuk geometri spesifik seperti kiub, kuboid, prisma
dan rombus
4. Different crystals have different geometrical shapes
Hablur berbeza jenis mempunyai bentuk geometri berbeza
5. Some crystals of different sizes still have the same geometrical
shape
Hablur berbeza saiz masih mempunyai bentuk geometri yang
sama

Modul Kimia Onsoi! 261

How are the soluble salts 1. Dissolve soluble salt crystal in distilled water
Larutkan hablur garam terlarutkan dalam air suling
purified by 2. Heat the solution to speed up dissolving the salt
Panaskan larutan supaya garam cepat larut
recrystallisation? 3. Filter the hot solution to remove insoluble impurities
Turaskan larutan yang panas untuk menyingkirkan bendasing
Bagaimanakah garam takterlarutkan
4. Pour the filtrate in evaporating dish
terlarutkan ditulenkan Tuangkanhasil larutan ke dalam mangkuk penyejat
5. Heat the filtrate until saturated
melalui proses Panaskan hasil turasan hingga tepu
6. Cool down until a salt crystal formed
penghabluran? Sejukkan sehingga hablur terbentuk
7. Filter the saturated salt solution
Turaskan larutan garam tepu
8. Rinse the salt with little distilled water
Bilas hablur garam dengan sedikit air suling
9. Dry the salts by pressing them between 2 filter papers
Keringkan garam dengan menekannya antara 2 kertas turas

Name classification of Soluble salt // Garam terlarutkan
Insoluble salt // Garam tak terlarutkan
salts?

Namakan pengelasan

garam?

Complete the table to

classify salt correctly based Type of salts Solubility of salts
Jenis garam Keterlarutan garam
on their solubility in water.

Lengkapkan jadual untuk

mengelaskan garam

berdasarkan Potassium, sodium All soluble
Semua larut
keterlarutannya dalam air. dan ammonium salts

Garam natrium,

kalium dan

ammonium

Nitrate salts All soluble
Garam nitrat Semua larut

Modul Kimia Onsoi! 262

Carbonate salts All insoluble except for potassium
Garam karbonat
carbonate , sodium carbonate

and ammonium carbonate salts

Semua tak terlarutkan kecuali

garam kalium karbonat ,natrium

karbonat dan ammonium

karbonat

Chloride salts All soluble except lead(II) chloride ,
Garam klorida silver chloride and mercury
chloride salt
Semua larut kecuali garam
plumbum(II) klorida, argentum
klorida dan merkuri klorida

Sulphate salts All soluble except lead(II) sulphate ,
Garam sulfat calcium sulphate and barium
sulphate salt
Semua larut kecuali garam
plumbum(II) sulfat, kalsium sulfat
dan kalsium sulfat

What are method to 1. Titration acid-base
Pentitratan asid-bes
prepare soluble salts? 2. Dissolve excess metal/ metal oxide/ metal carbonate in acid
Melarutkan hablur logam/ logam oksida/ logam karbonat dalam
Apakah kaedah untuk asid

menyediakan garam

terlarutkan?

What are method to Double decomposition // Precipitation
prepare insoluble salts? Penguraian ganda dua // Pemendakan
Apakah kaedah untuk
menyediakan garam tak
terlarutkan?

Modul Kimia Onsoi! 263

PREPARATION SALT TYPE SODIUM, POTASSIUM & AMMONIUM
PENYEDIAAN GARAM JENIS NATRIUM, KALIUM & AMMONIUM

Reaction Acid + Alkali
Tindakbalas

Method Titration
Kaedah Pentitratan

Procedure 1. Titrate and pour 25 cm3 of 1.0 mol dm-3 of alkali
Prosedur solutions in conical flask.
Titrat dan tuang 25 cm3 larutan alkali berkepekatan
1.0 mol dm-3 kedalam kelalang kon
2. Add a few drops phenolpthalein.
Tambahkan beberapa titis fenolftalein
3. 1.0 mol dm-3 acid is tittrated to the alkali until
phenoplthalein change pink to colourless.
1.0 mol dm-3 asid dititratkan ke dalam alkali
sehingga warna fenolftalein berubah dari merah
jambu kepada tidak berwarna
4. Record the volume of acid used.
Rekodkan isipadu asid digunakan
5. Repeat titration step without using phenolpthalein.
Ulang langkah pentitratan tanpa menggunakan
penunjuk fenolftalein
6. Filter the salt solution and pour into evaporating
dish
Turaskan larutan garam dan tuang ke mangkuk
penyejat
7. Heat the filtrate until saturated
Panaskan hasil turasan sehingga tepu
8. Cool it until salt crystal formed
Sejukkan sehingga hablur garam terbentuk
9. Filter the saturated salt solution
Turas larutan garam tepu
10. Rinse the salt crystal with a little distilled water
Bilas hablur garam dengan sedikit air suling
11. Dry the salt by pressing between 2 filter papers
Keringkan garam dengan menekannya antara 2
keping kertas turas

Modul Kimia Onsoi! 264

Why first round of titration the alkali need To get exactly volume of acid needed to completely
to be added with acid-base indicator? neutralised alkali
Kenapa penunjuk asid-bes ditambah ke Untuk mendapat isipadu asid yang tepat untuk
dalam alkali dalam prosedur titratan meneutralkan alkali
pertama?

Why second round of titration the alkali To get pure salt solution
need not to be added with acid-base Mendapat larutan garam yang tulen
indicator?
Kenapa penunjuk asid-bes tidak
ditambah ke dalam alkali dalam prosedur
titratan seterusnya?

PREPARATION SOLUBLE SALT NOT TYPE SODIUM, POTASSIUM & AMMONIUM
PENYEDIAAN GARAM TERLARUTKAN BUKAN JENIS NATRIUM, KALIUM & AMMONIUM

Reaction 1. Acid + Metal
Tindakbalas 2. Acid + Metal oxide
3. Acid + Metal carbonate

Procedure Modul Kimia Onsoi! 265
Prosedur
1. Pour 20 cm3 of 2.0 mol dm-3 acid solution into
beaker
Tambahkan 20 cm3 larutan asid 2.0 mol dm-3
dalam bikar
2. Add metal powder// metal oxide powder //
metal carbonate powder gradually into acid.
Tambahkan hablur logam// logam oksida dan
logam karbonat beransur-ansur ke dalam bikar
3. Stir the mixture
Kacau campuran
4. Added the metal powder// metal oxide
powder // metal carbonate powder until it more
more dissolved
Tambahkanhablur logam// logam oksida dan
logam karbonat sehingga ia tidak lagi larut
dalam asid
5. Filter the excess metal powder// metal oxide
powder // metal carbonate powder and pour
salt solution into evaporating dish
Turaskan lebihan halur logam/ logam oksida
dan logam karbonat dan tuang larutan garam ke
mangkuk penyejat
6. Heat the filtrate until saturated
Panaskan hasil turasan sehingga tepu
7. Cool it until salt crystal formed
Sejukkan sehingga hablur garam terbentuk
8. Filter the saturated salt solution
Turas larutan garam tepi
9. Rinse the salt crystal with a little distilled
water
Bilas hablur garam dengan sedikit air suling
10. Dry the salt by pressing between 2 filter
papers
Keringkan garam dengan menekannya antar 2
keping kertas turas

Modul Kimia Onsoi! 266

Why the metal powder//metal oxide powder // To ensure complete reaction with asid
metal carbonate powder added excessively Untuk memastikan ia lengkap bertindakbalas
in acid? dengan asid
Kenapakah serbuk logam// logam oksida //
logam karbonat ditambah secara berlebihan
dalam asid?

Why the mixture of excess metal/ metal oxide To remove impurities from salt solution larutan
// metal carbonate and salt solution need to Untuk menyingkirkan bendasing dari
be filtered? garam
Kenapa campuran logam// logam oksida//
logam karbonat berlebihan dengan larutan
garam perlu dituras?

Write all possible reaction to produce 1. Mg + 2 HNO3 → Mg(NO3)2 + H2
magnesium nitrat salt and also its ionic Ionic equation : Mg +2 H+→ Mg2+ + H2
equation
Tuliskan semua tindak balas yang mungkin 2. MgO + 2 HNO3 → Mg(NO3)2 + H2O
untuk menghasilkan garam magnesiun nitrat
dan tulis juga persamaan ion terlibat

3. MgCO3 + 2 HNO3 → Mg(NO3)2 + H2O + CO2
Ionic equation : CO32- +2 H+→ H2O + CO2

Modul Kimia Onsoi! 267

Write all possible reaction to produce 1. CuO + H2SO4→ CuSO4 + H2O
copper(II) sulphate salt and also its ionic
equation 2. CuCO3 + H2SO4→ CuSO4 + H2O + CO2
Tuliskan semua tindak balas yang mungkin Ionic equation : CO32- +2 H+→ H2O + CO2
untuk menghasilkan garam kuprum(II) sulfat
dan tulis juga persamaan ion terlibat

PREPARATION INSOLUBLE SALT
PENYEDIAAN GARAM TAK TERLARUTKAN

Method Double decomposition // Precipitation
Kaedah Penguraian ganda dua // Pemendakan

Procedure 1. Pour 10 cm3 of 1.0 mol dm-3 of sodium sulphate solution into a
Prosedur beakar
Tuangkan 10 cm3 larutan natrium sulfat 1.0 mol dm_3 ke dalam
bikar
2. Add 10 cm3 of 1.0 mol dm-3 of barium chloride solution into the
sodium sulphate solution.
Campurkan 20 cm3larutan barium klorida 1.0 mol dm-3 ke dalam
larutan natrium sulfat
3. Filter the mixture
Turas campuran
4. Rinse the residue with distilled water
Bilas baki turasan dengan air suling
5. Dry the salt by pressing it between 2 filter papers
Keringkan garam dengan menekannya antara 2 kertas turas

Modul Kimia Onsoi! 268

Write possible chemical Chemical equation :
1. Pb(NO3)2 + Na2CO3→ PbCO3 + 2 NaNO3
equation to prepare lead(II) 2. Pb(NO3)2 + K2CO3→ PbCO3 + 2 KNO3
3. Pb(NO3)2 + (NH4)2CO3→ PbCO3 + 2 NH4NO3
carbonate salt
Ionic equation :
Write its ionic equation Pb2+ + CO32-→ PbCO3

Tulis beberapa

kemungkinan persamaan

kimia untuk menyediakan

garam plumbum(II)

karbonat.

Tulis persamaan ion

EFFECT OF HEAT ON SALT
KESAN HABA KE ATAS GARAM

Complete method and observation for the following gases
Lengkapkan kaedah dan pemerhatian bagi gas berikut

Gas Colour // Odour Chemical test // Ujian kimia
Warna// Bau

Method Observation
Langkah Pemerhatian

Hydrogen Colourless Put burning splinter near Pop sound produce
Hidrogen Tidak test tube Bunyi pop dihasilkan
berwarna Letakkan kayu uji menyala
dekat tabung uji

Oxygen Colourless Put glowing splinter into Splinter ignites
Oksigen Tidak test tube Kayu uji menyala
berwarna Letakkan kayu uji berbara
dalam tabung uji

Carbon dioxide Colourless Flow the gas into Limewater turns cloudy
limewater Air kapur menjadi keruh
Karbon Tidak Alirkan gas ke dalam air
kapur
dioksida berwarna

Ammonia Colourless Put moist red litmus paper Red litmus paper turns blue
Ammonia Tidak
near test tube Kertas litmus merah menjadi biru

Modul Kimia Onsoi! 269

berwarna Letakkan kertas litmus biru
Pungent smell lembap pada mulut tabung
Bau tajam uji

Chlorine Greenish Placed moist blue litmus Blue litmus paper turns red then
Klorin yellow gas paper near test tube bleached
Gas kuning Letakkan kertas litmus biru Kertas litmus biru berubah merah
kehijauan lembap pada mulut tabung kemudian dilunturkan
uji

Hydrogen Colourless Dip glass rod into White fume formed
chloride Tidak concentrated concentrated Wasap putih terhasil
Hidrogen berwarna ammonia solution and put it
klorida near test tube
Celup rod kaca ke dalam
larutan ammonia pekat
dan letakkannya ke mulut
tabung uji

Sulphur Colourless Flow the gas into acidified Purple colour of potassium
manganate(VII) turns colourless//
dioxide Tidak potassium manganate (VII) Orange colour of potassium
dichromate(VI) turns green
Sulfur dioksida berwarna solution // acidified Letakkan kertas litmus biru lembap
pada mulut tabung uji
potassium dichromate(VI)

solution

Alirkan gas ke dalam

larutan kalium

manganat(VII) // Larutan

kalium dikromat(VI)

Nitrogen Brown gas Placed moist blue litmus Blue litmus paper turns red
dioxide Gas berwarna paper near test tube Kertas litmus biru berubah merah
Nitrogen perang Letakkan kertas litmus biru
dioksida lembap pada mulut tabung
uji

Complete the table with effect heat on carbonate and nitrate salt
Lengkapkan jadual bagi kesan haba ke atas garam karbonat dan nitrat.

Cation Nitrate Carbonate
Kation Nitrat Karbonat

Modul Kimia Onsoi! 270

Decompose into metal nitrite and Decompose into metal oxide and carbon
oxygen gas dioxide gas
Terurai kepada logam nitrit dan gas Terurai kepada logam oksida dan gas
oksigen karbon dioksida

K+ 2 KNO3 → 2 KNO3 + O2 -

White White

Colourless

gas

Na+ 2 NaNO3 → 2 NaNO3 + O2 -

White White

Colourless

gas

Decompose into metal oxide, oxygen gas
and nitrogen gas
Terurai kepada logam oksida, gas
oksigen dan gas nitrogen dioksida

Ca2+ 2 Ca(NO3)2 → 2 CaO + 4 NO2 + CaCO3 → CaO + CO2
White White Turns lime water
O2 cloudy
Putih Putih Air kapur menjadi
White White Brown keruh

Colourless

gas

gas

Mg2+ 2 Mg(NO3)2 → 2 MgO + 4 NO2 + MgCO3 → MgO + CO2
White White Turns lime water
O2 cloudy
Putih Putih Air kapur menjadi
White White Brown keruh

Colourless

gas

gas

Al3+ 4 Al2(NO3)3 → 2 CaO + 12 NO2 + 3 Al2(CO3)3 → Al2O3 + 3 CO2

O2 White White Turns lime water

White White Brown cloudy

Colourless Putih Putih Air kapur menjadi

gas keruh

gas

Zn2+ 2 Zn(NO3)2 → 2 ZnO + 4 NO2 + O2 ZnCO3 → ZnO + CO2

White Hot : Yellow Brown White Hot : Yellow Turns lime water

Colourless cloudy

Modul Kimia Onsoi! 271

Cold : White gas Putih Cold : White Air kapur menjadi
keruh
gas

Pb2+ 2 Pb(NO3)2 → 2 PbO + 4 NO2 + PbCO3 → PbO + CO2
White
O2 cloudy Hot : Brown Turns lime water
Putih
White Hot : Brown Brown keruh

Colourless Cold : Yellow Air kapur menjadi

Cold : Yellow gas

gas

Cu2+ 2 Cu(NO3)2 → 2 CuO + 4 NO2 + CuCO3 → CuO + CO2
Green Black Turns lime water
O2 cloudy
Hijau Hitam Air kapur menjadi
Blue Black Brown keruh

Colourless

gas

gas

KERTAS 1 11. C 21. D 31. B 41. B
12. C 22. B 32. C 42. D
1. A 13. C 23. B 33. B 43. D
2. C 14. D 24. B 34. D 44. C
3. B 15. C 25. C 35. C 45. C
4. C 16. C 26. D 36. B 46. D
5. D 17. A 27. C 37. C 47. C
6. C 18. D 28. C 38. B 48. C
7. A 19. B 29. C 39. D 49. C
8. C 20. B 30. B 40. C 50. A
9. A
10 D Scheme Mark
Skema
KERTAS 2

Question
Soalan

Modul Kimia Onsoi! 272

1. (a) (i) ethanoic acid // methanoic acid// oxalic acid 1

(ii)1. Acid X is a weak acid while hydrochloric acid is strong acid // 1

The concentration of H+ion in ethanoic acid is lower

Asid X ialah asid lemah manakala asid hidroklorik ialah asid kuat

// Kepekatan ion hidrogen dalam asid etanoik rendah 1
2. The concentration of H+ ion higher, the lower pH value

Kepekatan ion hydrogen tinggi, nilai pH rendah

(iii) 1. pH increase // pH meningkat 1

2. Concentration of hydrogen ion decrease / kepekatan ion 1

hidrogen berkurang

(b) (i) A solution with known concentration// Larutan yang diketahui 1

kepekatan

(ii) Dillution 1

(iii) Volumetric flask measure more accurate than a beaker // 1

kelalang volumetrik mengukur lebih tepat daripada bikar

(iv) M1V1 = M2V2 1

(2.0) X V1 = (1.0) X 100 1
V1 = 20 cm3

2. (a) Standard solution // 1

(b) (i) Volumetric flask measured volume of solution correctly 1

Kelalang volumetrik menyukat isipadu larutan lebih tepat 1
(ii) prevent evaporation of solution

Halang penyejatan larutan

(c)(i) number of molen = MV /1000 = 0.25 1
mass = mol X RFM = 1.2 g 1

(ii) M1V1 = M2V2 1
1(V1) = 0.1( 250) 1
V1 = 25 cm3

(d)(i) 1. Hydrochloric acid a monoprotic acid and sulphuric acid is a 1
diprotic acid 1
Asid hidroklorik adalah asid monoprotik manakala asid sulfurik adalah 1
asid driprotik
2. Number of hydrogen ion in sulphuric acid is twice / double than
hydrochlorc acid
Bilangan ion hidrogen dalam asid sulfurik dua kali ganda dalam asid
hidroklorik
(ii) 1. Pour sodium hydroxide solution into conical flask with a few drops
of phenolpthalein

Modul Kimia Onsoi! 273

Tambahkan larutan natrium hidroksida ke dalam kelalang kon dengan 1
beberapa titis fenolftalein
2. Add hydrochloric caid into conical flask until the colour turn colourless
Tambahkan asid hiidroklorik ke dalam kelalang kon sehingga warna
merah muda fenolftalein berubah tidak berwarna

3. (a) Acid is a chemical that ionise in water to produce hydogen ion // asid

ialah bahan kimia yang mengion dalam air menghasilkan ion hidrogen 1

(b) S

(c) (i) P

(ii) strong acid has high concentration of hydrogen ion //asid kuat 1

mempunyai kepekatan ion hidrogen tinggi

(d) (i) 15cm3 1

(ii) 2 NaOH + H2SO4 →Na2SO4 + 2 H2O 1

(iii) 1
Step 1: calculate Mol H2SO4, n = MV/1000 = 15(1.0)/1000 = 0.015 mol 2
Step 2 : Compare ratio

1 mol H2SO4 → 2 mol NaOH

0.015 mol H2SO4 → 0.03 mol NaOH

Step 3 : concentration of NaOH 1
n = MV/1000
0.03 = M (20)/ 1000
M = 1.5 mol dm-3

1

1

4. (a) (i) Beaker A : Water/ Air 1

Beaker B : Methylbenzene

(ii) 1

1. pH of ethanoic acid in beaker is 4.0, shows it is a weak acid

pH asid etanoik dalam bikar ialah 4.0, menunjukkan ia asid lemah

2. Ethanoic acid molecule in beaker A ionises partially in water 1

Modul Kimia Onsoi! 274

Asid etanoik molekul dalam bikar A mengion separa dalam air 1
3. Concentration of hydrogen ion is low 1
1
Kepekatan ion hidrogen rendah
(iii)

1. Ethanoic acid in beaker B consist of molecule which is neutral
Asid etanoik dalam bikar B mengandungi molekul neutral

2. The ethanoic acid molecules are not ionise to form hydrogen ion
Molekul asid etanoik tidak mengion kepada ion hidrogen

(b)(i) T
Solution T strong alkali Larutan T alkali kuat
T ionises completely in water produce high concentration of hydroxide ion
T mengion lengkap dalam air menghasilkan kepekatan ion hidroksida
tinggi

1

1

1

1

1

5. (a) Acid that ionised completely in water 1

Asid mengion lengkap dalam air

(b) (i) Solution Q // Larutan Q

(ii) has high concentration of hydrogen ion

Mempunyai kepekatan ion hidrogen yang tinggi 1

(c) (i) Phenolpthalein // Methyl orange // universal indicator

(ii)Neutralisan // Peneutralan

(iii) Pink colour of phenolpthalein turns colourless 1

Warna merah muda fenoftalein berubah tidak berwarna

(d) (i) 2 KOH +2H2SO4 → K2SO4 + 2 H2O

(ii) MaVa/ MbVb = ½ 1
1
Ma = 25 x 1 / 10 x 2
-3

= 1.25 mol dm

Modul Kimia Onsoi! 275
1
2

1

1

6. (a) Pb(NO3)2 1
(b) (i) nitrogen dioxide // oxygen

(ii) 2 Pb(NO3)2 → 2 PbO + 4 NO2 + O2 1
(iii) Step 1 : mol of Pb(NO3)2 = 33.1/ 331 = 0.1 mol 2
Step 2 : ratio 2 mol Pb(NO3)2 → 2 mol PbO

0.1 mol Pb(NO3)2 → 0.1 mol PbO

Step 3 : mass of PbO = 0.1 X RFM PbO 1
= 0.1 X (207 +16) 1
= 22.3 g

(c) (i) lead(II) carbonate // plumbum (II) karbonat
(ii) double decomposition // precipitation // penguraian ganda dua //

pemendakan

(iii) Pb(NO3)2 + Na2CO3 → PbCO3 + 2 NaNO3

1

1

2

7. (a)(i) magnesium nitrate // Copper(II) sulphate 1

(ii) Mg(NO3)2 // CuSO4 1

(iii) Mg(NO3)2 : white 1

CuSO4 : blue

(b)(i) the residue brown when hot and yellow when cold// gas released will 1

turn limewater cloudy

Baki selepas pemanasan berwarna perang semasa panas dan kuning

Modul Kimia Onsoi! 276

semasa sejuk // gas
dibebaskan menyebabkan air kapur menjadi keruh

(ii) PbCO3 → PbO + CO2 2

(iii) step 1 : mol of PbCO3 = 26.7 / 267 = 0.1 mol 1
Step 2 : 1 mol PbCO3 → 1 mol CO2
1
0.1 mol PbCO3 → 0.1 mol CO2 1
Step 3 : Volume of CO2 = 0.1 X 24 dm3 = 2.4 dm3 1
1
(c)1. Pour 2 cm3 of salt solution into test tube
2.Add 1 cm3 dilute nitric acid
3.Add 2 cm3 silver nitrate solution
4.White precipitate formed
Confirmed sulphate salt present

8. (a) Precipitate V : Lead(II) chloride/ PbCl2 // plumbum(II) klorida 1
2
Solution A : Sodium nitrate // NaNO3 // natrium nitrat 1
(b) (i) Double decomposition // precipitation
2
Penguraian ganda dua // pemendakan 1
(ii) Pb(NO3)2 + 2 NaCl → PbCl2 + 2 NaNO3
(c) Mol lead (II) nitrate

n Pb(NO3)2 = (1.0) (50) / 1000 = 0.05 mol
1 mol Pb(NO3)2 → 1 mol PbCl2

0.05 mol Pb(NO3)2 → 0.05 mol PbCl2

Mass of PbCl2 = 0/05 x 242 = 12.1 g

(d) 1. Add diluted sulphuric acid followed by iron(II) sulphate solution 1
Tambahkan asid sulfurik cair diikuti dengan larutan ferum(II) sulfat 1
2. Slowly add concentrated sulphuric acid 1

Perlahan-lahan titis asid sulfurik pekat
3. Brown ring form indicate nitrate ion presence //

Cincin perang terhasil membuktikan ion nitrat hadir
(e) Same // 12.1 g

Modul Kimia Onsoi! 277
1

1

1
1

9. (a) CH3COOH + NaOH → CH3COONa + H2O 1
Mol of CH3COOH = (50X 0.1)/ 1000 = 0.005 mol 1
1 mol CH3COOH → 1 mol NaOH
1
0.005 mol CH3COOH → 0.005 mol NaOH
Mass of x = 0.005 x 40 = 0.2 g

(b) 1. Solvent L : Water 1

Pelarut L : Air 1
1
2. Solvent M : Methylbenzene // propanone // acetone // chloroform // 1

tetrachlromethane // toluene

Pelarut M : metilbenzena// propanon // aseton // kloroform //

tertklorometana // tuloune
3. Ethanoic acid ionised partially in water , H+ ion present
Asid etanoik mengion separa dalam air, ion H+ hadir
4. Ethanoic caid do not ionised in organic solvent, no H+ ion present
Asid etanoik tidak mengion dalam pelarut organic, tiada ion H+ hadir

2 CH3COOH + Mg → CH3( COO)2Mg + H2

(c) (i) 1. pH value for 0.001 mol dm_3 of HCl is higher than pH valur 1
for 0.1 mol dm-3 of HCl 1
Nilai pH untuk 0.001 mol dm_3 HCl lebih tinggi daripada nilai pH untuk
0.1 mol dm-3HCl

1. Concentration of H+ ion in 0.001 mol dm-3 of HCl is lower than
0.1 mol dm-3 HCl.

Modul Kimia Onsoi! 278

Kepekatan ion H+ dalam 0.001 mol dm_3 HCl lebih rendah 2
daripda 0.1 mol dm_3 HCl 1

(ii)1. HCl is a strong acid // HCl asid kuat
2. Ionized completely in water // mengion dalam air
3. to produce higher concentration of H+ ion // menghasilkan

kepekatan ion H+ tinggi
4. pH value is lower // Nilai pH rendah
5. CH3COOH is a weak acid // CH3COOH adalah asid lemah
6. ionised partially in water // mengion separa dalam air
7. to produce lower concentration of H+ ion // menghasilkan

kepekatan ion H+ ion rendah
8. pH value is higher // nilai pH tinggi

1

10. (a) 1. Acid A : hydrochloric acid // nitric acid// sulphuric acid 1
2. Acid B : Ethanoic acid 1
3. Acid A is a strong acid // ionised completely in water
Asid A ialah asid kuat // mengion lengkap dalam air 1

1
1
1
1
1
1
1
1

Modul Kimia Onsoi! 279

4. Acid B is a weak acid// ionised partially in water 1
Asid B ialah asid lemah // mengion separa dalam air 1
5. Acid A produced high concentration of hydrogen ion and acid B
produce low concentration of hydrogen ion 1
Asid A menghasilkan kepekatan ion hidrogen tinggi makala asidd
B menghasilkan ion hidrogen yang rendah
6. The higher concentration of hydrogen ion the lower pH value
Semakin tinggi kepekatan ion hidrogen, semakin rendah nilai pH

(b) 1. Vinegar// tamarind juice// lime Cuka/ air limau / jus asam jawa 1
2. The acid neutralised the sting which is alkaline. 1
Asid meneutralkan bisa yang beralkali
3.The acid do not burn Asid tidak membakar kulit 1

(c) (i) 1. S : Copper(II) carbonate// iron(II) carbonate // iron(III) carboate 1
2. T : Copper(II) oxide // iron(II) oxide // iron(II) oxide
3. U : carbon dioxide 1
4. W: Copper(II) sulphate // Iron(II) sulphate// Iron(III) sulphate 1
1
(ii)
2
(iii) 1. Pour of the 2 cm3 salt solution into 2 different test tubes
2. Add sodium hydroxide solution into one test tube 1
3. Blue/ Green/ Brown precipitate form to show present of 1
Copper(II) ion, Iron (II) ion and iron(III) ion 1
4. Add barium chloride solution into another test tube
5. White precipitate formed to show present of sulphate ion // 1
Mendakan putih terhasil menunjukkan kehdiran ion sulfat
1

10. (a) (i) Solvent X : tetrachloromethane // methylbenzene

Solvent Y : water 1

(ii) Procedure :

1. Add 1 spatula of zinc/ magnesium/ aluminium 1

powder or any metal carbonate powder into 2

beakers containing hydrogen chloride in solvent X

and solvent Y 1

2. No changes in beaker A

3. Gas bubble formed in beaker B

4. Hydrogen chloride in solvent X/ tetreachlromethane/

methylbenzene does not show acidic property// H+

Modul Kimia Onsoi! 280

ion absent 1
5. Hydrogen chloride in water shows acidic property/ 1
1
H+ ion present 1

1. Tambahkan 1 spatula serbukk logam zink/
magnesium/ aluminium // atau logam karbonat ke
dalam 2 bikar yang mengandungi hydrogen klorida
dalam pelarut X dan pelarut Y

2. Tiada perubahan dalam bikar A
3. Ada gelembung gas terhasil dalam bikar B
4. Hidrogen klorida dalam pelarut X/ tetraklorometana/

metilbenzena tidak menunjukkkan sifat asid// tiada
ion hydrogen hadir
5. Hidrogen klorida dalam air menunjukkan sifat asid/
ada ion hydrogen hadir

(b) (i) 1. Hydrogen chloride in solvent Y / water

2. Ionise to hydrogen ion to show acidic property

1.Hidrogen klorida dalam larutan Y/ Aair

2.Mengion lengkap kepada ion hydrogen

(ii) Able to describe an exeperiment to prepare dry zinc chloride

Procedure

1. Measure (20-100 cm3) of (0.1-2.0 mol dm-3) hydrochloric

caid using measuring cylinder

2. Pour the solution into a beaker and heat it

3. Add zinc oxide// Zinc carbonate excessively

4. Strir the mixture

5. Filter the mixture

6. Heated the filtrate until saturated / 1/3 from original

volume

7. Cool the saturated salt solution 1

8. Filter the salt formed

9. Rinse the salt with a little distilled water 1

10. Dry the salt by pressing between 2 filter papers

1. Ukur (20-100 cm3) (0.1-2.0 mol dm-3) asid hidroklorik
menggunakan silinder penyukat

2. Tuang asid ke dalam bikar dan panaskan
3. Tambah serbuk zink oksida// zink karbonat sehingga

berlebihan
4. Kacau campuran
5. Turaskan campuran

Modul Kimia Onsoi! 281

6. Panaskan hasil turasa sehingga menjadi tepu / 1/3 dari 1
isipadu asal
1
7. Sejukkan larutan garam yang tepu 1
8. Turas larutan garam tepu
9. Bilas garam dengan sedikit air suling
10. Keringkan garam sambil menekannya antar 2 keping

kertas turas

1

1

1

1

1

1

1

KERTAS 3 Marks
Markah
Question Answer
Soalan Jawapan 3
1(a)

PQRST UV
0.8 1.6 2.6 3.4 4.2 4.2 4.2

1(b) (i) Height of precipitation increase test tube P to T the start to constant 3
from T to V
Tinggi mendakan bertambah dari P ke T tetapi mulai kekal sama
bermula dari T, U dan V

Modul Kimia Onsoi! 282

1(b)(ii) Sulphate ions in test tube P,Q,R,S not completely react but in test 3
1 (c) tube T,U and V completely react. 3
Ion sulfat dalam tabung uji P,Q,R dan S tidak bertindakbalas dengan
1 (d) lengkap tetapi dalam tabung uji T, U dan V lengkap bertindakbalas.
1 (e)
Mv : Volume of barium chloride solution
Isipadu larutan barium klorida

Rv : Height of precipitate
Tinggi mendakan

Cv : Volume and concentration of potassium sulphate solution
Isipadu dan kepekatan larutan kalium sulfat

Volume of barium chloride increase until it reach a constant height 3

Isipadu larutan barium klorida bertambah tinggi mendakan bertambah

sehingga mencapai ketinggian yang sama

1. Both axis are labelled // kedua-dua paksi dilabel 3
2. Consistent scale at both axes // Skala seragam setiap paksi
3. All points transferred // semua titik dipindahkan
4. Curve cover at least 50% of graph paper // saiz graf sekurang-

kurang 50%
5. Correct curve // lengkung yang betul

Modul Kimia Onsoi! 283

1 (f) 5cm3 3
1(g) (i) 3
Moles of barium ion = 0.0025
1 g (ii) Moles of sulphate ion = 0.0025 3
Moles of barium chloride = 1

2+ 2-
Ba + SO4 → BaSO4

1(h) Add barium chloride solution into potassium sulphate and white 3

precipitate form

Tambahkan larutan barium klorida ke dalam larutan kalium sufat dan

mendakan putih terhasil

1 (i) 1. The reactants react thoroughly // faster 3
B ahan tindakbalas bertindakbalas sepenuhnya// cepat
2.precipitate sink at the bottom
Mendakan terbentuk di dasar tabung uji

1(j) 3
cation
anion

Potassium ion Sulphate ion
Barium ion Chloride ion

Modul Kimia Onsoi! 284

Chapter 7: RATE OF REACTION
Bab 7: KADAR TINDAK BALAS

1.0 Set Induction
Set Induksi

Observe the diagrams below that show the chemical reactions that take place in our daily life
Perhatikan gambarajah di bawah yang menunjukkan tindak balas kimia yang berlaku dalam
kehidupan seharian.

Is the reaction faster or higher? Adakah tindak balas in cepat atau tinggi?
Is the rate of reaction faster or higher? kadar tindak balas cepat atau tinggi?

Fill in the blanks with a (√) Isikan tempat kosong dengan tanda (√)

Fast Reaction Phenomena Slow Reaction
Tindak balas cepat Fenomena Tindak balas perlahan

Rusting of Iron
Pengaratan besi

Photosynthesis
Fotosintesis

Boiling of water
Pendidihan air

Dissolving salt in water
Melarutkan garam dalam air

Modul Kimia Onsoi! 285

2.0 Check for understanding
Menyemak kefahaman

No. Terms/ Concepts Define/ explain Remark
Bil. istilah/ konsep Takrifkan / terangkan s
Catatan

1 Rate of reaction
Kadar tindak balas

2 Average rate of reaction
Kadar tindak balas purata

3 Rate of reaction at given time
Kadar tindak balas pada masa diberi

4 Collision theory
Teori pelanggaran

5 Frequency of collision
Frekuensi pelanggaran

6 Frequency of effective collision
Frekuensi pelanggaran berkesan

7 Catalyst
Mangkin

8 Activation energy
Tenaga pengaktifan

3.0 Easy to do questions (baby steps)

Meaning of Rate of Reaction
Maksud Kadar Tindak Balas

1. Rate of reaction is a measurement of the change in quantity of reactant or product in a certain
range of time.
Kadar tindak balas adalah suatu pengukuran perubahan kuantiti bahan tindak balas atau
hasil tindak balas dalam selang masa tertentu.

Rate of reaction = change in quantity of reactant or product / time taken
Kadar tindak balas = perubahan kuantiti bahan tindak balas atau hasil tindak balas / masa

Modul Kimia Onsoi! 286

2. It is the speed at which reactants are converted into the products in a chemical reaction.
Kelajuan bahan tindak balas ditukarkan kepada hasil tindak balas dalam tindak balas kimia.

3. A rate of reaction is…………. if the reaction occurs fast within a short period of time.
Kadar tindak balas ialah ……… jika tindak balas berlaku cepat dalam masa yang singkat.

4. A rate of reaction is…………. if the reaction occurs slowly within a long period of time.
Kadar tindak balas ialah ……… jika tindak balas berlaku perlahan dalam masa yang lama.

5. Rate of reaction is inversely proportional with time
Kadar tindak balas berkadar songsang dengan masa

Rate of reaction ∝ 1 / time taken

Kadar tindak balas ∝ 1/ masa yang diambil

The shorter the time taken, the …………. the rate of reaction.
Masa tindak balas semakin rendah, kadar tindak balas …………..

The longer the time taken, the …………. the rate of reaction.
Masa tindak balas semakin panjang, kadar tindak balas …………..
6. The change in amount of reactant or product that can be measured by :

● increasing in volume of gas released
Peningkatan isipadu gas yang dibebaskan

● increasing in mass/concentration of product
Peningkatan jisim/ kepekatan bahan tindak balas

● decreasing in mass/concentration of reactant
Pengurangan jisim/ kepekatan bahan tindak balas

● Formation of precipitate
Pembentukan mendakan

● changes in pH, temperature or electrical conductivity
Perubahan nilai pH, suhu atau kekonduksian elektrik

Modul Kimia Onsoi! 287

Activity/ Aktiviti
Study the reaction between 2.0 g of CaCO3 with excess of 1.0 mol dm-3 HCl.
Kaji tindak balas antara 2.0 g CaCO3 dengan 1.0 mol dm-3 HCl berlebihan

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

Sketch the graph below for:
Lakarkan graf untuk perkara berikut:

Mass of CaCO3 against of time Concentration of HCl against time
Jisim CaCO3 lawan masa Kepekatan HCl lawan masa

Concentration of CaCl2 against time Volume of CO2 against time
Kepekatan CaCl2 lawan masa Isipadu CO2 lawan masa

Application of Rate of Reaction Modul Kimia Onsoi! 288
Aplikasi Kadar tindak balas
Collision Theory of Particles
Teori Pelanggaran Zarah

Factor/ Faktor
…………………………………………………..
causes/ memberikan kesan kepada
……………………………………………………
……………………………………………………
…………………………………………………...

Frequency of Collision/ Frekuensi
pelanggaran
………………………………………………….
Frequency of Effective Collision/ Frekuensi
pelanggaran Berkesan
………………………………………………….

Factor/ Faktor
…………………………………………………..
causes/ memberikan kesan kepada
……………………………………………………
……………………………………………………
…………………………………………………...

Frequency of Collision/ Frekuensi
pelanggaran
………………………………………………….
Frequency of Effective Collision/ Frekuensi
pelanggaran Berkesan
………………………………………………….

Factor/ Faktor
…………………………………………………..
causes/ memberikan kesan kepada
……………………………………………………
……………………………………………………
…………………………………………………...

Frequency of Collision/ Frekuensi
pelanggaran
………………………………………………….
Frequency of Effective Collision/ Frekuensi
pelanggaran Berkesan
………………………………………………….

Modul Kimia Onsoi! 289

Factor/ Faktor
…………………………………………………..
causes/ memberikan kesan kepada
……………………………………………………
……………………………………………………
…………………………………………………...

Frequency of Collision/ Frekuensi
pelanggaran
………………………………………………….
Frequency of Effective Collision/ Frekuensi
pelanggaran Berkesan
………………………………………………….

4.0 SPM Format Questions
Soalan -soalan format SPM

Paper 1/ Kertas 1
1.

Modul Kimia Onsoi! 290
2.

Paper 2/ Kertas 2

An experiment was conducted to study the effect of temperature on the rate of reaction between 50 cm3 of
sodium thiosulphate solution 0.2 mol dm-3 and 5 cm3 of sulphuric acid 1.0 mol dm-3 to form a yellow
precipitate. Diagram 5 shows the set-up of apparatus for the experiment.
Satu eksperimen dijalankan untuk mengkaji kesan suhu terhadap kadar tindak balas antara 50 cm3
larutan natrium tiosulfat dengan kepekatan 0.2 mol dm-3 dan 5 cm3 asid sulfurik berkepekatan 1.0 mol
dm-3 bagi membentuk mendakan kuning. Rajah 5 menunjukkan susunan radas untuk eksperimen
tersebut.

The experiment was repeated five times at different temperatures. Table 4 shows the temperature and
time taken for mark „X‟ to disappear from view. Eksperimen ini diulangi lima kali pada suhu yang berbeza.
Jadual 4 menunjukkan suhu dan masa yang diambil untuk tanda “X” hilang.

Modul Kimia Onsoi! 291

(a) Draw a graph of temperature against time for this experiment. Lukiskan graf suhu lawan masa bagi
eksperimen ini.

[3 marks]

Modul Kimia Onsoi! 292

Modul Kimia Onsoi! 293

(b) Compare the rate of reaction between Experiment 2 and Experiment 4. Explain your answer by using
the Collision Theory. Bandingkan kadar tindak balas di antara Eksperimen 2 dan Eksperimen 4.
Terangkan jawapan anda dengan menggunakan teori perlanggaran jirim.
………….…………………………………………………………………………………………………
…………………….………………………………………………………………………………………
……………………………….……………………………………………………………………………
………………………………………….…………………………………………………………………
……………………………………………………..………………………………………………………

[5 marks]
(c) Write the chemical equation for the reaction. Tuliskan persamaan kimia bagi tindak balas tersebut.
………………….………………………………………………………………………………………………

[1 mark]
(d) Name the yellow precipitate formed. Namakan mendakan kuning yang terbentuk.
……………..……………………………………………………………………………………………………

[1 mark]
(e) State one other factor that can affect the rate of reaction for this experiment. Nyatakan satu faktor
yang mempengaruhi kadar tindak balas bagi eksperimen ini.
……………..…………………………………………………………………………………………………….

[1 mark]

Paper 3/ Kertas 3

Two experiments are carried to study the effect of the size of calcium carbonate on the rate of reaction.
Experiment I: 1 g of calcium carbonate chips react with 20.0 cm3 of 0.2 mol dm-3 hydrochloric acid
Experiment II: 1 g of calcium carbonate powder react with 20.0 cm3 of 0.2 mol dm-3 hydrochloric acid
The rate of reaction is determined by measuring the volume of of carbon dioxide gas given off against
time using the downward displacement of water method. The volume of gas released is recorded in Table
8 and Table 9 below.

Dua eksperimen dijalankan untuk mengkaji kesan saiz kalsium karbonat terhadap kadar tindak balas.
Eksperimen I: 1g ketulan kalsium karbonat bertindak balas dengan 20.0 cm3 of 0.2 mol dm-3 asid
hidroklorik
Eksperimen II: 1g serbuk kalsium karbonat bertindak balas dengan 20.0 cm3 of 0.2 mol dm-3 asid
hidroklorik

Experiment 1/ Eksperimen 1

Time/ s 0 60 120 180 240 300 360
Masa/ s 50.00
0.00 9.50 8.00 8.00
Burette reading/ cm3
Bacaan buret/ cm3

Volume of gas/ cm3
Isipadu gas/ cm3

Table 8/ Jadual 8

Modul Kimia Onsoi! 294

Experiment 2/ Eksperimen 2

Time/ s 0 60 120 180 240 300 360
Masa/ s 9.50 8.00 8.00
50.00 22.00 13.50 9.00
Burette reading/ cm3
Bacaan buret/ cm3 0.00

Volume of gas/ cm3
Isipadu gas/ cm3

Table 9/ Jadual 9

Diagram 17 shows the burette reading for Experiment I at 60, 120 and 180 seconds
Rajah 17 menunjukkan bacaan buret bagi eksperimen I pada 60, 120 dan 180 saat.

Diagram 17/ Rajah 17

(a) Record the burette reading and the volume of gas released at 60, 120 and 180 seconds in Table
8. Rekodkan bacaan buret dan isipadu gas terbebas pada 60, 120 dan 180 saat dalam jadual 8.

[3 marks]
(b) Calculate the volume of gas released in both experiments in Table 8 and Table 9. Kirakan isipadu gas
yang dibebaskan dalam kedua-dua eksperimen dalam jadual 8 dan 9.

[3 marks]
(c) By using the same axes, draw a graph of the volume of gas released against time for Experiment
I and II. Dengan menggunakan paksi yang sama, lukiskan graf isipadu gas melawan masa bagi kedua-
dua eksperimen.

[3 marks]

Modul Kimia Onsoi! 295

Modul Kimia Onsoi! 296

(d) Based on the graph in (c), which experiment has a higher rate of reaction? Explain your answer.
Berdasarkan graf di (c), eksperimen yang manakah mempunyai kadar tindak balas yang lebih tinggi?
Jelaskan jawapan anda.

………………………………………………………………………………………
.............…………………………………………………. ....................................

[3 marks]
(e)
(i) State the variables involved in this experiment. Nyatakan pemboleh ubah yang terlibat.

Manipulated variable Pemboleh ubah dimanipulasikan
………………………………………………………………………………………………………
Responding variable Pemboleh ubah bergerak balas
…………………………………………………….…………………………………………………
Constant variable Pemboleh ubah yang dimanipulasikan
……….………………………………………………………………………………………………

[3 marks]
(ii) State the hypothesis for these experiments. Nyatakan hipotesis untuk eksperimen tersebut.
………..…....…………………………………………...……………………………………………
………. ..…..……………………...…………………………………………………………………
………….……….……………………………………………………………………………………

[3 marks]
(f) Predict the volume of gas released at 420 seconds in Experiment II.
Ramalkan isipadu gas terbebas pada 420 saat di Eksperimen II.
……………..…………………………………………………………………………………………...…

[3 marks]
(g) Classify the following reaction into fast reaction and slow reaction
Kelaskan tindak balas yang berikut kepada tindak balas cepat dan perlahan.

Rusting, Fermentation, Neutralization, Photosynthesis, Combustion,
Precipitation, Corrosion, Displacement

Pengaratan, Penapaian, Peneutralan, Fotosintesis, Pembakaran,
Mendakan, Penghakisan, Penyesaran

[3 marks]

Modul Kimia Onsoi! 297

5.0 Answers
Jawapan

Set Induksi

Is the reaction fast or slow? Adakah tindak balas in cepat atau perlahan?
fast fast slow fast slow

Is the rate of reaction low or high? kadar tindak balas rendah atau tinggi?
high high low high low

Fill in the blanks with a (√) Isikan tempat kosong dengan tanda (√)

Fast Reaction Phenomena Slow Reaction
Tindak balas cepat Fenomena Tindak balas perlahan

/ Rusting of Iron /
Pengaratan besi
/ /
Photosynthesis
Fotosintesis

Boiling of water
Pendidihan air

Dissolving salt in water
Melarutkan garam dalam air

Modul Kimia Onsoi! 298

2.0 Check for understanding/ Menyemak kefahaman

No. Terms/ Concepts Define/ explain Remarks
Bil. istilah/ konsep Takrifkan / terangkan Catatan

1 Rate of reaction Rate at which reactants are converted to
Kadar tindak balas products over time/ Measure of how quickly
a chemical reaction happen over time
Kadar bahan tindak balas ditukarkan
kepada hasil tindak balas pada suatu
tempoh masa/ pengukuran berapa cepat
suatu tindak balas kimia berlaku dalam
suatu tempoh masa

2 Average rate of reaction The rate of reaction over a period of time
Kadar tindak balas purata interval
Kadar tindak balas yang berlaku dalam satu
tempoh masa

3 Rate of reaction at given time Rate of reaction at any given time
Kadar tindak balas pada masa Kadar tindak balas pada masa tertentu
diberi

4 Collision theory Colliding particles must have enough energy
Teori pelanggaran and the reacting particles must collide at a
correct orintation
Zarah-zarah yang berlanggar mesti
mempunyai tenaga yang cukup dan
perlanggaran zarah-zarah bahan mesti
pada orientasi yang betul

5 Frequency of collision Rate of collision between the reacting
Frekuensi pelanggaran particles
Kadar perlanggaran zarah-zarah bahan

6 Frequency of effective collision A qualitative explanation of chemical

Frekuensi pelanggaran reactions and the rates at which they occur.

berkesan Penerangan secara kualitatif untuk tindak

balas kimia dan kadar

7 Catalyst A chemical substance that speeds up the
Mangkin rate of chemical reaction
Sebatian kimia yang dapat meningkatkan
kadar tindak balas suatu tindak balas kimia

8 Activation energy Modul Kimia Onsoi! 299
Tenaga pengaktifan
The minimum energy required for reactants
to start reaction
Tenaga minima yang diperlukan oleh zarah-
zarah bahan untuk memulakan tindak balas
kimia

3.0 Easy to do questions (baby steps)
10.1 : Meaning of Rate of Reaction Maksud Kadar Tindak Balas
3. A rate of reaction is HIGH if the reaction occurs fast within a short period of time.

Kadar tindak balas ialah TINGGI jika tindak balas berlaku cepat dalam masa yang singkat.
4. A rate of reaction is LOW if the reaction occurs slowly within a long period of time.

Kadar tindak balas ialah RENDAHjika tindak balas berlaku perlahan dalam masa yang
lama.
5. Rate of reaction is inversely proportional with time
Kadar tindak balas berkadar songsang dengan masa

Rate of reaction ∝ 1 / time taken

Kadar tindak balas ∝ 1/ masa yang diambil

The shorter the time taken, the HIGHER the rate of reaction.
Masa tindak balas semakin rendah, kadar tindak balas semakin TINGGI

The longer the time taken, the LOWER. the rate of reaction.
Masa tindak balas semakin panjang, kadar tindak balas semakin RENDAH

Sketch the graph below for the below Lakarkan graf untuk perkara berikut:

Mass of CaCO3 against of time Concentration of HCl against time
Kepekatan HCl lawan masa

Jisim CaCO3 lawan masa