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99 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Meaning of salt Soluble salt Insoluble salt Anion Cation SALT 1 In food preparation 2 In agriculture 3 In medical substances 1 Characteristics of salt crystals 2 To grow salt crystal Some salts decompose when they are heated: Salt → metal oxide + gas (Colour of residue refers to certain cation) (Gas identification refers to certain anion/cation) Reaction of acids with: (a) Alkali (b) Metal oxide (c) Metal carbonate (d) Metal (a)Silver nitrate and nitric acid for chloride ion (b)Barium nitrate and nitric acid for sulphate ion (c) Iron(II) sulphate and sulphuric acid for nitrate ion (d)Nitric acid and limewater for carbonate ion (a)Sodium hydroxide solution (b)Aqueous ammonia (c) Specific reagents Double decomposition through precipitation Confirmatory test using reagent Solubility Qualitative analysis Determining formula Preparation Preparation Action of heat on salt Confirmatory test using reagent Salt crystals Uses in daily life Concept Map Continuous variation method 06 U6a Chemistry F4(p98-131)csy2p.indd 99 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 100 UNIT 6 MODULE • Chemistry FORM 4 Define acid LS 6.1.1 • Acid is a chemical substance which ionises in water to produce hydrogen ion, H+. Hydrogen chloride gas is a covalent compound exist in the form of molecule. When the gas is bubbled into water, hydrochloric acid is produced. Explain. • 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. − + I When dissolve in water + I • Ionisation equation: HCl(aq) H+ (aq) + Cl– (aq) Hydrochloric acid Hydrogen ion Chloride ion • 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+ . HCl(g) +H2O(l) H3O+ (aq) + Cl– (aq) Hydrogen Hydroxonium Chloride chloride ion ion H3O+ H+ (aq) + H2O Hydroxonium ion Hydrogen ion The ionisation of hydrochloric acid is represented as: H2O HCl(aq) H+ (aq) + Cl– (aq) H+ Remark: Hydroxonium ion, H3O+ is formed by the transfer of a hydrogen ion (H+) from the hydrogen chloride to the lone pair of electrons on the water molecule. The type of bond formed is of dative bond (studied in Unit 5, chemical bond) What are the physical properties of acid? • Acid tastes sour, corrosive, turns moist blue litmus paper to red and conduct electricity in aqueous solution state. What is basicity of acid? LS 6.1.2 • Basicity of an acid is the number of ionisable hydrogen atom per molecule of an acid molecule in an aqueous solution. • Monoprotic: One acid molecule ionises to one hydrogen ion. • Diprotic: One acid molecule ionises to two hydrogen ions. • Triprotic: One acid molecule ionises to three hydrogen ions. • Hydrochloric acid is monoprotic acid because one molecule of hydrochloric acid ionises to one hydrogen ion. Examples of acid and their basicity Ionisation of acid Number of hydrogen ions produced per molecule of acid Basicity of acid HNO3(aq) H+ (aq) + NO3 – (aq) Nitric acid Hydrogen ion Nitrate ion One Monoprotic H2SO4(aq) 2H+ (aq) + SO4 2–(aq) Sulphuric acid Hydrogen ion Sulphate ion Two Diprotic H3PO4(aq) 3H+ (aq) + PO4 3–(aq) Phosphoric acid Hydrogen ion Phosphate ion Three Triprotic *CH3COOH(aq) CH3COO– (aq) + H+ (aq) Ethanoic acid Ethanoate ion Hydrogen ion One Monoprotic Not all hydrogen atoms in ethanoic acid are ionisable 6.1 THE ROLE OF WATER IN SHOWING ACIDIC AND ALKALINE PROPERTIES CS 6.1 PERFORMANCE LEVEL (PL) Mastered Not mastered PL1 Recall knowledge and basic skills on acids, bases and salts. 06 U6a Chemistry F4(p98-131)csy2p.indd 100 21/12/2022 3:50 PM

101 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Define base • Bases is a chemical substance that reacts with acid to produce salt and water only. Example of base • 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. (a) Copper(II) oxide (a base) reacts with sulphuric acid to produce copper(II) sulphate (a salt) and water. CuO + H2SO4 CuSO4 + H2O (b) Zinc hydroxide (a base) reacts with hydrochloric acid to produce zinc chloride (a salt) and water. Zn(OH)2 + 2HCl ZnCl2 + 2H2O Define alkali • Alkali is a base that is soluble in water and ionises to hydroxide ion, OH– . Examples of alkali (a) Sodium hydroxide dissolves in water and ionises to hydroxide ion. NaOH(aq) Na+ (aq) + OH– (aq) (b) Ammonia solution is obtained by dissolving ammonia molecule in water, ionisation occur to produce a hydroxide ion, OH– . NH3(g) + H2O(l) NH4 + (aq) + OH– (aq) (c) Other examples of alkalis are barium hydroxide and calcium hydroxide. What are the physical properties of alkali? • Alkali tastes bitter, slippery and turns moist red litmus paper to blue and conduct electricity in aqueous solution state. Complete the following table: PL2 Soluble base (alkali) Insoluble base Names Formula Ionisation equation Names Formula Sodium oxide Na2O Na2O(s) + H2O 2NaOH(aq) NaOH(aq) Na+ (aq) + OH– (aq) Copper(II) oxide CuO Potassium oxide K2O K2O(s) + H2O 2KOH(aq) KOH(aq) K+ (aq) + OH– (aq) Copper(II) hydroxide Cu(OH)2 Ammonia NH3 NH3(g) + H2O NH4 + (aq) + OH– (aq) Zinc hydroxide Zn(OH)2 Sodium hydroxide NaOH NaOH(aq) Na+ (aq) + OH– (aq) Aluminium oxide Al2O3 Potassium hydroxide KOH KOH(aq) K+ (aq) + OH– (aq) Lead(II) hydroxide Pb(OH)2 Barium hydroxide Ba(OH)2 Ba(OH)2(aq) Ba2+ (aq) + 2OH– (aq) Magnesium hydroxide Mg(OH)2 Bases that can dissolve in water (soluble bases) are known as alkali Exercise LS 6.1.1 LS 6.1.1 PERFORMANCE LEVEL (PL) Mastered Not mastered PL2 Understand and explain acids, bases and salts. 06 U6a Chemistry F4(p98-131)csy2p.indd 101 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 102 UNIT 6 MODULE • Chemistry FORM 4 How do you explain the role of water in the formation of hydrogen ion which causes acidic properties? • Acid molecules ionise in water to form hydrogen ions. • The presence of hydrogen ions is needed for the acid to show its acidic properties. Acid does not show acidic properties without water or dissolved in organic solvent. Explain. • Acid will remain in the form of molecules in two conditions: (a) Without the presence of water for example dry hydrogen chloride gas and *glacial ethanoic acid. (b) Acid is dissolved in *organic solvent for example solution of hydrogen chloride in methylbenzene and ethanoic acid in propanone. * Glacial ethanoic is pure ethanoic acid * Organic solvent is covalent compound that exist as liquid at room temperature such as propanone, methylbenzene and trichloromethane Role of Water and the Properties of Acid 1 Uses of acid Acid Uses Sulphuric acid • To make fertiliser, detergents, paint, synthetic polymer and electrolyte in lead-acid accumulator Hydrochloric acid • As a cleansing agent in a toilet cleaner, to clean metal before electroplating • Hydrochloric acid react with oxide layer on the surface of metal Nitric acid • To manufacture fertilisers, explosive substances, dyes and plastic Ethanoic acid • To make vinegar and as a preservative for pickles Benzoic acid • To preserve food 2 Uses of base and alkali Alkali / Base Uses Ammonia • To make fertiliser, nitric acid, grease remover and keep latex in liquid state Calcium hydroxide (lime) • To neutralise acidic soil, to make cement and limewater Magnesium hydroxide • To make toothpaste and antacid (to neutralise excess acid in the stomach) Sodium hydroxide • To make soap and detergents Aluminium hydroxide • To make antacid (to neutralise excess acid in the stomach) Uses of Acid, Bases and Alkali in Daily Life PERFORMANCE LEVEL (PL) Mastered Not mastered PL3 Apply knowledge on acids, bases and salts to explain the natural occurrences or phenomena and be able to carry out simple tasks. Example 1: Hydrogen chloride gas dissolved in propanone and in water. LS 6.1.3 Hydrogen chloride gas dissolved in propanone Propanone Hydrogen chloride molecules in propanone do not ionise . Hydrogen chloride gas dissolved in water Water Hydrogen chloride molecule in water ionises to hydrogen ion and chloride ion: HCl(aq) ➝ H+ (aq) + Cl– (aq) 06 U6a Chemistry F4(p98-131)csy2p.indd 102 21/12/2022 3:50 PM

103 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Example 2: Reaction between zinc powder with hydrogen chloride in propanone and hydrogen chloride in water. Zinc powder with hydrogen chloride in propanone Hydrogen chloride gas dissolved in propanone Zinc powder • No bubbles. • Hydrogen chloride in propanone does not react with zinc. • Hydrogen chloride molecules in propanone do not ionise. • Hydrogen chloride exist as molecule only, there are no hydrogen ions present. • Hydrogen chloride in propanone does not show acidic properties. Zinc powder with hydrogen chloride in water Hydrogen chloride gas dissolved in water Zinc powder • Bubbles are released. • Hydrogen chloride in water react with zinc. • Hydrogen chloride molecule in water ionises . • Hydrogen ions present. • Hydrogen chloride in water (hydrochloric acid) shows acidic properties. Example 3: Glacial ethanoic acid and ethanoic acid dissolved in water. Glacial ethanoic acid • Glacial ethanoic acid molecules do not ionise . • Glacial ethanoic acid exist as molecule only, no hydrogen ions present: (i) Glacial ethanoic acid does not react with metal, base or metal carbonate. (ii) Glacial ethanoic acid does not turn blue litmus paper to red. • There are no free moving ions, glacial ethanoic acid cannot conduct electricity (non electrolyte). Ethanoic acid dissolved in water Water • Ethanoic acid molecules ionise partially in water to produce hydrogen ion: CH3COOH(aq) CH3COO– (aq) + H+ (aq) • Hydrogen ions are present. (i) Ethanoic acid react with metal, base or metal carbonate. (ii) Ethanoic acid turn blue litmus paper to red . • There are free moving ions, ethanoic acid can conduct electricity (electrolyte). Role of Water and Properties of Alkali How do you explain the role of water in the formation of hydroxide ion which cause alkaline properties? • Alkali dissolves and ionises in water to produce hydroxide ions. Alkaline does not show alkaline properties without water or dissolved in organic solvent. Explain. • Without water or in organic solvents, no hydroxide ions are produced, so the alkaline properties are not shown . 06 U6a Chemistry F4(p98-131)csy2p.indd 103 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 104 UNIT 6 MODULE • Chemistry FORM 4 Example 1: Sodium hydroxide without water and sodium hydroxide dissolved in water. LS 6.1.3 Solid sodium hydroxide without water • Sodium ions and hydroxide ions are attracted by strong electrostatic force in solid sodium hydroxide. • Sodium hydroxide do not ionise to hydroxide ion. (i) Solid sodium hydroxide does not react with ammonium salt. (ii) Solid sodium hydroxide does not turn red litmus paper to blue . • There are no free moving ions, solid sodium hydroxide cannot conduct electricity (non electrolyte). Sodium hydroxide dissolved in water (sodium hydroxide solution) Water • Sodium hydroxide ionise in water to produce hydroxide ion. NaOH(aq) ➝ Na+ (aq) + OH– (aq) • Hydroxide ions are present: (i) Sodium hydroxide solution reacts with acid and ammonium salt. (ii) Sodium hydroxide solution turns blue litmus paper to red . • There are free moving ions, sodium hydroxide solution can conduct electricity (electrolyte). Example 2: Ammonia gas dissolved in propanone and ammonia dissolved in water. Ammonia gas dissolved in propanone Propanone • Ammonia molecules in propanone do not ionise . • Ammonia in propanone exist as molecule only, no hydroxide ions present. (i) Ammonia in propanone does not react with ammonium salt. (ii) Ammonia in propanone does not turn red litmus paper to blue. • There are no free moving ions, ammonia in propanone cannot conduct electricity (non electrolyte). Ammonia gas dissolved in water Water • Ammonia molecules ionise partially in water to produce hydroxide ions. NH3(g) + H2O(l) NH4 + (aq) + OH– (aq) • Hydroxide ions are present. (i) Ammonia aqueous reacts with acid and ammonium salt. (ii) Ammonia aqueous turns red litmus paper to blue . • There are free moving ions, ammonia aqueous can conduct electricity (electrolyte). 06 U6a Chemistry F4(p98-131)csy2p.indd 104 21/12/2022 3:50 PM

105 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Aim : To show that water is needed for an acid to show its acidic properties. Problem statement : Does an acid need water to show its acidic properties? Manipulated variable : Presence of water Responding variable : Color change of blue litmus paper Constant variable : Oxalic acid Hypothesis : Oxalic acid without water do not shows acidic properties/ do not change blue changes blue litmus paper to red. Oxalic acid in water shows acidic properties/changes blue litmus paper to red Materials : Oxalic acid powder, distilled water and blue litmus paper Apparatus : Test tube, test tube rack Oxalic acid powder Blue litmus paper Oxalic acid + water Blue litmus paper I II Procedures: 1 Two test tubes are labelled and placed on a test tube rack. 2 About 1 spatula of oxalic acid powder is poured into each test tube. 3 About 2 cm3 distilled water is added to test tube II and the mixture is shaken. 4 A piece of dry blue litmus paper is dipped into each test tube. 5 Any changes are observed and recorded. Observations: Test tube Effect on blue litmus paper Inference I No change Oxalic acid without water does not show acidic property II Blue litmus paper turns red Oxalic acid in water shows acidic property Conclusion: Hypothesis is accepted . Aim : To show that water is needed for an alkali to show its alkaline properties. Problem statement : Does an alkali need water to show its alkaline properties? Manipulated variable : Presence of water Responding variable : Colour change of red litmus paper Constant variable: Barium hydroxide Hypothesis : Barium hydroxide without water does not change colour of red litmus paper/does not show alkaline properties. Barium hydroxide added with water changes red litmus paper to blue/ shows alkaline properties. Materials : Barium hydroxide, distilled water, red litmus paper Apparatus : Watch glass, dropper Red litmus paper Barium hydroxide Procedures: 1 One spatula of dry barium hydroxide powder is placed in a dry watch glass. 2 A piece of dry red litmus paper is dipped into the dry barium hydroxide powder. 3 Colour change of red litmus paper is observed and recorded. 4 A few drops of distilled water are added to dissolve dry barium hydroxide powder in the watch glass. 5 Colour change of red litmus paper is observed and recorded. Observations: Barium hydroxide Effect on red litmus paper Inference Without water No change Dry barium hydroxide does not show alkaline property Aqueous solution Red litmus paper turns blue Aqueous solution of barium hydroxide shows alkaline property Conclusion: Hypothesis is accepted . Experiment to Show That the Presence of Water is Essential for the Acid and Alkali to Show the Acidity and Alkaline Properties LS 6.1.3 PERFORMANCE LEVEL (PL) Mastered Not mastered PL4 Analyse knowledge on acids, bases and salts in the context of problem solving the natural occurrences or phenomena. 06 U6a Chemistry F4(p98-131)csy2p.indd 105 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 106 UNIT 6 MODULE • Chemistry FORM 4 What is meant by strong acid and weak acid? Explain with examples. (a) A strong acid is an acid that ionises completely in water to produce high concentration of hydrogen ion, H+ . Example: Hydrochloric acid is a strong acid ionises completely in water to produce high concentration of hydrogen ions and chloride ions, HCl(aq) H+ (aq) + Cl– (aq) Water Hydrochloric acid ionises completely to produce high concentration of hydrogen ions, H+ Other examples of strong acid: Strong acid Ionisation equation Particles present Nitric acid, HNO3 HNO3(aq) H+ (aq) + NO3 – (aq) H+ and Cl– Sulphuric acid, H2SO4 H2SO4(aq) 2H+ (aq) + SO4 2–(aq) H+ and SO4 2– (b) A weak acid is an acid that partially ionises in water to produce low concentration of hydrogen ion, H+ . Example: Ethanoic acid is a weak acid ionises partially in water to produce low concentration of hydrogen ions and ethanoate ions, CH3COOH(aq) H+ (aq) + CH3COO– (aq) Water An aqueous solution ethanoic acid ionises partially to produce low concentration of hydrogen ions, H+ . The majority remains as molecules, CH3COOH. Other example of weak acid: Weak acid Ionisation equation Particles present Carbonic acid, H2CO3 H2CO3(aq) 2H+ (aq) + CO3 2–(aq) H2CO3, H+ and CO3 2– 6.2 THE STRENGTH OF ACID AND ALKALI CS 6.3 LS 6.3.1 LS 6.3.2 06 U6a Chemistry F4(p98-131)csy2p.indd 106 21/12/2022 3:50 PM

107 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. What is meant by strong alkali and weak alkali? Explain with examples. (a) A strong alkali is an alkali that ionises completely in water to produce high concentration of hydroxide ion, OH– . Example: Sodium hydroxide solution is a strong alkali ionises completely in water to produce high concentration of hydroxide ions and sodium ions: NaOH(aq) Na+ (aq) + OH– (aq) All sodium hydroxide ionises completely to Water produce high concentration of hydroxide ion Other examples of strong alkali: Strong alkali Ionisation equation Particles present Potassium hydroxide, KOH KOH(aq) K+ (aq) + OH– (aq) K+ and OH– Barium hydroxide, Ba(OH)2 Ba(OH)2(aq) Ba2+(aq) + 2OH– (aq) Ba2+ and OH– (b) A weak alkali is an alkali that partially ionises in water to produce low concentration of hydroxide ion, OH– . Example: Ammonia aqueous is a weak alkali ionises partially in water to produce low concentration of hydroxide ions: NH3(g) + H2O(l) NH4 + (aq) + OH– (aq) An aqueous ammonia ionises partially to produce low concentration of hydroxide ions, OH– and ammonium ions, NH4 + . The majority remains as ammonia molecules, NH3 Water LS 6.3.1 LS 6.3.2 06 U6a Chemistry F4(p98-131)csy2p.indd 107 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 108 UNIT 6 MODULE • Chemistry FORM 4 1 Describe chemical properties of acid: LS 6.4.1 Chemical properties Examples of experiment Observations Remarks 1 Acid + Metal Salt + Hydrogen * Acids react with metals that are more electropositive than hydrogen in Electrochemical Series, acids do not react with copper and silver (The type of reaction is metal displacement that will be studied in the redox topic, Form 5) * Application of the reaction: • Preparation of soluble salt (Topic: Salt) • Preparation of hydrogen gas in determination of the empirical formula of copper(II) oxide (Topic: Chemical Formula and Equation) Magnesium + Hydrochloric acid Magnesium powder Hydrochloric acid Burning wooden splinter (a) About 5 cm3 of dilute hydrochloric acid is poured into a test tube. (b) One spatula of magnesium powder is added to the acid. (c) A burning wooden splinter is placed at the mouth of the test tube. (d) The observations are recorded. • The grey solid dissolves. • Colourless gas bubbles are released. • When a burning wooden splinter is placed at the mouth of the test tube, ‘pop sound’ is produced. Chemical equation: Mg + 2HCl MgCl2 + H2 Inference: • Magnesium reacts with hydrochloric acid. • Hydrogen gas is released. 2 Acid + Metal carbonate Salt + Water + Carbon dioxide * Application of the reaction: • Preparation of soluble salt (Topic: Salt) • Confirmatory test for anion carbonate ion in qualitative analysis of salt (Topic: Salt) Calcium carbonate + Hydrochloric acid Limewater Calcium carbonate Hydrochloric acid (a) About 5 cm3 of dilute hydrochloric acid is poured into a test tube. (b) One spatula of calcium carbonate powder is added into the acid. (c) The gas released is passed through limewater as shown in the diagram. (d) The observations are recorded. • The white solid dissolves. • Colourless gas bubbles are released. • When the gas is passed through limewater, the limewater turns chalky. Chemical equation: CaCO3 + 2HCl CaCl2 + H2O + CO2 Inference: • Calcium carbonate reacts with hydrochloric acid. • Carbon dioxide gas is released. 6.3 CHEMICAL PROPERTIES OF ACID AND ALKALI CS 6.4 06 U6a Chemistry F4(p98-131)csy2p.indd 108 21/12/2022 3:50 PM

109 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 3 Acid + Base / Alkali Salt + Water * Acid neutralises base/alkali * Application of the reaction: • Preparation of soluble salt (Topic Salt) Copper(II) oxide + Sulphuric acid Sulphuric acid Copper(II) oxide (a) Sulphuric acid is poured into a beaker until half full. (b) The acid is warmed gently. (c) One spatula of copper(II) oxide powder is added to the acid. (d) The mixture is stirred with a glass rod. (e) The observations are recorded. • The black solid dissolves. • The colourless solution turns blue. Chemical equation: CuO + H2SO4 CuSO4 + H2O Inference: • Copper(II) oxide reacts with sulphuric acid. • The blue solution is copper(II) sulphate . 2 Write the chemical formula for the following compounds: Compounds Chemical formula Compounds Chemical formula Hydrochloric acid HCl Magnesium oxide MgO Nitric acid HNO3 Calcium oxide CaO Sulphuric acid H2SO4 Copper(II) oxide CuO Ethanoic acid CH3COOH Lead(II) oxide PbO Sodium hydroxide NaOH Sodium nitrate NaNO3 Potassium hydroxide KOH Potassium sulphate K2SO4 Calcium hydroxide Ca(OH)2 Barium hydroxide Ba(OH)2 Sodium carbonate Na2CO3 Sodium chloride NaCl Magnesium hydroxide Mg(OH)2 Magnesium Mg Ammonium sulphate (NH4)2SO4 Zinc Zn Hydroxide ion OH– Sodium Na Sodium sulphate Na2SO4 Calcium carbonate CaCO3 Carbon dioxide CO2 Hydrogen gas H2 Copper(II) carbonate CuCO3 Sodium oxide Na2O Water H2O Magnesium nitrate Mg(NO3)2 06 U6a Chemistry F4(p98-131)csy2p.indd 109 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 110 UNIT 6 MODULE • Chemistry FORM 4 3 Ionic equation: What is ionic equation? • Ionic equation is an equation that shows particles that change during chemical reaction. How to write ionic equation? Examples: (i) Reaction between sulphuric acid and sodium hydroxide solution: Write balanced equation: H2SO4 + 2NaOH Na2SO4 + 2H2O Write the formula of all the particles in the reactants and products: 2H+ + SO4 2– + 2Na+ + 2OH– 2Na+ + SO4 2– + 2H2O Remove all the particles in the reactants and products which remain unchanged: 2H+ + SO4 2– + 2Na+ + 2OH– 2Na+ + SO4 2– + 2H2O Ionic equation: 2H+ + 2OH– 2H2O H+ + OH– H2O (ii) Reaction between zinc and hydrochloric acid : Write balanced equation: 2HCl + Zn ZnCl2 + H2 Write the formula of all the particles in the reactants and products: 2H+ + 2Cl– + Zn Zn2+ + 2Cl– + H2 Remove all the particles in the reactants and products which remain unchanged: 2H+ + 2Cl– + Zn Zn2+ + 2Cl– + H2 Ionic equation: 2H+ + Zn Zn2+ + H2 4 Write the chemical equations and ionic equations for the following reactions: Reactants Chemical equations Ionic equations Hydrochloric acid and # magnesium oxide MgO + 2HCl MgCl2 + H2O 2H+ + MgO Mg2+ + H2O Hydrochloric acid and sodium hydroxide HCl + NaOH NaCl + H2O H+ + OH– H2O Hydrochloric acid and magnesium 2HCl + Mg MgCl2 + H2 2H+ + Mg Mg2+ + H2 Hydrochloric acid and # calcium carbonate 2HCl + CaCO3 CaCl2 + CO2 + H2O 2H+ + CaCO3 Ca2+ + CO2 + H2O Sulphuric acid and zinc H2SO4 + Zn ZnSO4 + H2 2H+ + Zn Zn2+ + H2 Sulphuric acid and # zinc oxide H2SO4 + ZnO ZnSO4 + H2O 2H+ + ZnO Zn2+ + H2O Sulphuric acid and # zinc carbonate H2SO4 +ZnCO3 ZnSO4 + CO2 + H2O 2H+ + ZnCO3 Zn2+ + CO2 + H2O Nitric acid and # copper(II) oxide 2HNO3 + CuO Cu(NO3)2 + H2O 2H+ + CuO Cu2+ + H2O Nitric acid and sodium hydroxide HNO3 + NaOH NaNO3 + H2O H+ + OH– H2O # 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. PERFORMANCE LEVEL (PL) Mastered Not mastered PL5 Evaluate knowledge on acids, bases and salts in context of problem solving and decision-making to carry out a task. 06 U6a Chemistry F4(p98-131)csy2p.indd 110 21/12/2022 3:50 PM

111 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 5 Describe chemical properties of alkali: LS 6.4.2 Chemical properties Write the balanced chemical equation for the reaction 1 Alkali + Acid Salt + Water * Alkali neutralises acid. * Application of the reaction: • Preparation of soluble salt (Topic Salt) (a) Potassium hydroxide and sulphuric acid: H2SO4 + 2KOH K2SO4 + 2H2O (b) Barium hydroxide and hydrochloric acid: 2HCl + Ba(OH)2 BaCl2 + H2O 2 Alkali + Ammonium salt Salt + Water + Ammonia gas * Ammonia gas is released when alkali is heated with ammonium salt. Ammonia gas has pungent smell and turns moist red litmus paper to blue. * Application of the reaction: • Confirmatory test for cations ammonium in qualitative analysis of salt (Topic Salt) (a) Ammonium chloride and potassium hydroxide: KOH + NH4Cl KCl + H2O + NH3 (b) Ammonium sulphate and sodium hydroxide: 2NaOH + (NH4)2SO4 Na2SO4 + 2H2O + 2NH3 3 Alkali + Metal ion Insoluble metal hydroxide * Most of the metal hydroxides are insoluble. * Hydroxides of transition element metals are coloured. * Application of the reaction: • Confirmatory test for cations in qualitative analysis of salt (Topic Salt) (a) 2OH– (aq) + Mg2+(aq) Mg(OH)2(s) (white precipitate) (b) 2OH– (aq) + Cu2+(aq) Cu(OH)2(s) (blue precipitate) 1 The table below shows two experiments for the reaction between magnesium with hydrogen chloride in solvent X and solvent Y. Set-up of apparatus Experiment I: Magnesium ribbon with hydrogen chloride in solvent X. Experiment II: Magnesium ribbon with hydrogen chloride in solvent Y. Observation Bubbles of colourless gas are released and magnesium ribbon dissolves No bubble of gas (a) Name possible substances that can be solvent X and solvent Y. Solvent X: Water Solvent Y: Propanone / methylbenzene / trichloromethane (b) Name the gas released in Experiment I. Hydrogen gas (c) Compare observation in Experiment I and Experiment II. Explain your answer. • Hydrogen chloride in solvent X in Experiment I reacts with magnesium. • Hydrogen chloride in solvent Y in Experiment II does not react with magnesium. • Hydrogen chloride in solvent ionises to H+ : HCl H+ + Cl– . • H+ ions react with magnesium atom to produce hydrogen molecule: Mg + 2H+ Mg2+ + H2. • Hydrogen chloride in solvent Y remains in the form of molecule . No hydrogen ion present. PL5 PL4 Exercise PL1 Tree Map on Role of Water in Acid or Alkali 06 U6a Chemistry F4(p98-131)csy2p.indd 111 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 112 UNIT 6 MODULE • Chemistry FORM 4 STRONG ACID & WEAK ACID : (a) A strong acid is an acid that ionises completely in water to produce high concentration of hydrogen ion, H+ Example: (i) Hydrochloric acid : HCl (aq) → H+ (aq) + Cl– (aq) (ii) Nitric acid : HNO3 (aq) → H+ (aq) + NO3– (aq) (iii)Sulphuric acid : H2SO4 (aq) → 2H+ (aq) + SO42– (aq) (b) A weak acid is an acid that partially ionises in water to produce low concentration of hydrogen ion, H+ Example: Ethanoic acid : CH COOH (aq) 3 CH COO 3 – (aq) + H+ (aq) STRONG ALKALI & WEAK ALKALI : (a) A strong alkali is an alkali that ionises completely in water to produce high concentration of hydroxide ion, OHExample: (i) Sodium hydroxide: NaOH (aq) → Na+ (aq) + OH– (aq) (ii) Potassium hydroxide: KOH (aq) → K+ (aq) + OH– (aq) (b) A weak alkali is an alkali that partially ionises in water to produce low concentration of hydroxide ion, OH– Example: Ammonia solution : NH3 (g) + H O (l) 2 NH4+ (aq) + OH– (aq) MEANING : Acid is Example : a chemical substance which ionises in HCl (aq) → H+ (aq) + Cl– (aq) Hydrochloric acid → Hydrogen ion + Chloride ion water to produce hydrogen ion. MEANING : • Base is • Most bases are metal oxide or metal hydroxide Example : • Alkali is Example: Sodium hydroxide dissolves in water and ionises to hydroxide ion. a chemical substance that reacts a base that is soluble in water and CuO, MgO, Pb(OH)2 NaOH (aq) → Na+ (aq) + OH– (aq) with acid to produce salt and water only. ionises to hydroxide ion. CHARACTERISTICS : 1. Change blue litmus paper to red 2. pH is less than 7 3. Taste sour CHARACTERISTICS : 1. Change red litmus paper to blue 2. pH is more than 7 3. Taste bitter CHEMICAL PROPERTIES : 1. 2. 3. Base/alkali + Acid → Salt + Water Base/alkali + Ammonium salt → Base/alkali + Metal ion → Metal Salt + Water + Ammonia gas hydroxide CHEMICAL PROPERTIES : 1. 2. 3. Acid + Metal carbonate → Salt + Acid + Metal → Salt + Hydrogen Acid + Base/alkali → Salt + Water Water + Carbon dioxide Acid + Base/Alkali ➝ Salt + Water Acid Base/ Alkali Complete the following To Compare the Acid and Base/Alkali 06 U6a Chemistry F4(p98-131)csy2p.indd 112 21/12/2022 3:50 PM

113 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. What is a solution? • 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). What is concentration? LS 6.5.1 • Concentration of a solution is the quantity of solute in a given volume of solution which is usually 1 dm3 of solution. (a) Mass of solute in gram per 1 dm3 solution, g dm–3. Concentration of solution (g dm–3) = Mass of solute in gram (g) Volume of solution (dm3 ) (b) Number of moles of solute in 1 dm3 solution, mol dm–3. Concentration of solution (mol dm–3) = Mass mole of solute (mol) Volume of solution (dm3 ) What is molarity? • The concentration in mol dm–3 is called molarity or molar concentration. • The unit mol dm–3 can be represented by ‘M’. What is the relationship between number of moles with molarity and volume of solution? Molarity = Number of mole of solute (mol) Volume of solution (dm3 ) Number of mole of solute (mol) = Molarity × Volume (dm3 ) n = MV n = Mv 1 000 n = Number of moles of solute M = Concentration in mol dm–3 (molarity) V = Volume of solution in dm3 v = Volume of solution in cm3 Can concentration in mol dm–3 be converted to g dm–3 and vice versa? The concentration of a solution can be converted from mol dm–3 to g dm–3 and vice versa. mol dm–3 g dm–3 × molar mass of the solute ÷ molar mass of the solute How to solve numerical problems involving molarity of acid and alkali? LS 6.5.2 Example: What is the concentration of the following solution in g dm–3 and mol dm–3? (a) 10 g sodium hydroxide, NaOH in 500 cm3 solution. (b) 0.5 mol potassium hydroxide, KOH in 2.5 dm3 solution. Solution: (a) Volume of solution = 500 cm3 1 000 = 0.5 dm3 Concentration of solution = 10 g 0.5 dm3 = 20 g dm–3 Relative formula mass of NaOH = 23 + 16 + 1 = 40, molar mass of NaOH = 40 g mol–1 Concentration of solution = 20 g dm–3 40 g mol–1 = 0.5 mol dm–3 (b) Concentration of solution = 0.5 mol 2.5 dm3 = 0.2 mol dm–3 Relative formula mass of KOH = 39 + 16 + 1 = 56, Molar mass of KOH = 56 g mol–1 Concentration of solution = 0.2 mol dm–3 × 56 g mol–1 = 11.2 g mol–1 6.4 CONCENTRATION OF AQUEOUS SOLUTION CS 6.5 06 U6a Chemistry F4(p98-131)csy2p.indd 113 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 114 UNIT 6 MODULE • Chemistry FORM 4 1 The molarity of sodium hydroxide solution is 2 mol dm–3. What is the concentration of the solution in g dm–3? [Relative atomic mass: Na = 23, O = 16, H = 1] Answer: 80 g dm–3 2 Calculate the molarity of the solution obtained when 14 g of potassium hydroxide is dissolved in distilled water to make up 500 cm3 solution. [Relative atomic mass: K = 39, H = 1, O = 16] Answer: 0.5 mol dm–3 3 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. Answer: 2 mol dm–3 4 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? [Relative atomic mass: Na = 23, O = 16, H = 1] Answer: 10 g Exercise LS 6.5.2 PL2 What is a standard solution? LS 6.6.1 • Standard solution is a solution in which its concentration is accurately known. How to prepare a standard solution? LS 6.6.2 • The steps taken in preparing a standard solution are: 1 Calculate the mass of solute needed to produce the required volume and molarity. 2 The solute is weighed. 3 The solute is completely dissolved in distilled water and then transferred to a volumetric flask that is partially filled with distilled water. 4 Distilled water is added until the calibration mark of the volumetric flask and the flask is inverted to make sure thorough mixing. 6.5 STANDARD SOLUTION CS 6.6 06 U6a Chemistry F4(p98-131)csy2p.indd 114 21/12/2022 3:50 PM

115 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Example: Preparing 100 cm3 of 2.0 mol dm–3 standard sodium hydroxide solution. LS 6.6.2 1 Calculate mass sodium hydroxide Number of mol NaOH = 2 × 100 1 000 = 0.2 mol Mass of NaOH = 0.2 mol × 40 g mol–1 = 8.0 g 2 8.0 g of solid sodium hydroxide is weighed in a dry weighing bottle. 3 8.0 g of solid sodium hydroxide in the weighing bottle is transferred into a beaker containing 25 cm3 of distilled water. The mixture is stirred to dissolve the solid. g ON OFF g ON OFF Solid sodium hydroxide 4 The solution from the beaker is then carefully poured into a 100 cm3 volumetric flask through a filter funnel . 5 The weighing bottle and the beaker are rinsed with a small amount of distilled water and poured into the volumetric flask. Filter funnel Glass rod 100 cm3 volumetric flask 6 Distilled water is poured into the volumetric flask until the calibration mark . Distilled water Calibration mark 7 The volumetric flask is then closed with a stopper and inverted a few times to get homogenous solution. Calibration mark 2 mol dm–3 What is dilution method? • Dilution is a process of diluting a concentrated solution by adding a solvent such as water to obtain a diluted solution. How to prepare a standard solution by dilution? • 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: • Number of mol of solute before dilution = Number of mole of solute after dilution M1V1 1 000 = M2V2 1 000 Therefore, M1V1 = M2V2 M1 = Initial concentration of the solute V1 = Initial volume of the solution in cm3 M2 = Final concentration of the solute V2 = Final volume of the solution in cm3 Preparation of a Solution by Dilution LS 6.6.3 06 U6a Chemistry F4(p98-131)csy2p.indd 115 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 116 UNIT 6 MODULE • Chemistry FORM 4 Example: Preparing 100 cm3 of 0.2 mol dm–3 of sodium hydroxide solution from 2 mol dm–3 of sodium hydroxide solution using dilution method. LS 6.6.3 1 Calculate the volume of 2 mol dm–3 sodium hydroxide solution required. M1V1 = M2V2 2 mol dm–3 × V1 = 0.2 mol dm–3 × 100 cm3 V1 = 10 cm3 2 Use pipette to draw 10 cm3 of sodium hydroxide solution from stock solution (2 mol dm–3 sodium hydroxide solution). 3 Transfer 10 cm3 of 2 mol dm–3 sodium hydroxide solution into 100 cm3 volumetric flask. 4 Add distilled water until the calibration mark . 5 The volumetric flask is then closed with a stopper and inverted a few times to get homogenous solution. 1 If 300 cm3 water is added to 200 cm3 hydrochloric acid, 1 mol dm–3. What is the resulting molarity of the solution? Answer: 0.4 mol dm–3 2 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. Answer: 50 cm3 Exercise LS 6.6.3 PL2 What is pH scale? LS 6.2.1 • The pH is a scale of numbers ranging between 0 to 14 to measure the degree of acidity and alkalinity of an aqueous solution. What is measured by each pH values? • Each pH value is a measure of the concentration of hydrogen ions, H+ or hydroxide ions, OH– . 6.6 PH VALUE CS 6.2 Distilled water Calibration mark Calibration mark 2 mol dm–3 Volumetric flask 06 U6a Chemistry F4(p98-131)csy2p.indd 116 21/12/2022 3:50 PM

117 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Example 1: Calculate pH value for: (a) An acid with the concentration of hydrogen = 3.5 × 10–3 mol dm–3 (b) 0.01 mol dm–3 of hydrochloric acid. LS 6.2.2 (a) pH = –log [H+ ] = –log (3.5 × 10–3) = 2.46 (b) Hydrochloric acid is a strong acid, ionises completely in water to produce hydrogen ion. HCl → H+ + Cl– 0.01 mol dm–3 0.01 mol dm–3 pH = –log [H+ ] = –log (0.01) = –log (10–2) = 2 Example 2: pH value of nitric acid is 3.30. Calculate the concentration of hydrogen ion in the acid. pH = –log [H+ ] 3.3 = –log [H+ ] [H+ ] = antilog(–3.3) [H+ ] = 5.0 × 10–4 What is measured by each pOH value? • Each pOH value is a measure of concentration of hydroxide ions, OH– in a solution. pOH = –log [OH– ] where [OH– ] = concentration of the hydroxide ion, OH– in mol dm–3. Example 3: Calculate the pOH value of an alkali if the concentration of OH– ion is 0.0001 mol dm–3. pOH = –log [OH– ] pOH = –log (0.0001) = –log (10–4) pOH = 4 Remark: Here it helps to rewrite the concentration as 1.0 × 10–4 What is the relationship of pH and pOH? • The sum of pH and pOH is always 14. pH + pOH = 14 Example 4: What are the pOH and the pH value of potassium hydroxide, 0.0125 mol dm–3? • Potassium hydroxide is a strong alkali, ionises completely to produce hydroxide ion. KOH → K+ + OH– 0.0125 mol dm–3 0.0125 mol dm–3 pOH = −log [OH− ] = −log 0.0125 = −(−1.903) = 1.903 • The pH can be obtained from the pOH value: pH + pOH = 14.00 pH = 14.00 − pOH = 14.00 − 1.903 = 12.10 What is the relationship of the concentration of hydrogen and hydroxide ions with pH value? pH = 7 Neutral pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [H+ ] 100 10–1 10–2 10–3 10–4 10–5 10–6 10–7 10–8 10–9 10–10 10–11 10–12 10–13 10–14 [OH– ] 10–14 10–13 10–12 10–11 10–10 10–9 10–8 10–7 10–6 10–5 10–4 10–3 10–2 10–1 100 pH < 7: • Acidic solution. • The higher the concentration of hydrogen ion, H+ , the lower the pH value. pH > 7: • Alkaline solution. • The higher the concentration of hydroxide ion, OH– , the higher the pH value. Remark: [H+] = concentration of hydrogen ion in mol dm–3 [OH– ] = concentration of hydroxide ion in mol dm–3 pH = –log(10–x) = –(–x)log10 = x(1) = x pOH = –log(10–y) = –(–y)log10 = y(1) = y 06 U6a Chemistry F4(p98-131)csy2p.indd 117 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 118 UNIT 6 MODULE • Chemistry FORM 4 pH of solutions used in daily life. Complete the table. Solutions pH value pOH [H+ ] (mol dm–3) [OH– ] (mol dm–3) Acidic/alkaline/ neutral Gastric juice 1 13 10–1 10–13 Acidic Lime juice 2 12 10–2 10–12 Acidic Carbonated drink 3 11 10–3 10–11 Acidic Vinegar 3 11 10–3 10–11 Acidic Orange juice 4 10 10–4 10–10 Acidic Coffee 5 9 10–5 10–9 Acidic Tea 5 9 10–5 10–9 Acidic Milk 6 8 10–6 10–7 Acidic Distilled water 7 7 10–7 10–7 Neutral Toothpaste 8 6 10–8 10–6 Alkaline Blood 8 6 10–8 10–6 Alkaline Detergent 10 4 10–10 10–4 Alkaline Household cleaner 11 3 10–11 10–3 Alkaline How to measure the pH value of a solution? • The pH of an aqueous solution can be measured by: (a) pH meter (b) Acid-base indicator State the colours of the acid-base indicators in acidic, neutral and alkaline solution. Indicators Colours Acid Neutral Alkali Litmus solution Red Purple Blue Methyl orange Red Orange Yellow Phenolphthalein Colourless Colourless Pink Universal indicator Red Green Purple 06 U6a Chemistry F4(p98-131)csy2p.indd 118 21/12/2022 3:50 PM

119 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 1 Stomach acid is a solution of hydrochloric. The concentration of hydrogen ion in the acid is 1.2 × 10–3 mol dm–3. What is the pH value of stomach acid? pH = −log [H+ ] = −log [1.2 × 10–3] = −(−2.92) = 2.92 Answer: 2.92 2 What is the hydroxide ion concentration in a solution that has a pOH value of 5.70? 5.70 = –log [OH– ] –5.70 = log [OH– ] [OH– ] = 10–5.70 = 2.00 × 10–6 mol dm–3 Answer: 2 × 10–6 mol dm–3 3 Blood has a pH of 7.3 (slightly alkaline). Calculate the concentration of hydrogen ions and hydroxide ions in the blood. pH = −log [H+ ] = 7.3 log [H+ ] = −7.3 [H+ ] = antilog −7.3 [H+ ] = 5 × 10–8 mol dm–3 pOH + pH = 14 pOH = 14 – 7.3 = 6.7 pOH = –log [OH– ] = 6.7 log [OH– ] = –6.7 [OH– ] = 2.00 × 10–7 mol dm–3 Answer: 4 Water exposed to air contains carbonic acid, H2CO3 due to the reaction between carbon dioxide and water: CO2(aq) + H2O(l) H2CO3(aq) The concentration of hydrogen ion air-saturated water caused by the dissolved CO2 is 2.0 × 106 mol dm–3. Calculate the pH of the solution. pH = −log [H+ ] = −log (2.0 × 10–6) = 5.70 Answer: 5.70 [H+ ] = 5 × 10–8 mol dm–3 [OH– ] = 2.00 × 10–7 mol dm–3 Exercise PL2 How do you relate the pH with the molarity of an acid and an alkali? • The pH value of an acid or an alkali depends on the concentration of hydrogen ions or hydroxide ions: The higher the concentration of hydrogen ions in acidic solution, the lower the pH value. The higher the concentration of hydroxide ions in alkaline solution, the higher the pH value. What are the factors that can affect the concentration of hydrogen and hydroxide ions of acid and alkali? • The pH value of an acid or an alkali depends on: (a) The strength of acid or alkali – the degree of ionisation or dissociation of the acid and alkali in water. (b) Molarity of acid or alkali – the concentration of acid or alkali in mol dm–3. (c) Basicity of an acid – the number of ionisable hydrogen atom per molecule of an acid molecule in an aqueous solution. The Relationship between pH Value and Concentration of Acid and Alkali 06 U6a Chemistry F4(p98-131)csy2p.indd 119 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 120 UNIT 6 MODULE • Chemistry FORM 4 Experiment to investigate the relationship between pH values with the molarity of solution: LS 6.2.3 Problem statement : How does the concentration of an acid and an alkali affect the pH value? Manipulated variable : Concentration of sodium hydroxide solution and hydrochloric acid. Responding variable : pH value Constant variable : Sodium hydroxide solution and hydrochloric acid Hypothesis : The higher the concentration of sodium hydroxide solution, the higher the pH value. The higher the concentration of hydrochloric acid, the lower the pH value. Materials : Hydrochloric acid of 0.1 mol dm–3, 0.01 mol dm–3, 0.001 mol dm–3 and 0.0001 mol dm–3 Sodium hydroxide solution of 1 mol dm–3, 0.1 mol dm–3, 0.01 mol dm–3, 0.001 mol dm–3 and 0.0001 mol dm–3 Apparatus : 100 cm3 beakers, pH meter 1.21 pH meter Hydrochloric acid Procedures: 1 About 30 cm3 of 1.0 mol dm–3 hydrochloric acid is poured into a dry beaker. 2 A clean and dry pH meter is dipped into the hydrochloric acid as shown in the diagram. 3 The value of pH meter is recorded. 4 Steps 1 to 3 are repeated by replacing 1.0 mol dm–3 hydrochloric acid with 0.1 mol dm–3, 0.01 mol dm–3, 0.001 mol dm–3 and 0.0001 mol dm–3 hydrochloric acid. 5 The experiment is repeated by replacing hydrochloric acid with sodium hydroxide solutions of different concentration. Results: 1 Hydrochloric acid Molarity of HCl (mol dm–3) 1.0 0.1 0.01 0.001 0.0001 pH value 0 1 2 3 4 2 Sodium hydroxide solution Molarity of NaOH (mol dm–3) 1.0 0.1 0.01 0.001 0.0001 pH value 14 13 12 11 10 Conclusion: Hypothesis is accepted. The higher the concentration of sodium hydroxide solution, the higher the pH value. The higher the concentration of hydrochloric acid, the lower the pH value. PERFORMANCE LEVEL (PL) Mastered Not mastered PL6 Invent by applying knowledge and skills about acids, bases and salts in the context of problem solving and decision-making or when carrying out an activity/task in new situations creatively and innovatively; giving due considerations to the social/ economic/cultural values of the community. 06 U6a Chemistry F4(p98-131)csy2p.indd 120 21/12/2022 3:50 PM

121 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Example 1: 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. Experiments I II III pH meter reading 1.00 0.1 mol dm–3 HCl 2.00 0.01 mol dm–3 HCl 1.00 0.05 mol dm–3 H2SO4 1.30 0.05 mol dm–3 HCl 1.00 0.1 mol dm–3 HCl 3.45 0.1 mol dm–3 CH COOH 3 Compare and explain the concentration of ion hydrogen, H+ and pH value • Hydrochloric acid is a strong acid ionises completely in water to hydrogen ion. • 0.1 mol dm–3 of hydrochloric acid ionises completely to form 0.1 mol dm–3 hydrogen ion: HCl H+ + Cl– 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: HCl H+ + Cl– 0.01 mol dm–3 0.01 mol dm–3 • The concentration of hydrogen ion in 0.1 mol dm–3 hydrochloric acid is higher than 0.01 mol dm–3 hydrochloric acid. • The pH value of 0.1 mol dm–3 of hydrochloric acid is lower than 0.01 mol dm–3 of hydrochloric acid. • Sulphuric acid is a strong diprotic acid. • 0.05 mol dm–3 of sulphuric acid ionises completely in water to form 0.1 mol dm–3 hydrogen ion: H2SO4 2H+ + SO42– 0.05 mol dm–3 0.1 mol dm–3 • Hydrochloric acid is a strong monoprotic acid. • 0.05 mol dm–3 of hydrochloric acid ionises completely in water to form 0.05 mol dm–3 hydrogen ion: HCl H+ + Cl– 0.01 mol dm–3 0.05 mol dm–3 • The concentration of hydrogen ion in 0.05 mol dm–3 sulphuric acid is double of (higher than) 0.05 mol dm–3 hydrochloric acid. • The pH value of 0.05 mol dm–3 of sulphuric acid is lower than 0.05 mol dm–3 of hydrochloric acid. • Hydrochloric acid is a strong acid ionises completely in water to produce higher concentration hydrogen ion. • 0.1 mol dm–3 of hydrochloric acid ionises completely in water to form 0.1 mol dm–3 hydrogen ion: HCl H+ + Cl– 0.1 mol dm–3 0.1 mol dm–3 • Ethanoic acid is a weak acid ionises partially in water to produce lower concentration of hydrogen ion. • 0.1 mol dm–3 of ethanoic acid ionises to produce less than 0.1 mol dm–3 hydrogen ion: CH COOH( 3 aq) H+ + CH COO 3 – (aq) 0.1 mol dm–3 less than 0.1 mol dm–3 • The concentration of hydrogen ion in 0.1 mol dm–3 of hydrochloric acid is higher than of 0.1 mol dm–3 of ethanoic acid. • The pH value of 0.1 mol dm–3 of hydrochloric acid lower than of 0.1 mol dm–3 of ethanoic acid. 06 U6a Chemistry F4(p98-131)csy2p.indd 121 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 122 UNIT 6 MODULE • Chemistry FORM 4 Example 2: The diagram below shows the reading of pH meter for different types and concentration of alkali. Compare the concentration of hydroxide ions and the pH value of the following alkalis. Explain your answer. Experiment I II pH meter reading 14.0 1 mol dm–3 NaOH(aq) 12.0 0.01 mol dm–3 NaOH(aq) 14.0 1 mol dm–3 NaOH(aq) 11.60 1 mol dm–3 NH3(aq) Compare and explain the concentration of ion hydrogen, OH– and pH value • Sodium hydroxide is a strong alkali ionises completely in water hydroxide ion. • 1 mol dm–3 of sodium hydroxide ionises to completely to 1 mol dm3 of hydroxide ion: NaOH(aq) Na+ (aq) + OH– (aq) 1 mol dm–3 1 mol dm–3 • 0.01 mol dm–3 of sodium hydroxide ionises completely to 0.01 mol dm–3 hydroxide ion. NaOH(aq) Na+ (aq) + OH– (aq) 0.01 mol dm–3 0.01 mol dm–3 • The concentration of hydroxide ions in 1 mol dm–3 of sodium hydroxide is higher than 0.01 mol dm–3 of sodium hydroxide solution. • The pH value of 1 mol dm–3 of sodium hydroxide solution higher than 0.01 mol dm–3 of sodium hydroxide solution. • Sodium hydroxide is a strong alkali ionises completely in water to hydroxide ion. • 1 mol dm–3 of sodium hydroxide ionises to completely to 1 mol dm3 of hydroxide ion: NaOH(aq) Na+ (aq) + OH– (aq) 1 mol dm–3 1 mol dm–3 • Aqueous ammonia a weak alkali ionises partially in water to produce lower concentration of hydroxide ion. • Aqueous ammonia 1 mol dm–3 ionises partially to less than 1 mol dm3 hydroxide ion. NH3(g) + H2O(l) NH4 + (aq) + OH– (aq) 1 mol dm–3 less than 1 mol dm–3 • The concentration of hydroxide ions in 1 mol dm–3 is higher than 0.01 mol dm–3 of sodium hydroxide solution. • The pH value of 1 mol dm–3 of sodium hydroxide solution is higher than 0.01 mol dm–3 of sodium hydroxide solution. 06 U6a Chemistry F4(p98-131)csy2p.indd 122 21/12/2022 3:50 PM

123 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 1 The table below shows the pH value of a few substances. Substances pH value Ethanoic acid 0.1 mol dm–3 3 Hydrochloric acid 0.1 mol dm–3 1 Glacial ethanoic acid 7 (a) (i) What is meant by weak acid and strong acid? Weak acid : An acid that partially ionises in water to produce lower concentration of hydrogen ion, H+ . Strong acid : An acid that ionises completely in water to produce higher concentration of hydrogen ion, H+ . (ii) Between ethanoic acid and hydrochloric acid, which acid has the higher concentration of H+ ion? Explain your answer. • Hydrochloric acid has higher concentration of H+ ions than ethanoic acid. • Hydrochloric acid is a strong acid which ionises completely in water to produce higher concentration of H+ ions: HCl(aq) H2O H+ (aq) + Cl– (aq) • Ethanoic acid is a weak acid which ionises partially in water to produce lower concentration of H+ ions: CH3COOH(aq) H2O CH3COO– (aq) + H+ (aq) (iii) Why do ethanoic acid and hydrochloric acid have different pH value? • The higher the concentration of H+ ions, the lower the pH value. • The concentration of H+ ions in hydrochloric acid is higher , the pH value is lower . • The concentration of H+ ions in ethanoic acid is lower , the pH value is higher . (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. • Glacial ethanoic acid molecules do not ionise . Glacial ethanoic acid consists of only CH3COOH molecules . The CH3COOH molecules are neutral . No hydrogen ions present. The pH value of glacial ethanoic acid is 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. 2 Emission of sulphur dioxide from factories cause the formation of acid rain and this led to the decreasing pH value of lake. Table below shows the pH value of a lake in a year from January to June. Months pH value January 6.0 February 5.0 March 4.0 April 5.5 May 6.0 June 6.5 Exercise PL1 PL4 PL4 PL2 STEM Example 2: The diagram below shows the reading of pH meter for different types and concentration of alkali. Compare the concentration of hydroxide ions and the pH value of the following alkalis. Explain your answer. Experiment I II pH meter reading 14.0 12.0 0.01 mol dm–3 14.0 1 mol dm–3 NaOH(aq) 11.60 1 mol dm–3 NH3(aq) Compare and explain the concentration of ion hydrogen, OH– and pH value • Sodium hydroxide is a strong alkali ionises completely in water to hydroxide ion. • 1 mol dm–3 of sodium hydroxide ionises to completely to 1 mol dm3 of hydroxide ion: NaOH(aq) Na+ (aq) + OH– (aq) 0.1 mol dm–3 1 mol dm–3 • 0.01 mol dm–3 of sodium hydroxide ionises to completely to 0.01 mol dm–3 ion hydroxide. NaOH(aq) Na+ (aq) + OH– (aq) 0.1 mol dm–3 0.01 mol dm–3 • The concentration of hydroxide ions in 1 mol dm–3 is higher than 0.01 mol dm–3 of sodium hydroxide solution. • The pH value of 1 mol dm–3 of sodium hydroxide solution higher than 0.01 mol dm–3 of sodium hydroxide solution. • Sodium hydroxide is a strong alkali ionises completely in water to hydroxide ion. • 1 mol dm–3 of sodium hydroxide ionises to completely to 1 mol dm3 of hydroxide ion: NaOH(aq) Na+ (aq) + OH– (aq) 0.1 mol dm–3 1 mol dm–3 • 0.01 mol dm–3 of sodium hydroxide ionises to completely to 0.01 mol dm–3 ion hydroxide. NaOH(aq) Na+ (aq) + OH– (aq) 0.1 mol dm–3 0.01 mol dm–3 • The concentration of hydroxide ions in 1 mol dm–3 is higher than 0.01 mol dm–3 of sodium hydroxide solution. • The pH value of 1 mol dm–3 of sodium hydroxide solution higher than 0.01 mol dm–3 of sodium hydroxide solution. 06 U6a Chemistry F4(p98-131)csy2p.indd 123 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 124 UNIT 6 MODULE • Chemistry FORM 4 (a) What is meant by pH scale? The pH is a scale of numbers ranging between 0 to 14 to measure the degree of acidity and alkalinity of an aqueous solution. (b) What is measured by pH value? Each pH value is a measure of the concentration of hydrogen ions, H+ or hydroxide ions, OH– (c) From the table above, (i) State the month that has; Highest concentration of hydrogen ions: March Lowest concentration of hydrogen ions: June (ii) Calculate the concentration of hydrogen ions in the lake for April: 3.16 × 10–6 (iii) Suggest method that can be done in order to increase the pH value from March to June. Add calcium oxide into the lake. (d) If you are going for fishing at this lake, which month is the best for you to obtain the healthier fish? Explain your answer. • June. • The pH value of the lake is almost to neutral. • The fish grow healthier at this month. 3 (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. Acid 50 cm3 of 1 mol dm–3 of sulphuric acid 50 cm3 of 1 mol dm–3 of hydrochloric acid Calculate number of hydrogen ion, H+ Number of mol of sulphuric acid = 50 × 1 1 000 = 0.05 mol H2SO4 2H+ + SO4 2– From the equation, 1 mol H2SO4 : 2 mol H+ 0.05 mol H2SO4 : 0.1 mol H+ Number of mol of hydrochloric acid = 50 × 1 1 000 = 0.05 mol HCl H+ + Cl– From the equation, 1 mol HCl : 1 mol H+ 0.05 mol HCl : 0.05 mol H+ Compare the number of hydrogen ions • The number of H+ ion in 50 cm3 of 1 mol dm–3 of sulphuric acid is twice the number of H+ in 50 cm3 of 1 mol dm–3 of hydrochloric acid. Explanation • Sulphuric acid is diprotic acid whereas hydrochloric acid is monoprotic acid. • 1 mol of sulphuric ionises to 2 mol of H+ ions whereas 1 mol of hydrochloric acid ionises to 1 mol of H+ ions. • The number of H+ ions in both acid with the same volume and concentration 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. 100 cm3 PL2 PL4 Additional Question 06 U6a Chemistry F4(p98-131)csy2p.indd 124 21/12/2022 3:50 PM

125 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. What is neutralisation? LS 6.7.1 • Neutralisation is the reaction between an acid and a base to form only salt and water: Acid + Base Salt + Water Examples: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 2HNO3(aq) + MgO(s) Mg(NO3)2(aq) + H2O(l) What happens during neutralisation reaction? • During 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: H+ (aq) + OH– (aq) H2O(l) Application of neutralisation in daily life: LS 6.7.3 Applications Examples Agriculture 1 Acidic soil is treated with powdered soda lime (calcium oxide, CaO), limestone (calcium carbonate, CaCO3) or ashes of burnt wood. 2 Basic soil is treated with compost. The acidic gas from the decomposition of compost neutralises the alkalis in basic soil. 3 The acidity of water in farming is controlled by adding soda lime (calcium oxide, CaO). 4 Neutralisation reaction between acid and alkali can produce fertiliser. For example, ammonium nitrate, ammonium sulphate and urea. Industries 1 Acidic gases emitted by industries are neutralised with soda lime (calcium oxide, CaO) before the gases are released into the air. 2 Organic acid produced by bacteria in latex is neutralises by ammonia solution/ ammonium hydroxide and prevents coagulation. 3 Effluents from the electroplating industry contain acids such as sulphuric acid. It is treated by adding lime to neutralise it before it is discharged into rivers and streams. Health 1 Excess acid in the stomach is neutralised with its anti-acids that contain bases such as aluminium hydroxide and magnesium hydroxide . 2 Toothpastes contain bases (such as magnesium hydroxide) to neutralise the acid produced by bacteria in the mouth. 3 Baking powder (sodium hydrogen carbonate) is used to cure acidic bee stings. 4 Vinegar (ethanoic acid) is used to cure alkaline wasp sting. 6.7 NEUTRALISATION CS 6.7 Neutralisation Quiz 06 U6a Chemistry F4(p98-131)csy2p.indd 125 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 126 UNIT 6 MODULE • Chemistry FORM 4 The diagram below shows two different fertilisers used by the farmers onto the crops so that the crops can grow faster and bigger and thus, their crop yields can be increased. By using your knowledge in chemistry, determine the best fertiliser to be used by the farmers. Answer: Urea, (NH2)2CO Percentage of N = 28 60 × 100% = 46.67% Ammonium nitrate, NH4NO3 Percentage of N = 28 80 × 100% = 35% Urea, (NH2)2CO is the best fertiliser because it contains higher percentage of nitrogen by mass. Exercise FERTILISER A Urea, (NH2)2CO FERTILISER B Ammonium nitrate, NH4NO3 PL5 Aspirin is a modern medicine. Aspirin is a type of analgesic that is used as a painkiller. The molecular structure of aspirin consists of carboxyl group which is acidic. Taking excessive aspirin can cause gastric due to excess acid in stomach. Excess acid can be neutralised by taking antacid medicine or by drinking magnesia milk. Both antacid dan magnesia milk contain base such as magnesium hydroxide. Neutralisation is a reaction between an acid with an alkali. You are required to conduct an experiment to determine the type of acid P used to neutralise a sodium hydroxide solution. Diagram 1 shows the apparatus set up to conduct the experiment. Diagram 1 Acid P 0.5 mol dm–3 Sodium hydroxide solution + phenolphthalein You are provided with 0.5 mol dm–3 of sodium hydroxide solution, NaOH labelled as Solution R1, 0.5 mol dm–3 of acid P labelled Solution R2, phenolphthalein, pipette and pipette filler, 250 cm3 conical flask, burette, retort stand and white tile 1 Conduct the experiment by using the following procedures: (i) By using a pipette , measure 25.0 cm3 of R1 solution and transfer into a conical flask. Add three drops of phenolphthalein into the solution in the conical flask. (ii) Fill a burette with R2 solution and record the initial burette reading. (iii) Carry out the titration by adding R2 solution into R1 solution. Continue addition of R2 solution until there is a colour change at the end point. (iv) Record the final burette reading. (iv) Repeat the titration twice. Eksperimen Pentitratan Asid dan Bes SPM K3 Experiment: Titration of Acid and Base S P O T S P M K 3 06 U6a Chemistry F4(p98-131)csy2p.indd 126 21/12/2022 3:50 PM

127 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 2 Construct a table to record burette reading and volume of R2 solution used. Titration I II III Initial reading of burette (cm3 ) V1 V3 V5 Final reading of burette (cm3 ) V2 V4 V6 Volume of acid P (cm3 ) V2 – V1 = x V4 – V3 = y V6 – V5 = z [4 marks] 3 Based on the experiment, calculate (a) (i) the average volume of R2 solution used. Average volume R2 = x + y + z 3 = _____________ cm3 [2 marks] (ii) By using your answer in 3(a)(i), name an acid P. Explain your answer. Acid P is sulphuric acid. This is because sulphuric acid has two moles of hydrogen ions. [2 marks] (iii) Write the chemical equation for the reaction between the named acid P and sodium hydroxide. H2SO4 + 2NaOH→Na2SO4 + 2H2O [2 marks] (b) What is the colour change at the end point? Pink to colourless [1 mark] (c) State the operational definition for end point for this experiment. When acid P is titrated into a conical flask containing sodium hydroxide and phenolphthalein as indicator, the colour of mixture change from pink to colourless shows that end point is reached. [2 marks] (d) Predict the volume of acid P if the experiment is repeated by replacing sodium hydroxide with magnesium hydroxide solution Same as the volume of acid P in this experiment. [2 marks] Titration Calculation S P O T S P M K 3 06 U6a Chemistry F4(p98-131)csy2p.indd 127 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 128 UNIT 6 MODULE • Chemistry FORM 4 What is salt? LS 6.8.1 • 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. Write the formulae of the salts in the Table of Salts below by replacing hydrogen ion in sulphuric acid, hydrochloric acid, nitric acid and carbonic acid with metal ions or ammonium ion: 6.8 SALT, CRYSTALS AND USES OF SALT IN DAILY LIFE CS 6.8 Table of Salts Metal ion Sulphate salt (from H2SO4) Chloride salt (from HCl) Nitrate salt (from HNO3) Carbonate salt (from H2CO3) Na+ Na2SO4 NaCl NaNO3 Na2CO3 K+ K2SO4 KCl KNO3 K2CO3 NH4 + (NH4 )2SO4 NH4Cl NH4NO3 (NH4 )2CO3 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 Ba2+ BaSO4 BaCl2 Ba(NO3 )2 BaCO3 How are salt crystals formed? • Salt crystals are formed when a saturated salt solution is cooled down. Example of crystals Sodium chloride crystals Copper(II) sulphate crystals Calcium fluoride crystals What are the characteristics of crystals? • Crystals of the same type of salt have the following characteristics: (a) Fixed geometrical shape. Their sizes could be different. Fast crystallisation will produce small crystals, slow crystallisation will produce bigger crystals (b) Flat faces, straight edges and sharp angles (c) Fixed angle between adjacent faces Remark: Crystals of the different salts have different geometric shape like cubes, rhombus, cuboid and prism. Physical Characteristics of Salt Crystals PERFORMANCE LEVEL (PL) Mastered Not mastered PL1 PL2 Recall knowledge and basic skills on acids, bases and salts. Understand and explain acids, bases and salts. 06 U6a Chemistry F4(p98-131)csy2p.indd 128 21/12/2022 3:50 PM

129 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. How are the soluble salts purified by recrystallisation? • The insoluble impurities in salt solution are removed by filtration whereas the soluble are removed by recrystallisation. • Steps in recrystallisation process includes: (i) The soluble salt is dissolved in a suitable solvent, usually water. (ii) The aqueous solution is then heated to evaporate until the solution saturated. (iii) When the hot saturated solution is allowed to cool, the salt reappears as pure crystals, leaving behind the impurities in the solvent. (iv) The crystals are obtained as a residue through filtration. How to grow copper(II) sulphate crystals? • Growing copper(II) sulphate crystals is done by hanging a small copper(II) sulphate crystal in a saturated copper(II) sulphate solution. • Let the water evaporate slowly to get a larger copper(II) sulphate crystal. Glass rod Nylon thread Small copper(II) sulphate crystals Saturated copper(II) sulphate solution Applications Salts Uses Food preparation Sodium chloride, NaCl As food preservative in salted fish Monosodium glutamate (MSG) To enhance the taste of food Sodium hydrogen carbonate (NaHCO3) As baking powder in cake and bread Sodium benzoate (C6H5COONa) Preservative in food such as tomato sauce, oyster sauce and jam Sodium nitrate (NaNO3) Preservative in processed meat such as burger and sausages Agriculture • Nitrate salt such as potassium nitrate (KNO3), sodium nitrate (NaNO3) • Ammonium salt such as ammonium sulphate [(NH4)2SO4], ammonium nitrate (NH4NO3) Fertiliser Copper(II) sulphate (CuSO4) and iron(II) sulphate (FeSO4) Pesticides Medical substances Calcium carbonate (CaCO3) and calcium hydrogen carbonate (CaHCO3) Antacid to reduce acidity in the stomach of gastric patient Calcium sulphate (CaSO4) To make plaster of Paris that is used to support fractured bones Barium sulphate, BaSO4 Enables the intestine of suspected stomach cancer patients to be seen clearly in X-ray film Potassium manganate (KMnO4) As antiseptic to kill germ Other uses Silver bromide (AgBr) To produce black and white photographic film Tin(II) fluoride, SnF2 Added to toothpaste to prevent tooth decay Uses of Salt in Daily Life LS 6.8.3 Examples: How to grow copper(II) sulphate crystal PERFORMANCE LEVEL (PL) Mastered Not mastered PL3 Apply knowledge on acids, bases and salts to explain the natural occurrences or phenomena and be able to carry out simple tasks. 06 U6a Chemistry F4(p98-131)csy2p.indd 129 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 130 UNIT 6 MODULE • Chemistry FORM 4 6.9 PREPARATION OF SALTS CS 6.9 What are the classification of salt? (i) Soluble salt (ii) Insoluble salt State general rules of solubility for salt in water. (i) All salts K+ , Na+ and NH4 + are soluble. (ii) All nitrate salts are soluble. (iii) All carbonate salts are insoluble except K2CO3, Na2CO3 and (NH4)2CO3. (iv) All sulphate salts are soluble except CaSO4, PbSO4 and BaSO4. (v) All chloride salts are soluble except PbCl2 and AgCl. * Based on the solubility of the salts in water, shade the insoluble salts in the Table of Salts in page 128. Plan an experiment to study the solubility of various salt in water. Aim: To study solubility of various salts in water. Problem statement: Are all salts of sulphate, chloride, nitrate and carbonate soluble in water? Manipulated variable: Sulphate salts, chloride salts, nitrate salts, carbonate salts Responding variable: Solubility of salts in water Constant variable: Volume of distilled water, quantity of salt Hypothesis: (i) All nitrate salts are soluble. (ii) All carbonate salts are insoluble except K2CO3, Na2CO3 and (NH4)2CO3. (iii) All sulphate salts are soluble except CaSO4, PbSO4 and BaSO4. (iv) All chloride salts are soluble except PbCl2 and AgCl. Materials: Powder of nitrate, sulphate, chloride and carbonate salts for sodium, potassium, ammonium, barium, silver, lead, calcium and magnesium, distilled water Apparatus: Test tubes, Bunsen burner, test tube holder, spatula, measuring cylinder Procedures: 1 Measure and pour 5 cm3 of distilled water into a test tube. 2 Add half spatula of sodium nitrate powder into the test tube. Shake the test tube. 3 Record the observation. 4 Repeat steps 1 – 3 for all salts stated above. Observation: (Draw table to record the observations) What are the methods to prepare soluble salt? LS 6.9.2 I Using titration method II By adding excess of solid reactant to acid What is the method to prepare insoluble salt? LS 6.9.3 III By the precipitation method Quiz 06 U6a Chemistry F4(p98-131)csy2p.indd 130 21/12/2022 3:50 PM

131 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 1 Preparing Soluble Salt I Using titration method When do we use this method? • When both reactants are aqueous solutions: Acid(aq) + Alkali(aq) ➝ Salt(aq) + Water • Salts of Na+ , K+ and NH4 + are prepared by using titration method. Identify these salts from the Table of Salts on page 128. • Using this method, acid is added to alkali until the end point with the help of an indicator. • At the end point, a neutral salt solution is produced without excess acid or alkali (Excess acid and alkali cannot be separated because both are in aqueous form) II By adding excess of solid reactant to acid When do we use this method? • One of the reactants is insoluble metal, carbonate or insoluble base. (i) Acid(aq) + Metal(s) ➝ Salt + Hydrogen (ii) Acid(aq) + Metal oxide(s) ➝ Salt + Water (iii) Acid(aq) + Metal carbonate(s) ➝ Salt + Water + Carbon dioxide • Soluble salts which are not salts of Na+ , K+ and NH4 + are prepared by adding excess of solid reactant to acid. Identify these salts from the Table of Salts on page 128. • When all the acid reacts, a neutral salt solution is produced while the excess solid unreacted substance is precipitated. • The precipitate is separated by filtration. LS 6.9.2 Quiz Choose pair of reactants that require the use of titration method or by adding excess solid. Reactants Titration method (3 / 7) By adding excess of solid reactant (3 / 7) CuO(s) + H2SO4(aq) 7 3 NaOH(aq) + HCl(aq) 3 7 KOH(aq) + HNO3(aq) 3 7 Zn(s) + HCl(aq) 7 3 Exercise 06 U6a Chemistry F4(p98-131)csy2p.indd 131 21/12/2022 3:50 PM

© Nilam Publication Sdn. Bhd. 132 UNIT 6 MODULE • Chemistry FORM 4 2 Preparing Insoluble Salt LS 6.9.3 III By precipitation method State the name of the reaction to produce an insoluble salt. • Double decomposition reactions. What type of reactants are needed for this method? • The precipitate of insoluble salt is formed when two different solutions that contain the cation and anion of the insoluble salts are mixed. • The precipitate is then obtained by filtration. Give the chemical equation for the precipitation of lead(II) sulphate. Pb(NO3)2(aq) + Na2SO4(aq) ➝ PbSO4(s) + 2NaNO3(aq) How to deduce the correct reactants from a given insoluble salt: lead(II) sulphate? SO4 2– Sulphate of NH4 + / Na+ / K+ because all are soluble Pb2+ Nitrate Pb2+ because all nitrate salts are soluble. • It is advisable to suggest the reactants in which one is nitrate salt and the other one salt of ammonium/potassium/sodium as they are always soluble. Write the ionic equation for the reaction. Pb2+ + SO4 2– ➝ PbSO4 Suggest the suitable reactants and write the ionic equation for the formation of the following insoluble salts. Insoluble salts Solutions of cation Solutions of anion Ionic equation Lead(II) iodide, PbI2 Lead(II) nitrate Potassium iodide Pb2+ + 2I– PbI2 Barium sulphate, BaSO4 Barium nitrate Ammonium sulphate Ba2+ + SO4 2– BaSO4 Silver chloride, AgCl Silver nitrate Sodium chloride Ag+ + Cl– AgCl Calcium carbonate, CaCO3 Calcium nitrate Potassium carbonate Ca2+ + CO3 2– CaCO3 Remark: Lead(II) iodide and lead(II) chloride are exceptional insoluble salts. 1 Lead (II) iodide is insoluble yellow solid but dissolves in hot water and forms a yellow solid again when cooled. 2 Lead (II) chloride is insoluble white solid but dissolves in hot water and forms a white solid again when cooled. How Salt is Made 06 U6b Chemistry F4(p132-160)csy2p.indd 132 21/12/2022 3:54 PM

133 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 1 Salts are prepared based on its solubility as shown in the flow chart below: LS 6.9.2 LS 6.9.3 PREPARATION OF SALT Soluble salt Insoluble salt Salts K+, Na+, NH4+ Other than K+, Na+, NH4+ The salt is prepared by titration method of acid and alkali using an indicator. • Acid + Alkali Salt + Water (Neutralisation Reaction) The salt is prepared by reacting acid with insoluble metal/metal oxide/metal carbonate: • Acid + Metal Salt + Hydrogen (Displacement reaction) • Acid + Metal oxide Salt + Water (Neutralisation reaction) • Acid + Metal carbonate Salt + Water + Carbon dioxide The salt is prepared by precipitation method. (Double decomposition reaction) • Mix two solutions containing cations and anions of insoluble salts. • Stir with glass rod. • Filter using filter funnel. • Rinse the residue with distilled water. • Dry the residue by pressing it between filter papers. • A titration is conducted to determine the volume of acid needed to neutralise a fixed volume of an alkali with the aid of an indicator. • The same volume of acid is then added to the same volume of alkali without any indicator to obtain pure and neutral salt solution. • Evaporate the filtrate until it becomes a saturated solution. • Dip in a glass rod, if crystals are formed, the solution is saturated. • Cooled at room temperature. • Filter and dry the salt crystals by pressing them between filter papers. • Add metal/metal oxide/metal carbonate powder until excess into a fixed volume of the heated acid. • Filter the mixture to remove the excess metal/metal oxide/ metal carbonate. Method III Method II Method I Describe the Preparation of Soluble and Insoluble Salt 06 U6b Chemistry F4(p132-160)csy2p.indd 133 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 134 UNIT 6 MODULE • Chemistry FORM 4 2 Steps to Prepare Soluble Salt • The salt solution is poured into evaporating dish . • Evaporate the salt solution until saturated solution is formed. • Cool it at room temperature until salts crystals are formed. • Measure and pour 100 cm3 of 1 mol dm–3 of any acid and pour into a beaker. • Add metal/metal oxide/metal carbonate powder into the acid and heat gently . • Filter the miture to separate the salt crystals . Method II: Soluble salt except K+, Na+ and NH4+ • Stir the mixture with a glass rod . • Add metal/metal oxide/metal carbonate powder to the acid until excess . Excess of metal/ metal oxide/ metal carbonate Heat • Filter the mixture to separate excess metal/ metal oxide/metal carbonate with the salt solution . The residue is metal/metal oxide/ metal carbonate . The filtrate is salt solution . Method I: Soluble salt of K+, Na+ and NH4 + Acid Alkali Heat Acid Residue is salt crystals Heat Saturated salt solutions Salt crystals • Dry the salt crystals by pressing them between filter papers. Salt crystals ➀ ➁ ➍ ➎ ➂ ➏ ➃ ➋ ➊ ➌ • Measure and pour 50 cm3 of 1 mol dm–3 of any alkali into a conical flask. Add a few drops of phenolphthalein. • 1 mol dm–3 of any acid is titrated to the alkali until neutral by using an indicator. The volume of acid used is recorded. • Repeat the titration without indicator to get pure and neutral salt solution. 06 U6b Chemistry F4(p132-160)csy2p.indd 134 21/12/2022 3:54 PM

135 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 3 Steps to Prepare Insoluble Salt Method III: Preparation of Insoluble XnYm Salt by Double Decomposition Reaction LS 6.9.3 Precipitate of XnYm salt is formed. 3 Mix both solutions and stir the mixture with glass rod . Salt XnYm 5 Press the precipitate between filter papers to dry it. The residue is XnYm salt. 4 Filter the mixture and rinse the precipitate with distilled water . The residue is XnYm salt. 2 Measure and pour 100 cm3 of 1 mol dm–3 of aqueous solution contains Yn– anion into another beaker. 1 Measure and pour 100 cm3 of 1 mol dm–3 of aqueous solution contains Xm+ cation into a beaker. 06 U6b Chemistry F4(p132-160)csy2p.indd 135 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 136 UNIT 6 MODULE • Chemistry FORM 4 1 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. Salts “S” / “IS” Chemical equations Zinc chloride S Zn + 2HCl ZnCl2 + H2 ZnCO3 + 2HCl ZnCl2 + CO2 + H2O ZnO + 2HCl ZnCl2 + H2O Sodium nitrate S NaOH + HNO3 NaNO3 + H2O Silver chloride IS AgNO3 + KCl AgCl + KNO3 AgNO3 + NaCl AgCl + NaNO3 Copper(II) sulphate S CuO + H2SO4 CuSO4 + H2O CuCO3 + H2SO4 CuSO4 + CO2 + H2O Lead(II) sulphate IS Pb(NO3)2 + K2SO4 PbSO4 + 2KNO3 Pb(NO3)2 + Na2SO4 PbSO4 + 2NaNO3 Aluminium nitrate S 2Al + 6HNO3 2Al(NO3)3 + 3H2 Al2O3 + 6HNO3 2Al(NO3)3 + 3H2O Al2(CO3)3 + 6HNO3 2Al(NO3)3 + 3CO2 + 3H2O Lead(II) chloride IS Pb(NO3)2 + 2KCl PbCl2 + 2KNO3 Pb(NO3)2 + 2NaCl PbCl2 + 2NaNO3 Magnesium nitrate S Mg + 2HNO3 Mg(NO3)2 + H2 MgO + 2HNO3 Mg(NO3)2 + H2O MgCO3 + 2HNO3 Mg(NO3)2 + CO2 + H2O Potassium chloride S KOH + HCl KCl + H2O Lead(II) nitrate S PbO + 2HNO3 Pb(NO3)2 + H2O PbCO3 + 2HNO3 Pb(NO3)2 + CO2 + H2O Barium sulphate IS BaCl2 + K2SO4 BaSO4 + 2KCl BaCl2 + Na2SO4 BaSO4 + 2NaCl PL2 Exercise PERFORMANCE LEVEL (PL) Mastered Not mastered PL4 Analyse knowledge on acids, bases and salts in the context of problem solving the natural occurrences or phenomena. 06 U6b Chemistry F4(p132-160)csy2p.indd 136 21/12/2022 3:54 PM

137 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 2 The diagram below shows the set-up of apparatus to prepare soluble salt Y. Nitric acid 25 cm3 of 1 mol dm–3 potassium hydroxide solution + phenolphthalein Phenolphthalein is used as an indicator in a titration between nitric acid and potassium 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. (a) Name salt Y. Potassium nitrate (b) Write a balanced equation for the reaction that occurs. HNO3 + KOH KNO3 + H2O (c) Calculate the concentration of nitric acid. Number of moles of KOH = 1 × 25 1 000 = 0.025 mol From the equation, 1 mol KOH : 1 mol HNO3 0.025 mol KOH : 0.025 mol HNO3 Concentration of HNO3, M 0.025 = M × 25 1 000 M = 1 mol dm–3 (d) Why the experiment is repeated without phenolphthalein? To get pure and neutral salt solution Y. (e) Describe briefly how a crystal of salt is obtained from the salt solution. • The salt solution is poured into an evaporating dish. • The solution is heated to evaporate the solution until one third of 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. Potassium/sodium/ammonium salt. Example: potassium nitrate, sodium sulphate. (g) State the type of reaction in the preparation of the salts. Neutralisation PL2 PL3 PL6 PL2 PL1 PL3 PL1 06 U6b Chemistry F4(p132-160)csy2p.indd 137 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 138 UNIT 6 MODULE • Chemistry FORM 4 3 The following are the steps in preparation of dry copper(II) sulphate crystals. 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. Step II: The mixture is filtered. Step IV: The salt solution is allowed to cool at room temperature for the crystallisation to take place. Step V: The crystals formed are filtered and dried by pressing them between filter papers. Step III: The filtrate is poured into an evaporating dish and heated to evaporate the solution until one third of its original. (a) (i) State two observations during Step I. • Black solid dissolve • Colourless solution turns blue (ii) Write a balance chemical equation for the reaction that occur in Step I. CuO + H2SO4 CuSO4 + H2O (iii) State the type of reaction in the preparation of the salts. Neutralisation (b) Why is copper(II) oxide powder added until some of it no longer dissolve in Step I? To make sure that all sulphuric acid has reacted. (c) What is the purpose of heating in Step III? To evaporate the water and make the copper(II) sulphate solution to become saturated. (d) What is the colour of copper(II) sulphate? Blue (e) What is the purpose of filtration in (i) Step II? • To remove the excess copper(II) oxide. • To obtain copper(II) sulphate solution as a filtrate. (ii) Step V? To obtain copper(II) sulphate crystals as a residue. (f) Draw a labelled diagram to show the set-up of apparatus used in Step II and Step III. Excess of copper(II) oxide Filter paper Copper(II) sulphate solution Copper(II) sulphate solution Heat PL4 PL2 PL2 PL3 PL3 PL5 06 U6b Chemistry F4(p132-160)csy2p.indd 138 21/12/2022 3:54 PM

139 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. (g) Can copper powder replace copper(II) oxide in the experiment? Explain your answer. 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. Substance : Copper(II) carbonate Balance equation : CuCO3 + H2SO4 CuSO4 + H2O + CO2 4 The diagram below showsthe flow chart for the preparation of zinc carbonate and zinc sulphate through reactions I and II. Zinc nitrate Zinc carbonate Zinc sulphate Reaction I Reaction II (a) Based on the flow chart above, classify the above salt to soluble salt and insoluble salt. Soluble salt : Zinc nitrate, zinc sulphate Insoluble salt : Zinc carbonate (b) (i) State the reactant for the preparation of zinc carbonate from zinc nitrate in reaction I. Sodium carbonate solution/potassium carbonate solution/ammonium carbonate solution (ii) State the type of reaction that occurs in reaction I. Double decomposition (iii) Describe the preparation of zinc carbonate from zinc nitrate in the laboratory through reaction I. • 50 cm3 of 1 mol dm–3 zinc nitrate solution is measured and poured into a beaker. • 50 cm3 of 1 mol dm–3 sodium carbonate solution is measured and poured into another beaker. • The mixture is stirred with a glass rod and a white solid, zinc carbonate (ZnCO3) is formed. • The mixture is filtered and the residue is rinsed with distilled water. • The white precipitate is dried by pressing it between filter papers. (iv) Write the chemical equation for the reaction in 4(b)(iii). Zn(NO3)2 + Na2CO3 ZnCO3 + 2NaNO3 (c) (i) State the reactant for the preparation of zinc sulphate from zinc carbonate in reaction II. Sulphuric acid (ii) Describe laboratory experiment to prepare zinc sulphate from zinc carbonate through reaction II. • 50 cm3 of 1 mol dm–3 of sulphuric acid is measured and poured into a beaker and is heated. • The white precipitate from reaction I/zinc carbonate powder is added to the acid until in excess. • The mixture is stirred with a glass rod. The excess white precipitate is filtered out. • The filtrate is poured into an evaporating dish • The salt solution is gently heated until saturated. • The hot saturated salt solution is allowed to cool for crystals to form. • The crystals formed are filtered and dried by pressing it between sheets of filter papers. (iii) Write the chemical equation for the reaction in 4(c)(ii). ZnCO3 + H2SO4 ZnSO4 + H2O + CO2 PL5 PL3 PL2 PL2 PL2 PL6 PL3 PL3 PL6 PL3 06 U6b Chemistry F4(p132-160)csy2p.indd 139 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 140 UNIT 6 MODULE • Chemistry FORM 4 How to construct ionic equation for the formation of insoluble salt? • The ionic equation for the formation of insoluble salt can be constructed if the number of moles of anion and cation to form 1 mol of insoluble salt are known. State the type of the experiment to determine the ionic equation for the formation of insoluble salt. • The continuous variation method. Example of experiment: Problem statement : How to construct ionic equation for the formation of lead(II) chromate precipitate? Aim : To construct ionic equation for the formation of lead(II) chromate precipitate Manipulated variable : Volume of lead(II) nitrate solution Responding variable : Height of precipitate Constant variable : Volume and concentration of potassium chromate(II) solution, concentration of lead(II) nitrate solution, size of test tube Hypothesis : When the volume of lead(II) nitrate solution increases, the height of precipitate will increase until all potassium chromate(VI) has reacted. Apparatus : Test tubes of same size, measuring cylinder//burette, test tube rack, stopper, ruler Materials : 0.5 mol dm–3 of lead(II) nitrate and potassium chromate(VI) solution 5 cm3 of 0.5 mol dm–3 of potassium chromate(VI) solution is poured into every test tube as shown in the following diagram: 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 1 cm3 2 cm3 3 cm3 4 cm3 5 cm3 6 cm3 7 cm3 8 cm3 Precipitate 1 cm3 of 0.5 mol dm–3 lead(II) nitrate solution is added to the first test tube, 2 cm3 to the second test tube and so on until 8 cm3 to the eighth test tube as shown in the following diagram: 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 1 cm3 2 cm3 3 cm3 4 cm3 5 cm3 6 cm3 7 cm3 8 cm3 Precipitate Constructing Ionic Equation for the Formation of Insoluble Salt LS 6.9.4 06 U6b Chemistry F4(p132-160)csy2p.indd 140 21/12/2022 3:54 PM

141 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. Procedures: (a) Eight test tubes of same size are labelled from 1 to 8 and placed on a test tube rack. (b) 5.00 cm3 of 0.5 mol dm-3 potassium chromate(VI) solution is measured and poured into each of the test tubes. (c) 1 cm3 of 0.5 mol dm-3 lead(II) nitrate solution is added to the first test tube, 2 cm3 to the second test tube and so on until 8 cm3 to the eighth test tube. (d) The test tubes are stoppered and shaken well. (e) The test tubes are left aside for one hour on a test tube rack. (f) The height of precipitate in each test tube is measured. (g) The colour of the solution above the precipitate in each test tube is recorded. (h) A graph of height of precipitate against volume of lead(II) nitrate solution added is plotted. Results: Test tubes 1 2 3 4 5 6 7 8 Volume of 0.5 mol dm–3 of potassium chromate(VI) solution / cm3 5.00 5.00 5.00 5.00 5.00 5.00 5.00 5.00 Volume of 0.5 mol dm–3 of lead(II) nitrate solution / cm3 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 Height of precipitate / cm 0.5 0.8 1.2 1.6 2.0 2.0 2.0 2.0 Colour of solution above precipitate Yellow Yellow Yellow Yellow Colourless Colourless Colourless Colourless Chemicals in the solution above precipitate Excess of potassium chromate(VI) and potassium nitrate formed. Excess of potassium chromate(VI) and potassium nitrate formed. Excess of potassium chromate(VI) and potassium nitrate formed. Excess of potassium chromate(VI) and potassium nitrate formed. Potassium nitrate formed. Excess of lead(II) nitrate and potassium nitrate formed. Excess of lead(II) nitrate and potassium nitrate formed. Excess of lead(II) nitrate and potassium nitrate formed. Ions present in the solution above the precipitate NO3 - , K+ CrO4 2- NO3 - , K+ CrO4 2- NO3 - , K+ CrO4 2- NO3 - , K+ CrO4 2- NO3 - , K+ Pb2+, NO3 - , K+ Pb2+, NO3 - , K+ Pb2+, NO3 - , K+ 06 U6b Chemistry F4(p132-160)csy2p.indd 141 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 142 UNIT 6 MODULE • Chemistry FORM 4 1 Complete the following table: Observations Inferences The height of precipitate in test tubes 1 to 4 increases. The solution above precipitate is yellow in test tubes 1 to 4. • The increase in volume of lead(II) nitrate added increases the mass of precipitate formed in test tubes 1 to 4. • There are excess of potassium chromate(VI) solution in test tubes 1 to 4. The height of precipitate remains constant in test tubes 5 to 8. The solution above precipitate is colourless in test tubes 5 to 8. • In test tube 5, potassium chromate(VI) solution has completely reacted with lead(II) nitrate solution. All chromate(VI) ions and all lead(II) ions have reacted. The solution above precipitate is potassium nitrate . • In test tubes 6 to 8, there are excess of lead(II) ions as more lead(II) nitrate solution is added. 2 The graph below is obtained when the height of precipitate is plotted against the volume of lead(II) nitrate solution. 0 1 2 3 4 5 6 7 8 2 Height of precipitate / cm Volume of lead(II) nitrate / cm3 9 (a) What is the colour of lead(II) nitrate solution and potassium chromate(VI) solution? Lead(II) nitrate solution : colourless Potassium chromate(VI) solution : yellow (b) (i) State the name of the precipitate formed. Lead(II) chromate (ii) What is the colour of the precipitate? Yellow (c) (i) Based on the above graph, what is the volume of lead(II) nitrate solution needed to completely react with potassium chromate(VI) 5 cm3 of solution? 5 cm3 (ii) Calculate the number of moles of lead(II) ions in 5.0 cm3 of 0.5 mol dm–3 lead (II) nitrate solution. Ionisation equation of lead(II) nitrate solution: Pb(NO3)2 Pb2+ + 2NO3 – Number of moles of Pb(NO3)2 = 5.0 × 0.5 1 000 = 0.0025 mol From the ionisation equation, 1 mol Pb(NO3)2 : 1 mol Pb2+ 0.0025 mol Pb(NO3)2 : 0.0025 mol Pb2+ PL4 PL4 06 U6b Chemistry F4(p132-160)csy2p.indd 142 21/12/2022 3:54 PM

143 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. (iii) Calculate the number of mol of chromate(VI) ions in 5.0 cm3 of 0.5 mol dm–3 potassium chromate(VI) solution. Ionisation equation of potassium chromate(VI) solution: K2CrO4 2K+ + CrO4 2- From the ionisation equation, 1 mol K2CrO4: 1 mol CrO4 2- 0.0025 mol K2CrO4: 0.0025 mol CrO4 2- (iv) Calculate the number of moles of chromate(VI) ions reacts completely with one mol of lead(II) ions. 0.0025 mol Pb2+ : reacts completely with 0.0025 mol CrO4 2- 1 mol Pb2+ : 1 mol CrO4 2- (d) Write the ionic equation for the formation of the precipitate. Pb2+ + CrO4 2- ➝ PbCrO4 Solving Numerical Problems Involving the Salt Preparation Relationship between mol of substance with mass, volume of gas, volume of solution and concentration of solution: ÷ (RAM/RMM/RFM) g mol–1 × (RAM/RMM/RFM) g mol–1 n = MV 1 000 × 24 dm3 mol–1 ÷ 24 dm3 mol–1 Mass in gram Volume of Number of mol (n) gas in dm3 Solution concentration in mol dm–3 (M) and volume in cm3 (V) Example: 50 cm3 of 2 mol dm–3 sulphuric acid is added to excess of copper(II) oxide powder. Calculate the mass of copper(II) sulphate formed in the reaction. [Relative atomic mass: H = 1, O = 16, Cu = 64, S = 32] Write a balanced equation Write the information from the question above the equation M = 2 mol dm–3 V = 50 cm3 ? g CuO(aq) + H2SO4 CuSO4(aq) + 2H2O(l) Convert the given quantity into moles by using the method shown in the chart above. Number of moles of sulphuric acid = 2 × 50 1 000 = 0.1 mol Use the mole ratio of the substances involved to find the number of moles of other substance. Remark: The coefficient of each formula shows the number of moles of reactants react and products formed. From the equation, 1 mol H2SO4 : 1 mol CuSO4 0.1 mol H2SO4 : 0.1 mol CuSO4 Convert the mole into the quantity required in the question by using the method shown in the chart above. Mass of CuSO4 = 0.1 mol × [64 + 32 + (16 × 4)] g mol–1 = 16 g 06 U6b Chemistry F4(p132-160)csy2p.indd 143 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 144 UNIT 6 MODULE • Chemistry FORM 4 1 27.66 g of lead(II) iodide is precipitated when 2.0 mol dm–3 of aqueous lead(II) nitrate solution is added to an excess of aqueous potassium iodide solution. Calculate the volume of aqueous lead(II) nitrate solution used. [Relative atomic mass: I = 127, Pb = 207] M = 2 mol dm–3 V = ? cm3 27.66 g Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq) Number of moles of PbI2 = 27.66 (207 + 2 × 127) = 0.06 mol From the equation, 1 mol PbI2 : 1 mol Pb(NO3)2 0.06 mol PbI2 : 0.06 mol Pb(NO3)2 Volume of Pb(NO3)2 = n mol M mol dm–3 = 0.06 mol 2 mol dm–3 = 0.03 dm3 = 30 cm3 2 Zinc oxide powder is added to 100 cm3 of 2 mol dm–3 nitric acid to form zinc nitrate. Calculate (i) the mass of zinc oxide that has reacted. (ii) the mass of zinc nitrate produced. [Relative atomic mass: H = 1, O = 16, N = 14, Zn = 65] M = 2 mol dm–3 V = 100 cm3 (i) 2HNO3(aq) + ZnO(s) Zn(NO3)2(aq) + H2O(l) Number of moles of HNO3 = 100 × 2 1 000 = 0.2 mol From the equation, 2 mol HNO3 : 1 mol ZnO 0.2 mol HNO3 : 0.1 mol ZnO Mass of ZnO = 0.1 × [65 + 16] = 8.1 g (ii) From the equation, 2 mol HNO3 : 1 mol Zn(NO3)2 0.2 mol HNO3 : 0.1 mol Zn(NO3)2 Mass of Zn(NO3)2 = 0.1 mol × [65 + [14 + (16 × 3)] × 2] g mol–1 = 0.1 × 189 = 18.9 g 3 200 cm3 of 1 mol dm–3 barium chloride solution reacts with 100 cm3 of 1 mol dm–3 silver nitrate solution. Calculate the mass of precipitate produced. [Relative atomic mass Ag = 108, Cl = 35.5] M = 1.0 mol dm–3 M = 1.0 mol dm–3 V = 200 cm3 V = 100 cm3 ? g BaCl2 + 2AgNO3 2AgCl + Ba(NO3)2 Number of moles of barium chloride = 1 × 200 1 000 = 0.2 mol (excess) Number of moles of silver nitrate = 1 × 100 1 000 = 0.1 mol From the equation, 1 mol BaCl2 : 2 mol AgNO3 : 2 mol AgCl 0.2 mol BaCl2 (excess) : 0.1 mol AgNO3 : 0.1 mol AgCl Mass of AgCl = 0.1 mol × [108 + 35.5] g mol–1 = 14.35 g PL3 PL3 PL3 Exercise 06 U6b Chemistry F4(p132-160)csy2p.indd 144 21/12/2022 3:54 PM

145 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. What are the effect of heat on salts? LS 6.10.1 • Some salts decompose when they are heated: Salt Metal oxide (Colour of residue refer to certain cation) Gas (Gas identification refers to certain anion/cation) + 1 Common gas identification: LS 6.10.2 Gas Observations/Tests Inferences Nitrogen dioxide, NO2 • Brown gas. • Place a moist blue litmus paper at the mouth of the boiling tube, blue litmus paper turns red. • Nitrogen dioxide gas is produced by heating nitrate salt. • Nitrate ion, NO3 – present. Oxygen, O2 • Colourless gas. • Put a glowing wooden splinter into boiling tube, the glowing wooden splinter is relighted. • Oxygen gas is produced by heating nitrate or chlorate(V) salt. • Nitrate ion, NO3 – present or ClO3 – ion present. Carbon dioxide, CO2 • Colourless gas. • Pass the gas through lime water, lime water turns chalky. • Draw the set-up of apparatus to conduct the test: Calcium carbonate Heat Limewater • Carbon dioxide gas is produced by heating carbonate salt. • Carbonate ion, CO3 2– present. Ammonia, NH3 • Colourless gas with pungent smell. • Place a moist red litmus paper at the mouth of the boiling tube, red litmus paper turns blue. • Ammonia gas is produced by heating ammonium salt with alkali. • Ammonium ion, NH4 + present. Hydrogen gas, H2 • Colourless gas. • Placed burning wooden splinter at the mouth of the test tube. • The gas burns with “pop” sound. • Hydrogen gas is produced. Chlorine gas, Cl2 • Greenish yellow gas. • Place moist blue litmus paper at the mouth of test tube. • Blue litmus paper turns red, it is then bleached. • Chlorine gas is produced. Hydrogen chloride gas, HCl • Colourless gas. • Dip a glass rod into concentrated ammonia solution and place it near the mouth of the test tube. • White fumes are formed. • Hydrogen chloride gas is produced by heating of sodium chloride with concentrated sulphuric acid. Sulphur dioxide gas, SO2 • Colourless gas with pungent smell. • Bubble gas through acidified potassium dichromate(VI). • The orange solution turns green. • Sulphur dioxide gas is produced. 6.10 EFFECT OF HEAT ON SALT CS 6.10 Glass rod dip in concentrated ammonia Concentrated H2SO4 NaCl 06 U6b Chemistry F4(p132-160)csy2p.indd 145 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 146 UNIT 6 MODULE • Chemistry FORM 4 2 Action of heat on nitrate and carbonate salts. Cations Nitrate (NO3 – ) Carbonate (CO3 2–) Decompose to oxygen gas and metal nitrite when heated Does not decompose when heated K+ 2KNO3 2KNO2 + O2 White solid White solid – Na+ 2NaNO3 2NaNO2 + O2 White solid White solid – Decompose to oxygen gas, nitrogen dioxide gas and metal oxide when heated Decompose to carbon dioxide gas and metal oxide when heated Ca2+ 2Ca(NO3)2 2CaO + 4NO2 + O2 White White Brown solid solid fume CaCO3 CaO + CO2 White White Turn lime solid solid water chalky Mg2+ 2Mg(NO3)2 2MgO + 4NO2 + O2 White White Brown solid solid fume MgCO3 MgO + CO2 White White Turn lime solid solid water chalky Al3+ 4Al(NO3)3 2Al2O3 + 12NO2 + 3O2 White White Brown solid solid fume 2Al2(CO3)3 2Al2O3 + 6CO2 White White Turn lime solid solid water chalky Zn2+ 2Zn(NO3)2 2ZnO + 4NO2 + O2 White Yellow when Brown solid hot, white gas when cold ZnCO3 ZnO + CO2 White Yellow when Turn lime solid hot, white water chalky when cold Pb2+ 2Pb(NO3)2 2PbO + 4NO2 + O2 White Brown when Brown solid hot, yellow gas when cold PbCO3 PbO + CO2 White Brown when Turn lime solid hot, yellow water chalky when cold Cu2+ 2Cu(NO3)2 2CuO + 4NO2 + O2 Blue Black Brown solid solid fume CuCO3 CuO + CO2 Green Black Turn lime solid solid water chalky 3 Sulphate salts are more stable, they are not easily decompose when heated. 4 Chloride salts do not decompose except NH4Cl: NH4Cl(s) NH3(g) + HCl(g) LS 6.10.2 06 U6b Chemistry F4(p132-160)csy2p.indd 146 21/12/2022 3:54 PM

147 UNIT 6 MODULE • Chemistry FORM 4 © Nilam Publication Sdn. Bhd. 5 Complete the following table: LS 6.10.2 Observations Inferences/conclusions A white salt is heated. • Brown gas is released, the gas turns moist blue litmus paper red. • Residue is yellow when hot and white when cold. • Nitrogen dioxide gas released. Nitrate ion present. • The residue is zinc oxide. Zinc ion present. • The white salt is zinc nitrate . A green salt is heated. • Colourless gas released, the gas turns limewater chalky. • Residue is black. • Carbon dioxide gas released. Carbonate ion present. • The residue is copper(II) oxide. Copper(II) ion present. • The green salt is copper(II) carbonate . A white salt is heated. • Colourless gas released, the gas turns limewater chalky. • Residue is brown when hot and yellow when cold. • Carbon dioxide gas released. Carbonate ion present. • The residue is lead(II) oxide. Lead(II) ion present. • The white salt is lead(II) carbonate . A white salt is heated. • Colourless gas released, the gas turns limewater chalky. • Residue is yellow when hot and white when cold. • Carbon dioxide gas released. Carbonate ion present. • The residue is zinc oxide. Zinc ion present. • The white salt is zinc carbonate . A blue salt is heated. • Brown gas is released, the gas turns moist blue litmus paper red. • Residue is black. • Nitrogen dioxide gas released. Nitrate ion present. • The residue is copper(II) oxide. Copper(II) ion present. • The blue salt is copper(II) nitrate . A white salt is heated. • Brown gas is released, the gas turns moist blue litmus paper red. • Residue is brown when hot and yellow when cold. • Nitrogen dioxide gas released. Nitrate ion present. • The residue is lead(II) oxide. Lead(II) ion present. • The white salt is lead(II) nitrate . A white salt is heated. • Colourless gas released, the gas turns limewater chalky. • Residue is white • Carbon dioxide gas released. Carbonate ion present. • The possible residues are CaO/MgO/Al2O3. • From the above table, action of heat on salt can be used to identify lead(II) nitrate , lead(II) carbonate , zinc nitrate , zinc carbonate , copper(II) nitrate and copper(II) carbonate . • Confirmatory test for other cations and anions is carried out by Confirmatory Tests for Anions and Cations. 06 U6b Chemistry F4(p132-160)csy2p.indd 147 21/12/2022 3:54 PM

© Nilam Publication Sdn. Bhd. 148 UNIT 6 MODULE • Chemistry FORM 4 What is qualitative analysis of salt? • Qualitative analysis of a salt is a chemical technique to identify the ions present in a salt. What is the preliminary examination on salt? LS 6.11.2 • The preliminary examination is on the physical properties such as colour and solubility, indicate the possibility the presence of certain cations, anions or metal oxide. Solid Aqueous Salts/Cation/Metal oxide White Colourless K+ , Na+ , Ca2+, Mg2+, Al3+, Zn2+, Pb2+, NH4 + Green Insoluble CuCO3 Light green Light green Fe2+, example: FeSO4, FeCl2, Fe(NO3)2 Blue Blue Cu2+, example: CuSO4, Cu(NO3 )2 and CuCl2 Brown Brown Fe3+ Black Insoluble CuO Yellow when hot, white when cold Insoluble ZnO Brown when hot, yellow when cold Insoluble PbO How are qualitative analysis being done to identify salts? LS 6.11.2 • The qualitative analysis consists of the following steps: (a) Observe the physical properties on salt. (b) The action of heat on salts. (c) Prepare aqueous solution of salts and conduct confirmatory test for cation and anion that present. 6.11 QUALITATIVE ANALYSIS CS 6.11 Confirmatory Tests for Cations LS 6.11.1 List the cations. Ca2+, Mg2+, Zn2+, Al3+, Pb2+, Cu2+, Fe2+, Fe3+, NH4 + What are the two steps for testing cations? Step 1: Add a few drops of sodium hydroxide solution or ammonia solution to the aqueous solution of salt and observe. Step 2: Add excess of sodium hydroxide solution or ammonia solution and observe. What are the essential observations? • Three main observations: (i) Is there a precipitate formed in a few drops of testing reagent? (ii) If yes, what is the colour of the precipitate? (iii) Does the precipitate dissolve in excess testing reagent? 06 U6b Chemistry F4(p132-160)csy2p.indd 148 21/12/2022 3:54 PM