Suggested Teaching Scheme - STMGSSintranet.stmgss.edu.hk/~ted/ccy/new_edition/Book2Aanswer.pdf · 2014-10-22 · Suggested Teaching Scheme Suggested Teaching Scheme The following

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Suggested Teaching Scheme


The following suggested teaching schemes are for teachers’ reference only. Teachers may revise them based on the time-tabling arrangement of their own schools.

Scheme 1: Chemistry to be studied in Secondary 3, 4, 5 and 6In many schools, the Chemistry curriculum is studied in Secondary 3, 4, 5 and 6. Although the distribution of periods varies from school to school, the total number of periods for the curriculum is generally around 4�6. A possible distribution of periods is as follows:

A possible distribution of periods for S3, S4, S5 and S6

S3 S4 S5 S6

Number of teaching weeks per year 28 28 28 �6

Number of periods per week 2 5 5 5

Total number of periods per year 56 �40 �40 80

Total number of periods for the curriculum 4�6

Suggested teaching scheme for the curriculum

Level ContentSuggested number

of period(s)

S3(56 periods)

Topic � Planet Earth �2

Topic 2 Microscopic World I 44

S4(�40 periods)

Revision on laboratory safety �

Topic 3 Metals 39

Topic 4 Acids and Bases 45

Topic 5 Redox Reactions, Chemical Cells and Electrolysis 4�

Topic 6 Microscopic World II �4

S5(�40 periods)


Topic 7 Fossil Fuels and Carbon Compounds 32

Topic 8 Chemistry of Carbon Compounds 45

Topic 9 Chemical Reactions and Energy �3

Topic �0 Rate of Reaction �6

Topic �� Chemical Equilibrium �8

Topic �2 Patterns in the Chemical World �5

S6(80 periods)


Topic �3 Industrial Chemistry 39

Topic �4 Materials Chemistry 39

Topic �5 Analytical Chemistry 40

Schools taking investigative study need to allocate an extra of 30 periods for the curriculum.

Only 2 out of 3 Only 2 out of 3

2

Topic 4 Acids and Bases

Scheme 2: Chemistry to be studied in Secondary 4, 5 and 6In some schools, the Chemistry curriculum is studied in Secondary 4, 5 and 6. The total number of periods for the curriculum is generally around 360. A possible distribution of periods is as follows:

A possible distribution of periods for S4, S5 and S6

S4 S5 S6

Number of teaching weeks per year 28 28 �6

Number of periods per week 5 5 5

Total number of periods per year �40 �40 80

Total number of periods for the curriculum 360

Suggested teaching scheme for the curriculum

Level ContentSuggested number

of period(s)

S4(�40 periods)

Topic � Planet Earth 8

Topic 2 Microscopic World I 3�

Topic 3 Metals 32

Topic 4 Acids and Bases 36

Topic 5 Redox Reactions, Chemical Cells and Electrolysis 33

S5(�40 periods)


Topic 6 Microscopic World II �3

Topic 7 Fossil Fuels and Carbon Compounds 29

Topic 8 Chemistry of Carbon Compounds 4�

Topic 9 Chemical Reactions and Energy �2

Topic �0 Rate of Reaction �5

Topic �� Chemical Equilibrium �6

Topic �2 Patterns in the Chemical World �3

S6(80 periods)


Topic �3 Industrial Chemistry 39

Topic �4 Materials Chemistry 39

Topic �5 Analytical Chemistry 40

Schools taking investigative study need to allocate an extra of 30 periods for the curriculum.

Only 2 out of 3 Only 2 out of 3

3


Suggested number of periods for Topic 4

Chemistry forTotal number

of periodsSuggested number of periods for each unit

S3–S6(Scheme �)

45

Unit �4 Acids and alkalisUnit �5 Molarity, pH scale and strengths of acids and

alkalisUnit �6 Salts and neutralizationUnit �7 Concentration of solutions and volumetric

analysis

�3

5��

�6

S4–S6(Scheme 2)

36

Unit �4 Acids and alkalisUnit �5 Molarity, pH scale and strengths of acids and

alkalisUnit �6 Salts and neutralizationUnit �7 Concentration of solutions and volumetric

analysis

�0

49

�3

4


Acids and bases are involved in numerous chemical processes, from industrial processes to biological ones. Students have already studied acids and bases in junior science courses. Here they will further study the properties of acids and bases, and the molarity concept. They should also develop an awareness of the potential hazards associated with the handling of acids and bases.

Students will learn to use an instrumental method of pH measurement and acquire knowledge about strong / weak acids and alkalis. They will also learn to prepare salts by different methods and to perform volumetric analysis involving acids and alkalis. Through these experimental practices students should be able to demonstrate essential experimental techniques, to analyze data and to interpret experimental results. As an introduction to analytical chemistry, students will need to use the chemical reactions learnt to identify the species in a sample. All these serve as a basis for their further study in Topic �5 Analytical Chemistry (optional).

Organization of the topic

Teaching Plan

Acids and Bases

Unit 14Acids and alkalis

Unit 15Molarity, pH scale and strengths of acids and alkalis

Unit 17Concentration of

solutions andvolumetric analysis

Unit 16Salts and

neutralization

5

Teaching Plan

Section Key point(s)Suggested task(s) for

studentsRemark

Total number of period = 1

�4.� Acids in our daily lives

• Acids commonly found in our daily lives

• Students have studied ‘common acids used at home’ and ‘everyday uses of acids’ in Unit �0 of Science (secondary �–3).

�4.2 Acids in the laboratory

• Acids commonly used in the laboratory

• Proper procedure in diluting concentrated sulphuric acid

• Students have studied ‘proper procedures in diluting concentrated acids and alkalis’ in Unit �0 of Science (secondary �–3).

Total number of periods = 2

�4.3 Characteristics of dilute acids

• Taste• Effect on indicators• Reaction with metals• Reaction with

carbonates• Reaction with

hydrogencarbonates• Reaction with

hydroxides and oxides of metals

• Electrical conductivity

• Activity �4.� — Investigating the properties of dilute hydrochloric acid

• Practice �4.�

• Do you know — What is the colour of

hydrangea flowers in acidic and alkaline soils?

• Reaction of magnesium

with dilute hydrochloric acid

• Action of dilute

hydrochloric acid on sodium carbonate

• Students have studied ‘the use of indicators to classify solutions into acidic and alkaline’ in Unit �0 of Science (secondary �–3).

Unit 14 Acids and alkalis

Continued on next page

6



studentsRemark

Total number of periods = 2 (Scheme 1), total number of period = 1 (Scheme 2)

�4.4 The role of water for acids

• Water must be present for an acid to show its acidic properties

• Definition of an acid• Hydrogen ion in

aqueous solution

• Activity �4.2 — Comparing the properties of solid citric acid and its aqueous solution

• Practice �4.2

• Studying the role of

water in exhibiting the acidic properties of citric acid

• Do you know — Baking powder

• Dissolving hydrogen

chloride in water

�4.5 Basicity of an acid • What basicity of an acid means

• Basicity of some common acids


�4.6 Bases and alkalis • What bases and alkalis are

�4.7 Bases and alkalis in the home

• Bases and alkalis commonly used in our daily lives

• Students have studied ‘common alkalis used at home’ and ‘everyday uses of alkalis’ in Unit �0 of Science (secondary � – 3).

�4.8 Alkalis in the laboratory

• Alkalis commonly used in the laboratory

Total number of periods = 3 (Scheme 1), total number of periods = 2 (Scheme 2)

�4.9 Characteristics of solutions of alkalis

• Taste• Feel• Effect on indicators• Reaction with metal

ions• Reaction with

ammonium compounds• Reaction with acid• Electrical conductivity

• Activity �4.3 — Investigating the properties of dilute solutions of alkalis

• Refer to the video clip

‘Acids and Alkalis’: http://resources.hkedcity.

net/resource_detail.php?rid=�85�70�568

(accessed July 20�4)

• Action of solid

calcium hydroxide on ammonium chloride

�4.�0 The role of water for alkalis

• Properties of solutions of alkalis depend on the presence of hydroxide ions

• Dissolving ammonia in

water


7

Teaching Plan


studentsRemark


�4.�� An introduction to analytical chemistry

• Using the chemical reactions learnt to– identify species in a

sample – distinguish between

chemicals

• Practice �4.3• Problem Solving


�4.�2 Concentrated acids • Introduction to concentrated hydrochloric acid, nitric acid and sulphuric acid

• Refer to the following

website for the use, manufacturing, storage and safety information of sulphuric acid:

http://www.sulphuric-acid.org/


�4.�3 Corrosive nature of concentrated acids and alkalis

• How corrosive concentrated acids and alkalis are

• Find & Share — Corrosive nature of concentrated acids and alkalis

• Action of concentrated

sodium hydroxide solution on meat

• Students have studied ‘corrosive nature of acids’, ‘potential dangers in handling strong acids and alkalis’ and ‘emergency treatment involving acids and alkalis’ in Unit �0 of Science (secondary �–3).

�4.�4 Hygroscopic and deliquescent substances

• Examples of hygroscopic and deliquescent substances

• Drying agents

8



studentsRemark


�5.� Concentration of a solution

• Calculations involving molarity



�5.2 The pH scale • Definition of pH• Calculate concentration

of hydrogen ions from pH value and vice versa

• pH values of some common substances

• Practice �5.2 • Do you know — The origin of the term

pH• Do you know — pH of normal rainwater• Students have studied

‘pH scale as an indication of relative acidity’ in Unit �0 of Science (secondary �–3).

�5.3 Determining pH values of solutions

• Methods to determine pH values of solutions– universal indicator

solution– pH meter– data-logger

• Activity �5.� — Classifying substances as acidic, alkaline or neutral using indicators

• Do you know — Monitoring water

quality

• Refer to the following website for simulations of measuring the pH of solutions of acids, bases and salts: http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/acidbasepH/ph_meter.html

(accessed July 20�4)• Students have studied

‘measurement of pH values’ in Unit �0 of Science (secondary �–3).

Unit 15 Molarity, pH scale and strengths of acids and alkalis


9

Teaching Plan


studentsRemark


�5.4 Strong and weak acids

• Definitions• Comparing the

dissociation of strong and weak acids in water

• Activity �5.2 – Distinguishing between a strong acid and a weak acid

• Strong and weak acids

�5.5 Comparing the strengths of acids

• Comparing the strength of acids by– pH– electrical conductivity – reaction with

magnesium

• Discussion

• Comparing the pH, electrical conductivity and reaction with magnesium of acids with different strength


�5.6 Strong and weak alkalis

• Definitions• Comparing the

dissociation of strong and weak alkalis in water

• Strong and weak alkalis

�5.7 Comparing the strengths of alkalis

• Comparing the strength of alkalis by– pH– electrical conductivity

�5.8 Strength versus concentration

• Distinguishing between the strength and concentration of acids

• Practice �5.3

�0



studentsRemark


�6.� Acid-base reactions • Neutralization of an acid and an alkali

• Neutralization of an acid and an insoluble metal hydroxide

• Neutralization of an acid and an insoluble metal oxide

• Practice �6.� • Refer to the video clip

‘Neutralization’: http://resources.hkedcity.

net/resource_detail.php?rid=�065476039


�6.2 Heat change during neutralization

• Heat released during the neutralization between a strong acid and a strong alkali

• Measuring the temperature change of the neutralization reaction between dilute hydrochloric acid and dilute sodium hydroxide solution


�6.3 Formation of salts • Salt formation when hydrogen ions in acids are replaced by metal ions or ammonium ions

• Normal salts and acid salts

�6.4 Naming of salts • Rules for naming salts • Practice �6.2


�6.5 Soluble and insoluble salts

• Solubility of some common salts in water

�6.6 Preparing soluble salts (except sodium, potassium and ammonium salts)

• Preparing copper(II) sulphate crystals

• Action of acids on metal / insoluble bases / insoluble carbonates

– using the preparation of zinc sulphate as an example

• Activity �6.� — Preparing magnesium sulphate crystals from the reaction between an acid and an insoluble carbonate

• Preparing copper(II) sulphate crystals from the neutralization reaction between dilute sulphuric acid and copper(II) oxide

Unit 16 Salts and neutralization


��

Teaching Plan


studentsRemark

�6.7 Preparing sodium, potassium and ammonium salts

• Titration• Action of acids on

alkalis / soluble carbonates

– using the preparation of sodium sulphate crystals as an example

• Activity �6.2 — Preparing sodium chloride from an acid-alkali titration

• Practice �6.3

• Preparing sodium sulphate crystals from the neutralization reaction between dilute sulphuric acid and dilute sodium hydroxide solution

• Students have studied ‘neutralization of acids and alkalis to get salts’ in Unit �0 of Science (secondary �–3).


�6.8 Preparing insoluble salts

• Preparing insoluble salts by precipitation reaction

– using the preparation of lead(II) chloride as an example

• Activity �6.3 — Preparing silver chloride

by precipitation• Practice �6.4


�6.9 Uses of neutralization

• Soil treatment• Treatment of industrial

waste• Production of fertilizers• Treatment of an ache

caused by excess acid in the stomach

• Find & Share — Uses of neutralization

• Practice �6.5

• Students have studied ‘everyday uses of neutralization’ in Unit �0 of Science (secondary �–3).

�2



studentsRemark


�7.� Concentration of a solution

• Calculations involving concentration (in g dm–3)


�7.2 Dilution • Calculations involving dilution

• Practice �7.2


�7.3 Volumetric analysis • What volumetric analysis is

• Apparatus used in volumetric analysis

�7.4 Preparing a standard solution of an acid / alkali

• Dissolving a solid acid / alkali in water

• Diluting a concentrated acid / alkali of known concentration

• Activity �7.� — Preparing solutions of known concentrations

• Preparing 250.0 cm3 of about 0.� mol dm–3

ethanedioic acid solution by dissolving the solid acid

• Preparing 250.0 cm3 of 0.�0 mol dm–3 (M) sulphuric acid by diluting �.0 mol dm–3 (M) sulphuric acid


�7.5 Acid-alkali titration • Steps in titration• Using titration data

to calculate the concentration of an unknown solution

• Washing apparatus in volumetric analysis

• Discussion

• Titration of dilute hydrochloric acid against dilute sodium hydroxide solution

• Refer to the following website for simulations of acid-alkali titrations:

http://www.vias.org/simulations/simusoft_titration.html


Unit 17 Concentration of solutions and volumetric analysis


�3

Teaching Plan


studentsRemark

�7.6 pH change during a titration

• What equivalence point is

• Monitoring the pH change during an acid-alkali titration using a pH meter

• Titration curve

• Activity �7.2 — Following the pH change during an acid-alkali titration

• Following the pH

change during an acid-alkali titration

• A strong acid-strong alkali titration curve


�7.7 Using an indicator in an acid-alkali titration

• Choosing a suitable indicator for an acid-alkali titration

• Practice �7.3 • Titration curves

�7.8 Equivalence point detection by temperature change

• Detecting the equivalence point of a titration by temperature change of the solution mixture


�7.9 Steps to solve problems of acid-alkali titrations

• Steps to solve problems involving acid-alkali titrations

• Calculations involving acid-alkali titrations

• Activity �7.3 — Determining the concentration of citric acid in lemon squash

• Activity �7.4 —

Determining the concentration of ammonia in a household glass cleanser

• Practice �7.4• Practice �7.5


�4



studentsRemark


�7.�0 Back titration • Determining the percentage by mass of aluminium hydroxide in one drug tablet by back titration

• Activity �7.5 — Determining the mass of the active ingredient in an antacid tablet

• Activity �7.6 —

Determining the percentage by mass of calcium carbonate in eggshells by back titration

• Practice �7.6• Chemistry Magazine — Sulphur dioxide content

in wine

�5

Teaching Notes


N1 page 2

Fizzy drinks

Fizzy drinks contain carbon dioxide dissolved under pressure. Carbon dioxide exists mainly as hydrated molecules, CO2(aq). Only a small amount of the gas reacts with water to form carbonic acid.

CO2(aq) + H2O(l) H2CO3(aq)

Most fizzy drinks are made by dissolving carbon dioxide gas in the still drink under pressure. This method is used for soft drinks such as lemonade and Coca-Cola, for most bottled beers, and for some cheap sparking wines.

When the cap of a bottle of fizzy drink is taken off, gaseous carbon dioxide escapes and forms bubbles which rise to the surface.

Suppose a test tube with a magnesium ribbon is immersed in a beaker of freshly opened carbonated water.

When dilute hydrochloric acid is added to the magnesium ribbon, an exothermic reaction occurs. The temperature is increased. Solubility of carbon dioxide in the carbonated water decreases. Hence more gas bubbles are seen in the carbonated water outside the test tube.

N2 page 2

Toilet bowl cleansers

The discolourations and tight scale buildup that occur in toilet bowls are mostly calcium carbonate deposits from hard water. Calcium carbonate reacts with some acids to form water soluble substances. The solid type toilet bowl cleansers are mostly sodium hydrogensulphate while the liquid type contain hydrochloric acid.

In addition to the acidic substances, these cleansers may contain a carbonate or hydrogencarbonate to cause bubbling and therefore agitation to help loosen the scale and give the illusion of being a ‘powerful’ cleanser. Some cleansers also contain a blue colouring substance as blue is supposed to look ‘clean and pure’. This is purely the result of psychological conditioning by the advertizing industry.

Precaution: never use toilet bowl cleansers with chlorine bleaches as they react to give toxic chlorine gas.

Teaching Notes

�6


N8 page 9

Acid-base nature of oxides

In general, non-metals react with oxygen to form acidic oxides and metals react with oxygen to form basic oxides.

Carbon dioxide (CO2) and phosphorus pentoxide (P4O�0) are common acidic oxides. Carbon dioxide reacts with water to form an acid and reacts with bases to form salts.

CO2(g) + H2O(l) H2CO3(aq) carbonic acid

CO2(g) + 2NaOH(aq) H2O(l) + Na2CO3(aq) sodium carbonate

Phosphorus pentoxide reacts with water to form phosphoric acid.

P4O�0(s) + 6H2O(l) 4H3PO4(aq) phosphoric acid

Sodium oxide and copper(II) oxide are common basic oxides. Sodium oxide dissolves in water to form sodium hydroxide.

Na2O(s) + H2O(l) 2NaOH(aq) sodium hydroxide

Copper(II) oxide reacts with dilute sulphuric acid to form copper(II) sulphate.

CuO(s) + H2SO4(aq) CuSO4(aq) + H2O(l) copper(II) sulphate

There are some oxides, such as ZnO, PbO and Al2O3, that react with acids to form salts, but also react with alkalis. These are amphoteric oxides.

Aluminium oxide is insoluble in water. However, it reacts with acids to give salts.

Al2O3(s) + 6HCl(aq) 2AlCl3(aq) + 3H2O(l)

Aluminium oxide also dissolves in sodium hydroxide solution to give a complex salt.

Al2O3(s) + 2OH–(aq) + 3H2O(l) 2[Al(OH)4]–(aq)

tetrahydroxoaluminate ion

There is another group of oxides which do not react with either acids or bases. These are neutral oxides. Common neutral oxides are carbon monoxide (CO), dinitrogen oxide (N2O) and nitrogen monoxide (NO).

�7

Teaching Notes

Common acidic, basic, amphoteric and neutral oxides

Acidic oxides Basic oxides Amphoteric oxides Neutral oxides

CO2 Na2O ZnO CO

NO2 K2O Al2O3 NO

P4O6 MgO PbO N2O

P4O�0 CaO SnO

SO2 CuO

SO3 Fe2O3

Cl2O Ag2O

Look at the location of different types of oxides in the periodic table shown below:

• Oxides of non-metals (on the right of the periodic table) are acidic oxides. They are covalent compounds.

• Oxides of metals (on the left of the periodic table) are basic oxides. They are ionic compounds.

The acid-base nature of oxides will be discussed in Topic �2 Patterns in the Chemical World.

page 11N10

Citric acid

Examination questions often ask about citric acid.

• It is a weak acid.

• It is an electrolyte.

• It exists as a solid at room conditions.

• It contains ionizable hydrogen atoms.

• When citric acid is dissolved in water, citric acid molecules become mobile.

• When water is added to a solid mixture of citric acid and sodium hydrogencarbonate, a colourless gas (carbon dioxide) evolves.

�8


page 15N11

Kitchen cleansers

Many cleaning problems involve grease-bound dirt. Fats, grease and oil bind dirt to a surface, making the dirt difficult to be removed with water alone, because water and grease do not mix. The usual solution is to use a detergent, which helps grease and water mix. For very greasy surfaces, such as oven and kitchen surfaces, detergent alone is not very effective. In such cases, an alkali cleanser can be used to react with the grease.

The grease, fats and oil encountered in the kitchen are all of natural origin. They contain the triesters of long-chain carboxylic acids. They are hydrolyzed by alkalis. Such reactions are called saponification. The general equation is like this:

Alkalis not only break down the fats and oils but turn them into soap. Therefore they are very effective in removing fatty, greasy deposits. Oven cleansers usually contain sodium hydroxide.

Drain cleansers

Some drain cleansers contain solid sodium hydroxide together with a little aluminium powder. When this mixture is poured down the drain, it comes into contact with water and the following reaction occurs:

2Al(s) + 2OH–(aq) + 6H2O(l) 2[Al(OH)4]–(aq) + 3H2(g)

This reaction is very exothermic and at the high temperature reached, the grease blocking the drain melts. The cleansing action of the sodium hydroxide is much more effective. The process is aided by the hydrogen bubbles produced.

�9

Teaching Notes

page 17N12

Nickel(II) compounds

Examination questions often ask about nickel(II) compounds.

• Ni2+(aq) is green in colour.

• Nickel(II) hydroxide can be precipitated by adding dilute sodium hydroxide solution to a solution of nickel(II) salt.

Ni2+(aq) + 2OH–(aq) Ni(OH)2(s) green precipitate

• Nickel(II) carbonate can be precipitated by mixing Ni2+(aq) and CO32–(aq).

Ni2+(aq) + CO32–(aq) NiCO3(s)

• Ni is below Fe but above Pb in the electrochemical series.

page 21N17

Reagents or methods for distinguishing between different species

Examination questions often ask about the reagents (or methods) that can be used to distinguish between two species. Examples:

• aluminium sulphate solution and lead(II) ethanoate solution

distinguished by hydrochloric acid(only lead(II) ethanoate solution produces a white precipitate, PbCl2)

• solid sodium carbonate and solid calcium carbonate

distinguished by testing water solubility

• iron(II) sulphate solution and iron(III) sulphate solution

distinguished by colour / dilute aqueous ammonia / dilute sodium hydroxide solution (they give precipitates of different colours with dilute aqueous ammonia / dilute sodium hydroxide solution)

• magnesium nitrate solution and silver nitrate solution

distinguished by potassium chloride solution(only silver nitrate solution produces a white precipitate, AgCl)

• solid ammonium chloride and solid potassium chloride

distinguished by warming with sodium hydroxide solution / calciumhydroxide solution(only ammonium chloride gives an alkaline gas, NH3)

• dilute sulphuric acid and dilute nitric acid

distinguished by heating with copper(only dilute nitric acid gives gas bubbles)

• dilute sulphuric acid and dilute nitric acid

distinguished by barium chloride solution(only dilute sulphuric acid gives a white precipitate, BaSO4)

20


• dilute hydrochloric acid and dilute nitric acid

distinguished by heating with copper(only dilute nitric acid gives gas bubbles)

• dilute hydrochloric acid and dilute nitric acid

distinguished by silver nitrate solution(only dilute hydrochloric acid gives a white precipitate, AgCl)

• dilute sodium hydroxide solution and dilute aqueous ammonia

distinguished by a solution containing aluminium ions / lead(II) ions(only dilute sodium hydroxide solution gives a white precipitate which dissolves in excess alkali)

• solution of cane sugar and solution of sodium chloride

distinguished by electrical conductivity(only solution of sodium chloride can conduct electricity)

page 27N18

Choosing a suitable drying agent

Drying agent Applicable for Not applicable for

Anhydrous calcium chloride Saturated hydrocarbons, haloalkanes, ethers, most esters

Ammonia, amines, alcohols, aldehydes, some esters and ketones

Calcium oxide Ammonia, amines, alcohols Acids, aldehydes, ketones

Anhydrous calcium sulphate Universal applications —

Anhydrous magnesium sulphate Almost all compounds, including acids, acid derivatives, aldehydes, ketones

Acid sensitive compounds

Concentrated sulphuric acid For neutral and acidic gases All organic compounds, hydrogen sulphide, hydrogen iodide

Silica gel Gas drying, wide use for organic liquids

Hydrogen fluoride

Anhydrous sodium sulphate Haloalkanes, fatty acids, esters, aldehydes, ketones

—

References:

https://www.erowid.org/archive/rhodium/chemistry/equipment/dryingagent.html

http://www.emdmillipore.com/US/en/analytics-sample-prep/analytical-sample-preparation/drying-agents/dFKb.qB..QgAAAE_KPZ3.Lxi,nav

both accessed July 20�4.

2�

Teaching Notes


page 45N1

pH values of some common substances

The following table lists the approximate pH values of some common substances.

Approximate pH values of some common solutions

Solution pH Solution pH

Hydrochloric acid (4%) 0 Saliva 6.2–7.4

Gastric juice �.6–�.8 Pure water 7.0

Lemon juice 2.� Blood 7.4

Vinegar (4%) 2.5 Fresh egg white 7.6–8.0

Rainwater (thunderstorm) 3.5–4.2 Bile 7.8–8.6

Milk 6.3–6.6 Milk of magnesia �0.5

Urine 5.5–7.0 Washing soda �2.0

Normal rainwater 5.6 Sodium hydroxide (4%) �3.0

page 48N3

Weak acids arranged in decreasing order of acid strength

Name Formula Equation

Ethanedioic acid H2C2O4 H2C2O4(aq) H+(aq) + HC2O4–(aq)

Hydrogensulphate ion HSO4– HSO4

–(aq) H+(aq) + SO42–(aq)

Phosphoric acid H3PO4 H3PO4(aq) H+(aq) + H2PO4–(aq)

Hydrofluoric acid HF HF(aq) H+(aq) + F–(aq)

Ethanoic acid CH3COOH CH3COOH(aq) H+(aq) + CH3COO–(aq)

Carbonic acid H2CO3 H2CO3(aq) H+(aq) + HCO3–(aq)

Hydrogen sulphide H2S H2S(aq) H+(aq) + HS–(aq)

Dihydrogenphosphate ion H2PO4– H2PO4

–(aq) H+(aq) + HPO42–(aq)

Hypochlorous acid HOCl HOCl(aq) H+(aq) + OCl–(aq)

Hydrocyanic acid HCN HCN(aq) H+(aq) + CN–(aq)

acid strength

decreasing

22


page 50N4

Comparing strong acids with weak acids

Examination questions often ask students to compare strong acids with weak acids.

Examples:

• Comparing 20 cm3 of � mol dm–3 CH3COOH(aq) with 20 cm3 of � mol dm–3 HCl(aq)

– they contain different number of hydrogen ions;

– they have different pH values;

– they have different electrical conductivity;

– they react with magnesium at different rates;

– they give the same amount of hydrogen gas with equal mass of magnesium;

– both react with NH3(aq), each giving a salt;

– they require the same number of moles of NaOH for complete neutralization;

– they give different colour change with the same quantity of universal indicator solution;

– when completely neutralized by � mol dm–3 NaOH(aq), HCl(aq) gives a larger temperature rise than CH3COOH(aq)◀.

• Comparing 20 cm3 of � mol dm–3 CH3COOH(aq) with �0 cm3 of � mol dm–3 H2SO4(aq)

– they have different pH values;

– they have different electrical conductivity;

– they react with magnesium at different rates;

– they give the same amount of hydrogen gas with equal mass of magnesium;

– they require the same number of moles of NaOH for complete neutralization.

• Comparing 45 cm3 of �.2 mol dm–3 HCl(aq) with 60 cm3 of 0.9 mol dm–3 CH3COOH(aq)

– they require the same volume of � mol dm–3 NaOH(aq) for complete neutralization.

• Comparing �00 cm3 of � mol dm–3 HCl(aq) with �00 cm3 of � mol dm–3 H2SO4(aq)

– they contain different concentration of H+(aq) ions;

– they react with magnesium at different rates.

◀ We will further discuss the heat released during a neutralization reaction in Topic 9 Chemical Reactions and Energy.

23

Teaching Notes


page 65N2

Temperature rise upon mixing an acid and an alkali

Examination questions often ask students to compare the temperature rises upon mixing acids and alkalis.

ExampleTemperature

rise

Number of moles of

water formed

Heat released◀ Explanation

Mixture �:20 cm3 of � mol dm–3

HCl(aq) mixed with 20 cm3 of � mol dm–3 NaOH(aq)andMixture 2:40 cm3 of � mol dm–3 HCl(aq) mixed with 40 cm3 of � mol dm–3 NaOH(aq)

∆T� = ∆T2

0.02 mole

0.04 mole

�.�4 kJ

2.28 kJ

Mixture � (total volume 40 cm3) releases �.�4 kJ of heat while mixture 2 (total volume 80 cm3) releases 2.28 kJ of heat. Hence the two mixtures give the same temperature rise.

Mixture �:20 cm3 of � mol dm–3

HCl(aq) mixed with 20 cm3 of � mol dm–3 NaOH(aq)andMixture 2:20 cm3 of 2 mol dm–3

HCl(aq) mixed with 20 cm3 of 2 mol dm–3 NaOH(aq)

2∆T� = ∆T2

0.02 mole

0.04 mole

�.�4 kJ

2.28 kJ

The total volumes of the two mixtures are the same. Hence the temperature rise of mixture 2 is twice that of mixture �.

Mixture �:20 cm3 of � mol dm–3 CH3COOH(aq) mixed with 20 cm3 of � mol dm–3 NaOH(aq)andMixture 2:20 cm3 of � mol dm–3

HCl(aq) mixed with 20 cm3 of � mol dm–3 NaOH(aq)

∆T� < ∆T2

0.02 mole

0.02 mole

< �.�4 kJ

�.�4 kJ

For neutralization in which either the acid or alkali or both are weak, the heat released will be less than 57 kJ for � mole of water formed.This is because some energy is consumed when the weak acid and weak alkali dissociate to give hydrogen ions and hydroxide ions before neutralization.CH3COOH(aq) is a weak acid. Hence the temperature rise of the first mixture is less than that of the second mixture.

◀ For the reaction between a strong acid and a strong alkali, the heat released is 57 kJ for � mole of water formed.

The heat change in neutralization will be further discussed in Topic 9 Chemical Reactions and Energy.

24


page 70N3

Hydrolysis of salts

Besides dissolving, some salts may at the same time react with water to form new products. Such reactions are called hydrolysis.

Solutions of salts may be acidic, neutral or alkaline, depending on the strengths of acids and bases from which the salts are formed.

Salt of a strong acid and a strong base

Salts such as sodium chloride (NaCl) and potassium nitrate (KNO3) belong to this class. When sodium chloride is dissolved in water, it dissociates into sodium ions and chloride ions. The ions have no reaction with water and no hydrolysis takes place.

Salt of a strong acid and a weak base

Salts such as ammonium chloride (NH4Cl) and copper(II) sulphate (CuSO4) belong to this class. When ammonium chloride is dissolved in water, it dissociates into ammonium ions and chloride ions. Chloride ions have no reaction with water. The ammonium ions undergo hydrolysis to form ammonia molecules and hydroxonium ions.

NH4+(aq) + H2O(l) NH3(aq) + H3O

+(aq)

The solution will be acidic due to the formation of hydroxonium ions (H3O+).

Salt of a weak acid and a strong base

Salts such as sodium carbonate (Na2CO3) and sodium ethanoate (CH3COONa) belong to this class. When sodium ethanoate is dissolved in water, it dissociates into sodium ions and ethanoate ions. The ethanoate ions undergo hydrolysis to form ethanoic acid molecules and hydroxide ions.

CH3COO–(aq) + H2O(l) CH3COOH(aq) + OH–(aq)

The solution will be alkaline due to the formation of hydroxide ions.

The following table shows a summary of the hydrolysis of different types of salts.

Summary of hydrolysis of different types of salts

Strength of parent acid

Strength of parent base

ExamplesUndergo

hydrolysis?pH of salt solution

Strong Strong NaCl, KNO3 No 7 (neutral)

Strong Weak NH4Cl, CuSO4 Yes <7 (acidic)

Weak Strong Na2CO3, CH3COONa Yes >7 (alkaline)

25

Teaching Notes

page 76N5

Pairs of solutions (or ions) producing a precipitate when mixed

Examination questions often give pairs of solutions (or ions) and ask students to identify pairs which would produce a precipitate when mixed.

Examples:

NH3(aq) and Pb(NO3)2(aq)

Pb(NO3)2(aq)◀ and K2CO3(aq)

Pb(NO3)2(aq)◀ and HCl(aq)

Pb(NO3)2(aq)◀ and Na2SO4(aq) produce a white precipitate when mixed

K2CO3(aq) and CaCl2(aq)

(NH4)2CO3(aq) and CaCl2(aq)

Ca2+(aq) and SO42–(aq)

Ni2+(aq) and CO32–(aq) produce a green precipitate (nickel(II) carbonate) when mixed

Na2S2O3(aq) and H2SO4(aq) produce a yellow precipitate (sulphur) when mixed

Ba(NO3)2(aq) and ZnCl2(aq)

NH4Cl(aq) and K2SO4(aq)

CuSO4(aq) and MgCl2(aq) do NOT produce a precipitate when mixed

KI(aq) and NH3(aq)

Cu2+(aq) and NO3–(aq)

NH4+(aq) and OH–(aq)

page 78N8

Acids in the soil

The pH of soil can vary from about 4 to 8. However, most soils have a pH between 6.5 and 7.5. In chalk or limestone areas, the soil is usually alkaline. On the other hand, it is generally acidic in moorland, sandstone and forest areas. Peatbogs and clay soils are normally acidic also.

For general gardening and farming purposes, the best results are obtained from a neutral or slightly acidic soil of pH 6.5 to 7.0. Only a few plants can grow well in soils which are acidic.

Causes of soil acidity

Soil acidity may be caused by a combination of the following factors:

� Acid rain

This factor is more important in urban areas and industrial countries.

◀ Examination questions often ask about lead compounds.

26


2 Use of acidic fertilizers

Fertilizers such as ammonium sulphate and ammonium nitrate hydrolyze to produce H3O+(aq), causing

acidity in soil:

NH4+(aq) + H2O(l) NH3(aq) + H3O

+(aq)

3 Selective leaching

Leaching is the washing away of minerals from soil by rain or underground water. Salts of alkali metals and alkaline earth metals are usually alkaline and responsible for soil alkalinity. They are more soluble and thus more readily leached away than salts of other metals. The soil may become acidic due to hydrolysis of metal ions such as Al3+ and Fe3+ remaining in the soil:

Al3+(aq) + H2O(l) [Al(OH)]2+(aq) + H+(aq)

Fe3+(aq) + H2O(l) [Fe(OH)]2+(aq) + H+(aq)

4 Biological activities in soil

Aerobic oxidation of organic matter in soil uses up oxygen and produces carbon dioxide. Normal air contains about 2�% oxygen and 0.03% carbon dioxide. In soil the percentage of oxygen may drop to as low as �5% and that of carbon dioxide may rise above 5%. Carbon dioxide increases the acidity of soil by reacting with water:

CO2(g) + H2O(l) H+(aq) + HCO3–(aq)

The bacterial decomposition of organic matter also produces acidic substances other than carbon dioxide.


page 100N5

Experimental aspects of titration

Examination questions often ask about experimental aspects of titration, such as:

• liquid that should be used to rinse each piece of equipment before titration;

• procedure for preparing a standard solution from a solid acid / alkali;

• procedure for diluting a concentrated acid / alkali of known concentration;

• procedure for preparing a burette for titration;

• the filter funnel should NOT remain on top of the burette after using it to fill the burette with a solution. It is because the solution clinging onto the stem of the funnel may fall into the burette and affect the titre readings.

27

Teaching Notes

page 111N10

End point detection by electrical conductivity measurements

Reactions which show a significant change in electrical conductivity as they proceed can be studied by monitoring these changes, using a conductivity cell with dipping electrodes.

Conductivity measurement serves as another method for detecting the end point in acid-alkali titrations.

Consider the addition of �.0 mol dm–3 NaOH(aq) into a beaker containing 50.0 cm3 of 0.�0 mol dm–3 HCl(aq). �.0 mol dm–3 NaOH(aq) is added, � cm3 at a time. The resistance of the solution mixture is measured after each addition. The reason for having the acid much more dilute than the alkali is to minimize dilution effects, which would tend to mask any conductivity changes. The reciprocal of the measured resistance is proportional to the conductivity. Thus, it is not essential to calculate the conductivity values at each stage.

The changes in conductivity for the titration are shown in the following graph.

• As alkali is added, the highly mobile hydrogen ions are replaced by the slower sodium ions, since virtually unionized water is produced. The conductivity falls in a linear manner along the line XY.

28


• After the end point, the conductivity rises sharply along the line YZ as additional sodium ions and hydroxide ions are added. The sharp rise in conductivity in this region is primarily due to the presence of the highly mobile hydroxide ions.

• The slope of the line YZ is not as great as that of the line XY as hydroxide ions are not as mobile as hydrogen ions.

page 112N12

Which reactant is in excess?

Examination questions may not tell students which reactant is in excess in a chemical reaction. Students need to identify that.

• Consider two reactions involving magnesium ribbons:

Reaction Reaction mixture

� �.5 g of Mg + �00 cm3 of � mol dm–3 HCl(aq)

2 �.5 g of Mg + �00 cm3 of � mol dm–3 H2SO4(aq)

For Reaction �:

Number of moles of Mg present = �.5 g

24.3 g mol–�

= 0.06�7 mol

Number of moles of HCl present = � mol dm–3 x �00

� 000 dm3

= 0.� mol

� mole of Mg requires 2 moles of HCl for complete reaction. Hence 0.� mole of HCl would react with 0.05 mole of Mg. HCl is the limiting reactant. Mg does not react completely in Reaction �. Mg is in excess.

For Reaction 2:

Number of moles of H2SO4 present = � mol dm–3 x �00

� 000 dm3

= 0.� mol

� mole of Mg requires � mole of H2SO4 for complete reaction. Hence 0.06�7 mole of Mg would react with 0.06�7 mole of H2SO4. Mg is the limiting reactant. H2SO4 does not react completely in Reaction 2. H2SO4 is in excess.

The initial rate of Reaction 2 is higher than that of Reaction �. This is because H2SO4(aq) is a dibasic acid while HCl(aq) is a monobasic acid. � mol dm–3 H2SO4(aq) contains a higher concentration of hydrogen ions than � mol dm–3 HCl(aq).

• Besides calculations, we can deduce that the acid is in excess if magnesium ribbons disappear after reaction.

29

Teaching Notes

page 118N13

Determining the percentage by mass of calcium carbonate in a sample

Examination questions often ask students to determine the percentage by mass of calcium carbonate in a sample.

• Alternative method �:

– Let the sample react with excess dilute hydrochloric acid.

– Measure the volume of carbon dioxide given off with a syringe. Calculate the number of moles of CO2 evolved. (The relationship between the volume of a gas and the number of moles will be discussed in Topic �0 Rate of Reaction.)

• Alternative method 2:

– Convert insoluble CaCO3 to insoluble CaSO4 via soluble Ca(NO3)2.

excess HNO3(aq) excess Na2SO4(aq) CaCO3(s) Ca(NO3)2(aq) CaSO4(s)

– Calculate the percentage by mass of CaCO3 in the sample from the mass of CaSO4 obtained.

30


page 1

� Making use of pH paper (or meter) / litmus paper (or solution) / universal indicator solution

2 acid + alkali salt + water

3 One mole


Practice

P14.1 page 10

� a) The oxide dissolves in the acid.

b) ZnO(s) + 2HNO3(aq) Zn(NO3)2(aq) + H2O(l)

c) ZnO(s) + 2H+(aq) Zn2+(aq) + H2O(l)

2 a) Effervescence occurs.

b) (NH4)2CO3(aq) + 2HCl(aq) 2NH4Cl(aq) + H2O(l) + CO2(g)

c) CO32–(aq) + 2H+(aq) H2O(l) + CO2(g)

P14.2 page 13

When the tablet is added to water, the citric acid in the tablet dissociates in water to give hydrogen ions.

The sodium hydrogencarbonate in the tablet dissolves in water, releasing hydrogencarbonate ions.

The hydrogen ions react with the hydrogencarbonate ions, forming carbon dioxide gas.

H+(aq) + HCO3–(aq) H2O(l) + CO2(g)

P14.3 page 23

a) Add dilute hydrochloric acid to each substance separately.

Only calcium carbonate gives a gas that turns limewater milky (the gas is carbon dioxide).

b) Add dilute sodium hydroxide solution to each substance separately and warm.

Only ammonium chloride gives a gas that turns moist red litmus paper blue (the gas is ammonia).

Suggested Answers

3�

Suggested Answers

Problem Solving page 23

Carry out a flame test on all the substances.

Only sodium chloride gives a golden yellow flame.

Dissolve the remaining three substances separately in water. Divide each solution into two portions.

Add silver nitrate solution to one portion of each solution.

Only the zinc chloride solution gives a white precipitate.

Add dilute sodium hydroxide solution to the remaining two solutions.

The lead(II) nitrate solution gives a white precipitate which is soluble in excess alkali.

The magnesium nitrate solution gives a white precipitate which does not dissolve in excess alkali.

(Other possible tests are also acceptable.)

page 25Find & Share

Corrosive nature of concentrated acids and alkalis

Concentrated hydrochloric acid

Hazard warning symbol

Risk descriptions• Causes burns• Irritating to respiratory systemSafety precautions• In case of contact with eyes, rinse immediately with plenty of water and seek

medical advice.• In case of accident and if you feel unwell, seek medical advice immediately.

Concentrated nitric acid

Hazard warning symbols

Risk descriptions• Causes severe burns• Contact with combustible material may cause fireSafety precautions• Do not inhale fumes / vapour / spray.• In case of contact with eyes, rinse immediately with plenty of water and seek



32


Concentrated sulphuric acid


Risk descriptions• Reacts violently with water• Causes severe burns• Irritating to respiratory systemSafety precautions• Never add water to concentrated sulphuric acid.• In case of contact with eyes, rinse immediately with plenty of water and seek


Concentrated sodium hydroxide

solution


Risk description• Causes severe burnsSafety precautions• Wear protective gloves and eye / face protection.• In case of contact with eyes, rinse immediately with plenty of water and seek


Concentrated potassium hydroxide

solution


Risk descriptions• Causes severe burnsSafety precautions• Wear protective gloves and eye / face protection.• In case of contact with eyes, rinse immediately with plenty of water and seek


33

Suggested Answers

pages 32–39Unit Exercise

� a) sulphuric acid

b) nitric acid

c) dibasic acid

d) tribasic acid

e) sour

f) red

g) hydrogen

h) carbon dioxide

i) salt

j) water

2 a) sodium hydroxide solution

b) calcium hydroxide solution

c) aqueous ammonia

d) slippery

e) blue

f) ammonia

g) salt

h) water

i) electricity

3

IndicatorColour in

dilute hydrochloric acid dilute sodium hydroxide solution

Litmus solution red blue

Methyl orange red yellow

Phenolphthalein colourless red

34


4 Adding

NaOH(aq) to solution containing

Colour of precipitate

formed

Precipitate dissolves in excess

NaOH(aq)?(Yes or No)

Ionic equation(s)

Ca2+(aq) white No Ca2+(aq) + 2OH–(aq) Ca(OH)2(s)

Mg2+(aq) white No Mg2+(aq) + 2OH–(aq) Mg(OH)2(s)

Al3+(aq) white YesAl3+(aq) + 3OH–(aq) Al(OH)3(s)Al(OH)3(s) + OH–(aq) [Al(OH)4]

–(aq)

Pb2+(aq) white YesPb2+(aq) + 2OH–(aq) Pb(OH)2(s)Pb(OH)2(s) + 2OH–(aq) [Pb(OH)4]

2–(aq)

Zn2+(aq) white YesZn2+(aq) + 2OH–(aq) Zn(OH)2(s)Zn(OH)2(s) + 2OH–(aq) [Zn(OH)4]

2–(aq)

Fe2+(aq) green No Fe2+(aq) + 2OH–(aq) Fe(OH)2(s)

Fe3+(aq) reddish brown No Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)

Cu2+(aq) pale blue No Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)

5 Adding

NH3(aq) to solution

containing

Colour of precipitate

formed

Precipitate dissolves in

excess NH3(aq)?(Yes or No)

Ionic equation(s)

Mg2+(aq) white No Mg2+(aq) + 2OH–(aq) Mg(OH)2(s)

Al3+(aq) white No Al3+(aq) + 3OH–(aq) Al(OH)3(s)

Pb2+(aq) white No Pb2+(aq) + 2OH–(aq) Pb(OH)2(s)

Zn2+(aq) white YesZn2+(aq) + 2OH–(aq) Zn(OH)2(s)Zn(OH)2(s) + 4NH3(aq) [Zn(NH3)4]

2+(aq) + 2OH–(aq)

Fe2+(aq) green No Fe2+(aq) + 2OH–(aq) Fe(OH)2(s)

Fe3+(aq) reddish brown No Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)

Cu2+(aq) pale blue YesCu2+(aq) + 2OH–(aq) Cu(OH)2(s)Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]

2+(aq) + 2OH–(aq)

35

Suggested Answers

6

excess CO2(g)

solution F

solution Dsolution C

H2O(l)

solid A

liquid E

gas B

HCl(aq)

CaCO3

heat

+

+CaO CO2

Ca(OH)2

Ca(HCO3)2

CaCl2 H2O

7 D

8 C CuO(s) is insoluble in water and there is no separate Cu2+ ion or O2– ion in water.

9 A

�0 C H3PO4 is a tribasic acid.

�� A

�2 B FeSO4 is soluble in water. Addition of dilute aqueous ammonia to FeSO4(aq) gives a precipitate, iron(II) hydroxide.

Fe2+(aq) + 2OH–(aq) Fe(OH)2(s)

The precipitate does not dissolve in excess alkali.

�3 D

�4 B Iron(III) hydroxide is insoluble in aqueous ammonia.

Copper(II) hydroxide reacts with excess aqueous ammonia to form a soluble complex salt.

Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 2OH–(aq)

�5 D

�6 D

�7 A

36


�8 (�) a) Copper does not react with dilute hydrochloric acid.

b) Add magnesium to dilute hydrochloric acid.

(2) a) Insoluble calcium sulphate forms when calcium carbonate reacts with dilute sulphuric acid. The calcium sulphate covers the surface of calcium carbonate and prevents further reaction.

b) Add calcium carbonate to dilute hydrochloric acid / dilute nitric acid.

(3) a) Sodium hydroxide is very corrosive.

b) Use magnesium hydroxide / aluminium hydroxide as the active ingredient.

�9 a) Gas bubbles are given off. / The zinc granules dissolve in the acid.

Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)

or Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

b) Effervescence occurs. / Solid copper(II) carbonate dissolves in the acid. / A blue solution forms.

CuCO3(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l) + CO2(g)

or CuCO3(s) + 2H+(aq) Cu2+(aq) + H2O(l) + CO2(g)

c) Effervescence occurs. / Solid sodium hydrogencarbonate dissolves in the acid.

NaHCO3(s) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g)

or NaHCO3(s) + H+(aq) Na+(aq) + H2O(l) + CO2(g)

20 Answers for the HKASLE question are not provided.

2� a) An acid is a hydrogen-containing substance that gives hydrogen ions (H+(aq)) as the only type of positive ions when dissolved in water.

b) i) copper(II) oxide and hydrochloric acid

CuO(s) + 2HCl(aq) CuCl2(aq) + H2O(l)

or CuO(s) + 2H+(aq) Cu2+(aq) + H2O(l)

sodium carbonate and hydrochloric acid

Na2CO3(s) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)

or Na2CO3(s) + 2H+(aq) 2Na+(aq) + H2O(l) + CO2(g)

ii) Effervescence

The solid disappears / dissolves.

A blue / green solution is formed.

37

Suggested Answers

22 a) A blue precipitate forms;

the precipitate dissolves in excess dilute aqueous ammonia to give a deep blue solution.

Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)

Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 2OH–(aq)

b) A white precipitate forms.

Al3+(aq) + 3OH–(aq) Al(OH)3(s)

c) A white precipitate forms;

the precipitate dissolves in excess dilute sodium hydroxide solution to give a colourless solution.

Zn2+(aq) + 2OH–(aq) Zn(OH)2(s)

Zn(OH)2(s) + 2OH–(aq) [Zn(OH4)]2–(aq)

d) Ammonia gas is given off.

2NH4Cl(aq) + Ca(OH)2(aq) 2NH3(g) + CaCl2(aq) + 2H2O(l)

23 a) dilute sodium hydroxide solution / calcium hydroxide

b) ammonia

c) ammonium

d) silver chloride

e) chloride

24 a) A brick-red flame is observed.

b) Effervesence

c) carbon dioxide

d) carbonate

e) A white precipitate is formed.

f) calcium hydroxide

25 a) Warm with dilute sodium hydroxide solution.

A gas that turns moist red litmus paper blue (ammonia) is given off.

b) Add dilute hydrochloric acid.

A gas that turns limewater milky (carbon dioxide) is given off.

38


26 a) Any one of the following:

• Add dilute aqueous ammonia to each solution.

Only the aluminium nitrate solution gives a white precipitate.

• Add dilute sodium hydroxide solution to each solution.

The aluminium nitrate solution gives a white precipitate which dissolves in excess alkali.

The calcium nitrate solution gives a white precipitate which does not dissolve in excess alkali.

b) Add dilute sodium hydroxide solution / dilute aqueous ammonia to each solution.

The iron(II) sulphate solution gives a green precipitate.

The iron(III) sulphate solution gives a reddish-brown precipitate.

27 a) Ag+(aq) + Cl–(aq) AgCl(s)

b) Filtration

c) i) Copper(II) hydroxide

ii) Pale blue

d) Warm E. A gas that turns moist red litmus paper blue is evolved. The gas is ammonia. This shows the presence of ammonium ions in E.

e) Blue; the solution contains blue Cu2+(aq) ions and the other ions are colourless.

28 a) Add dilute hydrochloric acid to the salt.

A gas that can turn limewater milky is evolved.

b) A white precipitate

c) The presence of potassium cannot be shown. As in a flame test, the lilac flame of potassium will be masked by the brillant golden yellow flame of sodium.

39

Suggested Answers


Practice

P15.1 page 42

� Molar mass of NH4NO3 = (2 x �4.0 + 4 x �.0 + 3 x �6.0) g mol–� = 80.0 g mol–�

Number of moles of NH4NO3 = mass

molar mass

= 26.0 g

80.0 g mol–�

= 0.325 mol

Molarity of ammonium nitrate solution = number of moles of NH4NO3

volume of solution

= 0.325 mol

( 500.0� 000 ) dm3

= 0.650 mol dm–3 (M)

2 Molar mass of K2CO3 = (2 x 39.� + �2.0 + 3 x �6.0) g mol–� =�38.2 g mol−�

Molarity of potassium carbonate solution = number of moles of K2CO3

volume of solution

0.200 mol dm–3 = number of moles of K2CO3

2.50 dm3

Number of moles of K2CO3 = 0.200 mol dm–3 x 2.50 dm3

= 0.500 mol

Mass of K2CO3 = number of moles of K2CO3 x molar mass of K2CO3

= 0.500 mol x �38.2 g mol–�

= 69.� g

P15.2 page 44

� pH of milk = –log�0(2.5� x �0–7) = 6.60

2 pH of acid rain = –log�0[H+]

= 3.44

i.e. log�0[H+] = –3.44

[H+] = �0–3.44

= 3.63 x �0–4 mol dm–3

40


P15.3 page 54

� mol dm–3 sulphuric acid < � mol dm–3 hydrochloric acid < 0.� mol dm–3 nitric acid < 0.� mol dm–3 ethanoic acid

Both sulphuric acid and hydrochloric acid are strong acids and dissociate completely.

H2SO4(aq) 2H+(aq) + SO42–(aq)

HCl(aq) H+(aq) + Cl–(aq)

Every HCl molecule dissociates to give one hydrogen ion while every H2SO4 molecule dissociates to give two hydrogen ions.

Thus, the concentration of hydrogen ions in � mol dm–3 sulphuric acid is higher than that in � mol dm–3 hydrochloric acid. Hence the pH of � mol dm–3 sulphuric acid is lower than that of � mol dm–3 hydrochloric acid.

The concentration of hydrogen ions in � mol dm–3 hydrochloric acid is higher than that in 0.� mol dm–3 nitric acid. Hence the pH of � mol dm–3 hydrochloric acid is less than that of 0.� mol dm–3 nitric acid.

Ethanoic acid is a weak acid. It only partially dissociates in water.

CH3COOH(aq) H+(aq) + CH3COO–(aq)

The concentration of hydrogen ions in 0.� mol dm–3 ethanoic acid is lower than that in 0.� mol dm–3 nitric acid, a strong acid. Hence the pH of 0.� mol dm–3 ethanoic acid is higher than that of 0.� mol dm–3 nitric acid.

Discussion page 51

No.

‘A is a stronger acid than B’ only means that the degree of dissociation of A is larger than that of B.

However, the pH of an aqueous solution of an acid depends on both its degree of dissociation and concentration.

Thus, an aqueous solution of the stronger acid A may have a higher pH than that of the weaker acid B if the concentration of acid B is higher than that of acid A by an adequate amount.


� a) universal

b) pH

c) acid

d) strong

e) hydrogen

f) weak

g) ethanoic

4�

Suggested Answers

h) hydroxide

i) strong

j) hydroxide

k) weak

l) ammonia

2 a) number of moles of solute (mol)

b) acidity; alkalinity

c) acidic

d) alkaline

e) neutral

f) strong

g) weak

3 a) C

b) B

c) A

d) D

e) E

4 C

5 A Molarity of the acid solution = number of moles of (COOH)2•2H2O

volume of solution

0.�20 mol dm–3 = number of moles of (COOH)2•2H2O

( 500.0� 000 ) dm3

Number of moles of (COOH)2•2H2O = 0.�20 mol dm–3 x 500.0� 000

dm3

= 0.0600 mol

Mass of (COOH)2•2H2O required = 0.0600 mol x �26.0 g mol–�

= 7.56 g

\ 7.56 g of ethanedioic acid crystals are required.

6 C

7 A Car battery acid is sulphuric acid.

42


8 C

9 C

�0 C

�� D (�) Citric acid is a weak acid. When citric acid crystals dissolve in water, only a few citric acid molecules dissociate to give hydrogen ions.

(3) Citric acid exists as a solid at room temperature. Its aqueous solution can conduct electricity.

�2 D

�3 a) Molar mass of KOH = (39.� + �6.0 + �.0) g mol–� = 56.� g mol–�

Number of moles of KOH = mass

molar mass

= �0.� g

56.� g mol–�

= 0.�80 mol

Molarity of KOH solution = number of moles of KOH

volume of solution

= 0.�80 mol

( 250.0� 000 ) dm3

= 0.720 mol dm–3 (M)

b) Molar mass of (NH4)2SO4 = [2 x (�4.0 + 4 x �.0) + 32.� + 4 x �6.0] g mol–� = �32.� g mol–�

Number of moles of (NH4)2SO4 = mass

molar mass

= 7.93 g

�32.� g mol–�

= 0.0600 mol

Molarity of (NH4)2SO4 solution = number of moles of (NH4)2SO4

volume of solution

= 0.0600 mol

( 750.0� 000 ) dm3

= 0.0800 mol dm–3 (M)

43

Suggested Answers

c) Molar mass of FeCl2 = (55.8 + 2 x 35.5) g mol–� = �26.8 g mol–�

Number of moles of FeCl2 = mass

molar mass

= 57.� g

�26.8 g mol–�

= 0.450 mol

Molarity of FeCl2 solution = number of moles of FeCl2

volume of solution

= 0.450 mol2.50 dm3

= 0.�80 mol dm3 (M)

�4 Molar mass of CuSO4 = (63.5 + 32.� + 4 x �6.0) g mol–� = �59.6 g mol–�

Molarity of copper(II) sulphate solution = number of moles of CuSO4

volume of solution

�.20 mol dm–3 = number of moles of CuSO4

( 200.0� 000 ) dm3

Number of moles of CuSO4 = �.20 mol dm–3 x ( 200.0� 000 ) dm3

= 0.240 mol

Mass of CuSO4 required = number of moles of CuSO4 x molar mass of CuSO4

= 0.240 mol x �59.6 g mol–�

= 38.3 g

�5 � mole of (NH4)2CO3 contains 2 moles of NH4+ ions.

Number of moles of NH4+ ions in ammonium carbonate solution = 2 x 0.400 mol dm–3 x

20.0� 000

dm3

= 0.0�60 mol

� mole of NH4NO3 contains � mole of NH4+ ion.

Number of moles of NH4+ ions in ammonium nitrate solution = 0.500 mol dm–3 x

40.0� 000

dm3

= 0.0200 mol

Total number of moles of NH4+ ions = (0.0�60 + 0.0200) mol

= 0.0360 mol

Total volume of resulting solution = (20.0 + 40.0) cm3

= 60.0 cm3

Concentration of NH4+ ions in resulting solution =

0.0360 mol

( 60.0� 000 ) dm3

= 0.600 mol dm–3

44


�6 a) Hydrochloric acid dissociates completely according to the following equation:

HCl(aq) H+(aq) + Cl–(aq) �.00 x �0–4 mol dm–3 ? mol dm–3

According to the equation, � mole of HCl dissociates to give � mole of hydrogen ions.

i.e. concentration of hydrogen ions = �.00 x �0–4 mol dm–3

pH of acid = –log�0(�.00 x �0–4) = –(–4.00) = 4.00

b) Sulphuric acid dissociates completely according to the following equation:

H2SO4(aq) 2H+(aq) + SO42–(aq)

0.0480 mol dm–3 ? mol dm–3

According to the equation, � mole of H2SO4 dissociates to give 2 moles of hydrogen ions.

i.e. concentration of hydrogen ions = 2 x 0.0480 mol dm–3 = 0.0960 mol dm–3

pH of acid = –log�0(0.0960) = –(–�.02) = �.02

�7 a) pH of coffee = –log�0[H+] = 5.40

i.e. log�0[H+] = –5.40

[H+] = �0–5.40

= 3.98 x �0–6 mol dm–3

b) pH of saliva = –log�0[H+] = 6.70

i.e. log�0[H+] = –6.70

[H+] = �0–6.70

= 2.00 x �0–7 mol dm–3

�8 pH of effluent before treatment = 6.80

pH of effluent after treatment = 7.00

Concentration of hydrogen ions in effluent before treatment = �0–6.80

= �.58 x �0–7 mol dm–3

Concentration of hydrogen ions in effluent after treatment = �0–7.00 mol dm–3

Change in concentration of hydrogen ions = (�0–7 – �.58 x �0–7) mol dm–3

= –0.58 x �0–7 mol dm–3

\ Concentration of hydrogen ions in the effluent decreases by 0.58 x �0–7 mol dm–3 upon treatment.

45

Suggested Answers

�9 a) Bubbles stopped evolving.

b) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

c) The time required for the completion of Reaction II would be longer.

During the reaction between magnesium and the acids, magnesium would react with hydrogen ions in the acids.

Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)

Hydrochloric acid is a strong acid that completely dissociates in water.

Ethanoic acid is a weak acid that only partially dissociates in water.

Thus, the concentration of hydrogen ions in ethanoic acid is lower than that in hydrochloric acid. The reaction rate between magnesium and ethanoic acid is thus lower and the reaction takes a longer time to complete.

20 a) A weak acid is an acid that only partially dissociates in water.

b) Na2CO3(s) + 2HF(aq) 2NaF(aq) + H2O(l) + CO2(g)

c) Na2CO3(s) + 2H+(aq) 2Na+(aq) + H2O(l) + CO2(g)

d) In the reaction between Na2CO3 and an acid, Na2CO3 reacts with the hydrogen ions in the acid.

The rate of evolution of gas bubbles for hydrochloric acid is higher than that for hydrofluoric acid. / The solid disappears faster in hydrochloric acid than in hydrofluoric acid.

The concentration of the acids must be the same.

2� a) CH3COOH(aq) H+(aq) + CH3COO–(aq)

b) Dilute hydrochloric acid has a higher concentration of hydrogen ions than dilute ethanoic acid does.

c) Dilute hydrochloric acid has a higher concentration of mobile ions.

22 a) Hydroxide ion / OH–(aq)

b) Almost completely dissociate

c) Any one of the following:

• Measure the pH of each solution.

The sodium hydroxide solution has a higher pH.

• Measure the electrical conductivity of each solution.

The sodium hydroxide solution conducts better.

46



Practice

P16.1 page 65

� a) 2HCl(aq) + CaO(s) CaCl2(aq) + H2O(l)

2H+(aq) + CaO(s) Ca2+(aq) + H2O(l)

b) H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l)

H+(aq) + OH–(aq) H2O(l)

c) 2HNO3(aq) + Pb(OH)2(s) Pb(NO3)2(aq) + 2H2O(l)

2H+(aq) + Pb(OH)2(s) Pb2+(aq) + 2H2O(l)

2 a) ZnCO3(s) + H2SO4(aq) ZnSO4(aq) + H2O(l) + CO2(g)

b) No.

In neutralization, salt and water are the only products.

P16.2 page 69

a) NH4HSO4 ammonium hydrogensulphate

(NH4)2SO4 ammonium sulphate

b) LiCl lithium chloride

c) Mg(NO3)2 magnesium nitrate

P16.3 page 75

a) Methyl orange shows a colour change when enough dilute nitric acid is added to the sodium carbonate solution and their reaction is complete, which indicates the end point of titration.

b) Na2CO3(s) + 2HNO3(aq) 2NaNO3(aq) + H2O(l) + CO2(g)

c) Washing with distilled water can remove the water soluble impurities.

Using a small amount of cold distilled water helps reduce the loss of salt.

d) Absorb the water by filter paper. / Place in a desiccator.

47

Suggested Answers

P16.4 page 77

a) i) C

ii) Barium nitrate solution and excess dilute sulphuric acid / sodium sulphate solution / potassium sulphate solution

iii) Ba2+(aq) + SO42−(aq) BaSO4(s)

b) i) B

ii) Dilute aqueous ammonia and dilute hydrochloric acid

iii) NH3(aq) + HCl(aq) NH4Cl(aq)

c) i) A

ii) Dilute nitric acid and excess copper(II) oxide

iii) CuO(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l)

P16.5 page 80

a) The acidity of liquid waste is harmful to water life.

b) Ca(OH)2(s) + 2HCl(aq) CaCl2(aq) +2H2O(l)

or Ca(OH)2(s) + 2H+(aq) Ca2+(aq) +2H2O(l)

c) Calcium ion forms a number of insoluble salt. /

The liquid waste may contain sulphuric acid, which forms a layer of insoluble calcium sulphate on the surface of the slaked lime solid. Such layer hinders the reaction of slaked lime and thus decreases its efficiency in neutralization.

page 78Find & Share

Uses of neutralization

Besides the four areas suggested, students may search for other uses of neutralization, such as:

• Insect sting treatment

Stings by ants and bees are acidic. The uneasy feeling can be relieved by washing with weakly alkaline solution (e.g. sodium hydrogencarbonate solution).

To relieve a wasp sting, apply vinegar or lemon juice to neutralize it, as a wasp has an alkaline sting.

• Toothpaste

The liquid secreted in our mouths is called saliva. Bacteria present in our mouths can produce acids by the action on food. The acids dissolve the enamel on the teeth and can make holes. Toothpaste is alkaline. It contains alkali which helps neutralize acids in the mouth.

48


• Using scrubbers in power stations to reduce sulphur dioxide emission

Attaching scrubbers to smokestacks of power stations can remove up to 95% of sulphur dioxide from power station exhaust. In dry scrubbing (on the left of the following figure), powdered limestone is blown into the combustion chamber where it decomposes.

CaCO3(s) CaO(s) + CO2(g)

The calcium oxide is a base and it neutralizes the sulphur dioxide to form calcium sulphite.

CaO(s) + SO2(g) CaSO3(s)

In wet scrubbing (on the right of the above figure), calcium oxide reacts with water to produce calcium hydroxide solution. The calcium hydroxide solution neutralizes the sulphur dioxide gas.

CaO(s) + H2O(l) Ca(OH)2(aq)

Ca(OH)2(aq) + SO2(g) CaSO3(s) + H2O(l)

These products are washed away as a slurry — a mixture of solids and water.


� a) base

b) water

c) precipitation

d) metal

e) insoluble base

f) insoluble carbonate

g) alkali / soluble carbonate

49

Suggested Answers

2 a) acid alkali salt

sulphuric acid potassium hydroxide

sodium sulphate

hydrochloric acid ammonium hydroxide

potassium chloride

nitric acid sodium hydroxide

magnesium nitrate

phosphoric acid magnesium hydroxide

b) i) H+(aq)

ii) OH–(aq)

iii) H+(aq) + OH–(aq) H2O(l)

3 a) KOH(aq) + HNO3(aq) KNO3(aq) + H2O(l)

b) Ca(OH)2(s) + 2HCl(aq) CaCl2(aq) + 2H2O(l)

c) 2Fe(OH)3(s) + 3H2SO4(aq) Fe2(SO4)3(aq) + 6H2O(l)

d) 2NaOH(aq) + H2CO3(aq) Na2CO3(aq) + 2H2O(l)

e) CuO(s) + H2SO4(aq) CuSO4(aq) + H2O(l)

f) AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

4

CO2(g)

precipitate C

AgNO3(aq)

precipitate D

solution BBa(OH)2(aq)

H2O(l)

solid A

HCl(aq)

Ba(s)heat

airBaO(s)

BaCl2(aq)

AgCl(s)BaCO3(s)

5 B

6 B

7 A Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

ZnO(s) + 2HCl(aq) ZnCl2(aq) + H2O(l)

ZnCO3(s) + 2HCl(aq) ZnCl2(aq) + H2O(l) + CO2(g)

50


8 D Sodium salts are prepared by titration method.

9 D

�0 C

�� A

�2 D (�) Pb(NO3)2(aq) + 2HCl(aq) PbCl2(s) + 2HNO3(aq)

(2) Pb(NO3)2(aq) + MgSO4(aq) PbSO4(s) + Mg(NO3)2(aq)

(3) Pb(NO3)2(aq) + Na2CO3(aq) PbCO3(s) + 2NaNO3(aq)

�3 a) (�) H2; hydrogen

(3) H2O; CO2; water; carbon dioxide

b) Effervescence occurs.

c) Observation: white precipitate

Name of product: barium sulphate

�4 a) Magnesium hydroxide in the milk of magnesia neutralizes the excess hydrochloric acid in the stomach and so the pain can be relieved.

Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)

b) The sodium hydrogencarbonate and citric acid in the tablet react upon dissolution in water to give carbon dioxide gas.

c) The acid will react with the sodium hydrogencarbonate in the presence of moisture.

�5 a) Sodium carbonate solution Sodium sulphate solution

Barium chloride solution precipitate of barium carbonate precipitate of barium sulphate

Potassium chloride solution no precipitate no precipitate

Calcium chloride solution precipitate of calcium carbonate precipitate of calcium sulphate

b) Pb(NO3)2( aq ) + 2KBr( aq ) PbBr2( s ) + 2KNO3( aq )

c) i) To separate the solid and liquid in the reaction mixture, in order to obtain lead(II) bromide solid.

ii) To wash away any water soluble impurities that may adhere to the solid residue.

iii) Distilled water does not contain impurities.

5�

Suggested Answers

iv) Any one of the following:

• To evaporate the water.

• To dry the solid.

• To increase the rate of evaporation of water.

�6 a) i) Add magnesium oxide to dilute sulphuric acid until some of the oxide remain in the beaker.

Remove the excess magnesium oxide by filtration.

Heat the filtrate to evaporate about half of the water.

Set the concentrated solution aside to cool and crystallize.

Filter the crystals from the remaining solution.

Wash the crystals with a small amount of cold distilled water.

Dry using filter papers. / Put in an oven. / Put in a desiccator.

ii) Rinse with plenty of water. / Add a weak alkali (e.g. solid or aqueous sodium hydrogencarbonate).

b) Insoluble strontium sulphate forms on the strontium carbonate.

�7 a) The formation of a solid from the solutions of two soluble salts.

b) copper(II) carbonate

lead(II) sulphate

c) i)

ii) Molar mass of Ca(NO3)2 = [40.� + 2 x (�4.0 + 3 x �6.0)] g mol–�

= �64.� g mol–�

Number of moles of calcium ions in 8.20 g of Ca(NO3)2 = mass

molar mass

= 8.20 g

�64.� g mol–�

= 5.00 x �0–2 mol

Molar mass of CaSO4 = (40.� + 32.� + 4 x �6.0) g mol–�

= �36.2 g mol–�

Mass of CaSO4 produced = 5.00 x �0–2 mol x �36.2 g mol–�

= 6.8� g

iii) To make sure that all the calcium nitrate reacts.

iv) To wash off any soluble impurities that may adhere to the calcium sulphate.

52


�8 a) Add excess iron powder to dilute sulphuric acid and mix the mixture. Bubbles (hydrogen) evolve from the mixture.

Once bubbles stop evolving, which indicates that the reaction stops, filter the reaction mixture to remove excess iron powder.

Collect the filtrate with an evaporating dish and heat it with a steam bath to evaporate half of the water.


Filter the crystals from the remaining solution and wash them with a small amount of cold distilled water.

Dry them using filter papers. / Put in an oven. / Put in a desiccator.

Fe(s) + H2SO4(aq) FeSO4(aq) + H2(g)

b) Add excess copper(II) oxide solid to very dilute nitric acid and mix the mixture well. The colourless solution turns blue upon reaction.

Filter the reaction mixture to remove excess copper(II) oxide.

Collect the filtrate with an evaporating dish and heat it with a steam bath to evaporate half of the water.


Filter the crystals from the remaining solution and wash them with a small amount of cold distilled water.

Dry them using filter papers. / Put in an oven. / Put in a desiccator.

CuO(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l)

c) Mix silver nitrate solution and potassium chloride solution. A white precipitate, silver chloride, is then formed.

Filter the reaction mixture with filter paper. Wash the residue on the filter paper with distilled water.

Collect the residue, dry it using filter papers. / Put in an oven. / Put in a desiccator. Set it aside in the dark.

AgNO3(aq) + KCl(aq) AgCl(s) + KNO3(aq)

d) Add excess calcium carbonate solid to limited amount of dilute nitric acid and mix the mixture. Bubbles (carbon dioxide) evolve from the mixture.

Once bubbles stop evolving, which indicates that the reaction stops, filter the reaction mixture. Collect the filtrate, calcium nitrate solution.

Add sodium sulphate solution to the filtrate and a white precipitate, calcium sulphate, is formed.

Filter the reaction mixture with filter paper. Wash the residue on the filter paper with distilled water.

Collect the residue and dry it using filter papers. / Put in an oven. / Put in a desiccator. HNO3(aq) Na2SO4(aq) CaCO3(s) Ca(NO3)2(aq) CaSO4(s)

53

Suggested Answers

�9 a) 2NH3(aq) + H2SO4(aq) (NH4)2SO4(aq)

b) i) Burette

ii) Yellow

iii) Mix 25 cm3 of dilute aqueous ammonia and 22.4 cm3 of dilute sulphuric acid.

Heat the ammonium sulphate solution gently to obtain a concentrated solution.


Filter the crystals from the remaining solution.

Wash the crystals with a small amount of cold distilled water.

Dry the crystals using filter papers / put in an oven / put in a desiccator.

20 Answers for the HKDSE question are not provided.


Practice

P17.1 page 93

� Concentration of magnesium sulphate solution = mass of MgSO4•7H2O

volume of solution

= 5.80 g

( 250.0� 000 ) dm3

= 23.2 g dm−3

2 a) Consider � 000 cm3 (i.e. �.00 dm3) of the sample.

Mass of � 000 cm3 of the sample = �.25 g cm–3 x � 000 cm3

= � 250 g

Mass of H3PO4 acid in � 000 cm3 of the sample = mass of � 000 cm3 of sample x percentage by mass of H3PO4 in the sample = � 250 g x 8�.8% = � 022.5 g

Concentration of H3PO4 in g dm−3 = mass of H3PO4

�.00 dm3

= � 022.5 g�.00 dm3

= � 022.5 g dm–3

54


b) Molar mass of H3PO4 = (3 x �.0 + 3�.0 + 4 x �6.0) g mol–�

= 98.0 g mol–�

Number of moles of H3PO4 in �.00 dm3 of the sample = mass

molar mass

= � 022.5 g

98.0 g mol–�

= �0.4 mol

Molarity of phosphoric acid = number of moles of H3PO4

volume of solution

= �0.4 mol�.00 dm3

= �0.4 mol dm–3

P17.2 page 95

� (MV) before dilution = (MV) after dilution, where M = molarity, V = volume

6.60 x 40.0� 000

= M x (40.0 + 80.0)

� 000 M = 2.20 (mol dm–3)

\ the molarity of the diluted sulphuric acid is 2.20 mol dm–3.

2 Suppose V cm3 of diluted solution is obtained.

(MV) before dilution = (MV) after dilution, where M = molarity, V = volume

3.60 x 50.0� 000

= �.50 x V

� 000 V = �20 (cm3)

Volume of diluted solution = �20 cm3

\ volume of water added = (�20 – 50.0) cm3

= 70.0 cm3

P17.3 page 110

Methyl orange is a suitable indicator as the indicator changes colour within the pH range of the vertical part of the titration curve.

P17.4 page 114

a) Pipette

b) Ca(OH)2(aq) + 2HCl(aq) CaCl2(aq) +2H2O(l)

c) Methyl orange

55

Suggested Answers

d) � Ca(OH)2(aq) + 2HCl(aq) CaCl2(aq) + 2H2O(l)

2 ? mol dm–3 0.�00 mol dm–3

25.0 cm3 �2.� cm3

3 Number of moles of HCl in �2.� cm3 solution = molarity of solution x volume of solution

= 0.�00 mol dm–3 x �2.�

� 000 dm3

= 0.00�2� mol

4 According to the equation, � mole of Ca(OH)2 requires 2 moles of HCl for complete neutralization.

i.e. number of moles of Ca(OH)2 = 0.00�2� mol

2 = 0.000605 mol

5 Molarity of Ca(OH)2 solution = number of moles of Ca(OH)2

volume of solution

= 0.000605 mol

( 25.0� 000 ) dm3

= 0.0242 mol dm–3

\ the concentration of the calcium hydroxide solution is 0.0242 mol dm–3.

P17.5 page 117

� Let n be the basicity of aicd T, so we can represent the acid by HnX.

HnX(s) + nNaOH(aq) NanX(aq) + nH2O(l)

Number of moles of HnX = mass

molar mass

= 0.960 g�92.0 g mol–�

= 0.00500 mol

Number of moles of NaOH in 25.0 cm3 solution = molarity of solution x volume of solution

= 0.400 mol dm–3 x 25.0

� 000 dm3

= 0.0�00 mol

Number of moles of NaOHNumber of moles of acid

= n

n = 0.0�00 mol

0.00500 mol

n = 2

\ the basicity of acid T is 2.

56


2 a) Wash the burette with distilled water and then with the sulphuric acid.

Use burette to contain the sulphuric acid.

Add the indicator to the flask, and then the acid from the burette until the indicator changes colour.

b) Yellow to orange

c) � Na2CO3(aq) + H2SO4(aq) Na2SO4(aq) + H2O(l) + CO2(g)

2 �0.3 g 0.�20 mol dm3

250.0 cm3 30.0 cm3

(used) 25.0 cm3

3 Molar mass of Na2CO3•nH2O = (�06.0 + �8n) g mol–�

Number of moles of Na2CO3•nH2O in 250.0 cm3 solution = mass of Na2CO3•nH2O

molar mass of Na2CO3•nH2O

= �0.3 g

(�06.0 + �8.0n) g mol–�

Number of moles of Na2CO3•nH2O in 25.0 cm3 solution

= �0.3 g

(�06.0 + �8.0n) g mol–� x 25.0 cm3

250.0 cm3

Number of moles of H2SO4 in 30.0 cm3 solution = molarity of solution x volume of solution

= 0.�20 mol dm–3 x 30.0

� 000 dm3

= 0.00360 mol

4 According to the equation, � mole of Na2CO3 requires � mole of H2SO4 for complete reaction.

i.e. number of moles of Na2CO3•nH2O in 25.0 cm3 solution = 0.00360 mol

5 Number of moles of Na2CO3•nH2O in 25.0 cm3 solution

= �0.3 g

(�06.0 + �8.0n) g mol–� x 25.0 cm3

250.0 cm3 = 0.00360 mol

n = �0

\ the number of molecules of water of crystallization which combine with one formula unit of sodium carbonate is �0.

P17.6 page 120

a) Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)

b) By filtration

57

Suggested Answers

c) i) � H+(aq) + OH–(aq) H2O(l)

2 0.200 mol dm–3 20.5 cm3

3 Number of moles of H+ ions in 20.5 cm3 solution = molarity of solution x volume of solution

= 0.200 mol dm–3 x 20.5

� 000 dm3

= 0.004�0 mol

4 According to the equation, � mole of H+ ions requires � mole of OH– ions for complete neutralization.

i.e. number of moles of OH– ions left over = 0.004�0 mol

ii) Number of moles of NaOH added in Step 1 = molarity of solution x volume of solution

= 0.660 mol dm–3 x 25.0

� 000 dm3

= 0.0�65 mol

iii) Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)

Number of moles of OH– ions reacted with Cu2+ ions = (0.0�65 – 0.004�0) mol = 0.0�24 mol

According to the equation, � mole of Cu2+ ions requires 2 moles of OH– ions for complete reaction.

i.e. number of moles of Cu2+ ions in copper(II) nitrate solution = 0.0�24

2 mol = 0.00620 mol

Concentration of Cu2+ ions in copper(II) nitrate solution = number of moles of Cu2+ ions

volume of solution

= 0.00620 mol

( 25.0� 000 ) dm3

= 0.248 mol dm–3

\ the concentration of copper(II) ions in the copper(II) nitrate solution was 0.248 mol dm–3.

Discussion page 104

Step 1 A standard sodium hydroxide solution should not be prepared using the method described.

Explanation: Sodium hydroxide pellets absorb moisture and carbon dioxide from the air.

Step 2 The conical flask should not be rinsed with the sample of vinegar.

Explanation: This would increase the amount of sample in the flask.

Step 5 Calculation should not be based on the result of one titration only.

Explanation: There may be errors in the titration.

58


Chemistry Magazine page 121

Sulphur dioxide content in wine

� Examples:

• determining the ethanoic acid content in vinegar;

• determining the citric acid content in fruit juice;

• determining the lactic acid content in milk.

2 a) Food additives are grouped into classes according to their functions:

• colourings add or restore colour to foods;

• acidity regulators (including acids and alkalis) adjust the acid or alkaline level in food or maintain a sour taste;

• antioxidants help stop oils and fats from deteriorating and developing rancid flavours; they also slow down colour and flavour changes so foods made using oils and fats can be kept for longer;

• emulsifiers help prevent oil and water mixtures separating into layers;

• stabilizers make it possible for two or more ingredients (which usually do not stay mixed) to stay together;

• thickeners and vegetable gums improve texture and maintain uniform consistency;

• anti-caking agents keep powdered products (such as salt) flowing freely when poured;

• humectants prevent foods such as dried fruits from drying out;

• raising agents are used in bakery products to make them rise;

• flavour enhancers improve the existing flavour and / or taste of food;

• flavourings restore taste losses due to processing, maintain uniformity and make food more palatable;

• sweeteners replace the sweetness normally provided by sugars in foods;

• mineral salts improve the texture of foods, such as processed meats;

• nutrients (such as vitamin C) increase the nutritive value of food.

b) Advantages of using food additives:

• they make the food stay fresh longer;

• they make the food look nicer;

• they make the food taste better;

• they make the food more nutritious;

• consumers can be offered a wide choice of food;

• sometimes they are essential for making the food.

59

Suggested Answers

Disadvantages of using food additives:

• they may disguise the poor quality of the food;

• they may deceive consumers;

• they may make the food less nutritious.


� a) electronic balance

b) pipette and pipette filler

c) burette

d) dissolving a solid acid / alkali in water

e) diluting a concentrated acid / alkali of known concentration

f) titration

g) methyl orange

h) phenolphthalein

2 a) Volumetric flask

b) Pipette

c) Conical flask

d) Burette

3 B Consider � 000 cm3 (i.e. �.00 dm3) of the sample.

Mass of � 000 cm3 of the sample = �.93 g cm–3 x � 000 cm3 = � 930 g

Mass of H2SO4 in � 000 cm3 of sample = mass of � 000 cm3 of sample x percentage by mass of H2SO4 in sample = � 930 g x 93.5% = � 805 g

Molar mass of H2SO4 = (2 x �.0 + 32.� + 4 x �6.0) g mol–� = 98.� g mol–�

Number of moles of H2SO4 in �.00 dm3 of sample = mass

molar mass

= � 805 g98.� g mol–�

= �8.4 mol

60


Concentration of sulphuric acid = number of moles of H2SO4

volume of solution

= �8.4 mol�.00 dm3

=�8.4 mol dm–3

4 B Suppose V cm3 of 2.50 mol dm–3 dilute acid is obtained.

(MV) before dilution = (MV) after dilution, where M = molarity, V = volume

8.25 x 30.0

� 000 = 2.50 x

V� 000

V = 99.0 (cm3)

Volume of dilute acid obtained = 99.0 cm3

\ volume of water added = (99.0 – 30.0) cm3 = 69.0 cm3

5 D (MV) before dilution = (MV) after dilution, where M = molarity, V = volume 5.00 x �.24 = M x 20.0 M = 0.3�0 (mol dm–3)

\ molarity of the diluted potassium hydroxide solution is 0.3�0 mol dm–3.

Molar mass of KOH = (39.� + �6.0 + �.0) g mol–�

= 56.� g mol–�

Mass of KOH in �.00 dm3 diluted solution = number of moles x molar mass = 0.3�0 mol x 56.� g mol–�

= �7.4 g

Concentration of diluted solution = mass of KOH

volume of solution

= �7.4 g

�.00 dm3

= �7.4 g dm–3

\ the concentration of the diluted solution is �7.4 g dm–3.

6 C

7 A We can represent the dibasic acid by H2X.

H2X(aq) + 2NaOH(aq) Na2X(aq) + 2H2O(l) ? mol dm–3 0.240 mol dm–3

�0.0 cm3 25.0 cm3


= 0.240 mol dm–3 x 25.0

� 000 dm3

= 0.00600 mol

6�

Suggested Answers

According to the equation, � mole of H2X requires 2 moles of NaOH for complete neutralization.

i.e. number of moles of H2X = 0.00600

2 mol

= 0.00300 mol

Concentration of dibasic acid = number of moles of acid

volume of solution

= 0.00300 mol

( �0.0� 000 ) dm3

= 0.300 mol dm–3

\ the concentration of the acid is 0.300 mol dm–3.

8 A

9 C

�0 A

�� C

�2 A

�3 a) pH = –log�0[H+]

b) pH = –log�0(0.�54) = –(–0.8�2) = 0.8�2

c) (MV) before dilute = (MV) after dilute, where M = molarity, V = volume

0.�54 x �0.0

� 000 = M x

�0.0 + 990.0� 000

M = 0.00�54 (mol dm–3)

\ molarity of the diluted acid = 0.00�54 mol dm–3

pH = –log�0(0.00�54) = –(–2.8�) = 2.8�

�4 Answers for the HKCEE question are not provided.

62


�5 a)

Curve starting from pH below 7 and finishing at pH above 7.

Curve includes an almost vertical section centred around pH 7.

b) Number of moles of nitric acid = molarity of nitric acid x volume of nitric acid

= 0.�50 mol dm–3 x 25.0

� 000 dm3

= 0.00375 mol

�6 a) Test ammonia with moist red litmus paper.

The litmus paper turns blue.

b) i) Methyl orange

ii) Number of moles of H2SO4 in 32.0 cm3 solution = molarity of solution x volume of solution

= 0.0500 mol dm–3 x 32.0

� 000 dm3

= 0.00�60 mol

According to the equation, � mole of H2SO4 requires 2 moles of NH3 for complete neutralization.

i.e. number of moles of NH3 = 2 x 0.00�60 mol = 0.00320 mol

Concentration of ammonia solution in mol dm–3 = number of moles of NH3

volume of solution

= 0.00320 mol

( 25.0� 000 ) dm3

= 0.�28 mol dm–3

\ the concentration of the ammonia solution is 0.�28 mol dm–3.

iii) Concentration of ammonia solution in g dm–3 = 0.�28 mol dm–3 x �7.0 g mol–�

= 2.�8 g dm–3

63

Suggested Answers

�7 a) To obtain consistent results.

b) To show the end point of the reaction.

c) 9.70 + 9.80

2 cm3 = 9.75 cm3

d) • Wash the burette with acid. / Wash the pipette with alkali.

• Deliver �0.0 cm3 of the alkali to a conical flask using the pipette. / Fill the burette with the acid.

• Add a few drops of methyl orange to the alkali.

• Run the acid from the burette into the alkali and swirl.

• Add the acid until the indicator turns orange.

e) 0.080 mol dm–3 x 40.0 g mol–� = 3.2 g dm–3

�8 a) The results are more consistent.

b) CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l) ? g dm–3 0.�00 mol dm–3

25.0 cm3 �2.5 cm3

Number of moles of NaOH in �2.5 cm3 solution = 0.�00 mol dm–3 x �2.5

� 000 dm3

= 0.00�25 mol

According to the equation, � mole of CH3COOH requires � mole of NaOH for complete neutralization.

i.e. number of moles of CH3COOH in 25.0 cm3 vinegar = 0.00�25 mol

Number of moles of CH3COOH in each dm3 of vinegar = 0.00�25 mol

( 25.0� 000 ) dm3

= 0.0500 mol dm–3

Mass of CH3COOH in each dm3 of vinegar = number of moles x molar mass = 0.0500 mol dm–3 x 60.0 g mol–�

= 3.00 g dm–3

\ the mass of ethanoic acid in each dm3 of the vinegar is 3.00 g.

�9 a) Burette

b) H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) ? mol dm–3 0.0500 mol dm–3

�0.0 cm3 26.4 cm3

250.0 cm3

(used) 20.0 cm3

64



= 0.0500 mol dm–3 x 26.4

� 000 dm3

= 0.00�32 mol

According to the equation, � mole of H2SO4 requires 2 moles of NaOH for complete neutralization.

i.e. number of moles of H2SO4 in 20.0 cm3 of diluted acid = 0.00�32

2 mol

= 0.000660 mol

Number of moles of H2SO4 in 250.0 cm3 of diluted acid = 0.000660 mol x 250.0 cm3

20.0 cm3

= 0.00825 mol

Concentration of sulphuric acid = number of moles of H2SO4

volume of solution

= 0.00825 mol

( �0.0� 000 ) dm3

= 0.825 mol dm–3

\ the concentration of the sulphuric acid in the cave water is 0.825 mol dm–3.

20 a) i) Thermal decomposition of limestone

ii) Reduce the acidity of soil.

b) i) Ca(OH)2(aq) + 2HNO3(aq) Ca(NO3)2(aq) + 2H2O(l)

ii) Number of moles of HNO3 used = molarity of solution x volume of solution

= 0.0�05 mol dm–3 x 22.45� 000

dm3

= 2.36 x �0–4 mol

iii) Number of moles of Ca(OH)2 reacted with the HNO3 = 2.36 x �0–4

2 mol

= �.�8 x �0–4 mol

iv) Concentration of Ca(OH)2 in limewater = number of moles of Ca(OH)2

volume of solution

= �.�8 x �0–4 mol

( 25.0� 000 ) dm3

= 0.00472 mol dm–3

v) Let Ca(NO3)2•nH2O be the formula of the hydrated salt.

Molar mass of Ca(NO3)2•nH2O = [40.� + 2 x (�4.0 + 3 x �6.0) + n(2 x �.0 + �6.0)] g mol–�

272.� = �64.� + �8n n = 6

\ the formula of the hydrated salt is Ca(NO3)2•6H2O.

65

Suggested Answers

2� Answers for the HKDSE question are not provided.

22 a) Place a sample of the lime scale remover in a conical flask with a few drops of a suitable indicator.

Record the initial burette reading.

Run the sodium hydroxide solution from the burette to the lime scale remover until the indicator changes colour.

Record the final burette reading.

b) To identify and discard any outliers.

To obtain results to calculate a mean titre.

c) Any one of the following:

• For quality control

• To match the information on the label.

• To ensure that the product is safe to use.

• To ensure that the product is effective.

• To ensure that the product does not cause damage to kettles.

d) i) Mass of NaOH in 25.0 cm3 solution = 60.0 g dm–3 x 25.0

� 000 dm3

= �.50 g

ii) H3PO4(aq) + 3NaOH(aq) Na3PO4(aq) + 3H2O(l) �0.0 g of lime �.50 g scale remover 25.0 cm3

Number of moles of NaOH in 25.0 cm3 solution = mass

molar mass

= �.50 g

(23.0 + �6.0 + �.0) g mol–�

= 0.0375 mol

According to the equation, � mole of H3PO4 requires 3 moles of NaOH for complete neutralization.

i.e. number of moles of H3PO4 in �0.0 g of lime scale remover = 0.0375

3 mol

= 0.0�25 mol

Mass of H3PO4 in �0.0 g of lime scale remover = number of moles x molar mass = 0.0�25 mol x (3 x �.0 + 3�.0 + 4 x �6.0) g mol–�

= �.23 g

\ the mass of phosphoric acid in �0.0 g of the lime scale remover is �.23 g.

66


23 a) i) Any two from H+, SO42– and HSO4

–

ii) Effervescence occurs.

The potassium carbonate dissolves / a colourless solution is formed.

H2SO4(aq) + K2CO3(s) K2SO4(aq) + H2O(l) + CO2(g)

b) H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) 0.�00 mol dm–3 2.00 g

24.60 dm3

250.0 cm3

(used) 25.0 cm3

Number of moles of H2SO4 in 24.60 cm3 solution = molarity of solution x volume of solution

= 0.�00 mol dm–3 x 24.60� 000

dm3

= 0.00246 mol

According to the equation, � mole of H2SO4 requires 2 moles of NaOH for complete neutralization.

i.e. number of moles of NaOH in 25.0 cm3 solution = 2 x 0.00246 mol = 0.00492 mol

Number of moles of NaOH in 250.0 cm3 solution = 0.00492 mol x 250.0 cm3

25.0 cm3

= 0.0492 mol

Mass of NaOH in 2.00 g of impure caustic soda = number of moles x molar mass = 0.0492 mol x 40.0 g mol–�

= �.97 g

Percentage by mass of NaOH in impure caustic soda = �.97 g2.00 g

x �00%

= 98.5%

\ the percentage by mass of NaOH in the impure caustic soda is 98.5%.

24 a) • Add a few drops of phenolphthalein to the champagne.

• Fill a burette with the standard sodium hydroxide solution. Record the initial burette reading.

• Run the standard sodium hydroxide solution into the champagne until the reaction mixture changes from colourless to pink. Record the final burette reading.

• Repeat the titration to obtain three consistent titres (titres that are within 0.� cm3).

b) Number of moles of NaOH in �3.5 cm3 solution = molarity of solution x volume of solution

= 0.�00 mol dm–3 x �3.5

� 000 dm3

= 0.00�35 mol

67

Suggested Answers

� mole of NaOH reacts completely with � mole of acid.

i.e. number of moles of acid in 25.0 cm3 of champagne = 0.00�35 mol

Concentration of acid in champagne = 0.00�35 mol

( 25.0� 000 ) dm3

= 0.0540 mol dm–3

\ the concentration of acid in the champagne is 0.0540 mol dm–3.

c) Any one of the following answer:

• Yes — the acid concentration is low compared with the acid secreted by the stomach.

• No — acid level too high / other substances or bacteria may be present.

• Cannot decide — insufficient evidence to decide / depends on whether the acid level is considered safe or not.

d) Methyl orange changes colour before the pH range of the vertical part of the curve.

25 Answers for the HKCEE question are not provided.

26 Answers for the HKDSE question are not provided.


pages 134–140Topic Exercise

� A

2 A

3 D

4 A

5 C

6 D

7 D

8 A

68


9 C

�0 D

�� Answers for the HKCEE question are not provided.

�2 a) Observation: Effervescence occurs.

Product: Carbon dioxide gas

Equation: Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)

b) Observation: A white precipitate forms.

Product: Barium sulphate

Equation: Na2SO4(aq) + BaCl2(aq) BaSO4(s) + 2NaCl(aq)

�3 a) Hydrogen ion / H+(aq)

b) Any one of the following:

• Measure of pH of each acid.

The weaker acid has a higher pH.

• Measure the electrical conductivity of each acid.

The weaker acid conducts less.

• Add magnesium / a metal carbonate to each acid separately.

The weaker acid gives fewer gas bubbles.

�4 a) To neutralize acidic soils.

Excess calcium hydroxide will make the soil too alkaline for growing crop.

b) According to the equation, � mole of Ca(s) forms 2 moles of OH–(aq) ions.

i.e. number of moles of OH–(aq) ions in solution = 2 x number of moles of Ca(s) reacted

Number of moles of OH–(aq) ions in solution = 2 x 0.00�3� mol = 0.00262 mol

Concentration of OH–(aq) ions in solution = 0.00262 mol

( 250.0� 000 ) dm3

= 0.0�05 mol dm–3

\ the concentration of hydroxide ions in the solution is 0.0�05 mol dm–3.

69

Suggested Answers

�5 Answers for the HKDSE question are not provided.

�6 a) Iron(II) hydroxide / Fe(OH)2

b) Ammonia / NH3

c) Barium sulphate / BaSO4

d) Fe2+, NH4+, SO4

2–

�7 Answers for the HKCEE question are not provided.

�8 a) Add an excess of calcium hydroxide solid to water until no more will dissolve.

Filter the undissolved calcium hydroxide solid.

b) i) �0.0 cm3 pipette

ii) Indicator: methyl orange

Colour change from yellow to orange

c) i) 8.95

ii) The first titre is not consistent with the other. / The first titre is not within the difference of 0.� cm3 when compared with other runs.

iii) Mean titre = 8.85 + 8.95

2 cm3

= 8.90 cm3

iv) Ca(OH)2(aq) + 2HCl(aq) CaCl2(aq) + 2H2O(l) ? mol dm–3 0.0500 mol dm–3

�0.0 cm3 8.90 cm3

Number of moles of HCl in 8.90 cm3 solution = molarity of solution x volume of solution

= 0.0500 mol dm–3 x 8.90

� 000 dm3

= 4.45 x �0–4 mol

According to the equation, � mole of Ca(OH)2 requires 2 moles of HCl for complete neutralization.

i.e. number of moles of Ca(OH)2 in �0.0 cm3 solution = 4.45 x �0–4

2 mol

= 2.23 x �0–4 mol

Concentration of Ca(OH)2 solution in mol dm–3 = number of moles of Ca(OH)2

volume of solution

= 2.23 x �0–4 mol

( �0.0� 000 ) dm3

= 0.0223 mol dm–3

70


v) Molar mass of Ca(OH)2 = (40.� + 2 x �6.0 + 2 x �.0) g mol–�

= 74.� g mol–�

Concentration of Ca(OH)2 solution in g dm–3 = 74.� g mol–� x 0.0223 mol dm–3

= �.65 g dm–3

�9 Answers for the HKDSE question are not provided.

20 a) i) C

ii) A

iii) D

b) i) Bromocresol green

ii) Purple to yellow

2� HxA(aq) + xNaOH(aq) NaxA(aq) + xH2O(l) 0.0500 mol dm–3 0.200 mol dm–3

25.00 cm3 �2.50 cm3

Number of moles of HxA used = molarity of solution x volume of solution

= 0.0500 mol dm–3 x 25.00� 000

dm3

= 0.00�25 mol

Number of moles of NaOH used = molarity of solution x volume of solution

= 0.200 mol dm–3 x �2.50� 000

dm3

= 0.00250 mol

number of moles of NaOHnumber of moles of HxA

= x�

0.00250 mol0.00�25 mol

= x�

\ x = 2

\ in the formula of sodium tartrate HxA, the value of x is 2.




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Suggested Teaching Scheme - STMGSSintranet.stmgss.edu.hk/~ted/ccy/new_edition/Book2Aanswer.pdf · 2014-10-22 · Suggested Teaching Scheme Suggested Teaching Scheme The following