Physical Sciences Acids and Bases

Physical Sciences GRAD E 12 SELF STUDY GUIDE ACIDS AND BASES

TABLE OF CONTENTS Page 1. Introduction 2 2. How to use this self-study guide 3 3. Acid and basis 3.1 Extract from the examination guidelines 4–6 3.2 Mind map 7 3.3 Definitions 8 3.4 Arrhenius and Lowrey-Brønsted models 9 3.5 Strong acids/bases and weak acid/bases 9 – 10 3.6 Concentrated/Dilute acids/bases 11 3.7 Conjugate acid-base pairs 12 – 13 3.8 Hydrolysis 13 – 15 3.9 pH and the pH scale and pH calculations 15 – 20 3.10 Titrations and stoichiometric calculations 21 – 27 3.11 Ka and Kb values 28 – 29 4. Exercises 4.1 Multiple choice questions 30 - 32 4.2 Structured questions 33 4.3 Typical examination questions 34 - 36 5. Solutions to exercises 37 – 43 6. Examination guidance 44 7. Study and examination tips 45 8. References 46 9. Acknowledgement 46 1

1 Introduction The declaration of COVID-19 as a global pandemic by the World Health Organisation led to the disruption of effective teaching and learning in many schools in South Africa. The majority of learners in various grades spent less time in class due to the phased-in approach and rotational/ alternate attendance system that was implemented by various provinces. Consequently, the majority of schools were not able to complete all the relevant content designed for specific grades in accordance with the Curriculum and Assessment Policy Statements in most subjects. As part of mitigating against the impact of COVID-19 on the current Grade 12, the Department of Basic Education (DBE) worked in collaboration with subject specialists from various Provincial Education Departments (PEDs) developed this Self-Study Guide. The Study Guide covers those topics, skills and concepts that are located in Grade 12, that are critical to lay the foundation for Grade 12. The main aim is to close the pre-existing content gaps in order to strengthen the mastery of subject knowledge in Grade 12. More importantly, the Study Guide will engender the attitudes in the learners to learning independently while mastering the core cross-cutting concepts. 2

2 How to use the booklet This booklet is meant to help you improve your understanding of the subject Physical Sciences. It summarises the work that must be studied for examination purposes. This guide does not give full explanations of all the concepts. Its aim is to assist you in the understanding of the important facts as highlighted in the examination guidelines, it gives tips and suggested methods of solving problems, and on how to answer certain questions, your textbook will explain the work in depth, thus it does not replace your textbook, it should be used in conjunction with your preferred textbook. The authors wrote this booklet using their experience in a classroom situation. This guide aims to help you improve your performance, or to help you score marks in Acids and Bases. Examples are given with solutions/answers, some explanations are provided next to the solution/answer to enhance your understanding. After studying a certain example, using a blank paper/cardboard shield the solution/answer for that example and try to solve it on your own without looking at the answer. Use the provided solutions to mark your own work. If you fail to get it right on the first attempt, do not give up, keep on trying until you can solve it successfully. Solutions to the exercises are provided in this booklet. Attempt the exercises without looking at the solutions. After doing an exercise compare your solution to the one provided and go through the provided solutions carefully and make sure you understand the steps taken to solve the problem/question. If you cannot do a certain exercise, go back to the relevant section/theory and study it again. 3

͵Ǥ Acids and Bases 3.1 Extracts from the Examination Guidelines Acid-base reactions x Define acids and bases according to Arrhenius and Lowry-Brønsted: Arrhenius theory: Acids produce hydrogen ions (H+) in solution. Bases produce hydroxide ions (OH-) in solution. Lowry-Brønsted theory: An acid is a proton (H+ ion) donor. A base is a proton (H+ ion) acceptor. x Distinguish between strong acids/bases and weak acids/bases with examples. Strong acids ionise completely in water to form a high concentration of H3O+ ions. Examples of strong acids are hydrochloric acid; sulphuric acid and nitric acid. Weak acids ionise incompletely in water to form a low concentration of H3O+ ions. Examples of weak acids are ethanoic acid and oxalic acid. Strong bases dissociate completely in water to form a high concentration of OH- ions. Examples of strong bases are sodium hydroxide and potassium hydroxide. Weak bases dissociate/ionise incompletely in water to form a low concentration of OH- ions. Examples of weak bases are ammonia, calcium carbonate, potassium carbonate, calcium carbonate, sodium hydrogen carbonate. x Distinguish between concentrated and dilute acids/bases. Concentrated acids/bases contain a large amount (number of moles) of acid/base in proportion to volume of water. Dilute acids/bases contain a small amount (number of moles) of acid/base in proportion to volume of water. x Write down the reaction equations of aqueous solutions of acids and bases. Examples: HCℓ(g) + H2O(ℓ) → H3O+(aq) + Cℓ-(aq) (HCℓ is a monoprotic acid.)  NH3(g) + H2O(ℓ) → NH 4 (aq) + OH-(aq) H2SO4(aq) + 2H2O(ℓ) → 2H3O+(aq) + SO 24  (aq) (H2SO4 is a diprotic acid.) x Identify conjugate acid-base pairs for given compounds. When the acid, HA, loses a proton, its conjugate base, A-, is formed. When the base, A-, accepts a proton, its conjugate acid, HA, is formed. These two are a conjugate acid-base pair. x Describe a substance that can act as either acid or base as ampholyte. Water is a good example of an ampholyte substance. Write equations to show how an ampholyte substance can act as acid or base. 4

x Write down neutralisation reactions of common laboratory acids and bases. Examples: HCℓ(aq) + NaOH(aq)/KOH(aq) → NaCℓ(aq)/KCℓ(aq) + H2O(ℓ) HCℓ(aq) + Na2CO3(aq) → NaCℓ(aq) + H2O(ℓ) + CO2(g) HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(ℓ) H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(ℓ) (COOH)2(aq) + NaOH(aq) → (COO)2Na2(aq) + H2O(ℓ) CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(ℓ) Note: The above are examples of equations that learners will be expected to write from given information. However, any other neutralisation reaction can be given in a question paper and used to assess e.g. stoichiometry. Hydrolysis x Define hydrolysis as the reaction of a salt with water. x Determine the approximate pH (equal to, smaller than or larger than 7) of salts in salt hydrolysis. o Hydrolysis of the salt of a weak acid and a strong base results in an alkaline solution i.e. the pH > 7. Examples of such salts are sodium ethanoate, sodium oxalate and sodium carbonate. o Hydrolysis of the salt of a strong acid and a weak base results in an acidic solution i.e. the pH < 7. An example of such a salt is ammonium chloride. o The salt of a strong acid and a strong bases does not undergo hydrolysis and the solution of the salt will be neutral i.e. pH = 7. Acid-base titrations x Motivate the choice of a specific indicator in a titration. Choose from methyl orange, phenolphthalein & bromothymol blue. x Define the equivalence point of a titration as the point at which the acid /base has completely reacted with the base/acid. x Define the endpoint of a titration as the point where the indicator changes colour. x Perform stoichiometric calculations based on titrations of a strong acid with a strong base, a strong acid with a weak base and a weak acid with a strong base. Calculations may include percentage purity. x For a titration e.g. the titration of oxalic acid with sodium hydroxide: o List the apparatus needed or identify the apparatus from a diagram. o Describe the procedure to prepare a standard oxalic acid solution. o Describe the procedure to conduct the titration. o Describe safety precautions . 5

o Describe measures that need to be in place to ensure reliable results. o Interpret given results to determine the unknown concentration. pH and pH scale x Explain the pH scale as a scale of numbers from 0 to 14 used to express the acidity or alkalinity of a solution. x Calculate pH values of strong acids and strong bases using pH = -log[H3O+]. x Define Kw as the equilibrium constant for the ionisation of water or the ionic product of water, i.e. Kw= [H3O+][OH-]=1 × 1014 at 298 K. x Explain the auto-ionisation of water i.e. the reaction of water with itself to form H3O+ ions and OH- ions. x Interpret Ka values of acids to determine the relative strength of given acids. Interpret Kb values of bases to determine the relative strength of given bases. x Compare strong and weak acids by looking at: o pH (monoprotic and diprotic acids) o Conductivity o Reaction rate 6

3.2 Lowry-Brønsted Models Arrhenius Models Conjugate Acid-Base Pairs x Lowry-Brønsted acid- Acids produce hydrogen is a proton donor. Mind Map x Lowry-Brønsted base- ions (H+) in solution. is a proton acceptor. Bases produce hydroxide ions (OH-) in solution. pH Scale and pH calculations Strong Acid-ionises completely in water. Strong base-dissociates completely in water. Weak Acid- ionises ACIDS AND BASES incompletely in water. Hydrolysis- the reaction of Weak Base- a salt with water dissociates/ionises incompletely in water. Neutralisation Reactions of Concentrated Acids/Bases Titration Calculations- Common laboratory Acids Dilute Acids/Bases stoichiometric calculations and Bases 7

3.3 Definitions TERM DEFINITION/EXPLANATION Arrhenius Acid Acids produce hydrogen ions (H+) in solution Arrhenius Base Bases produce hydroxide ions (OH-) in solution Lowry-Brønsted Acid Is a proton (H+ ion) donor Lowry-Brønsted Base Is a proton (H+ ion) acceptor. Strong acids ionise completely in water to form a high concentration of Strong Acid H3O+ ions. Weak acids ionise incompletely in water to form a low concentration of H3O+ Weak Acid ions Strong bases dissociate completely in water to form a high concentration Strong Base of OH- ions Weak bases dissociate/ionise incompletely in water to form a low Weak Base concentration of OH- ions Concentrated acids contain a large amount (number of moles) of acid in Concentrated Acid proportion to volume of water Concentrated bases contain a large amount (number of moles) of base in Concentrated Base proportion to volume of water Dilute acids contain a small amount (number of moles) of acid in Dilute Acid proportion to volume of water. Dilute acids contain a small amount (number of moles) of acid in Dilute Base proportion to volume of water. Ampholyte A substance that can act as acid or base. Hydrolysis Is the reaction of a salt with water Equivalence Point of a Is the point at which the acid /base has completely reacted with the titration base/acid End Point of a titration Is the point where the indicator changes colour Ionisation Constant of Is the equilibrium constant for the ionisation of water Water (Kw) Standard solution A solution of known concentration 8

3.4 Arrhenius and Lowry-Brønsted Models Examination Guidelines Arrhenius theory: Acids produce hydrogen ions (H+) in solution. Bases produce hydroxide ions (OH-) in solution. Lowry-Brønsted theory: An acid is a proton (H+ ion) donor. A base is a proton (H+ ion) acceptor. According to the Arrhenius theory Acids produce hydrogen ions (H+) in solution. e.g. HNO3(aq) → H+(aq) + NO3-(aq), H+ was produced thus it’s an Arrhenius acid, Bases produce hydroxide ions (OH-) in solution. e.g. KOH(aq) → K+(aq) + OH-(aq), OH- was produced, thus it’s an Arrhenius base. According to the Lowry-Brønsted theory, an acid is a proton (H+ ion) donor, and a base is a proton (H+ ion) acceptor. H+ e.g HNO3(aq) + H2O(ℓ) ֖ H3O+(aq) + NO3-(aq) HNO3 donated a proton to H2O, it is a Lowry-Brønsted acid, and H2O accepted the proton, so it is a Lowry-Brønsted base. 3.5 Strong acids/bases and weak acids/bases Examination Guidelines Strong acids ionise completely in water to form a high concentration of H3O+ ions. Weak acids ionise incompletely in water to form a low concentration of H3O+ ions. Strong bases dissociate completely in water to form a high concentration of OH- ions. Weak bases dissociate/ionise incompletely in water to form a low concentration of OH- ions. 9

Acids Strong acids ionise completely in water to form a high concentration of H3O+ ions. HA The acid HA ionised completely. There are no HA molecules present. Only ions are A- present. H+ Weak acids ionise incompletely in water to form a low concentration of H3O+ ions. HA The acid HA ionised incompletely. A- HA molecules are still present. Very few ions are present. H+ Bases Strong bases dissociate completely in water to form a high concentration of OH- ions. (In a NaOH solution, all the base dissociate to form ions, there are now Na+ and OH- ions in the container) Weak bases dissociate/ionise incompletely in water to form a low concentration of OH- ions. (There are fewer ions in the solution compared to the base itself, e.g. in a Zn(OH)2 solution there is a low amount of Zn2+ and OH- ions compared to the Zn(OH)2 molecules) Please familiarise yourself with the list/table given below. STRONG ACIDS STRONG BASES HCℓ - Hydrochloric acid (monoprotic) NaOH – Sodium hydroxide HNO3 – Nitric acid (monoprotic) KOH – Potassium hydroxide H2SO4 – Sulphuric acid (diprotic) LiOH – Lithium hydroxide H3PO4 – Phosphoric acid (triprotic) Ba(OH)2 – Barium hydroxide WEAK ACIDS WEAK BASES CH3COOH – Acetic acid NH3 - Ammonia (COOH)2 – Oxalic acid Zn(OH)2 – Zinc hydroxide Note: A monoprotic acid is an acid that can donate one proton only. e.g. H Cℓ Only ONE proton is available/can be donated. A diprotic acid is an acid that can donate a maximum of two protons. e.g. H2 SO4 TWO protons are available/can be donated. 10

3.6 Concentrated/Dilute Acids/Bases Examination Guidelines Concentrated acids/bases contain a large amount (number of moles) of acid/base in proportion to volume of water. Dilute acids/bases contain a small amount (number of moles) of acid/base in proportion to volume of water. Concentrated acids/bases contain a large amount (number of moles) of acid/base in proportion to volume of water. Water There is ‘more’ acid than water in the solution, i.e., there is a large amount of acid in proportion to the volume of water. Acid Dilute acids/bases contain a small amount (number of moles) of acid/base in proportion to volume of water. Water There is ‘less’ acid than water in the solution, i.e., there is a small amount of acid in proportion to the volume of water. Acid Note: The diagrams just show the proportion of acid to volume of water in the solution, not that the water is above the acid, or they have different densities and they cannot mix. Note: A dilute acid is not a weak acid, a weak acid can either be concentrated or dilute. A concentrated acid is not a strong acid, a strong acid can be concentrated or dilute. A strong acid/base conducts electricity better than a weak acid/base because of the greater number of ions in solutions, since the strong acid ionises/ the strong base dissociates completely, provided the strong acid/base and the weak acid/base have equal concentrations. A strong acid/base reacts very fast due to the high concentration of ions (the acid ionises or the base dissociates completely in water) 11

3.7 Conjugate acid-base pairs Examination Guidelines Identify conjugate acid-base pairs for given compounds. When the acid, HA, loses a proton, its conjugate base, A-, is formed. When the base, A-, accepts a proton, its conjugate acid, HA, is formed. These two are a conjugate acid-base pair. Describe a substance that can act as either acid or base as ampholyte. Water is a good example of an ampholyte substance. Write equations to show how an ampholyte substance can act as acid or base. You must be able to write balanced equations of aqueous solutions of acids and bases and be able to identify conjugate acid-base pairs. Conjugate acid-base pairs are compounds that differ from each other by a proton (H+ ion) To form a conjugate acid, add a proton. Example The conjugate acid of: (a) NH3 is NH+4 (A proton (H+) was added to NH3) This may assist especially when answering multiple choice (b) HCO-3 is H2CO3 (A proton (H+) was added to HCO-3 ) questions. To form a conjugate base, remove a proton. Example The conjugate base of: (a) HCO-3 is COଶି 3 (A proton was removed from HCO3) Forward reaction (b) H2O is OH- (A proton was removed from H2O) Example NH3 is a base because it accepts a proton from H2O, and H2O is an acid Identify conjugate acid base pairs. because it donates a proton to NH3. Base 1 Acid 1 Reverse reaction NH3(aq) + H2O ֖ NH (aq) + OH-(aq) NH4+ is an acid because it donates a proton to OH-, and OH- is a base Acid 2 Base because it accepts a proton from NH4+. 2 The conjugate acid-base pairs are: NH3 and N 4+ and, H2O and H3O+. 12

Exercise Identify conjugate acid-base pairs in the following balanced equations. Hydroxide ion (a) HCO-3 (aq) + H2O(ℓ) ֖H2CO3(aq) + OH-(aq) (b) HCO-3 (aq) + H2O(ℓ) ֖COଶ- 3 ሺaq) + H3O+(aq) Hydronium ion Solution (a) H2CO3 is the conjugate acid of HCO3- (base) In both equations, it can be noted that in (a) HCO3- is a base and in (b) it is an acid, OH- is the conjugate base of H2O (acid) thus HCO3- is an ampholyte, i.e., it can (b) CO-3 is the conjugate base of HCO-3 - (acid) act as either an acid or a base. H2O is also an ampholyte as can be seen in both H3O+ is the conjugate acid of H2O (base) (a) and (b). 3.8 Hydrolysis Examination Guidelines Define hydrolysis as the reaction of a salt with water. Determine the approximate pH (equal to, smaller than or larger than 7) of salts in salt hydrolysis. o Hydrolysis of the salt of a weak acid and a strong base results in an alkaline solution i.e. the pH > 7. Examples of such salts are sodium ethanoate, sodium oxalate and sodium carbonate. o Hydrolysis of the salt of a strong acid and a weak base results in an acidic solution i.e. the pH < 7. An example of such a salt is ammonium chloride. o The salt of a strong acid and a strong bases does not undergo hydrolysis and the solution of the salt will be neutral i.e. pH = 7. Hydrolysis is the reaction of a salt with water. The salt of a strong acid and a weak base is acidic, pH < 7. The salt of a weak acid and a strong base is basic, pH > 7. The salt of a strong acid and a strong base does not undergo hydrolysis. 13

EXAMPLE 1 STEPS NH4Cℓ x Split the salt into ions. x Identify the acid and the base that formed the salt. NH4+ Cℓ- i.e. NH3(aq)+HCℓ(aq) → NH4Cℓ(aq) x Classify the acid and the base as NH3 HCℓ strong or weak. Weak base Strong acid x The ion of the weak species OR acid/base will undergo hydrolysis. Hydrolysis No Hydrolysis x It forms the original acid/base and a NH+4 (aq)+H2O(ℓ) ֖ H3O+(aq) + NH3(aq) hydronium or hydroxide ion when it reacts with water. The salt is acidic, a strong acid reacted with a weak base OR H3O+ which is an acid is formed. EXAMPLE 2 STEPS CH3COONa x Split the salt into ions. CH3COO- Na+ x Identify the acid and the base that formed the salt. i.e. CH3COOH(aq) + NaOH(aq) → CH3COOH(aq) + H2O(ℓ) CH3COOH NaOH x Identify the strength of the acid and the base. Weak acid Strong base x The ion of the weak species OR acid/base will undergo hydrolysis. Hydrolysis No hydrolysis x It forms the original acid/base and a CH3COO-(aq) + H2O(ℓ) ֖ CH3COOH(aq)+ OH-(aq) hydronium or hydroxide ion when it reacts with water The salt is basic, a strong base reacted with a weak acid OR OH- which is a base is formed. Exercise 14

Is the salt KHCO3 acidic or basic? Explain with the aid of a balanced chemical equation. Solution KHCO3 K+ HCO3- KOH H2CO3 Strong base Weak acid HCO3-(aq) + H2O(ℓ) ֖H2CO3(aq) + OH-(aq) The salt is basic, OH- is formed which is a base (A strong base reacted with a weak acid) 3.9 pH and the pH scale and pH calculations Examination Guidelines x Explain the pH scale as a scale of numbers from 0 to 14 used to express the acidity or alkalinity of a solution. x Calculate pH values of strong acids and strong bases using pH = -log[H3O+]. x Define the concept of Kw as the equilibrium constant for the ionisation of water – the ionic product of water (ionisation constant of water). x Explain the auto-ionisation of water i.e. the reaction of water with itself to form H 3O+ ions and OH- ions. The pH scale is a scale of numbers from 0 to 14 used to express the acidity or alkalinity of a solution. Acids have a pH less than 7 (pH<7), a neutral solution has a pH equal to 7 (pH=7), bases have a pH greater than 7 (pH>7). 15

Acid-base indicators are used to test the pH of solutions, the common indicators are: x Methyl orange (red in an acidic solution and yellow in a basic solution) x Bromothymol blue (yellow in an acidic solution and blue in a basic solution) x Phenolphthalein (colourless in an acidic solution and pink in a basic solution) Calculation of pH To calculate the pH of a solution the concentration of H 3O+ must be known or be determined. The reaction of water with itself produces H3O+ and OH- ions, it is called the auto-ionisation of water. The equation is given below: H2O(ℓ) + H2O(ℓ) ֖ H3O+(aq) + OH-(aq) The expression for the equilibrium constant of the chemical equation given above is: ൣH3 O+ ൧[OH- ] kw = 2 [H2 O] the concentration of water is very large (constant) compared to the concentrations of H3O+ and OH- /water is a pure liquid so its concentration will be 1 mol.dm-3, it can therefore be omitted from the ൣH3 O+ ൧[OH- ] equilibrium expression, thus kw = , this expression is called the ionic product of water, the 1 symbol Kw is used, Kw = [H3O+][OH-]. Kw =1×10-14 at 25°C. For a neutral solution [H3O+] = [OH-] = 1×10-7 mol·dm-3. For an acidic solution [H3O+] > 1×10-7 mol·dm-3 and [H3O+] > [OH-]. For a basic solution [H3O+] < 1×10-7and [H3O+] < [OH-]. Copy and complete the following table, use Kw = [H3O+][OH-] = 1×10-14 pH 2 3 5 6 7 9 10 12 13 [H3O+] mol·dm-3 1×10-2 1×10-3 1×10-6 1×10-9 1×10-12 1×10-13 [OH-] mol·dm-3 1×10-9 1×10-8 1×10-7 1×10-5 1×10-4 Solution Use the expression [H3O+][OH-]=1×10-14 to calculate the unknown value, e.g. [H3O+][OH-]=1×10-14 [H3O+](1×10-13)=1×10-14 Therefore, [H3O+]=1×10-1 mol·dm-3 pH 2 3 5 6 7 9 10 12 13 [H3O+] mol·dm-3 1×10-2 1×10-3 1×10-5 1×10-6 1×10-7 1×10-9 1×10-10 1×10-12 1×10-13 [OH-] mol·dm-3 1×10-12 1×10-11 1×10-9 1×10-8 1×10-7 1×10-5 1×10-4 1×10-2 1×10-1 16

For pH calculations the formula pH = -log[H3O+] is used. You are expected to calculate the pH of solutions of strong acids and strong bases only. Steps to follow when calculating the pH of an acid. x Write down a balanced equation for the ionisation reaction of the acid (reaction with water) x Use ratios to determine the concentration of [H3O+]. x Substitute the concentration of [H3O+] in the formula pH = -log[H3O+]. Examples 1. Calculate the pH of a 0,2 mol·dm-3 HNO3 solution. HNO3(aq) + H2O(ℓ) ֖ H3O+(aq) + O-3 (aq) 0,2 mol·dm-3 0,2 mol·dm-3 (1:1 ratio) (monoprotic acid) ‫[ ׵‬H3O+] = 0,2 mol·dm-3 pH = -log [H3O+] = -log (0,2) = 0,699 (Note: pH does not have a unit) 2. Calculate the pH of a 0,25 mol·dm-3 H2SO4 solution. H2SO4(aq) + H2O(ℓ) ֖ 2H3O+(aq) + SO42-(aq) 0,25 mol·dm-3 2(0,25) mol·dm-3 (1:2 ratio) (diprotic acid) [H3O+] = 0,5 mol·dm-3 pH = -log [H3O+] = -log (0,5) = 0,301 Steps to follow when calculating the pH of a base. x Write down a balanced equation for the dissociation reaction of the base. x Use ratios to determine the concentration of [OH-]. x Use kw to calculate the concentration of [H3O+] x Substitute the concentration of [H3O+] in the formula pH = -log[H3O+]. 17

Examples 1. Calculate the pH of a 0,4 mol·dm-3 NaOH solution. NaOH(aq) ֖ Na+(aq) + OH-(aq) 0,4 mol·dm-3 0,4 mol·dm-3 (1:1 ratio) ‫[ ׵‬OH-] = 0,4 mol·dm-3 . [H3O+][OH-]=1×10-14 [H3O+](0,4)=1×10-14 ‫[ ׵‬H3O+] = 2,5 ×10-14 mol·dm-3 pH = -log [H3O+] = -log (2,5 ×10-14) = 13,602 2. Calculate the pH of a 0,01 mol·dm-3 Ba(OH)2 solution. Ba(OH)2(aq) ֖ Ba2+(aq) + 2OH-(aq) 0,01 mol·dm-3 2(0,01) mol·dm-3 (1:2 ratio) ‫[ ׵‬OH-] = 0,02 mol·dm-3 [H3O+][OH-]=1×10-14 [H3O+](0,02)=1×10-14 ‫[ ׵‬H3O+] = 5 ×10-13 mol·dm-3 pH = -log [H3O+] = -log (5 ×10-13) = 12,301 18

Calculating the concentration of an acid or base if given the pH 1. Calculate the concentration of the acid HCℓ with a pH = 4,5 Solution x Write down a balanced equation for the ionisation reaction of the acid (reaction with water) x Substitute the pH value in the formula pH = -log[H3O+] and calculate the concentration of [H3O+] using antilog, use the button 10 on your calculator, i.e., press 2nd F/SHIFT then log x Use ratios to determine the concentration of the acid. HCℓ(aq) + H2O(ℓ) ֖ H3O+(aq) + Cℓ-(aq) pH = -log [H3O+] 4,5 = -log [H3O+] [H3O+] = 10-4,5 = 3,162 × 10-5 mol·dm-3 Therefore [HCℓ] = 3,162 × 10-5 mol·dm-3 (1:1 ratio) 19

2. Calculate the concentration of the base Ca(OH)2 with a pH of 11,2. Solution x Write down a balanced equation for the dissociation reaction of the base. x Substitute the pH value in the formula pH = -log[H3O+] and calculate the concentration of [H3O+] using antilog, use the button 10 on your calculator, i.e., press 2nd F/SHIFT then log x Use kw to calculate the concentration of OH-. x Use ratios to determine the concentration of the base. Ca(OH)2(aq) ֖ Ca2+(aq) + 2OH-(aq) pH = -log [H3O+] 11,2 = -log [H3O+] [H3O+] = 10-11,2 = 6,310 × 10-12 mol·dm-3 [H3O+][OH-]=1×10-14 (6,310 × 10-12) [OH-]=1×10-14 [OH-]=0,00158 mol·dm-3 Therefore [Ca(OH)2] = ½ (0,00158) = 0,0007924 (7,924 × 10-4) mol·dm-3 (ratio 2:1) 20

3.10 Titrations and Stoichiometric Calculations Examination Guidelines Write down neutralisation reactions of common laboratory acids and bases. Examples: HCℓ(aq) + NaOH(aq)/KOH(aq) → NaCℓ(aq)/KCℓ(aq) + H 2O(ℓ) HCℓ(aq) + Na2CO3(aq) → NaCℓ(aq) + H2O(ℓ) + CO2(g) HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(ℓ) H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(ℓ) (COOH)2(aq) + NaOH(aq) → (COO)2Na2(aq) + H2O(ℓ) CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(ℓ) Note: The above are examples of equations that learners will be expected to write from given information. However, any other neutralisation reaction can be given in a question paper and used to assess e.g. stoichiometry calculations. x Motivate the choice of a specific indicator in a titration. Choose from methyl orange, phenolphthalein & bromothymol blue. x The equivalence point of a titration as the point at which the acid /base has completely reacted with the base/acid. x The endpoint of a titration as the point where the indicator changes colour. x Perform stoichiometric calculations based on titrations of a strong acid with a strong base, a strong acid with a weak base and a weak acid with a strong base. Calculations may include percentage purity. Terms Neutralisation is a chemical reaction in which an acid and a base interact, with the formation of a salt and water, carbon dioxide will also be formed if a carbonate is used. (see table below) A titration is when a standard solution (solution of known concentration) is added to the sample solution (unknown concentration) until the end point (the point where the indicator changes colour) is reached. Examples HCℓ(aq) + NaOH(aq) → NaCℓ(aq) + H2O(ℓ) HCℓ(aq) + Na2CO3(aq) → NaCℓ(aq) + H2O(ℓ) + CO2(g) 21

The following setup is used in a titration: An acid-base indicator is used to determine the end point of a titration. You must be able to choose the correct indicator for a titration. pH colour change ACID BASE INDICATOR range Burette Bromothymol Strong Strong 6,0 – 7,6 blue Strong Weak Methyl Orange 3,2 – 4.4 Weak Strong Phenolphthalein 8,2 – 10,0 e.g., For a titration of NaOH (strong base) against HCℓ (strong Erlenmeyer acid), bromothymol blue is used. The equivalence point for the Flask reaction of a strong acid and a strong base is approximately 7. Burette Stand 3.10.1 Basic Calculations Tips 1. Calculate the number of moles, if given volume and concentration of a substance using n = m cV OR n= if given a mass. M 2. Use the mole ratio from the balanced equation to calculate the number of moles of the other substance required. 3. Simply calculate the concentration, volume or mass of the substance depending on what is n m asked, using c= OR n= . v M 22

Examples 1. Calculate the concentration of 4 g magnesium hydroxide dissolved in 24 ml of distilled water. x Calculate the molar mass of Mg(OH)2 M[Mg(OH)2] = 24 + 2(16) + 2(1) = 58 g·mol-1 x Calculate the number of moles using m n= , m M n= x Convert ml/cm3 to dm3 by dividing the M volume by 1000. Ͷ x Calculate the concentration using the ൌ n ͷͺ formula c= V ൌͲǡͲ͸ͺͻ͸‘Ž  ൌ 1 cm3 = 1 ml  ͲǡͲ͸ͺͻ͸ 1000 cm3 = 1 dm3 ൌ ͲǡͲʹͶ 1 dm3 = 1 L =2,873 mol∙dm-3 2. Calculate the mass of H2SO4 in 30 cm3 of a 0,02 mol·dm-3 solution.  ൌ   n x Convert ml/cm3 to dm3 by dividing the 0,02=  volume by 1000. 0,03 x Calculate the number of moles using n=0,0006 mol the formula c= n V m x Calculate the molar mass of H2SO4. n=  M x Calculate the mass using the formula m m n= , M 0,0006=  98 m=0,0588 g 3.10.2 Dilution Calculations Dilution is required when we prepare a certain concentration of a solution from a more concentrated solution. When diluting a solution, the number of moles does NOT CHANGE, the number of moles before dilution(n1) = number of moles after dilution(n2). From n1=c1V1 (before dilution) and n2=c2V2 (after dilution), it can be deduced that c1V1= c2V2 23

Example Calculate the volume of 0,5 mol·dm-3 H2SO4 that must be added to give 50 cm3 of the H2SO4 with concentration 0,025 mol·dm-3. Solution c1 V1 =c2 V2 (0,5)1 =(0,025)(50) ͳ ൌʹǡͷ ͵ 3.10.3 Stoichiometric Calculations m If given a mass use n= to calculate the number of moles, the same formula will be used to calculate a M required mass. n If given concentration and volume use c= to calculate the number of moles V Use mol ratios to calculate the number of moles of the required substance or solution. n Calculate the required concentration or volume by using c= V If initial or excess amounts are mentioned, the formulae nexcess = ninitial – nreacted, ninitial = nreacted + nexcess or nreacted = ninitial - nexcess should be used. 24

Example 1. 10 ml of a 0,25 mol·dm-3 sodium hydroxide solution reacts with 0,40 mol·dm-3 H2SO4 according to the balanced chemical equation given below. Calculate the volume of acid needed to react with the base. H2SO4 + 2NaOH → Na2SO4 + 2H2O ͳͲ Solution ൌ ͲǡͲͳ ͳͲͲͲ Calculate the number of moles of NaOH first. (1000 ml/cm3 = 1 L/dm3) nሺNaOHሻ=cV ൌ ሺͲǡʹͷሻሺͲǡͲͳͲሻ Calculate the number of moles using =0,0025 mol n c= . V n(H2 SO4 ) 1 =  n(NaOH) 2 Use mole ratios to calculate the n(H2 SO4 ) 1 =  number of moles. 0,0025 2 1 n(H2 SO4 )= ×0,0025 2 ൌͲǡͲͲͳʹͷ‘Ž n Calculate the volume using c= . V n c= V 0,00125 0,40=  V V=0,003125 dm3  =3,125 cm3 OR ca Va na = cb Vb nb 0,4×Va 1 = 0,25×0.010 2 Va =0,003125 dm3 A learner adds a sample of calcium carbonate to 50,0 cm3 of hydrochloric acid of concentration 1,0 mol.dm-3. The hydrochloric acid is in excess. The balanced equation for the reaction that takes place is: CaCO3 + 2HCε Ѝ CaCε2 + CO2 + H2O The excess HCε is now neutralised by 28,0 cm3 of a 0,5 mol.dm-3 sodium hydroxide solution. The balanced equation for this reaction is: HCε + NaOH Ѝ NaCε + H2O Calculate the mass of calcium carbonate in this sample. 25

Solution The information given will indicate from which side to start. In this one we will start HCℓ from NaOH reacting with the acid, then move to CaCO3. Note: HCℓ reacts with both. CaCO3 NaOH HCℓ(aq) + NaOH(aq) → NaCℓ(aq) + H2O(ℓ) cb = 0,5 Write down the data/values below each Vb = 0,028 reagent, it will become easy to see which reagent has complete/incomplete information/data. Step 1: Calculate the total number of moles of HCℓ (initial) nሺHClሻinitial =cV =ሺ1,0ሻሺ0,050ሻ =0,05 mol Step 2: Calculate the number of moles of NaOH. HCℓ(aq) + NaOH(aq) → NaCℓ(aq) + H2O(ℓ) Write down the data/values below each cb = 0,5 reagent, it will become easy to see which reagent has complete/ Vb = 0,028 incomplete information/data. nሺNaOHሻ=cV =ሺ0,5ሻሺ0,028ሻ =0,014 mol 26

Step 3: Calculate the number of moles HCℓ in excess (reacted with NaOH). n(HCl) 1 = n(NaOH) 1 n(HCl) 1 = 0,014 1 n(HCl)excess =0,014 mol Step 4: Calculate the total number of moles of HCℓ reacted (reacted with CaCO3) n(HCℓ) reacted = n(HCℓ) initial - n(HCℓ) excess = 0,05 – 0,014 = 0,036 mol Step 5: Calculate the number of moles of CaCO3 reacted with HCℓ. CaCO3 + 2HCℓ → CaCℓ2 + CO2 + H2O na = 0,036 n(CaCO3 ) 1 =  n(HCl) 2 nሺCaCO3 ሻ 1 =  0,036 2 1 = ×0,036 2 =0,018 mol Step 6: Calculate the mass of CaCO3. m n= M Calculate the molar mass of m CaCO3. 0,018= 100 M[CaCO3] = 40+12+3(16) =0,018×100 = 100 g·mol-1 =1,8 g 27

3.11 Ka and Kb values x Interpret Ka values of acids to determine the relative strength of given acids. Interpret Kb values of bases to determine the relative strength of given bases. x Compare strong and weak acids by looking at: o pH (monoprotic and diprotic acids) o Conductivity o Reaction rate Ka is the ionisation constant of an acid and kb is the dissociation/ionisation constant of a base. The ionisation constants are a measure of the relative strength of an acid or base. Strong acids have large ka values, because they ionise completely, and weak acids have small ka values since they ionise incompletely/partially. Strong bases have large kb values, because they dissociate completely, and weak bases have small kb values since they ionise/dissociate incompletely/partially. Strong acids have Ka values larger than 1 Weak acids have Ka values smaller than 1 Table: Ka values of some common acids Formula Name Ka values Type HBr Hydrobromic acid 1×109 Strong acid HCℓ Hydrochloric acid 1,3×106 Strong acid H2SO4 Sulphuric acid 1st H+ :1×103 Strong acid 2nd H+ :1×10-2 HNO3 Nitric acid 1×109 Strong acid (COOH)2 Oxalic acid 1st H+ :5,5×10-2 Weak acid nd + -5 2 H :1×10 CH3COOH Ethanoic acid 1,7×10-7 Weak acid 28

Examples 1. Write down the ka or kb expressions of the following acids/bases.  (a) NH3 (aq) + H2O (ℓ) ֖ NH 4 (aq) + OH- (aq) (b) HNO3 (aq) + H2O (ℓ) ֖ H3O+ (aq) + NO-3 (aq) Use the same principles as used when writing kc in chemical equilibrium, i.e. liquids and solids Solution are not included in the expression. Since NH3 is a weak base, its kb ൣNH+4 ൧[OH- ] value will be small. (a) kb = [NH3 ] + - ൣH3 O ൧[NO3 ] (b) ka = Since HNO3 is a strong acid, its ka [HNO3 ] value will be large. 2. The Ka values for two weak acids are as follows: NAME FORMULA KA Oxalic acid (COOH)2 5,5×10-2 Carbonic acid H2CO3 4,5×10-7 Which acid, Oxalic acid or Carbonic acid, is stronger? Give a reason for the answer. Solution Oxalic acid has a higher Ka value/Carbonic acid has a lower Ka value. 29

4. Exercises 4.1 Multiple Choice Questions 1 Which ONE of the following solutions has the highest conductivity? A 0,1 mol∙dm-3 H2CO3 B 0,1 mol∙dm-3 (COOH)2 C 0,1 mol∙dm-3 HNO3 D 0,1 mol∙dm-3 CH3COOH (2) 2 Consider the following acids: ACID A ACID B 0,1 mol·dm-3 hydrochloric acid 1,0 mol·dm-3 ethanoic (acetic) acid Which ONE of the following statements about the acids is CORRECT? A Acid A is weaker and more dilute than acid B. B Acid A is stronger and more concentrated than acid B. C Acid A is stronger but more dilute than acid B D Acid A is weaker and more concentrated than acid B. 3 Ammonium sulphate (NH4)2SO4) is dissolved in water. Which ONE of the following statements regarding the solution which is formed, is CORRECT? A pH = 7 B [H3O+] ∙ [OH-] < 1 X 10-14 C [H3O+] > [OH-] D [H3O+] < [OH-] (2) 4 A solution of ethanoic acid (acetic acid) is titrated against a standard sodium hydroxide solution. Which ONE of the following indicators would be the most suitable for this titration? Indicator pH range of the indicator A Phenolphthalein 8,3 - 10 B Methyl orange 3,1 – 4,4 C Bromothymol blue 6,0 – 7,6 D Universal indicator Changes colour over a wide range of pH values (2) 30

5 Which ONE is the conjugated acid of HC2O4-? A C2O42- B OH- C H2C2O4 D H2C2O4- (2) 6 An aqueous solution that contains more hydronium ions than hydroxide ions is a(n) A Basic solution B Acidic solution C Neutral solution D None of the above (2) 7 According to the Brønsted-Lowry theory, a base …….. A dissociates in aqueous solution B raises the hydrogen ion concentration of an aqueous solution above 1.0 x 10-7 mol.dm-3. C tastes bitter and feels slippery D accepts a proton during a collision with an acid (2) 8 Which ONE of the following species CANNOT act as a Brønsted-Lowry acid and a Brønsted-Lowry base? A H2PO4- B H2O C HSO4- D CH3COOH (2) 31

9 A sodium hydroxide (NaOH) solution of concentration 0,1 mol·dm·3 is added dropwise to an ethanoic acid (CH3COOH) solution of concentration 0,1 mol·dm·3. Which ONE of the following substances will increase in concentration as sodium hydroxide is added dropwise? A H3O+ B OH- C CH3COO- D H2O (2) 10 Consider the reversible reaction represented below: CH3COOH(aq) + OH-(aq) ֖ CH3COO-(aq) + H2O(ℓ) The correct conjugate acid-base pair is: A CH3COOH and CH3COO- B CH3COOH and OH- C CH3COO- and H2O D CH3COOH and H2O (2) 11 The graph (not drawn to scale) for pH versus volume for the titration of an unknown acid with a base was obtained, which ONE of the following combinations of base and acid best fits the graph? 14 9 7 pH 5 3 0 0 10 20 30 40 50 60 70 80 90 100 110 Volume of acid added (cm3) A Strong acid – weak base B Weak acid – strong base C Strong acid – strong base D Weak acid – weak base (2) 32

4.2 Structured Questions 1 Classify each of the following as strong acid or weak acid. Sulphuric acid, hydrochloric acid, carbonic acid. 2 State whether each of the following solutions can act as a Lowry-Brønsted acid or Lowry- Brønsted base. Give a reason for the answer. (a) Ca(OH)2 (b) HBr 3 Give the formula of the conjugate acid or conjugate base for each of the following, indicate whether it is an acid (conjugate) or base (conjugate). (a) H3O+ (b) Cℓ- 4 The Ka value of benzoic acid (C6H5COOH) at 25°C is 6,3 × 10-5. Can the acid be classified as a STRONG ACID or WEAK ACID? Explain. 5 Is the salt K2CO3 acidic or basic? Use a balanced equation to explain/support the answer. 6 A student wishes to prepare 400 cm3 of a 0,1 mol·dm-3 LiOH solution from a 0,25 mol·dm-3 solution. What volume of the 0,25 mol·dm-3 solution must be diluted? 7 The pH of a blood sample was measured using a pH meter, the pH was found to be 7,45. Calculate the concentration of the hydroxide ions in the blood. 8 Sulphuric acid reacts with sodium hydroxide according to the balanced equation given below: H2SO4(aq) + 2NaOH(aq) ֖ Na2SO4(aq) + 2H2O(ℓ) 25 cm3 of 0,3 mol·dm-3 sulphuric acid is mixed with 25 cm3 0,3 mol·dm-3 sodium hydroxide. Calculate the pH of the resulting mixture. 9 A 0,48 g sample of calcium carbonate reacts with 50 cm3 of a 0,11 mol·dm-3 of nitric acid solution according to the balanced equation given below: CaCO3 + 2HNO3 → Ca(NO3)2 + H2O + CO2 Calculate the percentage purity of the CaCO3. 33

4.3 Typical Examination Questions QUESTION 1 1.1 A bottle in a laboratory contains dilute sulphuric acid of unknown concentration. Learners wish to determine the concentration of the sulphuric acid solution. To do this they titrate the sulphuric acid against a standard potassium hydroxide solution of concentration 0.2 mol·dm-3. The balanced equation for the reaction taking place is: 2KOH + H2SO4 → K2SO4 + 2H2O 1.1.1 What is a standard solution? (1) 1.1.2 Calculate the mass of KOH which he must use to make 300 cm3 of a 0.2 mol·dm-3 KOH solution. (3) 1.1.3 Calculate the pH of the 0.2 mol·dm-3 KOH solution. (5) 1.1.4 Which one of the indicators listed in the table below should he use in this titration? Explain your answer. INDICATOR pH Methyl orange 2.9 – 4.0 Methyl red 4.4 – 6.0 Bromothymol blue 6.0 – 10.0 phenolpthalein 8.3 – 10.0 (2) 1.1.5 During the titration the learners find that 15 cm3 of the KOH solution neutralises 20 cm3 of the H2SO4 solution. Calculate the concentration of the H2SO4 solution. (4) 1.2 An impure sample of calcium oxalate, CaC2O4, with a mass of 0.803 g, is titrated with 15.70 cm3 of a 0.101 mol·dm-3 KMnO4. The net reaction is… 2MnO4– + 5C2O42- + 16H+ → 2Mn2+ + 10CO2 + 8H2O Calculate the percentage purity of the CaC2O4 in the original sample. (6) [21] 34

QUESTION 2 2.1 Sulphuric acid is a strong diprotic acid. Define the term strong acid (2) 2.2 HSO-4 can behave either as an acid or a base. 2.2.1 Give the term that is used for substances such as HSO-4 (1) 2.2.2 Write down the balanced equation for the reaction of HSO-4 with H2O in which it acts as an acid. (2) 2.3 Calculate the pH of a Mg(OH)2 solution with a concentration of 0,2 mol∙dm-3. Assume that Mg(OH)2 dissociates completely. (4) 2.4 Vinegar is (3 – 5) % ethanoic acid by mass, with the remaining mass being a solvent. To determine the percentage of ethanoic acid in a vinegar sample, it is titrated against a NaOH solution. The balanced equation for the reaction is: CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + H20 (") A 10 g sample of vinegar is titrated against a 0,309 mol.dm -3 NaOH solution.19,57 cm3 of NaOH is required to reach end point. 2.4.1 Determine the mass of the ethanoic acid in the 10 g sample. (5) 2.4.2 Show by calculation that the percentage of the ethanoic acid is within the given range. (2) The table below provides information of three different indicators: INDICATOR COLOUR CHANGE COLOUR CHANGE pH RANGE Methyl orange Red – yellow 3,0 – 4,4 Bromothymol blue Yellow – blue 6,0 – 7,6 Phenolphthalein Colourless - pink 8,2 – 9,8 2.4.3 Use the information in the table above and choose a suitable indicator to use in the above titration. Give a reason for the answer. (2) 2.4.4 The ethanoate ions that form during the reaction, react with water according to the balanced equation below: CH3COO- (aq) + H2O (") ֖ CH3COOH (aq) + OH- (aq) Write down the name of the process described by the underlined phrase. (1) [19] 35

QUESTION 3 3.1 The acid HF ionises according to the following equation: HF(aq) + H2O(ℓ) ֖ H3O+(aq) + F− (aq) When a 0,10 mol·dm-3 solution of HF is prepared, it is found that the concentration of the F− (aq) ions is 0,018 mol·dm-3. The temperature of the solution is 25 °C. 3.1.1 Is HF a strong acid? (Write down only YES or NO.) (1) 3.1.2 Give a reason for the answer to QUESTION 3.1.1. (2) 3.2 0,50 dm3 of a HCℓ solution of concentration x mol.dm-3 is added to 0,80 dm3 of a 0,25 mol·dm-3 solution of NaOH. At the completion of the reaction, it is found that 0,12 mol of hydroxide ions (OH-) is present in the solution. The balanced equation for the reaction is: HCℓ(aq) + NaOH(aq) → NaCℓ(aq) + H2O(ℓ) 3.2.1 Name the apparatus you will use to measure out the acid solution. (2) 3.2.2 Calculate x, the concentration of the HCℓ solution. (6) 3.2.3 Calculate the concentration of the hydroxide ions (OH-) at the completion of the reaction. (3) 3.2.4 Calculate the pH of the solution at the completion of the reaction. (4) [18] QUESTION 4 4.1 During a lesson on acids and bases, a teacher wrote the following equations on the board: HCℓ(g) + H2O(g) ˢ H3O+(aq) + Cℓ-(aq)……………...………(I) NH3(aq) + H2O(g) ֎ NH4+(aq) + OH-(aq)…………………….(II) 4.1.1 Give the Lowry-Brönsted definition of a base. (2) 4.1.2 Which ONE of the compounds in reactions (I) and (II) is an ampholyte? (2) 4.2 Write down 4.2.1 the meaning of the term diprotic acid. (2) 4.2.2 the formula of a common diprotic acid. (2) [8] 36

5. Solutions to Exercises 5.1 Solutions to Multiple Choice Questions 1 C, It’s a strong acid it ionises completely, there will be more ions in the solution. 2 C, Acid A is a strong acid, Acid B is a weak acid and the concentration of Acid B is higher than the concentration of Acid A 3 C, a strong acid reacted with a weak base. (Hydrolysis) 4 A, a weak acid is titrated against a strong base. 5 C, H+ was ‘added’ to make it an acid 6 B, it has a high concentration of hydronium ions. 7 D, a base is a proton acceptor. 8 D, it can only donate a proton, it cannot accept a proton. 9 B, the hydroxide ions are added. 10 A, CH3COOH ֖ H+ + CH3COO- 11 B, the end point is above the pH of 7, a strong base reacted with a weak acid. 5.2 Solutions to Structured Questions 1 Sulphuric acid – strong acid 3 Hydrochloric acid – strong acid 3 Carbonic acid – weak acid 3 (3) 2 (a) It is a base3, it can accept a proton3, OH- + H+ ֖ H2O (b) It is an acid3, it can donate a proton3, HBr + H2O ֖ H3O+ + Br- (4) 3 (a) H2O3 (conjugate base) a proton was ‘removed’, i.e., can accept a proton3 (b) HCℓ3 (conjugate acid) a proton was ‘added’, i.e., can donate a proton. 3 (4) 4 Weak acid3, it has a low ka value3, it ionises incompletely. (2) 5 CO32- + H2O 3֖ CO3- + OH- 3 A strong base (OH-) was formed3 (A strong base reacted with a weak acid) (3) 6 c1 V1 =c2 V2  ሺ0,1ሻሺ400ሻ3=ሺ0,25ሻV2 3 ሺ0,1ሻሺ400ሻ V2 =  0,25 =160cm3 3 37

7 pH = -log [H3O+]3 7,453= -log [H3O+] [H3O+] = 10-7,45 = 3,548 × 10-8 mol·dm-33 [H3O+][OH-] = 1×10-14 (3,548 × 10-8) [OH-] = 1×10-143 [OH-] = 2,818 × 10-7 mol·dm-33 (5) 8 nሺNaOHሻ=cV3 nሺH2 SO4 ሻ=cV (10) =ሺ0,2ሻሺ0,025ሻ3 =ሺ0,3ሻሺ0,025ሻ3 =0,005 mol =0,0075 mol NaOH is the limiting reactant. n(H2 SO4 )reacted 1 = 3 nሺNaOHሻ 2 1 Use mol ratios to find the n(H2 SO4 )reacted = ×0,005 number of moles H2SO4 2 reacted. =0,0025 mol n(H2SO4)excess = 0,0075 – 0,00253 = 0,005 mol n cሺH2 SO4 ሻ= 3 V 0,005 =  The new volume is now 50 cm3, i.e. 0,053 =0,1 mol 25 + 25 =50 H2SO4 + 2H2O ֖ 2H3O+ + SO42- 0,1 2(0,1) [H3O+] = 0.2 mol·dm-3 pH = -log [H3O+]3 = -log (0,2) 3 = 0,73 9 nሺHଷ ሻ=cV3 =ሺ0,ͳͳሻሺ0,0ͷሻ3 =0,00ͷͷ mol n(CaCO ) 1  ሺ ሻ 38

1 n(CaCO3 )= 3×0,0055 2 =0,00275 mol m n=  M m 0,00275=  1003 m=0,275 g mass pure substance % purity= ×1003 mass impure substance 0,275 = 3×100 0,48 =57,292 %3 5.3 Solutions to Typical Examination Questions QUESTION 1 1.1.1 A solution of known concentration.3 (1) 1.1.2 OPTION 1 OPTION 2 ሺ ሻ ൌ ͷ͸‰ ή ‘Žିଵ  ሺ ሻ ൌ ͷ͸‰ ή ‘Žିଵ  ൌ 3 ൌ Ͳǡʹšͷ͸šͲǡ͵3 ݊ ൌ ܸܿ 3 ൌ ͵ǡ͵͸‰3 ൌ Ͳǡʹ ൈ Ͳǡ͵ ൌ ͲǡͲ͸ ݉ ൌ ݊‫ܯ‬ ൌ ͲǡͲ͸ ൈ ͷ͸3 ൌ ͵ǡ͵͸‰3 (3) 1.1.3 0,2 mol KOH yields 0,2 mol OH- Kw = [H3O+][OH-]3 10-14 = [H3O+](0,2) [H3O+] = 5 x 10-143 pH = -log[H3O+]3 = -log(5 x 10-14) 3 = 13,33 39

1.1.4 Bromothymol blue;3H2SO4 is a strong acid and KOH is a strong base3 The equivalence (2) point will be at approximately pH = 7 which is the endpoint of bromothymol blue. 1.1.5 OPTION 1 na Ca Va ൬ ൰ =ቆ 3ቇ nb Cb Vb 1 Ca ×20 ൬ ൰ 3= 3 2 0,2×15 Ca =0,075mol∙dm-3 3 OPTION 2 nሺNaOHሻ=cV 3 =0,2×0,015 =0,003 mol nሺH2 SO4 ሻ 1 = nሺNaOHሻ 2 1 nሺH2 SO4 ሻ= ሺ0,003ሻ3 2 =0,0015 mol n c= V 0,0015 = 3 0,020 =0,075 mol∙dm-3 3 (4) 1.2 n =cV3 Moles of MnO4- : ͳͷǡ͹Ͳ ൈ ͲǡͳͲͳ ൌ ͳͲͲͲ ൌ ͳǡͷͺͷ͹ ൈ ͳͲିଷ mol3 ହ Moles of C2O42- = ൈ ͳǡͷͺ͹ͷ ൈ ͳͲିଷ ଶ ൌ ͵ǡͻ͸Ͷ͵ ൈ ͳͲିଷ ݉‫݈݋‬3 n(CaC2O4) = n(C2O42-) m=nM Mass of CaC2O4 = ͵ǡͻ͸Ͷ͵ ൈ ͳͲିଷ  ൈ ͳʹͺ3 =0,50743g ଴ǡହ଴଻ସଷ Percentage of CaC2O4 = ൈ ͳͲͲ3 ଴ǡ଼଴ଷ ൌ ͸͵ǡͳͻΨ3 (6) [21] 40

QUESTION 2 2.1 A strong acid is an acid that ionises completely in water 3 to form a high concentration of H3O+ ions.3 (2) 2.2.1 Ampholyte 3 (1) 2.2.2 HSO4- + H2O ֖ H3O+ + SO42- 33 (2) 2.3 OPTION 1 [OH-] = 0,4 mol∙dm-3 3 (ratio 1:2) ( Mg(OH)2 → Mg2+ + 2OH- ) [H3O+] [OH-] = 1 x 10-14 3 [H3O+] (0,4) = 1 x 10-14 1×1014 [H3O+] =  0,4 = 2,5 x 10-14 mol.dm-3 pH = – log [H3O+] 3 = – log (2,5 x 10-14) = 13,6 3 OPTION 2 [OH-] = 0,4 mol∙dm-3 3 (ratio) pOH = -log[OH-]3= -log(0,4) = 0,4 pH + pOH = 14 3 (4) ‫׵‬pH = 13,6 3 2.4.1 CH3COOH + NaOH → CH3COONa + H2O c = 0,309 mol˜dm-3 V = 0,01957 dm-3 nNaOH = c x V3 = 0,309 x 0,01957 = 6,05 x 10-3 mol3 m = n x MCH3 COOH 3(ratio 1:1) = (6,05 x 10-3) x 60 3 = 0,363 g 3 (5) 41

2.4.2 % = 0,363 / 10 x 100 = 3,63% 3 Yes 3 (2) 2.4.3 Phenolphthalein3 A weak acid reacted with a strong base 3. OR 6  13 pH of solution ≈ ≈ 9,5 2 (2) 2.4.4 Hydrolysis3 (1) [19] QUESTION 3 3.1.1 No3 (1) 3.1.2 The acid did not ionise completely/it ionised incompletely, since 0,018 mol·dm-3 < 0,10 mol·dm-3.33 (2) 3.2.1 Burette 33 (2) 42

3.2.2 HCℓ(aq) + NaOH(aq) → NaCℓ(aq) + H2O(ℓ) ca= ? cb=0,25 Va=0,50 Vb=0,80 n(NaOH)initial = cV3 =(0,25)(0,80)3 = 0,2 mol n(NaOH)reacted= n(NaOH)initial - n(NaOH)excess = 0,20 – 0,123 = 0,08 mol n(HCℓ) = n(NaOH)reacted = 0,08 mol3 n c= V 0,08 = 3 0,50 =1,6 mol∙dmǦ3 3 (6) 3.2.3 n c= 3 V 0,12 = 1.33 =0,0923 mol∙dmǦ3 3 (3) 3.2.4 [H3O+][OH-] = 1×10-14 [H3O](0,0923) = 1×10-14 [H3O+] = 1,0834×10-13 mol·dm-3 pH = -log[H3O+] = -log (1,0834×10-13) = 12,965 (4) [16] QUESTION 4 4.1.1 Is a proton donor (2) 4.1.2 H2O (2) 4.2.1 An acid which can donate a maximum of two protons.33 (2) 4.2.2 H2SO433 (2) [8] 43

6. Examination guidance (Acids and Bases) x In Acids and Bases a lot of stoichiometry which was learned in Grades 10 and 11 is used, for instance, calculating the number of moles if given masses, concentration, number of particles or gas volumes and using ratios in calculations of product formed, m V N n reactant used, etc. You must be able to use the formulae, n= , n= , n= , c= to M Vm NA V calculate the number of moles. You must revise your work on limiting reactants because some questions might involve limiting reactants. x When doing multistep calculations e.g., when calculating the number of moles of NaOH, the formula should be as follows: n(NaOH) = cV NOT just n = cV x When there are excess number of moles, initial number of moles, number of moles used/reacted, number of moles remaining, you should write a descriptor/label, e.g., n(NaOH)initial, n(NaOH)reacted/used, n(NaOH) excess/remaining, this will make it easy to identify what you have and what must be calculated/determined. x When substituting values given in the question paper they should not be rounded off, they should be used as they were given in the question paper. E.g. if given 0,00687, substitute 0,00687 not 0,01. m x The formula c= should only be used when dealing with solutions, it should NOT be VM m n used for solids. Preferably use n= M first and then c= V (only for solutions). x Always calculate the number of moles using relevant formulae, use ratios to find the unknown number of moles, one will always earn a mark for calculating the number of moles and using the ratios (Easy to Score Marks). x To calculate the antilog, i.e., to calculate the concentration of [H3O+] in the pH formula, use 10 to the power of, e.g. 10-7,6. Most calculators have the button , which should be used, i.e., press 2nd F/SHIFT then log 10 44

7. Study and Examination Tips: Physical Sciences x Always copy formulae correctly from the data sheet, do not rely on your memory. x Substitute directly into the original formula, do not manipulate the formula before substituting. x Rounding off should only be done at the final answer of a calculation. One should not round off in each step as it leads to an incorrect answer. The instruction in the paper reads a ‘MINIMUM of TWO decimal digits’ and NOT a ‘MAXIMUM of TWO decimal digits’, you may leave your answer with more than TWO decimal digits. x Always start with a question you feel you will manage or in which you will score good marks. x Learn the definitions as given in the examination guidelines. x Practice definitions by writing all the terms down and then try to write the correct definitions without referring to your examination guidelines. Do that until you can write the definitions correctly without omitting key words. 45

8 References The following documents were used in the Development of this booklet: x Physical Sciences Grade 10-12 CAPS document (DBE) x Physical Sciences Examination Guidelines 2021 (DBE) x Previous Grade 12 Question Papers (DBE) x Grade 12 Preparatory Examination Papers (PEDs) 9 Acknowledgement The Department of Basic Education (DBE) gratefully acknowledges the following officials for giving up their valuable time and families and for contributing their knowledge and expertise to develop this resource booklet for the children of our country, under very stringent conditions of COVID-19: Writer: Mr Stanley M Mndawe (Subject Specialist, PED: Mpumalanga) Reviewers: Mrs Thandeka P Hlongwane, Mr Sanele G Mncwango, Mr Magezi T Baloyi, Miss Linda Thyalibhongo DBE Subject Specialist: Ms JSK Maharaj (Veena) The development of the Study Guide was managed and coordinated by Ms Cheryl Weston and Dr Sandy Malapile 46

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