acids and bases	

	Name:	
________________________	

Class:	
________________________	

Date:	
________________________	

 

 	
Time:	384 minutes	
Marks:	370 marks	
Comments:		

Q1.          Iodine and propanone react in acid solution according to the equation
I2   +   CH3COCH3   →   CH3COCH2I   +   HI
The rate equation for the reaction is found to be
rate = k [CH3COCH3][H+]
(a)     Deduce the order of reaction with respect to iodine and the overall order of reaction.
Order with respect to iodine .........................................................................
Overall order ................................................................................................
(2)

(b)     At the start of the experiment, the rate of reaction was found to be 2.00 × 10–5 mol dm–3 s–1 when the concentrations of the reactants were as shown below.
 
Reactant	Concentration / mol dm–3	
CH3COCH3	1.50	
I2	2.00 × 10–2	
H+	3.00 × 10–2	

Use these data to calculate a value for the rate constant and deduce its units.
Rate constant ...............................................................................................
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Units .............................................................................................................
(3)

(c)     How can you tell that H+ acts as a catalyst in this reaction?
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(2)

(d)     Calculate the initial rate of reaction if the experiment were to be repeated at the same temperature and with the same concentrations of iodine and propanone as in part (b) but at a pH of 1.25
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(3)
(Total 10 marks)

 

Q2.A student was given a task to determine the percentage purity of a sample of salicylic acid. The method used by the student to prepare a solution of salicylic acid is described below.
•        0.500 g of an impure sample of salicylic acid was placed in a weighing bottle.
•        The contents were tipped into a beaker and 100 cm3 of distilled water were added.
•        Salicylic acid does not dissolve well in cold water so the beaker and its contents were heated gently until all the solid had dissolved.
•        The solution was poured into a 250 cm3 graduated flask and made up to the mark with distilled water.
(a)     Give two additional instructions that would improve this method for making up the salicylic acid solution.
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(2)

(b)     The pH of this solution was measured and a value of 2.50 was obtained. 
Calculate the concentration of salicylic acid in this solution. 
Assume that salicylic acid is the only acid in this solution. The Ka for salicylic acid is 1.07 × 10–3 mol dm–3. You may represent salicylic acid as HA. 
Show your working.
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(3)

(c)     Use your answer to part (b) to calculate the mass of salicylic acid (Mr = 138.0) present in the original sample. 
(If you were unable to complete the calculation in part (b), assume that the concentration of salicylic acid is 8.50 × 10–3 mol dm–3. This is not the correct answer.)
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(2)

(d)     Use your answer to part (c) to calculate the percentage purity of the salicylic acid used to make the solution. 
(If you were unable to complete the calculation in part (c), assume that the mass of salicylic acid is 0.347 g. This is not the correct answer.)
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(1)
(Total 8 marks)

Q3.          The sketch below shows the change in pH when a 0.200 mol dm–3 solution of sodium hydroxide is added from a burette to 25.0 cm3 of a 0.150 mol dm-3 solution of the weak acid HA at 25 °C.

(a)     The volume of sodium hydroxide solution added at the equivalence point is x cm3.
Calculate the value of x.
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(2)

(b)     (i)      Define the term pH.
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(ii)     The pH at the equivalence point is c. Suggest a value for c.
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(iii)     Identify a suitable indicator for detecting the equivalence point of the titration.
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(3)

(c)     The value of Kc for the weak acid HA at 25 °C is 2.75 × 10–5 mol dm–3.
(i)      Explain the term weak as applied to the acid HA.
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(ii)     Write an expression for Ka for the acid HA.
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(iii)     Calculate the pH of the 0.150 mol dm–3 solution of acid HA before any sodium hydroxide is added, i.e. the pH at point a.
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(5)

(d)     Calculate the pH of the solution formed when cm3 of the 0.200 mol dm–3 solution of sodium hydroxide are added to 25.0 cm3 of the 0.150 mol dm–3 solution of HA, i.e. the pH at point b.
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(3)
(Total 13 marks)

 

Q4.          (a)     By reference to the forces between molecules, explain why ammonia is very soluble in water.
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(2)

(b)     Aqueous solutions of ammonia have a pH greater than 7.
(i)      Write an equation for the reaction of ammonia with water.
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(ii)     Explain why the pH of a solution containing 1.0 mol dm–3 of ammonia is less than 
14 at 298 K.
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(3)

(c)     An ammonium ion in aqueous solution can behave as a Brønsted–Lowry acid. State what is meant by the term Brønsted–Lowry acid.
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(1)

(d)     State what is meant by the term buffer solution. Identify a reagent which could be added to a solution of ammonia in order to form a buffer solution.
Buffer solution ..............................................................................................
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Reagent ........................................................................................................
(3)

(e)     An acidic buffer solution is obtained when sodium ethanoate is dissolved in aqueous ethanoic acid.
(i)      Calculate the pH of the buffer solution formed at 298 K when 0.125 mol of sodium ethanoate is dissolved in 250 cm3 of a 1.00 mol dm–3 solution of ethanoic acid.
The acid dissociation constant, Ka, for ethanoic acid is 1.70 × 10–5 mol dm–3 at 298 K.
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(ii)     Write an ionic equation for the reaction which occurs when a small volume of dilute hydrochloric acid is added to this buffer solution.
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(5)
(Total 14 marks)

 

Q5.          (a)     At 50°C, the ionic product of water, Kw, has the value 5.48 × 10–14 mol2 dm–6.
(i)      Define the term Kw
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(ii)     Define the term pH
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(iii)     Calculate the pH of pure water at 50 °C. Explain why pure water at 50 °C is still neutral even though its pH is not 7.
Calculation ..........................................................................................
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Explanation ..........................................................................................
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(5)

(b)     At 25°C, Kw has the value 1.00 × 10–14 mol2 dm–6. Calculate the pH at 25 °C of
(i)      a 0.150 mol dm–3 solution of sodium hydroxide,
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(ii)     the solution formed when 35.0 cm3 of this solution of sodium hydroxide is mixed with 40.0 cm3 of a 0.120 mol dm–3 solution of hydrochloric acid.
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(8)

(c)     In a 0.150 mol dm–3 solution of a weak acid HX at 25 °C, 1.80% of the acid molecules are dissociated into ions.
(i)      Write an expression for Ka for the acid HX.
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(ii)     Calculate the value of Ka for the acid HX at this temperature and state its units.
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(5)
(Total 18 marks)

 

Q6.          (a)     Titration curves labelled A, B, C and D for combinations of different acids and bases are shown below. All solutions have a concentration of 0.1 mol dm–3.

(i)      Select from A, B, C and D the curve produced by the addition of
ammonia to 25 cm3 of hydrochloric acid  ................................................
ethanoic acid to 25 cm3 of sodium hydroxide  .........................................
sodium hydroxide to 25 cm3 of hydrochloric acid  ...................................

(ii)     A table of acid–base indicators and the pH ranges over which they change colour is shown below.
 
 	Indicator	pH range	 	
 	Thymol blue	1.2 – 2.8	 	
 	Bromophenol blue	3.0 – 4.6	 	
 	Methyl red	4.2 – 6.3	 	
 	Cresolphthalein	8.2 – 9.8 	 	
 	Thymolphthalein	9.3 – 10.5 	 	

Select from the table an indicator which could be used in the titration which produces curve A but not in the titration which produces curve B.
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(4)

(b)     (i)      Write an expression for the term pH.
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(ii)     A solution of potassium hydroxide has a pH of 11.90 at 25°C. Calculate the concentration of potassium hydroxide in the solution.
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(4)

(c)     The acid dissociation constant, Ka, for propanoic acid has the value of 
1.35 × 10–5mol dm–3 at 25 °C.

In each of the calculations below, give your answer to 2 decimal places.
(i)      Calculate the pH of a 0.117 mol dm–3 aqueous solution of propanoic acid.
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(ii)     Calculate the pH of a mixture formed by adding 25 cm3 of a 0.117 mol dm–3 aqueous solution of sodium propanoate to 25 cm3 of a 0.117 mol dm–3 aqueous solution of propanoic acid.
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(5)
(Total 13 marks)

 

Q7.          The value of the acid dissociation constant, Ka, for the weak acid HA, at 298 K, is
1.45 × 10–4 mol dm–3.
(a)     Write an expression for the term Ka for the weak acid HA.
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(1)

(b)     Calculate the pH of a 0.250 mol dm–3 solution of HA at 298 K.
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(4)

(c)     A mixture of the acid HA and the sodium salt of this acid, NaA, can be used to prepare a buffer solution.
(i)      State and explain the effect on the pH of this buffer solution when a small amount of hydrochloric acid is added.
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(ii)     The concentration of HA in a buffer solution is 0.250 mol dm–3. Calculate the concentration of A– in this buffer solution when the pH is 3.59
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(6)
(Total 11 marks)

 

Q8.          The pH curve shown below was obtained when a 0.150 mol dm–3 solution of sodium hydroxide was added to 25.0 cm3 of an aqueous solution of a weak monoprotic acid, HA.

(a)     Use the information given to calculate the concentration of the acid.
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(2)

(b)     (i)      Write an expression for the acid dissociation constant, Ka, for HA.
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(ii)     Write an expression for pKa
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(iii)     Using your answers to parts (b)(i) and (b)(ii), show that when sufficient sodium hydroxide has been added to neutralise half of the acid,
pH of the solution = pKa for the acid HA
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(4)

(c)     Explain why dilution with a small volume of water does not affect the pH of a buffer solution.
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(2)

(d)     (i)      Calculate the change in pH when 0.250 mol dm–3 hydrochloric acid is diluted with water to produce 0.150 mol dm–3 hydrochloric acid.
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(ii)     Calculate the volume of water which must be added to 30.0 cm3 of 0.250 mol dm–3  hydrochloric acid in order to reduce its concentration to 0.150 mol dm–3.
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(4)
(Total 12 marks)

 

Q9.In this question, give all values of pH to two decimal places.
Calculating the pH of aqueous solutions can involve the use of equilibrium constants such as Kw and Ka
Kw is the ionic product of water. The value of Kw is 5.48 × 10–14 mol2 dm–6 at 50 °C.
(a)     (i)      Write an expression for pH.
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(1)

(ii)     Write an expression for Kw
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(1)

(b)     (i)      Calculate the pH of pure water at 50 °C.
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(2)

(ii)     Suggest why this pure water is not acidic.
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(1)

(iii)     Calculate the pH of 0.140 mol dm–3 aqueous sodium hydroxide at 50 °C.
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(3)

(c)     Calculate the pH of the solution formed when 25.0 cm3 of 0.150 mol dm–3 aqueous sulfuric acid are added to 30.0 cm3 of 0.200 mol dm–3 aqueous potassium hydroxide at 25 °C. Assume that the sulfuric acid is fully dissociated.
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(6)
(Total 14 marks)

Q10.In this question, give all pH values to 2 decimal places.
(a)     (i)      Write expressions for the ionic product of water, Kw, and for pH.
KW = ….................................................................................................
pH = ....................................................................................................

(ii)     At 318 K, the value of Kw is 4.02 × 10–14 mol2 dm–6 and hence the pH of pure water is 6.70
State why pure water is not acidic at 318 K.
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(iii)     Calculate the number of moles of sodium hydroxide in 2.00 cm3 of 0.500 mol dm–3 aqueous sodium hydroxide.
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(iv)    Use the value of Kw given above and your answer to part (a)(iii) to calculate the pH of the solution formed when 2.00 cm3 of 0.500 mol dm–3 aqueous sodium hydroxide are added to 998 cm3 of pure water at 318 K.
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(6)

(b)     At 298 K, the acid dissociation constant, Ka, for propanoic acid, CH3CH2COOH, has the value 1.35 × 10–5mol dm–3.
(i)      Write an expression for Ka for propanoic acid.
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(ii)     Calculate the pH of 0.125 mol dm–3 aqueous propanoic acid at 298 K.
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(4)

(c)     Sodium hydroxide reacts with propanoic acid as shown in the following equation.
NaOH + CH3CH2COOH → CH3CH2COONa + H2O
A buffer solution is formed when sodium hydroxide is added to an excess of aqueous propanoic acid.
(i)      Calculate the number of moles of propanoic acid in 50.0 cm3 of 0.125 mol dm–3 aqueous propanoic acid.
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(ii)     Use your answers to part (a)(iii) and part (c)(i) to calculate the number of moles of propanoic acid in the buffer solution formed when 2.00 cm3 of 0.500 mol dm–3 aqueous sodium hydroxide are added to 50.0 cm3 of 0.125 mol dm–3 aqueous propanoic acid.
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(iii)     Hence calculate the pH of this buffer solution at 298 K.
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(6)
(Total 16 marks)

Q11.          (a)     A sample of hydrochloric acid has a pH of 2.34
Write an expression for pH and calculate the concentration of this acid.
pH ................................................................................................................
Concentration ..............................................................................................
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(2)

(b)     A 0.150 mol dm–3 solution of a weak acid, HX, also has a pH of 2.34
(i)      Write an expression for the acid dissociation constant, Ka, for the acid HX.
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(ii)     Calculate the value of Ka for this acid and state its units.
Calculation .........................................................................................
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Units ....................................................................................................
(iii)     Calculate the value of pKa for the acid HX. Give your answer to two decimal places.
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(5)

(c)     A 30.0 cm3 sample of a 0.480 mol dm–3 solution of potassium hydroxide was partially neutralised by the addition of 18.0 cm3 of a 0.350 mol dm–3 solution of sulphuric acid.
(i)      Calculate the initial number of moles of potassium hydroxide.
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(ii)     Calculate the number of moles of sulphuric acid added.
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(iii)     Calculate the number of moles of potassium hydroxide remaining in excess in the solution formed.
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(iv)    Calculate the concentration of hydroxide ions in the solution formed.
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(v)     Hence calculate the pH of the solution formed. Give your answer to two decimal places.
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(6)
(Total 13 marks)

 

Q12.          The hydrolysis of methyl propanoate was studied in acidic conditions at 25°C and the rate equation was found to be
rate = k[CH3CH2COOCH3][H+]
(a)     Use the data below to calculate the value of the rate constant, k, at this temperature.
Deduce its units.
Initial rate of reaction / 
mol dm–3 s–1	Initial concentration of methyl propanoate / mol dm–3	Initial concentration of hydrochloric acid / mol dm–3	
1.15 × 10–4	0.150	0.555	

Rate constant ...............................................................................................
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Units .............................................................................................................
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(3)

(b)     The reaction in part (a) was repeated at the same temperature, but water was added so that the volume of the reaction mixture was doubled. Calculate the initial rate of reaction under these conditions.
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(1)

(c)     A third experiment was carried out at a different temperature. Some data from this experiment are shown in the table below.
Initial rate of reaction / mol dm–3 s–1	Value of rate constant at this different temperature	Initial methyl propanoate / mol dm–3	
4.56 × 10–5	8.94 × 10–4	0.123	

Calculate the initial pH of the reaction mixture. Give your answer to two decimal places.
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(3)
(Total 7 marks)

 

Q13.          Buffer solutions are important in biological systems and in industry to maintain almost constant pH values.
(a)     In the human body, one important buffer system in blood involves the hydrogencarbonate ion, , and carbonic acid, H2CO3, which is formed when carbon dioxide dissolves in water.
(i)      Use the following equation to explain how this buffer maintains a constant pH of 7.41 even if a small amount of acid enters the bloodstream.
H2CO3(aq)      H+(aq)   +    (aq)
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(ii)     In a sample of blood with a pH of 7.41, the concentration of (aq) ions is 2.50 × 10–2 mol dm–3 and the concentration of H2CO3(aq) is 1.25 × 10–3 mol dm–3.
Calculate a value for the acid dissociation constant, Ka, for carbonic acid at this temperature.
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(5)

(b)     In industry, the pH of a solution used to dye cloth must be controlled or else the colour varies.
A solution of dye in a beaker is buffered by the presence of ethanoic acid and sodium ethanoate. In the solution, the concentration of ethanoic acid is 0.15 mol dm–3 and the concentration of sodium ethanoate is 0.10 mol dm–3. The value of Ka for ethanoic acid is 1.74 × 10–5 mol dm–3 at 298 K.
(i)      A 10.0 cm3 portion of 1.00 mol dm–3 hydrochloric acid is added to 1000 cm3 of this buffered solution.
Calculate the number of moles of hydrochloric acid added.
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(ii)     Calculate the number of moles of ethanoic acid and the number of moles of sodium ethanoate in the solution after addition of the hydrochloric acid.
Mol of ethanoic acid after addition ......................................................
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Mol of sodium ethanoate after addition ..............................................
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(iii)     Hence calculate the pH of this new solution. Give your answer to 2 decimal places.
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(6)
(Total 11 marks)

 

Q14.          Indicators and pH curves can be used to determine the end point in a titration.
(a)     The pH curves labelled J, K, L and M for combinations of different acids and bases are shown below. All solutions have a concentration of 0.1 mol dm–3.

(i)      Select from J, K, L and M the curve produced by the addition of
ammonia to 25 cm3 of hydrochloric acid .............................................
ethanoic acid to 25 cm3 of sodium hydroxide ......................................
sodium hydroxide to 25 cm3 of hydrochloric acid ................................

(ii)     A table of acid–base indicators and the pH ranges over which they change colour is shown below.
 
        Indicator                      pH range	
        Thymol blue                  1.2 – 2.8
        Bromophenol blue        3.0 – 4.6
        Methyl red                     4.2 – 6.3
        Cresolphthalein             8.2 – 9.8
        Thymolphthalein           9.3 – 10.5	

Select from the list above an indicator which could be used in the titration which produces curve J but not in the titration which produces curve K.
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(4)

(b)     The acid dissociation constant, Ka, for the weak acid, ethanoic acid, has a value of 1.74 × 10–5 mol dm–3 at 25 °C.
Ka = 
(i)      Write an expression for the term pH.
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(ii)     Calculate the pH of a 0.15 mol dm–3 solution of ethanoic acid. Give your answer to 
2 decimal places.
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(4)
(Total 8 marks)

 

Q15.          This question is about the pH of some solutions containing potassium hydroxide and ethanoic acid.
Give all values of pH to 2 decimal places.
(a)     (i)      Write an expression for pH.
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(1)

(ii)     Write an expression for the ionic product of water, Kw
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(1)

(iii)     At 10 °C, a 0.154 mol dm–3 solution of potassium hydroxide has a pH of 13.72. Calculate the value of Kw at 10 °C.
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(2)

(b)     At 25 °C, the acid dissociation constant Ka for ethanoic acid has the value
1.75 × 10–5 mol dm–3.
(i)      Write an expression for Ka for ethanoic acid.
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(1)

(ii)     Calculate the pH of a 0.154 mol dm–3 solution of ethanoic acid at 25 °C.
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(3)

(c)     At 25 °C, the acid dissociation constant Ka for ethanoic acid has the value
1.75 × 10–5 mol dm–3.
(i)      Calculate the pH of the solution formed when 10.0 cm3 of 0.154 mol dm–3 potassium hydroxide are added to 20.0 cm3 of 0.154 mol dm–3 ethanoic
acid at 25 °C.
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(4)

(ii)     Calculate the pH of the solution formed when 40.0 cm3 of 0.154 mol dm–3 potassium hydroxide are added to 20.0 cm3 of 0.154 mol dm–3 ethanoic acid at 25 °C.
At 25 °C, Kw has the value 1.00 × 10–14 mol2 dm–6.
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(4)
(Total 16 marks)

Q16.          This question is about the pH of several solutions.
          Give all values of pH to 2 decimal places.
(a)     (i)      Write an expression for pH.
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(1)

(ii)     Calculate the pH of 0.154 mol dm–3 hydrochloric acid.
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(1)

(iii)     Calculate the pH of the solution formed when 10.0 cm3 of 0.154 mol dm–3 hydrochloric acid are added to 990 cm3 of water.
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(2)

(b)     The acid dissociation constant, Ka, for the weak acid HX has the value
4.83 × 10–5 mol dm–3 at 25 °C.
A solution of HX has a pH of 2.48
Calculate the concentration of HX in the solution.
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(4)

(c)     Explain why the pH of an acidic buffer solution remains almost constant despite the addition of a small amount of sodium hydroxide.
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(2)

(d)     The acid dissociation constant, Ka, for the weak acid HY has the value
1.35 × 10–5 mol dm–3 at 25 °C.
A buffer solution was prepared by dissolving 0.0236 mol of the salt NaY in
50.0 cm3 of a 0.428 mol dm–3 solution of the weak acid HY
(i)      Calculate the pH of this buffer solution.
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(4)

(ii)     A 5.00 × 10–4 mol sample of sodium hydroxide was added to this buffer solution.
Calculate the pH of the buffer solution after the sodium hydroxide was added.
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(4)
(Total 18 marks)

Q17.Iron(II) sulfate is used to kill weeds in garden lawns. It is a by-product of the manufacture of steel. 
When a lawn is treated with iron(II) sulfate, the iron(II) ions are oxidised to form iron(III) ions.
Iron(III) ions are acidic in aqueous solution as shown by the following equation.
[Fe(H2O)6]3+(aq)   [Fe(H2O)5(OH)]2+(aq) + H+(aq)
In an experiment, a calibrated pH meter was used to measure the pH of an iron(III) salt in solution. At 20 °C the pH of a 0.100 mol dm–3 solution of iron(III) sulfate was found to be 1.62.
(a)     Explain briefly why a pH meter should be calibrated before use.
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(1)

(b)     Write an expression for the equilibrium constant, Ka, for the dissociation of iron(III) ions in aqueous solution.
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(1)

(c)     Use your answer from part (b) to calculate the value of Ka for this reaction at 20 °C. 
Give your answer to the appropriate precision. Show your working.
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(4)

(d)     Name the substance that is most likely to oxidise the iron(II) ions when iron(II) sulfate is used as a weed killer.
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(1)

(e)     Suggest a value for the pH of a 0.100 mol dm–3 solution of iron(II) sulfate.
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(1)
(Total 8 marks)

Q18.This question is about alkalis and carboxylic acids.
In this question, all data are quoted at 25 °C.
(a)     Carboxylic acids are weak acids.
State the meaning of the term weak as applied to carboxylic acids.
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(1)

(b)     Write an equation for the reaction of propanoic acid with sodium carbonate.
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(1)

(c)     Calculate the pH of a 0.0120 mol dm–3 solution of calcium hydroxide.
The ionic product of water Kw = 1.00 × 10–14 mol2 dm–6.
Give your answer to 2 decimal places.
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(3)

(d)     The value of the acid dissociation constant Ka for benzenecarboxylic acid (C6H5COOH) is 6.31 × 10–5 mol dm–3.
(i)      Write an expression for the acid dissociation constant Ka for benzenecarboxylic acid.
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(1)

(ii)     Calculate the pH of a 0.0120 mol dm–3 solution of benzenecarboxylic acid.
Give your answer to 2 decimal places.
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(3)

(iii)     A buffer solution with a pH of 4.00 is made using benzenecarboxylic acid and sodium benzenecarboxylate.
Calculate the mass of sodium benzenecarboxylate (Mr = 144.0) that should be dissolved in 1.00 dm3 of a 0.0120 mol dm–3 solution of benzenecarboxylic acid to produce a buffer solution with a pH of 4.00
The value of the acid dissociation constant Ka for benzenecarboxylic acid (C6H5COOH) is 6.31 × 10–5 mol dm–3.
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(5)

(e)     Two solutions, one with a pH of 4.00 and the other with a pH of 9.00, were left open to the air.
The pH of the pH 9.00 solution changed more than that of the other solution.
Suggest what substance might be present in the air to cause the pH to change.
Explain how and why the pH of the pH 9.00 solution changes.
Substance present in air ................................................................................
Explanation ....................................................................................................
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(3)
(Total 17 marks)

Q19.This question involves calculations about two strong acids and one weak acid.
All measurements were carried out at 25 °C.
(a)     A 25.0 cm3 sample of 0.0850 mol dm–3 hydrochloric acid was placed in a beaker and 100 cm3 of distilled water were added.
Calculate the pH of the new solution formed.
Give your answer to 2 decimal places.
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(2)

(b)     HX is a weak monobasic acid.
(i)      Write an expression for the acid dissociation constant, Ka, for HX.
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(1)

(ii)     The pH of a 0.0850 mol dm–3 solution of HX is 2.79
Calculate a value for the acid dissociation constant, Ka, of this acid.
Give your answer to 3 significant figures.
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(3)

(c)     A 25.0 cm3 sample of 0.620 mol dm–3 nitric acid was placed in a beaker and 38.2 cm3 of 0.550 mol dm–3 aqueous sodium hydroxide were added.
Calculate the pH of the solution formed.
Give your answer to 2 decimal places.
The ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 at 25 °C.
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(6)

(Total 12 marks)

 

Q20.Nitric acid (HNO–3) is a strong acid. Ethanoic acid (CH3COOH) is a weak acid.
(a)     Write an equation to show how ethanoic acid behaves as a weak acid in its reaction with water.
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(1)

(b)     When pure ethanoic acid reacts with pure nitric acid, ethanoic acid acts as a base.
Write an equation for this reaction.
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(1)

(c)     Two beakers, A and B, each contain 100.0 cm–3 of 0.0125 mol dm−3 nitric acid.
(i)      Calculate the pH of the solution formed after 50.0 cm–3 of distilled water are added to beaker A.
Give your answer to 2 decimal places.
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(2)
(ii)     Calculate the pH of the solution formed after 50.0 cm3 of 0.0108 mol dm−3 aqueous sodium hydroxide are added to beaker B.
Give your answer to 2 decimal places.
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(4)

(d)     A third beaker, C, contains 100.0 cm3 of 0.0125 mol dm−3 ethanoic acid.
The acid dissociation constant Ka for ethanoic acid has the value 1.74 × 10−5 mol dm−3 at 25 °C.
(i)      Write an expression for Ka for ethanoic acid and use it to calculate the pH of the ethanoic acid solution in beaker C.
Show your working. Give your answer to 2 decimal places.
Ka ..............................................................................................................
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Calculation ................................................................................................
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(4)
(ii)     Aqueous sodium hydroxide is added to beaker C until the pH of the solution becomes 4.84.
Name the salt formed in the reaction of ethanoic acid with sodium hydroxide.
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(1)
(iii)     Calculate the value of  in the solution with the pH of 4.84.
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(3)

(e)     Explain why chloroethanoic acid is a stronger acid than ethanoic acid.
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(2)

(f)     Explain why data books do not usually contain values of Ka for strong acids.
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(2)
(Total 20 marks)

Q21.This question is about several Brønsted–Lowry acids and bases.
(a)     Define the term Brønsted–Lowry acid.
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(1)

(b)     Three equilibria are shown below. For each reaction, indicate whether the substance immediately above the box is acting as a Brønsted–Lowry acid (A) or a Brønsted–Lowry base (B) by writing A or B in each of the six boxes.
 
(i)	CH3COOH	+	H2O
		CH3COO–
	+	H3O+	

(1)

 
(ii)	CH3NH2	+	H2O
		CH3NH3+
	+	OH–	

(1)

 
(iii)	   HNO3
	+	H2SO4		H2NO3+
	+	HSO4 –	

(1)

(c)     A 25.0 cm3 sample of 0.0850 mol dm–3 hydrochloric acid was placed in a beaker.
Distilled water was added until the pH of the solution was 1.25.
Calculate the total volume of the solution formed. State the units.
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(3)

(d)     At 298 K, the value of the acid dissociation constant (Ka) for the weak acid HX in aqueous solution is 3.01 × 10–5 mol dm–3.
(i)      Calculate the value of pKa for HX at this temperature.
Give your answer to 2 decimal places.
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(1)

(ii)     Write an expression for the acid dissociation constant (Ka) for the weak acid HX.
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(1)

(iii)     Calculate the pH of a 0.174 mol dm–3 solution of HX at this temperature.
Give your answer to 2 decimal places.
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(3)

(e)     An acidic buffer solution is formed when 10.0 cm3 of 0.125 mol dm–3 aqueous sodium hydroxide are added to 15.0 cm3 of 0.174 mol dm–3 aqueous HX.
The value of Ka for the weak acid HX is 3.01 × 10–5 mol dm–3.
Calculate the pH of this buffer solution at 298 K.
Give your answer to 2 decimal places.
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(6)

(Total 18 marks)

 

Q22.In this question, give all values of pH to 2 decimal places.
(a)     The ionic product of water has the symbol Kw
(i)      Write an expression for the ionic product of water.
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(1)

(ii)     At 42°C, the value of Kw is 3.46 × 10−14 mol2 dm−6.
Calculate the pH of pure water at this temperature.
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(2)

(iii)    At 75 °C, a 0.0470 mol dm–3 solution of sodium hydroxide has a pH of 11.36.
Calculate a value for Kw at this temperature.
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(2)

(b)     Methanoic acid (HCOOH) dissociates slightly in aqueous solution.
(i)      Write an equation for this dissociation.
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(1)

(ii)     Write an expression for the acid dissociation constant Ka for methanoic acid.
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(1)

(iii)    The value of Ka for methanoic acid is 1.78 × 10−4 mol dm−3 at 25 °C.
Calculate the pH of a 0.0560 mol dm−3 solution of methanoic acid.
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(3)

(iv)    The dissociation of methanoic acid in aqueous solution is endothermic.
Deduce whether the pH of a solution of methanoic acid will increase, decrease or stay the same if the solution is heated. Explain your answer.
Effect on pH .........................................................................................
Explanation ...........................................................................................
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(3)

(c)     The value of Ka for methanoic acid is 1.78 × 10−4 mol dm−3 at 25°C. 
A buffer solution is prepared containing 2.35 × 10−2 mol of methanoic acid and 1.84 × 10−2 mol of sodium methanoate in 1.00 dm3 of solution.
(i)      Calculate the pH of this buffer solution at 25°C.
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(3)

(ii)     A 5.00 cm3 sample of 0.100 mol dm−3 hydrochloric acid is added to the buffer solution in part (c)(i).
Calculate the pH of the buffer solution after this addition.
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(4)
(Total 20 marks)

Q23.This question is about Brønsted-Lowry acids of different strengths.
(a)     State the meaning of the term Brønsted–Lowry acid.
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(1)

(b)    (i)      Write an expression for the acid dissociation constant Ka for ethanoic acid.
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(1)

(ii)     The value of Ka for ethanoic acid is 1.75 × 10−5 mol dm−3 at 25 °C.
Calculate the concentration of ethanoic acid in a solution of the acid that has a pH of 2.69
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(4)

(c)     The value of Ka for chloroethanoic acid (ClCH2COOH) is 1.38 × 10−3 mol dm−3 at 25 °C.
(i)      Write an equation for the dissociation of chloroethanoic acid in aqueous solution.
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(1)

(ii)     Suggest why chloroethanoic acid is a stronger acid than ethanoic acid.
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(2)

(d)     P and Q are acids. X and Y are bases. The table shows the strength of each acid and base.
 
 	Acids	Bases	
 	strong	weak	strong	weak	
 	P	Q	X	Y	

The two acids were titrated separately with the two bases using methyl orange as indicator.
The titrations were then repeated using phenolphthalein as indicator.
The pH range for methyl orange is 3.1 − 4.4
The pH range for phenolphthalein is 8.3 − 10.0
For each of the following titrations, select the letter, A, B, C, or D, for the correct statement about the indicator(s) that would give a precise end-point.
Write your answer in the box provided.
A Both indicators give a precise end-point.
B Only methyl orange gives a precise end-point.
C Only phenolphthalein gives a precise end-point.
D Neither indicator gives a precise end-point.
 
 	(i)	Acid P with base X		

(1)

 	(ii)	Acid Q with base X		

(1)

 	(iii)	Acid Q with base Y		

(1)

(e)     Using a burette, 26.40 cm3 of 0.550 mol dm−3 sulfuric acid were added to a conical flask containing 19.60 cm3 of 0.720 mol dm−3 aqueous sodium hydroxide.
Assume that the sulfuric acid is fully dissociated.
Calculate the pH of the solution formed.
Give your answer to 2 decimal places.
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(6)
(Total 18 marks)

Q24.The acid dissociation constant, Ka, for ethanoic acid is given by the expression
Ka = 
The value of Ka for ethanoic acid is 1.74 × 10−5 mol dm−3 at 25 °C.
(a)     A buffer solution is prepared using ethanoic acid and sodium ethanoate. In the buffer solution, the concentration of ethanoic acid is 0.186 mol dm−3 and the concentration of sodium ethanoate is 0.105 mol dm−3.
Calculate the pH of this buffer solution.
Give your answer to 2 decimal places.
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(3)

(b)     In a different buffer solution, the concentration of ethanoic acid is 0.251 mol dm−3 and the concentration of sodium ethanoate is 0.140 mol dm−3.
A sample of hydrochloric acid containing 0.015 mol of HCl is added to 1000 cm3 of this buffer solution.
Calculate the pH of the buffer solution after the hydrochloric acid has been added.
You should ignore any change in total volume.
Give your answer to 2 decimal places.
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(5)
(Total 8 marks)

Q25.Water dissociates slightly according to the equation:
H2O(I)      H+(aq)   +   OH−(aq)
The ionic product of water, Kw, is given by the expression
Kw = [H+][OH−]
Kw varies with temperature as shown in the table.
 
 	Temperature / °C	Kw / mol2 dm−6	
 	25	1.00 × 10−14	
 	50	5.48 × 10−14	

(a)     Explain why the expression for Kw does not include the concentration of water.
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(2)

(b)     Explain why the value of Kw increases as the temperature increases.
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(2)

(c)     Calculate the pH of pure water at 50 °C.
Give your answer to 2 decimal places.
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(3)

(d)     Calculate the pH of 0.12 mol dm−3 aqueous NaOH at 50 °C.
Give your answer to 2 decimal places.
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(3)
(Total 10 marks)

Q26.Titration curves, labelled E, F, G and H, for combinations of different aqueous solutions of acids and bases are shown below.
All solutions have concentrations of 0.1 mol dm−3.

(a)     In this part of the question, write the appropriate letter in each box.
From the curves E, F, G and H, choose the curve produced by the addition of
 
 	(i)	sodium hydroxide to 25 cm3 of ethanoic acid		

(1)

 
 	(ii)	ammonia to 25 cm3 hydrobromic acid		

(1)

 
 	(iii)	hydrochloric acid to 25 cm3 of potassium hydroxide		

(1)

(b)     The table shows information about some acid-base indicators.
 
 	Indicator	pH range	Lower pH colour	Higher pH colour	
 	pentamethoxy red	1.2−3.2	violet	colourless	
 	naphthyl red	3.7−5.0	red	yellow	
 	4−nitrophenol	5.6−7.0	colourless	yellow	
 	cresol purple	7.6−9.2	yellow	purple	

(i)      Which indicator in the table could be used for the titration that produces curve E but not for the titration that produces curve F?
Tick (✓) one box.
 
 	pentamethoxy red		
 	naphthyl red		
 	4−nitrophenol		
 	cresol purple		

(1)

(ii)     Give the colour change at the end point of the titration that produces curve H when naphthyl red is used as the indicator.
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(1)

(iii)    A beaker contains 25 cm3 of a buffer solution at pH = 6.0
Two drops of each of the four indicators in the table are added to this solution.
State the colour of the mixture of indicators in this buffer solution.
You should assume that the indicators do not react with each other.
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(1)
(Total 6 marks)

Q27.Ethanedioic acid is a weak acid. 
Ethanedioic acid acts, initially, as a monoprotic acid.

(a)     Use the concept of electronegativity to justify why the acid strengths of ethanedioic acid and ethanoic acid are different.
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(6)

(b)     A buffer solution is made by adding 6.00 × 10–2 mol of sodium hydroxide to a solution containing 1.00 × 10–1 mol of ethanedioic acid (H2C2O4). 
Assume that the sodium hydroxide reacts as shown in the following equation and that in this buffer solution, the ethanedioic acid behaves as a monoprotic acid.
H2C2O4(aq) + OH–(aq)  HC2O4–(aq) + H2O(l)
The dissociation constant Ka for ethanedioic acid is 5.89 × 10–2 mol dm–3.
Calculate a value for the pH of the buffer solution. 
Give your answer to the appropriate number of significant figures.

 

pH = ....................................
(5)

(c)     In a titration, the end point was reached when 25.0 cm3 of an acidified solution containing ethanedioic acid reacted with 20.20 cm3 of 2.00 ×10–2 mol dm–3 potassium manganate(VII) solution.
Deduce an equation for the reaction that occurs and use it to calculate the original concentration of the ethanedioic acid solution.

Equation .........................................................................................................
Calculation

 

 

Original concentration = ............................... mol dm-3
(4)
(Total 15 marks)

Q28.          The value of the acid dissociation constant, Ka, for ethanoic acid is 1.74 × 10–5 mol dm–3 at 298 K.
(a)     (i)      Write an expression for Ka for ethanoic acid.
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(ii)     Calculate the pH at 298 K of a 0.220 mol dm–3 solution of ethanoic acid.
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(5)

(b)     A sample of the 0.220 mol dm–3 solution of ethanoic acid was titrated against sodium hydroxide solution.
(i)      Calculate the volume of a 0.150 mol dm–3 solution of sodium hydroxide required to neutralise 25.0 cm3 of the ethanoic acid solution.
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(ii)     From the list below, select the best indicator for this titration and explain your choice.
Name of indicator       pH range
bromophenol blue         3.0 – 4.6
methyl red                     4.2 – 6.3
bromothymol blue         6.0 – 7.6
thymol blue                    8.0 – 9.6
Indicator ..............................................................................................
Explanation .........................................................................................
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(5)

(c)     A buffer solution is formed when 2.00 g of sodium hydroxide are added to 1.00 dm3 of a 0.220 mol dm–3 solution of ethanoic acid.
Calculate the pH at 298 K of this buffer solution.
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(6)
(Total 16 marks)

 

 

M1.          (a)     Order with respect to iodine: 0 (1)
Overall order: 2 (1)
2

(b)     Rate constant: k =  = 4.4(4) × 10–4 (1)
Units: mol–1 dm3 s–1 (1)
3

(c)     Appears in rate equation (1)
OR implied by mention of concentration or order
          does not appear in (stoichiometric / overall) equation (1)
2

(d)     pH = –log10 [H+] (1)
      = 1.25
[H+] = 0.056(2) (1)
           rate = (4.44 × 10–4) × (1.50) × (0.0562)
 = 3.75 × 10–5 (1) (mol dm3 s–1)
(3.7 — 3.8)
Can score all 3 conseq on k from part (b)
3
[10]

 

M2.(a)     Any two from:
Weigh by difference or rinse weighing bottle and add to beaker
Rinse beaker and add washings to graduated flask
Invert flask several times to ensure uniform solution
Use a funnel to transfer to the flask and rinse the funnel
Use a stirrer to prepare the solution and rinse the stirrer
If more than two answers apply the list rule.
Max 2

(b)     Ka = [H+]2 / [HA]
Allow any correct expression relating Ka, [H+] and [HA]
1

[HA] = (10–2.50)2 / 1.07 × 10–3
M2 also scores M1
1

= 9.35 × 10–3 (mol dm–3)
Do not allow 9.4 (answer is 9.346).
Correct answer only scores 1 mark.
Do not penalise precision but must be to at least two significant figures.
1

(c)     (b) × 138.0 / 4
1

= 0.322
Using 8.50 × 10–3 gives 0.293
Correct answer scores M1 and M2.
Do not penalise precision but must be to at least two significant figures.
1

(d)     (c) × 100 / 0.500 = 64.5%
Using 0.293 from (c) gives 58.7%
Using 0.347 gives 69.4%
Do not penalise precision.
1
[8]

M3.          (a)     moles HA =  × 0.150 = 3.75 × 10–3 (1)
 vol NaOH =  = 1.875 × 10–2 dm3 (1)
or 18.75 cm3
2

(b)     (i)      pH = –log10 [H+] (1)
(ii)     Value above 7 but below 11 (1)
(iii)     phenol red / thymol blue / phenolphthalein / thymolphthalein
i.e. indicator with  7 < pKin <11
3

(c)     (i)      Only slightly dissociated/ionised (1)
NOT “not fully dissociated / ionised”

(ii)     Ka =  (1)
NOT 

(iii)     For weak acid alone:
Ka =  (1)
 [H+] = 
          = 2.03 × 10–3 (1)
 pH = 2.69 (1)
pH should be given to 2 decimal places
penalise answer to 1 d.p. once in question
5

(d)     moles OH– added = 1.875 × 10–3 = moles A– = moles HA left (1)
or  [A–] = [HA]
 Ka = [H+] or pH = pKa (1)
 pH = 4.56 (1)
3
[13]

 

M4.          (a)     Hydrogen bonding (1)
between H2O and NH3 (1)
2

(b)     (i)      NH3 + H2O  NH4+ + OH– (1)
(ii)     Ammonia is weak base (1)
NOT partially ionised
         Equilibrium to left or incomplete reaction (1)
3

(c)     A proton donor (1)
1

(d)     Buffer solution: A solution which resists change in pH (1)
when small amounts of acid or base added or on dilution (1)
Reagent: NH4Cl (1)
Allow a correct strong acid
3

(e)     (i)      Ka = [H+] [A–] / [HA] (1)
     = [H+] [0.125 × 4] (1) / 1.00
[H+] = 1.70 × 10–5 / 0.125 × 4 = 3.40 × 10–5 (1)
pH = –log10 [H+] = 4.47 (1)
Allow pH conseq to [H+] if 2 place decimals given
(ii)     H+ + CH3COO– → CH3COOH (1)
5
[14]

 

M5.          Penalise pH given to 1 dp first time it would have scored only
(a)     (i)      Kw = [H+] [OH–] (1)
(ii)     pH = – log [H+] (1)
or in words or below unless contradiction
(iii)     Calculation:  [H+]  =  (1)
 = 2.34 × 10–7
    pH = 6.63 or 6.64 (1)
         Explanation: pure water  [H+] = [OH–] (1)
5

(b)     (i)      [OH–] = 0.150             [H+] = 10–14/0.15 = 6.66 × 10–14
or pOH = 0.82
 pH = 13.18 (1)
or pH= 13.17

(ii)     moles OH– = (35 × 10–3) × 0.150 = 5.25 × 10–3 (1)a
moles H+ = (40 × 10–3) × 0.120 = 4.8(0) × 10–3 (1)b
 excess moles of OH– = 4.5 × 10–4 (1)c
 [OH–] = (4.5(0) × 10–4) × 1000/75d (1)e
    = 6.0(0) x 10-3
[H+] =  = 1.66 × 10–12 or pOH = 2.22
 pH = 11.78 (1)f
or 11.77
8

(c)     (i)      Ka =  (1)
(ii)     [H+] = 1.80 × 10–2 × 0.150 = 2.70 × 10–3 (1)
Ka =  = 4.86 × 10–5 (1) mol dm–3 (1)
or  = 4.95 × 10–5
5

          Notes
(a)     If Kw includes H2O allow 6.63 if seen otherwise no marks likely
(b)     (ii)     If no vol, max 4 for a, b, c, f answer = 10.65
If wrong volume max 5 for a, b, c, e, f
If no substraction max 3 for a, b, d
If missing 1000 max 5 for a, b, c, d, f  answer = 8.78
If uses excess as acid, max 4 for a, b, d, f  answer = 2.22
If uses excess as acid and no volume, max 2 for a,
b answer = 3.35
(c)     If wrong Ka in (i) max 2 in part (ii) for [H+] (1) and conseq units (1)
but mark on fully from minor errors
eg no [ ] or charges missing

          Organic points
(1)     Curly arrows: must show movement of a pair of electrons,
i.e. from bond to atom or from lp to atom / space
e.g.

(2)     Structures
penalise sticks (i.e. ) once per paper

Penalise once per paper
          allow CH3– or –CH3 or  or CH3
    or   H3C–
[18]

 

M6.          (a)     (i)      B;
1
         C;
1
         A;
1

(ii)     cresolphthalein
OR
thymolphthalein;
1

(b)     (i)      –log[H+];
1
(ii)     [H+] = 1.259 × 10–12 (or 1.26 or 1.3)
OR
OH = 14 – pH;
1

OR
 = 2.10;
1
= 7.9(4) × 10–3;
(if [H+] is wrong allow 1 for [OH] =  KW/[H+] or as numbers)
1

(c)     (i)      Ka = [H+]2/[CH3CH2COOH]
OR
[H+]2/[HA]
OR
[H+] = [A–] etc;
1
[H+]    = √l.35×l0–5 × 0.117 or expression without numbers;
1
          = 1.257 × l0–3
pH      = 2.90;
1

(iii)     Ka = [H+]
OR
pKa    = pH;
1
pH = 4.87;
(penalise 1dp once)
1
[13]

 

M7.          (a)     Ka = 
(All three sets of square brackets needed, penalise missing brackets or missing charge once in the question)
(Don’t penalise extra [H+]2/[HA])
1

(b)     Ka    =  or [H+] = [A–]
[H+]    = 
         = 6.02 × 10–3pH = 2.22
(must be to 2dp)
(allow 4th mark consequential on their [H+])
1

(c)     (i)      pH (almost) unchanged
(Must be correct to score explanation)
1
         H+ removed by A– forming HA
or acid reacts with salt
or more HA formed
1

(ii)     [H+]    = 10–3.59 = 2.57 × 10–4 or 2.6 × 10–4
1
[A–]     = 
1
          = 
1
          = 0.141 (mol dm3)
(Allow 0.139 to 0.141 and allow 0.14)
(If not used 3.59, to find [H+] can only score M2 for working)
(If 3.59 used but [H+] is wrong, can score M2 for correct method and conseq M4)
If wrong method and wrong expression, can only score M1)
1

(ii)     Alternative scheme for first three marks of part (c)(ii)
         pH = pKa  log
1
         pKa = 3.84
1
         3.59 = 3.84 – log
1
[11]

 

M8.          (a)     Concentration of acid:      m1v1 = m2v2 hence 25 × m1 = 18.2 × 0.150
OR
moles NaOH = 2.73 ×10–3;
1
m1 = 18.2 × 0.150/25= 0.109;
1

(b)     (i)      Ka=[H+][A–]/[HA]          not     Ka = [H+]2 / [HA];
1
(ii)     pKa = –logKa;
1
(iii)     [A–] = [HA];
1
         hence Ka = [H+] [A–] / [HA] = [H+]
         and –logKa  = –log[H+];
1

(c)     ratio [A–] :         [HA] remains constant;
1
hence as [H+] = Ka [HA] / [A–];              [H+] remains constant;
1

(d)     (i)      pH of 0.250 mol dm-3 HCl              = 0.60
and pH of 0.150 mol dm-3 HCl       = 0.82;
1
pH change = 0.22;
1

(ii)     moles HCl = 30 × 0.250 × 10–3 = v × 0.150 × 10–3 = 7.50 × 10–3
OR
v = 30 × 0.250 × 10–3/ 0.150 × 10–3 =50;
1
water added = 50 – 30 = 20 cm3;
1
[12]

 

M9.(a)     (i)      –log[H+]
Penalise missing [ ] here and not elsewhere
1
(ii)     [H+][OH–]
1

(b)     (i)      [H+] = 2.34 × 10–7
1
pH = 6.63
Penalise fewer than 3 sig figs but allow more than 2 dp
1
(ii)     [H+] = [OH–]
1

(iii)     M1    [H+] = Kw/[OH–]
if upside down or CE, allow M3 only for correct use of their [H+]
1
M2    (= 5.48 × 10–14/0.140) = 3.91 × 10–13
1
M3    pH = 12.4(1)
not 12.40 (AE from 12.407)
1
Penalise fewer than 3 sig figs but allow more than 3 sfs
For values above 10, allow 3sfs - do not insist on 2 dp.
For values below 1, allow 2dp – do not insist on 3 sig figs
Not allow pH = 14 – pOH but can award M3 only for pH = 13.1(46)
Can award all three marks if pKw = 13.26 is used

(c)     M1    mol NaOH = mol OH– = (30 × 10–3) × 0.20 = 6.0 × 10–3
mark for answer
1
M2    mol H2SO4 = (25 × 10–3) × 0.15 = 3.75 × 10–3
mark for answer
1
M3    mol H+ = (25 × 10–3) × 0.15 × 2 = 7.5 ×10–3
OR XS mol H2SO4 = 0.75 × 10–3
if factor of 2 missed or used wrongly, CE - lose M3 and next mark gained. In this case they must then use Kw to score any more.
see examples below
1

M4    XS mol H+ = 1.5 × 10–3
1
M5    [H+] = (1.5 × 10–3) × (1000/55) = 0.0273
if no use or wrong use of volume, lose M5 and M6 except if 1000 missed
AE –1 (pH = 4.56)
1
M6    pH = 1.56
Penalise fewer than 3 sig figs but allow more than 3 sfs
For values above 10, allow 3sfs - do not insist on 2 dp.
For values below 1, allow 2dp – do not insist on 3 sig figs
1
[14]

M10.(a)     (i)      [H+][OH–]
1
– log [H+]
1

(ii)     [H+] = [OH–]
1
(iii)     (2.0 × 10–3) × 0.5 = 1.0 × 10–3
1

(iv)    [H+] =                     (= 4.02 × 10–11)
1
pH = 10.40
1

(b)     (i)      Ka  =  [H+][CH3CH2COO-]
          [CH3CH2COOH]
1
         =            [H+]
          [CH3CH2COOH]
1
[H+] = √(1.35 × 10–5) × 0.125      (= 1.30 × 10–3)
1
pH = 2.89
1

(c)     (i)      (50.0 × 10–3) × 0.125 = 6.25 × 10–3
1
(ii)     (6.25 × 10–3) – (1.0 × 10–3) = 5.25 × 10–3
1
(iii)     mol salt formed = 1.0 × 10–3
1
[H+] = Ka ×   [CH3CH2COOH]
                      [CH3CH2COO–)
1
= (1.35 × 10–5) × (= 7.088 × 10–5)
1
pH = 4.15
1
[16]

M11.          (a)     –log [H+]
ecf if [ ] wrong and already penalised 
1
4.57 × 10–3 
allow 4.6 × 10–3
ignore units
1

(b)     (i)      Ka =  allow HA etc
not  but mark on
If expression wrong allow conseq units in (ii) 
but no other marks in (ii)
1

(ii)     = 
If use 4.6 × 10–3
1
Ka = 1.4(1) × 10–4 and pKa = 3.85
= 1.39 × 10–4
allow 1.39 – 1.41 × 10–4 mol dm–3
1
(iii)     pKa = 3.86
Penalise dp of final answer < or > 2 in pH once in paper
1

(c)     (i)       × 0.480 = 0.0144 or 1.4(4) ×10–2
Mark is for answer (M1)
1
(ii)      × 0.350 = 0.0063 or 6.3 × 10–3
Mark is for answer (M2)
1
(iii)     0.0144 – 2(0.0063) = 1.80 × 10–3
M3 is for (i) – 2(ii)
If x 2 missed, CE i.e. lose M3 and the next mark gained
1

(iv)    1.80 × 10–3 ×  = 0.0375 (0.038)
M4 is for answer
If vol is not 48 × 10–3 (unless AE) lose M4 and next mark gained
If multiply by 48 - this is AE - i.e. lose only M4
If multiply by 48 × 10–3 this is AE - i.e. lose only M4
1

(v)     10–14 / 0.0375         (10–14 / 0.038)
M5 for Kw/[OH–]
1
(= 2.66 × 10–13)     (= 2.63 × 10–13)
or pOH
or pOH = 1.426    (or pOH = 1.420)
If no attempt to use Kw or pOH lose both M5 and M6
1
pH = 12.57           (12.58)                 M6
Allow M6 conseq on AE in M5 if method OK
1
[13]

 

M12.          (a)     k = rate/[CH3CH2COOCH3][H+]
1
or
 = 
 = 1.38 × 10–3 to 1.4 × 10–3
1
mol–1 dm3 s–1
1

(b)     ans = rate constant × (½ × 0.150) × (½ × 0.555)
ignore units
      = rate constant × 0.0208
2.88 × 10–5                                     (1.38 × 10–3 gives 2.87 × 10–5)
Allow 2.87 – 2.91 × 10–5 (1.4 × 10–3 gives 2.91 × 10–5)
1

(c)     [H+] = rate/ k[CH3COOCH2CH3] 
1
= 
= 0.415 (0.4146)
1
pH = 0.38 mark independently
[H+] = 0.41 gives pH = 0.39
1
[7]

 

M13.          (a)     (i)      addition of small amounts of acid send eqm to left or extra H+
removed by reaction with HCO3–
1
ratio [H2CO3]/[HCO3–] remains constant hence [H+] and
pH remain const
1

(ii)     pH = 7.41  [H+] = 3.89 × 10–8 mol dm–3
1
Ka = 
1
=  = 7.78 × 10–8 mol dm–3
allow error carried forward mark. Do not penalise twice.
1

(b)     (i)      moles H+ added = 10 × 10–3 × 1.0 = 0.01
1

(ii)     moles ethanoic acid after addition = 0.15 + 0.01 = 0.16
1
         moles ethanoate ions after addition = 0.10 – 0.01 = 0.09
1

(iii)     [H+] = 
1
= 1.74 × 10–5 × 
1
pH = 4.51
1
[11]

 

M14.          (a)     (i)      B                                                                                                             1
C
1
A
1
(ii)     cresolphthalein or thymolphthalein
1

(b)     pH = -log[H+]
1
Ka =  or [H+] = [A–]
1
[H+] = √1.74 × 10–5 × 0.15 (or 1.62 × 10–3)
1
pH = 2.79 (penalise 1 dp or more than 2dp once in the qu)
1
[8]

 

M15.          (a)     (i)      - log[H+]
penalise missing [ ] here and not elsewhere
1
(ii)     [H+][OH–]
Allow ( ) brackets, but must have charges
1

(iii)     Mark independently from a(ii)
[H+] = 10-13.72 = 1.905 × 10–14
If wrong no further mark
1
Kw = 1.905 × 10–14 × 0.154 = = (2.93 – 2.94) × 10–15
1

(b)     (i)      Ka = 
Must have charges and all brackets, allow ( )
Acid/salt shown must be CH3COOH not HA
and correct formulae needed
1

(ii)     In pH values penalise fewer than 3 sig figs each time
but allow more than 2 dp
For values above 10, allow 3sfs - do not insist on 2 dp
Ka = 
Allow HA
1
([H+]2 = 1.75 × 10–5 × 0.154 = 2.695 × 10–6 = 2.70 × 10–6)
If √ shown but not done gets pH = 5.57 (scores 2)
[H+] = 1.64 × 10–3
Allow mark for pH conseq to their [H+] here only
1
pH = 2.78 or 2.79
1

(c)     (i)      In pH values penalise fewer than 3 sig figs each time but
allow more than 2 dp
For values above 10, allow 3sfs - do not insist on 2 dp
M1 Initially
mol OH– = (10 × 10–3) × 0.154 and
mol HA = (20 × 10–3) × 0.154
or mol OH– = 1.54 × 10–3 and mol HA = 3.08 × 10–3
1

M2 [H+] = Ka 
or with numbers
Allow Henderson Hasselbach
pH = pKa + log 

M3 mol ethanoic acid left = (mol ethanoate ions) = 1.54 × 10–3
Ka = [H+] or pH = pKa scores M1, M2 and M3
1 If either mol acid in mixture or mol salt wrong
- max 2 for M1 and M2
Any mention of [H+]2 - max 2 for M1 and M3
1
M4 pH (= - log 1.75 × 10–5) = 4.76 or 4.757
Not 4.75
1
If no subtraction (so mol ethanoic acid in buffer = original mol)
pH = 4.46 scores 2 for M1 and M2
If [H+]2 used, pH = 3.02 scores 2 for M1 and M3

(ii)     In pH values penalise fewer than 3 sig figs each time but
allow more than 2 dp
For values above 10, allow 3sfs - do not insist on 2 dp
M1 XS mol KOH (= (20 × 10–3) × 0.154) = 3.08 × 10–3
If no subtraction: max 1 for correct use of volume
No subtraction and no use of volume scores zero
If wrong subtraction or wrong moles
Can only score M2 and M3 for process
1

M2 [OH-] = 3.08 × 10–3 ×  = 0.0513(3)
Mark for dividing their answer to M1 by correct volume (method mark)
If no volume or wrong volume or multiplied by volume, max 2 for M1 and M3 process
1

M3 [H+] =  (= 1.948 × 10–13 to 1.95 × 10–13)
or pOH = 1.29
Mark for Kw divided by their answer to M2
If pOH route, give one mark for 14 – pOH
1
M4 pH = 12.7(1) 
Allow 3sf but not 12.70
1
If no subtraction and no use of volume (pH = 11.79 scores zero)
If no subtraction, max 1 for correct use of volume, (60cm3)
(pH = 13.01 scores 1)
If volume not used, pH = 11.49 (gets 2)
If multiplied by vol, pH = 10.27 (gets 2)
[16]

 

M16.          (a)     (i)      – log[H+] or log 1/[H+]
penalise missing square brackets here only
1
(ii)     0.81
2dp required, no other answer allowed
1

(iii)     M1    mol H+ = 1.54 × 10–3
if wrong no further mark
if 1.5 × 10–3 allow M1 but not M2 for 2.82
1
M2    pH = 2.81
allow more than 2dp but not fewer
1

(b)     M1    [H+] = 3.31 × 10–3
1
M2    Ka = or  or using numbers
do not penalise ( ) or one or more missing [ ]
1
M3    [HX] = 
allow conseq on their [H+]2/(4.83 × 10–5) (AE)
if upside down, no further marks after M2
1
M4    [HX] = 0.227
allow 0.225 – 0.23
1

(c)     M1    extra/added OH– removed by reaction with H+ or the acid
1
M2    correct discussion of equn shift i.e. HX  H+ + X– moves to right
1
OR
ratio  remains almost constant

(d)     (i)      M1    mol HY = (50 × 10–3) × 0.428 = 0.0214
OR    [Y] = .0236 × 
mark for answer
1

M2    [H+] = 1.35 × 10–5 × 
OR    1.35 × 10–5 = [H+] × 
OR    [H+] = 1.35 × 10–5 × 
OR    1.35 × 10–5 = [H+] × 
must be numbers not just rearrangement of Ka expression
If either HY value or Y– value wrong, (apart from AE -1) lose M2 and M3
1

M3    [H+] = 1.22 ×10–5
mark for answer
1
M4    pH = 4.91
allow more than 2dp but not fewer
allow M4 for correct pH calculation using their [H+] (this applies in (d)(i) only)
1

If Henderson Hasselbalch equation used:
M1    mol HY = (50 × 10–3) × 0.428 = 0.0214
OR    [Y] = .0236 × 
mark for answer
1
M2    pKa = 4.87
1

M3    log = –0.043
          log = –0.043
If either HY value or Y– value wrong, (apart from AE-1) lose M3 and M4
1
M4    pH = 4.87 – (– 0.043) = 4.91
allow more than 2dp but not fewer
1

(ii)     Can score full marks for correct consequential use of their
HY and Y– values from d(i)
M1    Mol HY after adding NaOH = 0.0214 – 5.0 × 10–4 = 0.0209
AE in subtraction loses just M1
If wrong initial mol HY (i.e. not conseq to part d(i)) or no subtraction or subtraction of wrong amount, lose M1 and M3
1
M2    Mol Y– after adding NaOH = 0.0236 + 5.0 × 10–4 = 0.0241
AE in addition loses just M2
If wrong mol Y– (i.e. not conseq to part d(i)) or no addition or addition of wrong amount lose M2 and next mark gained
1

M3    [H+] = 1.35 × 10–5 ×  (= 1.17 × 10–5)
          if convert to concentrations
          [H+] = 1.35 × 10–5 ×  (= 1.17 × 10–5)
1
if HY/Y– upside down, no further marks
M4    pH = 4.93
allow more than 2dp but not fewer
NOT allow M4 for correct pH calculation using their [H+] (this allowance applies in (d)(i) only)
1

If Henderson Hasselbalch equation used:
Can score full marks for correct consequential use of their HY and Y– values from d(i)
M1    Mol HY after adding NaOH = 0.0214 – 5.0 ×10–4 = 0.0209
AE in subtraction loses just M1
If wrong initial mol HY (i.e. not conseq to part d(i)) or no subtraction or subtraction of wrong amount lose M1 and M3
1

M2    Mol Y– after adding NaOH = 0.0236 + 5.0 × 10–4 = 0.0241
AE in addition loses just M2
If wrong mol Y– (i.e. not conseq to part d(i)) or no addition or addition of wrong amount lose M2 and next mark gained
1
M3    log  = –0.062
if HY/Y–– upside down, no further marks
1
M4    pH = 4.87 – (– 0.062) = 4.93
allow more than 2dp but not fewer
1
[18]

 

M17.(a)     Idea that over time / after storage meter does not give accurate readings
Do not accept ‘to get an accurate reading’ without further qualification.
Allow ‘temperature variations affect reading’.
1

(b)     
 
Allow without (aq) symbols.
Need at least one set of square brackets around complex ions
1

(c)     pH = –log [H+]
1

[H+] = 0.0240
Do not penalise precision of [H+]
Correct answer scores M1 and M2.
1

Ka = (0.0240)2 / 0.1 = 5.75 ×10-3 or 5.76 ×10-3
Correct answer without working loses M1 and M2.
Allow 7.58 ×10-3
1

Answer, even if incorrect, given to 3 sig figs
1

(d)     Oxygen (in the air) / O2
Ignore ‘air’ or ‘the atmosphere’ or ‘chemicals in soil’.
List principle.
1

(e)     4.0 – 6.9
Do not penalise precision.
[7]

M18.(a)    (only) slightly or partially dissociated / ionised
Ignore ‘not fully dissociated’.
Allow low tendency to dissociate or to lose / donate a proton.
Allow shown equilibrium well to the left.
Otherwise ignore equations.
1

(b)     2CH3CH2COOH + Na2CO3  2CH3CH2COONa + H2O + CO2
OR
2CH3CH2COOH + CO32−  2CH3CH2COO− + H2O + CO2
OR
CH3CH2COOH + Na2CO3  CH3CH2COONa + NaHCO3
OR
CH3CH2COOH + CO32−  CH3CH2COO− + HCO3−
Must be propanoic acid, allow C2H5COOH.
Not molecular formulae.
Allow multiples.
Ignore reversible sign.
Not H2CO3.
1

(c)     [OH−] = 2 × 0.0120 = 0.0240                M1
Correct answer for pH with or without working scores 3.
1

[H+] =  = 4.166 × 10−13 OR pOH = 1.62        M2
 
If × 2 missed or used wrongly can only score M3 for correct calculation of pH from their [H+].
1

pH = 12.38                M3
Lose M3 if not 2 decimal places: 12.4 scores 2.
12.08 scores 1 (missing × 2) ; 12.1 scores 0.
11.78 scores 1 (dividing by 2) 11.8 scores 0.
1

(d)    (i)      Ka    = 
 
Ignore ( ) here but brackets must be present.
Must be correct acid and salt.
If wrong, mark part (ii) independently.
1

(ii)     M1        Ka    =         OR with numbers
 
Correct answer for pH with or without working scores 3.
Allow HX, HA and ignore ( ) here.
May score M1 in part (i).
1

M2        [H+] = √(6.31 × 10−5 × 0.0120) or √(Ka × [C6H5COOH])
            (= √(7.572 × 10−7 = 8.70 ×10×4)
pH = 6.12 may score 2 if correct working shown and they show the square root but fail to take it.
But if no working shown or wrong Ka = 

used which also leads to 6.12, then zero scored.
1

M3        pH = 3.06
Must be 2 decimal places ie 3.1 loses M3.
1

(iii)    M1        [H+] = 10−4.00 = 1.00 × 10−4
Correct answer for mass with or without working scores 5.
Allow 1 × 10−4.
1

M2        [X−] = 
 
Ignore ( ) here.
If [HX] / [X−] upside down, can score M1 plus
M4 for 5.26 × 10−7.
1

M3        = 
 
And M5 for 7.57 × 10−5 g.
1

M4            = 7.572 × 10−3
1

M5        Mass (C6H5COONa) = 7.572 × 10−3 × 144 =1.09 g
                                                                                  or 1.1 g
Wrong method, eg using [H+]2 may only score M1 and M5 for correct multiplication of their M4 by 144
(provided not of obviously wrong substance).
1

(e)     M1        CO2
Allow NOx and SO2.
1

M2        pH (It) falls / decreases
If M1 wrong, no further marks.
1

M3        mark M2 & M3 independently
          acidic (gas)
          OR reacts with alkali(ne solution) / OH−
          OR CO2 + 2OH−  CO32− + H2O
          OR CO2 + OH−  HCO3−
Not forms H2CO3 H2SO3 H2SO4 etc OR H+ ions.
1
[17]

M19.(a)     M1    [H+] = 0.0170
1
M2    pH = 1.77
2 dp
Allow M2 for correct pH calculation from theirwrong [H+] for this pH calculation only
1

(b)     (i)              Ignore 
Penalize missing [ ] here and not elsewhere
Allow HA instead of HX
1

(ii)     M1    [H+] = 10–2.79 OR 1.6218… ×10–3
If [H+] wrong, can only score M2
1

M2         OR     
Allow HA instead of HX
1

M3    Ka = 3.09 ×10–5   3sfs min         (allow 3.10 ×10–5 if 1.6218 rounded to 1.622)         Ignore units
If [HX] used as (0.0850 –1.62 ×10–3 )
this gives Ka = 3.15 ×10–5
(0.0016)2/0.085 = 3.01 ×10–5 scores 2 for AE
1

(c)     M1    mol OH– (= (38.2 × 10–3) × 0.550 )
= 2.10(1) × 10–2 or 0.0210(1)
Mark for answer
1

M2    Mol H+ (= (25.0 × 10–3) × 0.620 )
= 1.55 × 10–2 or 0.0155
Mark for answer
1

M3    excess mol OH– = 5.5(1) × 10–3
Allow conseq for M1 – M2
If wrong method e.g. no subtraction or use of √
can only score max of M1, M2, M3 and M4.
1

M4    [OH− ] = 5.51 × 10−3 ×     [ = 0.08718   (0.0872)]
OR    [OH− ] = 5.5 × 10−3 ×  = 0.0870(2)
(M1 – M2) / vol in dm3 mark for dividing by volume
(take use of 63.2 without 10–3 as AE so 9.94 scores 5)
If no use or wrong use of vol lose M4 & M6
Can score M5 for showing (10–14/ their XS alkali)
1

M5    [H+] =  = 1.147 × 10−13
OR     = 1.149 × 10−13
OR    pOH = 1.06
If no use or wrong use of Kw or pOH no further marks
1

M6    pH = 12.9(4)   allow 3sf
If vol missed score max 4 for 11.7(4)
If acid– alkali reversed max 4 for pH = 1.06
Any excess acid – max 4
1
[12]

M20.(a)     CH3COOH + H2O ⇌ CH3COO- + H3O+
OR
CH3COOH ⇌ CH3COO- + H+
Must show ⇌
Allow CH3CO2H, CH3CO2-
Ignore state symbols
1

(b)     (CH3COOH + HNO3 → CH3COOH2+ + NO3-)
IGNORE ⇌
Allow CH3CO2H, CH3CO2H2+, CH3C+(OH)2
1

(c)     (i)      ( new [HNO3] = [H+] =  × 0.0125 )
M1     [H+] = 8.3(3) × 10-3 (mol dm-3)
OR
new[HNO3] = 
1
M2     pH = − log M1 OR 2.08
Must be 2dp
Allow correct pH conseq to their [H+] concentration
1

(ii)     M1     mol NaOH (= 50 × 10-3 × 0.0108) = 5.40 × 10-4
1
M2     Subtraction of M1 from moles of HNO3 (1.25 × 10-3 or conseq from 1c(i))
Excess mol H+ = 7.10 × 10-4
M2 allow ecf for subtraction of mol
If no subtraction, no further marks
1
M3     [H+] =  = 4.73 × 10-3
M3 if no use of volume, no further marks (pH=3.15)
If incorrect volume used, can score M4
1
M4     pH = -log M3 OR 2.32
M4 Allow 2.33 Must be 2 dp
1

(d)     (i)      M1     Ka = 
Penalise ( ) once here Not [H+][A-] / [HA]
If Ka expression wrong – Allow correct pH conseq to their [H+] concentration M4 only
1
M2     Ka =  or with numbers or with HA
1
M3     [H+] = [ √(1.74 ×10-5 × 0.0125)] = 4.66 ×10-4
Mark for answer
1
M4     pH = 3.33
Must be 2dp
Allow correct pH conseq to their [H+] concentration
(pH = 3.83 can score M1, M2 and M4)
1

(ii)     Sodium ethanoate
Ignore formula
Allow sodium acetate
1
(iii)     M1     [H+] = 1.45 × 10-5
Accept 1.445 × 10-5 or 1.4 × 10-5
1
M2     
If M1 incorrect CE=0
Inclusion of 0.0125 in calculation can only score M1
1
M3     1.2(0)
Ignore units
1.4 × 10-5 gives 1.24
1

(e)     M1     (Electronegative) chlorine withdraws electrons
Allow Cl has negative inductive effect
1
M2     Stabilises/reduces charge on COO-
OR weakens O-H bond
OR makes O-H more polar
Ignore chloroethanoic acid dissociates more readily
Mark independently
1

(f)     M1     Strong acids (almost) completely dissociated/ionised
OR not an equilibrium
OR equilibrium lies far to the right
Cannot have Ka value for a reaction not in equilibrium scores both marks
1
M2     Ka value for strong acids tends to infinity/is very large
OR can’t divide by zero in Ka
1
[20]

M21.(a)     Proton donor or H+ donor
Allow donator
1

(b)     (i)      B B
Both need to be correct to score the mark
1

(ii)     A A
Both need to be correct to score the mark
1

(iii)     B A
Both need to be correct to score the mark
1

(c)     M1    [H+] = 10–1.25 OR 0.05623
1

M2    mol HCl = (25 × 10–3) × 0.0850 (= 2.125 × 10–3)
Mark for Working
1

M3    vol   = 0.0378 dm3 or 37.8 cm3
allow 0.0375 – 0.038 dm3 or 37.5 – 38 cm3
Units and answer tied
Lose M3 if total given as (25 + 37.8) = 62.8 cm3
Ignore “vol added = 12.8 cm3 ” after correct answer
1

(d)     (i)      4.52
Must be 2dp
1

(ii)     Ka =        ignore =  but this may score M1 in (d)(iii)
Must have all brackets but allow ( ) Allow HA etc
NO mark for 10–pKa
1

(iii)     M1    Ka =  or with numbers
Allow [H+] = √(Ka × [HA]) for M1
1

M2    [H+] = (√(3.01 ×10–5 × 0.174) = √(5.24 ×10–6) )
= 2.29 ×10–3 - 2.3 ×10–3
Mark for answer
1

M3    pH = 2.64             (allow more than 2dp but not fewer)
Allow 1 for correct pH from their wrong [H+]
If square root forgotten, pH = 5.28 scores 2 for M1 and M3
1

(e)     M1    mol OH– = (10.0 × 10-3) × 0.125 = 1.25 × 10–3
Mark for answer
1

M2    orig mol HX = (15.0 × 10–3) × 0.174 = 2.61 × 10×3
Mark for answer
1

M3    mol HX in buffer = orig mol HX – mol OH–
Mark for answer
= 2.61 × 10–3 – 1.25 × 10–3 = 1.36 × 10–3
Allow conseq on their (M2 – M1)
([HX] = 1.36 × 10–3/25 × 10–3 = 0.0544)
If no subtraction, max 3 for M1, M2 & M4 (pH = 4.20)
If [H+] = [X–] & √used, max 3 for M1, M2 & M3 (pH = 2.89)
1

M4    mol X– in buffer = mol OH– = 1.25 × 10–3
([X–] = 1.25 × 10–3/25 × 10–3 = 0.05)
May be scored in M5 expression
1

M5    [H+]   
If use Ka =  no further marks
=    OR 
(= 3.27 × 10–5)
If either value of HX or X– used wrongly or expression upside down, no further marks
1

M6    pH = 4.48 or 4.49   (allow more than 2dp but not fewer)
Do not allow M6 for correct calculation of pH using their [H+] - this only applies in (d)(iii) - apart from earlier AE
1
[18]

M22.(a)     (i)      [H+][OH−] OR [H3O+][OH−]
Ignore (aq)
Must have [ ] not ( )
1

(ii)     3.46 × 10−14 (= 1.86 × 10−7)
If no square root, CE=0
1

pH = 6.73
Must be 2dp
1

(iii)    [H+] = 10−11.36 (= 4.365 × 10−12 OR 4.37 × 10−12)
Mark for working
1

Kw = [4.365 × 10−12 OR 4.37 × 10−12 × 0.047] = 2.05 × 10−13
Allow 2.05 × 10−13 − 2.1 × 10−13
Mark for answer
Ignore units
1

(b)     (i)      HCOOH    HCOO− +   H+
Must have  but ignore brackets.
OR   HCOOH + H2O     HCOO− +   H3O+
Allow HCO2− or CHOO− ie minus must be on oxygen, so penalise COOH−
1

(ii)      
 
Must have all brackets but allow ( )
Must be HCOOH etc.
Allow ecf in formulae from (b)(i)
1

(iii)    M1
 
Allow HA or HX etc.
Allow [H+] = (Ka × [HA]) for M1
1

M2    [H+] = 3.16 × 10−3
Mark for answer
1

M3    pH = 2.50     allow more than 2 dp but not fewer
Allow correct pH from their wrong [H+] here only If square root shown but not taken, pH = 5.00 can score max 2 for M1 and M3
1

(iv)    M1    Decrease     Mark M1 independently
1

M2    Eqm shifts / moves to RHS     OR    more H+    OR    Ka increases    
OR more dissociation
1

M3    To reduce temperature or oppose increase / change in temperature
Only award M3 following correct M2
1

(c)     (i)      M1    
 
If [HX]/[X-] upside down, no marks
1

M2    
         (= 2.27 × 10−4)
1

M3    pH = 3.64      allow more than 2 dp but not fewer
pH calc NOT allowed from their wrong [H+] here
1

(ii)     M1    Mol H+ added = 5.00 × 10−4
Mark on from AE in moles of HCl (eg 5 × 10−3 gives pH = 3.42 scores 3)
1

M2    Mol HCOOH = 2.40 × 10−2   and   Mol HCOO– = 1.79 × 10−2
If either wrong no further marks except AE (−1) OR if ECF in mol acid and / or mol salt from (c)(i), can score all 4
1

M3    
 
If [HX]/[X-] upside down here after correct expression in (c)(i), no further marks
                
If [HX]/[X-] upside down here and is repeat error from (c)(i), max 3 (pH = 3.88 after 3.86 in (c)(i))
1

M4    pH = 3.62      allow more than 2 dp but not fewer
pH calc NOT allowed from their wrong [H+] here
1
[20]

M23.(a)    Proton donor or H+ donor
1

(b)     (i)      
 
If Ka wrong, can only score M1 below.
Must be ethanoic acid not HA
Must have square brackets (penalise here only) but mark on in (b)(ii).
1
(ii)     M1  [H+] = 10−2.69 OR 2.042 × 10−3 (mol dm−3)
1

M2  
Ignore ( )
Mark for correctly rearranged expression incl [H+]2
1
M3

If M2 wrong no further marks.
1

M4  = 0.238 (mol dm−3) Allow 0.229 − 0.24
1

(c)    (i)      ClCH2COOH      ClCH2COO− + H+
OR ClCH2COOH + H2O      ClCH2COO− + H3O+
Allow 
Allow ClCH2CO2H and ClCH2CO 2−
1

(ii)     M1  Cl is (more electronegative so) withdraws electrons
       OR negative inductive effect of Cl
Ignore electronegativity.
Ignore chloroethanoic acid has a lower Ka value.
Allow Cl reduces +ve inductive effect of methyl group.
1

M2  Weakens O–H bond
       OR O–H bond is more polar
       OR reduces negative charge on COO−
       OR stabilizes COO− (more)
M1 & M2 are independent marks.
Ignore H+ lost more easily.
1

(d)     (i)      A
1

(ii)     C
1

(iii)    D
1

(e)     M1  Mol NaOH = mol OH− = (19.6 × 10–3) × 0.720 = 1.41(1) × 10−2
Mark for answer.
1
M2  Mol H2SO4 = (26.4 × 10−3) × 0.550 = 1.45(2) × 10−2
Mark for answer.
1

M3  Mol H+ added = 2 × (1.452 × 10−2) = 2.90(4) × 10−2
       OR
       XS mol H2SO4 = 7.46(4) × 10−3
If factor × 2 missed completely (pH = 2.05)
or used wrongly later,
can score max 4 for M1, M2, M5 & M6
1

M4  XS mol H+ = 0.0149(3)
1

M5  For dividing by volume
       [H+ ] = 0.0149(3) × (1000 / 46.0) = 0.324 − 0.325 mol dm−3
If no use or wrong use of volume lose M5 and M6
ie can score 4 for pH = 1.83 (no use of vol)
Treat missing 1000 as AE (−1) & score 5 for pH = 3.49
1

M6  pH = 0.49
2dp (penalise more or less).
If × 2 missed & vol not used, pH = 3.39 scores M1 & M2 only.
1
[18]

M24.(a)     [H+] =  or = 1.74 × 10 −5 × 
 
Allow ( )
M1
1
=     3.08 × 10−5
If [HX] / [X-] or  upside down, or any addition or subtraction lose M1 & M2.
M2
1
pH = 4.51                 (correct answer scores 3)
Can score M3 for correct pH conseq to their [H+], so pH = 5.01 scores one
Must be to 2 dp
M3
1

Alternative using Henderson−Hasselbach Equation
pH = pKa − log[HX] / X−] = −log(1.74 × 10−5) − log()
Allow ( )
M1
pKa = 4.76 − 0.248
If [HX] / [X−] or  upside down, can only score 1
M2
pH = = 4.51
so pH = 5.01
Must be to 2 dp
M3

(b)     mol HX after addition (= 0.251 + 0.015) = 0.266
For HX, if no addition or error in addition (other than AE) (or subsequent extra add or sub) MAX 3
M1
1

mol X- after subtraction (= 0.140 − 0.015) = 0.125
For X− if no subtraction or error in subtraction (other than AE) (or subsequent extra add or sub) MAX 3
M2
1
[H+] = (  ) = 
 
If errors above in both addition AND subtraction can only score M3 for insertion of their numbers in rearranged expression. One exception, if addition and subtraction reversed then pH =4.58 scores 2
M3
1
[H+] = 3.703 × 10−5 (mol dm−3)
If [HX] / [X−] upside down, lose M3 & M4 (or next two marks) but can score M5 for correct pH conseq to their [H+], so if M1 & M2 correct, pH = 5.09 scores 3.
M4
1
pH = 4.43
Correct use of HX and X− values from (d) gives
pH= 4.41 and scores 4
If wrong method, e.g. √ or no use of rearranged Ka expression, may score M1 & M2 but no more.
Allow more but not fewer than 2dp here.
M5
1

Alternative using Henderson−Hasselbach Equation
mol acid after addition = 0.251 + 0.015 = 0.266
For HX, if no addition or error in addition (other than AE) (or subsequent extra add or sub) MAX 3
M1
mol salt after addition = 0.140 − 0.015 = 0.125
For X− if no subtraction or error in subtraction (other than AE) (or subsequent extra add or sub) MAX 3
M2
pH = (pKa − log[HX] / [X−]) = −log(1.74 × 10−5) − log(0.266 / 0.125)
If errors above in both addition AND subtraction can only score M3 for insertion of their numbers − except if addition and subtraction reversed then pH = 4.58 scores 2
M3
pH = 4.76 − 0.328
M4
pH = = 4.43
If [HX] / [X−] upside down, lose M3 & M4 (or next two marks) but can score M5 for correct pH conseq to their working, so if M1 & M2 correct, pH = 5.09 scores 3.
Allow more but not fewer than 2dp here.
M5
[8]

M25.(a)     [H2O] is very high (compared with [H+] and [OH−])
OR
Very few H+ and OH− ions
OR
Only / very slightly dissociates
OR
Equilibrium lies far to the left
Not partially dissociates
M1
1

[H2O] is (effectively) constant
OR is incorporated into the constant K
Allow changes by only a very small amount
M2
1

(b)     (Dissociation OR breaking bonds) is endothermic
1
∴ Equilibrium moves to RHS (at higher T) to absorb heat or to lower T or oppose increase in T
Allow to oppose change only if increase T mentioned
1

(c)     [H+]     = √Kw (or = √5.48 × 10−14)
Correct pH answer scores 3
1
           If wrong method no marks
Using alternative Kw (1.00 × 10−14) gives pH = 7.00 which scores 1
           = 2.34 × 10−7
1
pH     = 6.63
Final answer must have 2dp
1

(d)     [H+]     = Kw / [OH−] or (= 5.48 × 10−14 / 0.12)
Correct pH answer scores 3
1
           If wrong method no marks
If use alternative Kw (1.00 × 10−14) again, do not penalise repeat error so pH = 13.08 scores 3
           = 4.566 × 10−13
1
pH     = 12.34
If use alternative Kw (1.00 × 10−14) not as a repeat error, pH = 13.08 scores 1
If AE in Kw value made in part (c) is repeated here, do not penalise again.
Final answer must have 2dp, but if dp penalised in (c) allow more than 2dp here but not fewer.
1
[10]

M26.(a)     (i)      G
1
(ii)     F
1
(iii)    H
1

(b)     (i)      cresol purple
1
(ii)     yellow to red
both colours needed and must be in this order
1
(iii)    yellow or pale yellow
Not allow any other colour with yellow
1
[6]

M27.(a)     This question is marked using levels of response. Refer to the Mark Scheme Instructions for Examiners for guidance on how to mark this question.
All stages are covered and the explanation of each stage is generally correct and virtually complete.
Answer is communicated coherently and shows a logical progression from stage 1 and stage 2 to stage 3. Steps in stage 3 must be complete, ordered and include a comparison.
Level 3
5 – 6 marks
All stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies OR two stages are covered and the explanations are generally correct and virtually complete.
Answer is mainly coherent and shows a progression from stage 1 and stage 2 to stage 3.
Level 2
3 – 4 marks
Two stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies, OR only one stage is covered but the explanation is generally correct and virtually complete.
Answer includes some isolated statements, but these are not presented in a logical order or show confused reasoning.
Level 1
1 – 2 marks
Insufficient correct Chemistry to warrant a mark.
Level 0
0 marks
Indicative Chemistry content
Stage 1: difference in structure of the two acids
•        The acids are of the form RCOOH
•        but in ethanoic acid R = CH3
•        whilst in ethanedioic acid R = COOH
Stage 2: the inductive effect
•        The unionised COOH group contains two very electronegative oxygen atoms
•        therefore has a negative inductive (electron withdrawing)effect
•        The CH3 group has a positive inductive (electron pushing) effect
Stage 3: how the polarity of OH affects acid strength
•        The O–H bond in the ethanedioic acid is more polarised / H becomes more δ+
•        More dissociation into H+ ions
•        Ethanedioic acid is stronger than ethanoic acid
6

(b)     Moles of NaOH = Moles of HOOCCOO– formed = 6.00 × 10–2
Extended response
1

Moles of HOOCCOOH remaining = 1.00 × 10–1 – 6.00 × 10–2
= 4.00 × 10–2
1

Ka = [H+][A–] / [HA]
[H+] = Ka × [HA] / [A–]
1

[H+] = 5.89 × 10–2 × (4.00 × 10–2 / V) / (6.00 × 10–2 / V) = 3.927 × 10–2
1

pH = –log10(3.927 ×10–2) = 1.406 = 1.41
Answer must be given to this precision
1

(c)     5H2C2O4 + 6H+ + 2MnO4–   2Mn2+ + 10CO2 + 8H2O
OR 5C2O42– + 16H+ + 2MnO4–   2Mn2+ + 10CO2 + 8H2O
1

Moles of KMnO4 = 20.2 × 2.00 × 10–2 / 1000 = 4.04 × 10–4
1

Moles of H2C2O4 = 5 / 2 × 4.04 × 10–4 = 1.01 × 10–3
1

Concentration = moles / volume (in dm3)
= 1.01 × 10–3 × 1000 / 25 = 4.04 × 10–2 (mol dm–3)
If 1:1 ratio or incorrect ratio used, M2 and M4 can be scored
1
[15]

M28.          (a)     (i)      Ka =  (1)

(ii)     (1)     Ka =  (1)
(2)     [H+] =  = 1.96 × 10–3 (1)
(3)     pH = -log10[H+] (1)
can score independently
(4)     pH = 2.71 (1)
2 d.p. essential
If forget √ can score (1) and (3) for pH = 5.42
5

(b)     (i)      moles acid =  (1) = 5.50 ×10–3
= 
 x = 25 ×  or 5.50 × 10–3 × 
       = 36.7 (or 37) cm3 (or 36.6) (1)
          
NOT 36 NOR 37.0 units must match

(ii)     Indicator:                  thymol blue (1)
Explanation:             weak acid – strong base (1)
                                 equivalent at pH > 7 (1)
                                 or high pH
5

(c)     (1)     mol NaOH added =  = 0.050 (1)
If wrong Mr: CE  lose marks (1) and (2) then mark on
consequentially → max 4
(2)     mol CH3COOH left = 0.220 – 0.050 = 0.170 (1)
(3)     mol CH3COO– formed = 0.050 (1)

(4)     [H+] = Ka  OR pH = pKa +  etc (1)
If expression wrong no marks for 4 / 5 / 6
can score (1) to (4) in (5)
(5)     [H+] = 1.74 × 10–5 ×  OR pH = 4.76 + log (1)

(6)     pH = 4.23 (1)
Correct answer gets (1)(1)(1)(1)(1)(1)
Mark (5) is for use of correct values of (acid moles) and
(salt moles)
if one wrong allow pH conseq
if both wrong, no further marks
e.g. if candidate forgets substitution in (2)
he loses (2) and (5) but can score (1) (3) (4) (6) = max 4
for pH = 4.12 if upside down; answer 5.29 scores 3 for (1) (2) (3)
6
[16]

 

 

E1.          This question on rate equations was answered well by most candidates even those who scored less well elsewhere in the paper. In part (a) most candidates deduced the overall order correctly, but to deduce that the order with respect to iodine was zero proved more difficult. Part (b) was well answered by most students, but only the better candidates were able to gain both marks in part (c). Many gave a general description of the role of a catalyst rather than the evidence that hydrogen ions appear in the rate equation but not in the overall stoichiometric equation. The use of pH to show a change of hydrogen ion concentration was understood by most candidates although some were unable to convert a pH value into a hydrogen ion concentration. Full marks were available for correct use in part (d) of an incorrect rate constant calculated in part (b).

E2.(a)     Many incomplete answers were given credit in this part. When the components of a two-part answer are linked by the underlined word ’and’ both components are needed to earn the mark.
(b)     A pleasing majority of students were able to complete the calculation in this part. Occasionally, a centre was unduly harsh in penalising minor rounding in the intermediate steps.

E3.          Parts (a) and (b) were well answered. Most candidates correctly calculated the volume of sodium hydroxide needed and although some candidates attempted to answer part (b)(i) in words, most were able to define pH as –log10[H+] . The indicator expected in part (b)(iii) was the familiar phenolphthalein, but thymol blue (the correct indicator to the January 2002 pH question) was a common and correct answer.
In part (c), the answer that weak acids are only slightly dissociated in water was required, rather than not fully dissociated, since even strong acids may not be fully dissociated. In part(c)(ii), the general expression for Ka was required, i.e. [H+][A–]/[HA] rather than the more particular [H+]2/[HA] which only applies to solutions of weak acids alone. The latter expression could however gain credit in the calculation in part (iii). This was answered well by many candidates although some used the number of moles of HA in the calculation rather than its concentration. Not all candidates seemed to be aware that it is important that pH values are given to two decimal places; these two figures give the value to the appropriate precision. The numbers before the decimal point only give the powers of ten involved in the hydrogen ion concentration.
Part (d) discriminated well. The better candidates realised that point b was the pH at half equivalence, i.e. when [A–] = [HA] such that pH = pKa.

E4.          Only the best candidates correctly explained the solubility of ammonia in water by reference to hydrogen bonding between water and ammonia molecules. The equation for the reaction between ammonia and water was often given correctly but few stated that, because ammonia is a weak base, it reacts reversibly with water forming a solution containing a small concentration of hydroxide ions. Most, but certainly not all, candidates knew the meaning of the term Bronsted-Lowry acid and answered part (c) correctly. Answers to part (d) were, by contrast, rather disappointing with many vague or incorrect statements of the term buffer solution. The calculation in part (e)(i) was not well done. Many candidates used the relationship Ka = [H+]2/[HA] which can only be used for a solution of a weak acid alone. Others failed to calculate the concentration of ethanoate ions in mol dm-3 before calculating the hydrogen ion concentration. Part (c)(ii) proved surprisingly difficult with only the better candidates giving the correct ionic equation.

E5.          Definitions of the terms Kw and pH in parts (a)(i) and (ii) were usually well answered except by those who failed to write mathematical definitions and attempted to describe the terms in words. Provided there were no contradictions, the marks could be earned in the later sections. The most common error in the expression for Kw was to include water. However this water term was conveniently forgotten in part (iii) and correct answers for the pH of pure water at 50°C were common. As reported regularly before, too many candidates seem not to be aware that pH values should be given to two decimal places; these two figures give the value to the appropriate precision. The numbers before the decimal point only give the powers of ten involved in the hydrogen ion concentration. However, failure to give pH values to the correct precision was only penalised once in the question. Part (b)(i) was answered well but (b)(ii) was more discriminating. A common error was to forget that the moles of excess sodium hydroxide were present in a total volume of solution of 75 cm3 and needed to be converted into a molar concentration before the pH of the solution could be calculated.

A number of candidates tried to answer the question as if a buffer solution had been produced despite the use of a named strong acid. Several also gave pH values below 7 although there was an excess of alkali present. Part (c) was answered well by most candidates although some failed to give the general expression for Ka and instead wrote the more specific [H+]2/[HX].
This expression was however credited in part (c)(ii).

E6.          Only the weakest failed to recognise the titration curves in part (a) or to give either of the indicators which changed colour above pH 7. The definition of pH was, as expected, well known, but many candidates were unable to complete the calculation of the concentration of potassium hydroxide in part (b)(ii) by using either Kw or the expression pH + pOH =14. Part (c)(i) was well done but many failed to recognise that the solution formed in part(c) was a buffer solution of the particular type where the moles of salt and moles of acid present were equal, so that the pH of the mixture equals the pKa of the weak acid. It was pleasing that very few ignored the stated request to quote pH values to two decimal places.

E7.          High marks were scored in this question by the more able candidates but the question discriminated well and although parts (a) and (b) were well answered by most, the weaker candidates scored poorly in part (c).
In part (a), although the expression for Ka was well known, several candidates did not give the general expression for Ka and instead wrote the more specific [H+]2/[HA]. This expression was however credited in part (b). Candidates should be reminded that square brackets are essential in these expressions.
In part (b), many correct answers were given although, in the calculation, several candidates failed to take the square root correctly. A mark was awarded for the correct calculation of a pH value from their hydrogen ion concentration.
As reported regularly before, too many candidates seem unaware that pH values should be given to two decimal places; these two figures give the value to the appropriate precision. The numbers before the decimal point only give the powers of ten involved in the hydrogen ion concentration.
In part (c)(ii), most candidates were able to state that the pH would remain unchanged, but fewer explained clearly that the extra hydrogen ions were removed by reaction with the A- ions from the salt.  Only the most able candidates answered the calculation in part (c)(ii) correctly. Many gained a mark for the hydrogen ion concentration at pH = 3.59, but few were able to use or, more particularly, to rearrange the expression for Ka correctly to calculate the concentration of A-.

E8.          Most candidates scored full marks for a correct calculation in part (a). Parts (b)(i) and (ii) were also well answered but (b)(iii) proved to be more difficult and many candidates failed to state that, when half the acid had been neutralised, [A–] = [HA]. Only the very best candidates stated correctly that on dilution the ratio [A–] : [HA] remains constant and, as a consequence, [H+] also remains constant. The calculations in part (d) were generally well done although one mark was lost by candidates who, in (d)(ii), calculated the total final volume of acid, 50.0 cm3, and not the volume of water which must be added as required by the question.

E9.Most of this question proved to be straightforward and few candidates lost marks in part (a). Part (c) was the most discriminating section with only the better candidates scoring all six marks. Many others either forgot that sulfuric acid is diprotic and so missed that the final mixture in the questions was acidic or failed to convert the number of moles of H+ ions present into a concentration in order to find pH.

E10.The standard expressions in part (a)(i) were answered well by almost all candidates, although some quote values for Kw and pH instead of reading the question carefully.
In part (a)(ii) some missed the essential point that in pure water, the concentrations of H+ and OH–ions are equal at all temperatures even if the pH scale is based on a value of Kw different from 1.0 –14. In part (a)(iii) most were able to calculate the number of moles of OH– present as 1.0 × 10–3, but many failed to realise that, in part (iv), when this number of moles were added to 998 cm3 of water, the concentration of OH– in the solution formed would be 1.0 × 10–3 mol dm–3. A few failed to use the given value of Kw but many were able to calculate the correct answer of 10.40.

Part (b) was generally very well answered and a pH of 2.89 was the most common answer. Some candidates, gave the answer 5.77 by forgetting to take the necessary square root in their working.

Part (c) was a buffer calculation, an area which candidates traditionally find difficult. Disappointingly, this again proved the case despite the lead in to the question given in parts (i) and (ii). While most were able to calculate the initial number of moles of propanoic acid and almost as many could calculate the number of moles of acid present in the buffer solution formed, few gained many more marks. Many tried to calculate the concentrations of the acid and the salt formed failing to realise that the volume involved cancels. Many were unable to rearrange the expression for Ka and many confused the number of moles of propanoic acid with the number of moles of H+ present inn the solution.

E11.          Part (a) was a straightforward question and was answered well. High marks were also gained in part (b), although some candidates lost the mark in part (i) for the expression for Ka because they wrote the less general expression [H+]2/[HX] instead of [H+] [X–]/[HX]. The calculation in part (ii) was usually well done. The mark in part (a)(iii) was awarded for a value of pKa correctly calculated from their Kain part (ii).
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In part (c), only the most able realised that, because the reaction was between potassium hydroxide and sulphuric acid, the mole ratio of reactants was 2:1. Hence although the answers to part (c)(i) and (ii) were usually correct, very few gained the mark in part (iii). For many, this was their only error in part (b). Others however failed to use the total volume of solution, 48 cm3, correctly in calculating the concentration of hydroxide ions in part (iv) and/or were unable to use Kw or pOH to calculate the pH of the mixture. It was disappointing to see that a considerable number of candidates did not give the pH to two decimal places despite being asked to do so.

E12.          As usual with questions involving rate equations, this question was answered well, particularly parts (a) and (c). Part (b) however was poorly answered and only the very best candidates gained this mark. Many of the rest only divided the original rate by 2 instead of by 4 as the concentrations of both the ester and of If” ions were halved. Again it was disappointing to see the number of candidates who did not give the pH to two decimal places despite being asked to do so.

E15.          As expected, there were very few incorrect answers to parts (a)(i) and (ii). However, in part (a)(iii), some candidates assumed that [H+] = [OH–] instead of using all the data in the question. Part (b) was generally well attempted, but common mistakes in part (b)(i) included incorrect formulae for ethanoic acid and the ethanoate ion, missing brackets and giving [H+]2 as the numerator.
Part (c)(i) proved to be difficult. While most candidates were able to calculate the initial numbers of moles of the acid and alkali, a large proportion could not rearrange the Ka expression correctly or assumed that the amount of acid had not changed on addition of the alkali. Very few realised that this was a half neutralisation and their attempts to calculate the concentrations of acid and salt were often wrong and sometimes included the term [H+]2.
Part (c)(ii) was more successfully attempted, but still many candidates scored only 2 marks, either because they did not realise that some of the alkali would be neutralised by adding acid or because they failed to use the total volume to convert the number of moles of OH– into a concentration.

E16.          The early parts of this question were answered well, although in part (a)(iii), about a third of the candidates did not realise that adding water had no effect on the number of moles of acid present. Over two thirds of the candidates scored full marks in part 2(b); a common error for the others was to rearrange the expression for Ka incorrectly. In part 2(c), many answers stated general properties of a buffer solution, without giving any explanation specific to this situation. 
It was pleasing to see that over 44% of the candidates gained full marks in part (d)(i). Many of the rest thought that addition of the salt to the solution of the acid changed the amount of acid present. Part (d)(ii) discriminated well and only the better candidates were able to calculate the amounts in moles of acid and of salt present in the solution after the addition of sodium hydroxide and then use these values correctly. 

E17.Part (a) proved to be the most demanding question in the paper. The great majority of candidates appreciated that calibration improves accuracy but did not explain why it is necessary in the first place. At A2 candidates should be wary of providing an answer such as ’to get a more accurate result’ without an appropriate qualification.
The expression in part (b) was widely known and many candidates were able to complete the calculation in part (c) in full.
In part (c) the great majority of candidates remembered to record the equilibrium constant to three significant figures.
Part (d) and part (e) defeated many candidates. The likelihood of oxidation by oxygen in the air seemed to be elusive while most candidates thought the iron(II) aqua ion to be a strong acid and estimated the pH accordingly.

E18.This question was generally well attempted, with most students scoring well in the mathematical sections. In part (a), most students gave the meaning of a weak acid as ‘only partially ionised or dissociated into ions’, and this was credited. The preferred answer was ߢonly slightly ionised’. ‘Not fully ionised’ did not gain the mark. The equation in part (b) was poorly answered mainly because the formula for sodium carbonate was incorrect. In some cases, the formula of sodium hydroxide or sodium hydrogen carbonate was given. A common error in part (c) was to omit the factor of 2 when calculating the concentration of hydroxide ions from the concentration of calcium hydroxide. Parts (d)(i) and (d)(ii) were answered well but in part (d)(iii) many students used an expression for Kc with [H+]2 or could not rearrange the correct expression to find the concentration of sodium benzenecarboxylate. Students should be encouraged to use words in their answers to make it clear whether the numbers they write refer to amounts in moles or to concentrations and also to which substance they refer. Answers to part (e) included many wrong substances and showed that some students think that H2 (or H+) is present in the atmosphere or that carbonic acid can exist in an alkaline solution.

E19.All of the pH calculations in this question were answered well. In part (a), more students than expected had difficulty calculating the [H+] after dilution but a mark was allowed for a correct calculation of pH from their [H+]. In part (b)(ii), some students struggled to convert the pH into [H+] but over three quarters of the entry gained all three marks. In part (c), common mistakes included using the initial number of moles of OH– rather than those remaining after partial neutralisation and also not using the total volume to convert the number of moles of excess hydroxide into a concentration. The degree of clarity of some students’ working was also disappointing; many wrote strings of numbers with no explanation of what they were calculating. As usual, students who made errors in their calculations or gave a wrong answer with no explanation or working could gain few marks.

 

E20.Question (a) was generally well done, but a large number of students did not include a reversible sign in their equation for dissociation of the weak acid. Part (b) differentiated well as weaker students failed to show how ethanoic acid could act as a base and accept a proton.
Part (c)(i) was more straightforward but some students failed to calculate the concentration of the acid correctly and just used a number of moles. Part (c)(ii) differentiated well: many students realised a subtraction was needed in the calculation, but then failed to use the total volume to convert the amount, in moles, of H+ ions to a concentration to calculate the pH.
Part (d)(i) was completed very well with most students gaining either 3 or 4 marks. In part (d)(iii) a significant number incorrectly used the original amount in moles of acid (0.0125) in their ratio. Even some of the strongest students struggled with the maths skills required and a frequent wrong answer was the inverse of 1.20 due to incorrect rearrangement. A significant number did not work out the hydrogen ion concentration and so lost M1 when they couldn’t complete the rest of the calculation.
Part (e) differentiated well with full descriptions seen in about a third of the answers and one mark gained by another third. In part (f) the first mark proved very accessible but the second proved trickier; many students showed poor maths skills and thought that Ka would equal 1 or zero.

E21.          Part (a) was well known, but part (b) was much less well understood. In part (c), most students scored the first two marks but then failed to use their answers correctly to calculate the total volume. Some who did not read the question fully, added 25 cm3 to their correct answer and so lost the final mark. 
In part (d)(iii), a significant number could not correctly rearrange the simplified expression for Kc. A mark was allowed, in this part only, for a correct calculation of pH using their value of [H+] even if this was not the correct concentration.
In part (e), most scored the first two marks, but many then failed to recognise that, on adding alkali, the number of moles of acid would have decreased. Others were unable to rearrange the Ka expression correctly or attempted to use the equation Ka = [H+]2/[HX].
Moles and concentration were often confused and many responses were also badly presented with little explanation. Such omission makes it difficult for examiners to award marks in an extended calculation if the answer is wrong.

 

E22.Students found part (a) relatively easy. The most common errors were in part (a)(iii) where some students squared the hydrogen ion concentration.
The double arrow sign for an equilibrium was frequently missed in part (b)(i) and there were several instances where the wrong proton was lost from HCOOH. Only 33% of students scored all 3 marks in part (b)(iv). Some students lost M1 for confusing higher [H+] with higher pH. Some lost M3 because they wrote an explanation in general terms, “according to Le Chatelier’s principle” rather than suggesting that the position of equilibrium shifts to oppose a specific change, ie to reduce the temperature.
Part (c) was a good discriminator and the full range of marks was seen, although there were more than usual who gained high marks in what has, in the past, been seen as a difficult topic. In part (c)(i), some students couldn’t rearrange the Ka expression correctly and wrote the acid and salt concentrations the wrong way round. Many calculated an ‘excess’ of methanoic acid by subtraction.
The calculation in part (c)(ii) proved to be the most difficult calculation on the paper. Most students scored M1, but several wrote nothing further. Many also gained M2 although adding and subtracting the wrong way round caused many to lose any further marks. Incorrect rearrangement of the Ka expression was again a problem.

E23.Although parts (a) and (b) were answered well, a considerable number of students again had difficulty in rearranging the expression for Ka in part (b)(ii). Part (c)(i) was well done, although HCl was a common wrong product. Part (c)(ii) proved to be difficult and was poorly understood by many students. In part (d), the indicator for the strong acid – strong base combination was surprisingly least well answered.
Many students found part (e) difficult and the question discriminated well. Despite the instruction to assume that sulfuric acid was fully dissociated, the factor of 2 was often omitted completely or alternatively used in the wrong place. Also, the number of moles of H+ was often used instead of the concentration of H+ to find the pH.

E24.The first part of this question was answered well and few students did not score all three marks. The most common error was in the rearrangement of the Ka expression. Some students subtracted the ethanoate concentration from the ethanoic acid concentration to give a wrong ethanoic acid concentration, although they could gain M3 for a correct pH consequential on their H+ concentration.
Part (b) discriminated slightly better as there was scope for a variety of errors. These included using the values given in part (a), no addition or error in addition to give the new concentration of ethanoic acid, no subtraction or error in subtraction to give the new concentration of the ethanoate salt, wrong rearrangement of the Ka expression or a combination of these. Students were given credit for correct working where appropriate.

E25.The answers to part (a) were such that few scored both marks. Not only was the reason not well known, but a lot of wrong chemistry was written. Common errors included 'H₄O is completely ionised' and 'the concentration of water is 1'.
The answers to part (b) were slightly better and most students stated that the ionisation of water is an endothermic reaction, but a high number failed either to appreciate that this ionisation produces an equilibrium mixture or to explain that the equilibrium position moves to the RHS to oppose the increase in temperature.
Most students answered part (c) and part (d) well. A significant number of students used the Kw value for 25°C, but this was only penalised once in the two parts. A small number failed to give their answer to the accuracy specified in the question (2 dp).

E26.Students showed a good understanding of this topic and part (a) was answered very well.
In part (b)(ii) some answers only showed one colour rather than the two seen in a colour change. Part (b)(iii) was answered correctly by a narrow majority of students.

E28.          Many candidates scored well in parts (a) and (b) but very few seemed confident in dealing with quantitative buffer calculations in part (c). The main errors in part (a)(i) were to write the expression for Ka as
Ka =  or  rather than 
The first is not a general statement while the second does not show that the acid is ethanoic. Despite the loss of this first mark many candidates were able to gain all the remaining marks in part (a). Marks were lost by those who did not take the square root to obtain the hydrogen ion concentration and also by those who gave the pH to one decimal place only. It is important that pH values are given to two decimal places; these two figures give the value to the appropriate precision.
The numbers before the decimal point only give the powers of ten involved in the hydrogen ion concentration. Part (b)(i) was the best answered part of the question and most candidates gained both marks for the calculation. In part (b)(ii) most realised that the titration involved a strong base and a weak acid but the indicator chosen was often bromothymol blue rather than thymol blue, despite the correct explanation that the equivalence point must be above pH 7 for this combination of acid and base. Part (c) was not attempted by a large number of candidates. Of those who did attempt the question, most calculated the number of moles of sodium hydroxide but then used Kw in their calculation as if there was a mixture of strong acid and strong base present. Of those who did realise that an acid buffer had been produced by the reaction of an excess of ethanoic acid with sodium hydroxide, a large proportion forgot to subtract the number of moles of acid reacted from the initial number of moles of acid before substituting in the expression for Ka and gave a pH of 4.12. This answer received four marks out of the six available for the correct answer, 4.23.