S5 Chemistry Paper: Acid-Base PH, Titration And Buffers (100 Marks)

S5 Chemistry Paper: Acid-Base pH, Titration and Buffers (100 marks)

2026-11-11

S5 Chemistry – acids, bases and titration

S5 Chemistry paper: acid pH and dilution, weak acid Ka, titration stoichiometry, titration curves, salt hydrolysis and buffers, with full answers.

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Paper details　S5 Chemistry practice paper with six structured questions on strong acid dilution and pH, weak acid Ka and degree of dissociation, titration stoichiometry and concentration, titration curves and indicator choice, salt hydrolysis, and buffer pH calculations, with full answers and marking notes.

Part A: pH, dilution and neutralisation of strong acids

Part A (20 marks). Show all working; give pH to two decimal places.

#	Question	Answer and marking notes
1	25.0 cm3 of 0.100 mol/dm3 hydrochloric acid is diluted to 250.0 cm3 in a volumetric flask. (a) Calculate the amount of hydrogen ions in the diluted solution.	Answer: n(H+) = 0.100 x 25.0 / 1000 = 2.50 x 10-3 mol. (3 marks: formula 1, substitution 1, answer with unit 1)
2	(b) Calculate the pH of the diluted solution.	Answer: [H+] = 2.50 x 10-3 / 0.2500 = 1.00 x 10-2 mol/dm3, pH = 2.00. (3 marks)
3	(c) The diluted solution is further diluted ten times. State the change in pH and explain.	Answer: [H+] falls to one tenth, so pH rises by 1.00, from 2.00 to 3.00. (4 marks: value 2, explanation 2)
4	(d) Calculate the pH of 0.0050 mol/dm3 barium hydroxide solution.	Answer: [OH-] = 2 x 0.0050 = 0.0100 mol/dm3, pOH = 2.00, pH = 12.00. (5 marks)
5	(e) 25.0 cm3 of 0.100 mol/dm3 hydrochloric acid is mixed with 30.0 cm3 of 0.100 mol/dm3 sodium hydroxide. Calculate the pH of the mixture.	Answer: base in excess by 5.0 x 10-4 mol in 55.0 cm3, so [OH-] = 9.09 x 10-3 mol/dm3, pOH = 2.04, pH = 11.96. (5 marks)

Part B: Ka, pH and degree of dissociation of a weak acid

Part B (18 marks). The approximation Ka = [H+]2 / [HA] may be used; state the assumptions.

#	Question	Answer and marking notes
6	(a) Write the Ka expression for a weak acid HA and state the two assumptions made when calculating [H+].	Answer: Ka = [H+][A-] / [HA]; water self-ionisation is negligible, and [H+] is approximately equal to [A-], so the equilibrium [HA] is taken as the initial concentration. (3 marks)
7	(b) For 0.100 mol/dm3 HA with Ka = 1.8 x 10-5, calculate [H+] and the pH.	Answer: [H+] = sqrt(1.8 x 10-5 x 0.100) = 1.34 x 10-3 mol/dm3, pH = 2.87. (5 marks)
8	(c) Calculate the degree of dissociation of this solution as a percentage.	Answer: 1.34 x 10-3 / 0.100 x 100% = 1.34%. (4 marks)
9	(d) The solution is diluted ten times. Calculate the new pH and the new degree of dissociation, and explain the direction of each change.	Answer: [H+] = sqrt(1.8 x 10-6) = 4.24 x 10-4 mol/dm3, pH = 3.37; degree of dissociation 4.24%. Dilution shifts the equilibrium to the right, so the degree of dissociation rises while [H+] still falls and pH rises by 0.50. (6 marks)

Part C: Titration stoichiometry and concentration determination

Part C (18 marks). 25.0 cm3 of sodium hydroxide solution of unknown concentration is titrated with 0.120 mol/dm3 hydrochloric acid; the mean titre is 21.35 cm3.

#	Question	Answer and marking notes
10	(a) Write the ionic equation for the neutralisation and state the mole ratio of acid to alkali.	Answer: H+ + OH- -> H2O; the mole ratio is 1 : 1. (4 marks: equation 3, ratio 1)
11	(b) Calculate the concentration of the sodium hydroxide solution.	Answer: n(HCl) = 0.120 x 21.35 / 1000 = 2.56 x 10-3 mol; [NaOH] = 2.56 x 10-3 / 0.0250 = 0.102 mol/dm3. (5 marks)
12	(c) The alkali sample was originally 250.0 cm3. Calculate the mass of sodium hydroxide it contains (relative atomic masses: Na 23.0, O 16.0, H 1.0).	Answer: n = 0.1025 x 0.250 = 0.0256 mol; m = 0.0256 x 40.0 = 1.02 g. (5 marks)
13	(d) Explain why the conical flask is rinsed with distilled water but not with the sodium hydroxide solution.	Answer: residual water does not change the amount of alkali in the flask, so the result is unaffected; rinsing with the alkali would add extra alkali, consume more acid and give a concentration that is too high. (4 marks)

Part D: Titration curves and the choice of indicator

Part D (16 marks). Answer for both strong acid with strong base, and weak acid with strong base.

#	Question	Answer and marking notes
14	(a) Describe the shape of the titration curve for a strong acid with a strong base and the feature near the equivalence point.	Answer: the pH starts low, rises slowly, then shows a steep vertical section from about pH 3 to about pH 11 at the equivalence point, and finally approaches the pH of the base. (4 marks)
15	(b) State the pH at the equivalence point for a strong acid with a strong base and identify the main species present.	Answer: pH = 7.00 at 25 C; the main species are Na+, Cl- and H2O, with no hydrolysis. (4 marks)
16	(c) For 0.050 mol/dm3 sodium ethanoate at the equivalence point, calculate the pH of a weak acid with strong base titration (Ka = 1.8 x 10-5).	Answer: Kb = Kw / Ka = 5.56 x 10-10; [OH-] = sqrt(5.56 x 10-10 x 0.050) = 5.27 x 10-6 mol/dm3; pOH = 5.28, pH = 8.72. (5 marks)
17	(d) For the titration in (c), name the most suitable indicator and give the reason.	Answer: phenolphthalein (range about pH 8.3 to 10.0), because its range covers the steep section around pH 8.72; methyl orange changes in the acidic range and is unsuitable. (3 marks)

Part E: Salt hydrolysis and the acidity of solutions

Part E (14 marks). Consider 0.10 mol/dm3 ammonium chloride solution (Kb for ammonia = 1.8 x 10-5).

#	Question	Answer and marking notes
18	(a) Calculate the pH of the solution.	Answer: Ka(NH4+) = Kw / Kb = 5.56 x 10-10; [H+] = sqrt(5.56 x 10-10 x 0.10) = 7.45 x 10-6 mol/dm3; pH = 5.13. (5 marks)
19	(b) Use an ionic equation to explain why the solution is acidic.	Answer: NH4+ + H2O = NH3 + H3O+ produces hydrated hydrogen ions, while chloride ions do not hydrolyse, so the solution is acidic. (4 marks: equation 2, explanation 2)
20	(c) The solution is diluted ten times. Calculate the new pH and explain why the pH rises.	Answer: [H+] = sqrt(5.56 x 10-11) = 2.36 x 10-6 mol/dm3, pH = 5.63; dilution raises the degree of hydrolysis but [H+] still falls, so pH rises by 0.50. (5 marks)

Part F: Buffer calculations and buffer action

Part F (14 marks). 500.0 cm3 of a buffer contains 0.050 mol ethanoic acid and 0.030 mol sodium ethanoate (Ka = 1.8 x 10-5).

#	Question	Answer and marking notes
21	(a) Calculate the pH of the buffer.	Answer: pKa = 4.74; pH = 4.74 + log(0.060 / 0.100) = 4.52. (5 marks: pKa 2, ratio 1, pH 2)
22	(b) Explain why the pH falls only slightly when a small amount of strong acid is added.	Answer: added H+ combines with ethanoate ions to form ethanoic acid (CH3COO- + H+ -> CH3COOH), so H+ is removed, and the undissociated acid suppresses further ionisation; hence the pH changes very little. (4 marks)
23	(c) 0.0050 mol of hydrogen chloride is added and the volume change is negligible. Calculate the new pH and the change in pH.	Answer: the acid becomes 0.055 mol and the ethanoate ion 0.025 mol, so pH = 4.74 + log(0.025 / 0.055) = 4.40, a fall of 0.12. (5 marks)

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