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

2026-11-10

S5 Chemistry – Acid-base titration and buffers

S5 Chemistry paper: 6 questions, 22 parts, 100 marks, 60 min. pH, titration calculations, two-stage carbonate titration, indicators and buffers; mark schemes included.

Download　Question paper PDF　|　Answer key PDF

Section A – pH of strong acids, strong bases and weak acids

Take Kw = 1.0 x 10^-14 at 25 C. Show formula, substitution, value and unit; give answers to 3 significant figures.

#	Question	Answer and marking notes
1	Calculate the pH of a 0.050 M hydrochloric acid solution.	HCl is a strong acid and ionises completely: [H+] = 0.050 M; pH = -log(0.050) = 1.30. (4 marks: formula 1; substitution 1; value 1; significant figures 1)
2	Calculate the pH of a 0.020 M sodium hydroxide solution.	pOH = -log(0.020) = 1.70; pH = 14.00 – 1.70 = 12.30. (4 marks: pOH formula 1; pOH value 1; conversion 1; unit 1)
3	Calculate the pH of a 0.10 M ethanoic acid solution (Ka = 1.8 x 10^-5).	Let [H+] = x, Ka = x^2 / (0.10 – x); since x is much less than 0.10, take x^2 / 0.10 = 1.8 x 10^-5, so x = 1.34e-03 M; pH = -log(1.34e-03) = 2.87. (4 marks: Ka expression 1; approximation 1; [H+] 1; pH 1)
4	From the previous part, find the percentage ionisation of ethanoic acid and state whether the approximation used is justified.	Percentage ionisation = 1.34e-03 / 0.10 x 100% = 1.34%; since 1.34% is far less than 5%, the assumption 0.10 – x = 0.10 is justified and the error is negligible. (5 marks: definition 1; value 2; judgement 1; reason 1)

Section B – Dilution, mixing and neutralisation

Assume volumes are additive and the temperature stays at 25 C.

#	Question	Answer and marking notes
5	25.0 cm³ of 0.100 M hydrochloric acid is diluted with water to 250.0 cm³. Find the new concentration and the pH.	Moles unchanged: 2.50 x 10^-3 mol; concentration = 2.50 x 10^-3 / 0.2500 = 0.0100 M; pH = -log(0.0100) = 2.00. (4 marks: mole conservation 1; concentration 1; pH formula 1; value 1)
6	50.0 cm³ of 0.100 M hydrochloric acid is mixed with 30.0 cm³ of 0.100 M sodium hydroxide. Find the pH of the mixture.	n(HCl) = 5.00 x 10^-3 mol, n(NaOH) = 3.00 x 10^-3 mol; acid is in excess by 2.00 x 10^-3 mol in 80.0 cm³; [H+] = 0.0250 M; pH = 1.60. (5 marks: moles 1; excess 1; volume 1; concentration 1; pH 1)
7	A 0.020 M sodium hydroxide solution is diluted 100 times. Find the pH after dilution.	Concentration becomes 2.0e-04 M; pOH = 3.70; pH = 14.00 – 3.70 = 10.30. (4 marks: concentration 1; pOH 1; conversion 1; value 1)
8	20.0 cm³ of 0.100 M sulphuric acid is exactly neutralised by 0.100 M sodium hydroxide. Find the volume of alkali needed and the moles of water formed.	n(H2SO4) = 2.00 x 10^-3 mol giving 4.00 x 10^-3 mol H+; volume of NaOH = 40.0 cm³; water formed = 4.00e-03 mol. (5 marks: moles of H+ 1; 1:1 ratio 1; volume 2; water 1)

Section C – Titration calculations

25.0 cm³ of sodium hydroxide of unknown concentration is titrated with 0.100 M hydrochloric acid using methyl orange.

#	Question	Answer and marking notes
9	The three titre values are 24.60 cm³, 24.50 cm³ and 24.70 cm³. Find the mean titre.	Mean = (24.60 + 24.50 + 24.70) / 3 = 24.60 cm³. (4 marks: sum 1; divide by 3 1; value 1; unit 1)
10	Find the number of moles of hydrochloric acid used.	n = c x V = 0.100 x 24.60 / 1000 = 2.46e-03 mol. (5 marks: formula 1; dm³ conversion 1; substitution 1; value 2)
11	HCl reacts with NaOH in a 1:1 ratio. Find the concentration of the sodium hydroxide solution.	n(NaOH) = 2.46e-03 mol; c = 2.46e-03 / 0.0250 = 0.0984 M. (5 marks: 1:1 ratio 1; formula 1; substitution 1; value 2)
12	If 250.0 cm³ of this sodium hydroxide solution is used, find the mass of sodium hydroxide present (M(NaOH) = 40.0 g mol⁻¹).	n = 0.0984 x 0.2500 = 0.0246 mol; m = 0.0246 x 40.0 = 0.984 g. (6 marks: n = cV 1; moles 2; multiply by molar mass 1; value 1; unit 1)

Section D – Two-stage titration of sodium carbonate

25.0 cm³ of 0.100 M sodium carbonate is titrated with 0.100 M hydrochloric acid. Take the molar volume of gas as 24.0 dm³ mol⁻¹ at room conditions.

#	Question	Answer and marking notes
13	Write the ionic equation for the first stage and find the volume of acid needed to reach the first end point.	CO3^2- + H+ -> HCO3^-; n(Na2CO3) = 2.50e-03 mol, reacting 1:1 with H+, so volume = 25.0 cm³. (4 marks: ionic equation 2; moles 1; volume 1)
14	Find the total volume of acid used by the second end point (methyl orange end point).	HCO3^- + H+ -> H2CO3 -> H2O + CO2; the second stage also needs 25.0 cm³, total 50.0 cm³. (4 marks: second stage 2; total 1; unit 1)
15	Find the volume of carbon dioxide released over the whole titration at room conditions, in cm³.	n(CO2) = 2.50e-03 mol; V = 0.060 dm³ = 60.0 cm³. (5 marks: mole ratio 1; formula 1; dm³ value 2; conversion 1)
16	Explain why phenolphthalein is chosen for the first stage but methyl orange must be used for the second.	At the first end point the solution is mainly HCO3^- with pH about 8.3, inside the phenolphthalein range 8.3-10.0; at the second end point the CO2-saturated solution has pH about 3.9, only inside the methyl orange range 3.1-4.4. Phenolphthalein is already colourless after the first end point, so it cannot show the second. (4 marks: first stage 1; pH 1; range 1; failure of phenolphthalein 1)

Section E – Titration curves and choice of indicator

Indicator ranges: methyl orange 3.1-4.4; phenolphthalein 8.3-10.0.

#	Question	Answer and marking notes
17	0.100 M hydrochloric acid is titrated with 0.100 M sodium hydroxide. Find the pH at the equivalence point and state whether both indicators are suitable.	Only NaCl remains, so pH = 7.00. The sharp pH change runs from about 3 to 11 and spans both ranges, so both indicators are suitable. (4 marks: pH 1; neutral salt 1; sharp change 1; conclusion 1)
18	0.10 M ethanoic acid is titrated with 0.10 M sodium hydroxide (Ka = 1.8 x 10^-5). Find the pH at the equivalence point and choose a suitable indicator.	Sodium ethanoate of 0.050 M is formed; Kb = 5.56e-10; [OH-] = 5.27e-06 M; pOH = 5.28, pH = 8.72, so phenolphthalein is suitable (methyl orange would change colour far too early). (5 marks: salt concentration 1; Kb 1; [OH-] 1; pH 1; indicator 1)
19	0.10 M aqueous ammonia is titrated with 0.10 M hydrochloric acid (Kb(NH3) = 1.8 x 10^-5). Find the equivalence pH and explain why phenolphthalein cannot be used.	NH4Cl of 0.050 M is formed; Ka(NH4+) = 5.56e-10; [H+] = 5.27e-06 M; pH = 5.28, acidic, so methyl orange is suitable. Phenolphthalein stays colourless below pH 8.3, so no end point can be seen. (5 marks: Ka 1; [H+] 1; pH 1; indicator 1; failure 1)

Section F – Buffer solutions and the ionic product of water

For ethanoic acid Ka = 1.8 x 10^-5 at 25 C; for a buffer use [H+] = Ka x [acid] / [salt].

#	Question	Answer and marking notes
20	Equal volumes of 0.10 M ethanoic acid and 0.10 M sodium ethanoate are mixed. Find the pH.	After mixing both are 0.050 M, still 1:1; [H+] = Ka x 0.050 / 0.050 = 1.80e-05 M; pH = 4.74 = pKa. (5 marks: concentrations 1; ratio 1; formula 1; [H+] 1; pH 1)
21	0.010 mol of hydrogen chloride gas is added to 1.00 dm³ of that buffer (volume unchanged). Find the new pH and the change in pH.	The added H+ converts 0.010 mol of ethanoate into acid: [acid] = 0.11 M, [salt] = 0.09 M; [H+] = 2.20e-05 M; pH = 4.66, a drop of only 0.087. The same amount of acid in pure water would give pH 2, so the buffer works. (4 marks: new concentrations 2; pH 1; change 1)
22	At 50 C, Kw = 5.5 x 10^-14. Find the pH of pure water at that temperature and state whether water is still neutral.	[H+] = [OH-] = 2.35e-07 M; pH = 6.63. Since [H+] still equals [OH-], water remains neutral even though pH is below 7; the rule that pH 7 means neutral only holds at 25 C. (5 marks: square root 1; [H+] 1; pH 1; neutrality 1; temperature 1)

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.