DSE Chemistry Practice Paper: Acids, Bases And Titration (12 Questions, 50 Marks, 75 Minutes)

DSE Chemistry Practice Paper: Acids, Bases and Titration (12 Questions, 50 Marks, 75 Minutes)

2026-11-02

DSE Chemistry: Acids, Bases and Titration

DSE Chemistry paper: 12 questions, 50 marks, 75 minutes on pH, titration sums, weak acid Ka, buffers, purity analysis and enthalpy of neutralisation. Full mark schemes.

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Paper details　Level: S6 | Subject: Chemistry | Time: 75 minutes | Marks: 50 | Questions: 12

Section A (2 marks each, 12 marks)

Answer all questions. Show your working and give units.

#	Question	Answer and marking notes
1	A strong acid solution has a hydrogen ion concentration of 0.01 mol dm-3 at 25 C. (a) Find its pH. (b) Find the pH after the solution is diluted ten times.	(a) pH = -log(0.01) = 2. (b) After ten-fold dilution [H+] = 1.0 x 10-3 mol dm-3, so pH = 3, a rise of one unit. Marks: correct negative logarithm (1); both pH values (1). Common error: dropping the minus sign or assuming dilution lowers the pH.
2	At 25 C a solution has a hydroxide ion concentration of 2.5 x 10-3 mol dm-3. Find its pOH and pH, and state whether it is acidic, alkaline or neutral.	pOH = -log(2.5 x 10-3) = 2.60; pH = 14.00 – 2.60 = 11.40, so the solution is alkaline since pH is above 7. Marks: pOH (1); pH and the acidic/alkaline conclusion (1). Common error: forgetting pH + pOH = 14.00 at 25 C.
3	25.0 cm3 of 0.200 mol dm-3 hydrochloric acid is diluted with water to a total volume of 500.0 cm3. (a) Find the amount of HCl before dilution. (b) Find the pH of the diluted solution.	(a) n(HCl) = 0.200 x 25.0 / 1000 = 0.005 mol. (b) [H+] = 0.005 / 0.500 = 0.01 mol dm-3, so pH = 2. Marks: amount (1); pH (1). Common error: using 25.0 cm3 instead of 500.0 cm3 for the concentration.
4	25.0 cm3 of a sodium hydroxide solution contains 3.00 x 10-3 mol of NaOH. (a) Find its molar concentration. (b) How many moles of hydrogen ions are needed to neutralise it?	(a) [NaOH] = 3.00 x 10-3 / 0.0250 = 0.12 mol dm-3. (b) NaOH reacts with H+ in a 1 : 1 ratio, so 3.00 x 10-3 mol of H+ is required. Marks: concentration (1); amount of H+ (1). Remember to convert cm3 to dm3 first.
5	25.0 cm3 of dilute hydrochloric acid is titrated with 0.100 mol dm-3 sodium hydroxide; 21.40 cm3 is required to reach the end point. (a) Write the equation. (b) Find the concentration of the acid.	(a) NaOH + HCl -> NaCl + H2O. (b) n(NaOH) = 0.100 x 21.40 / 1000 = 0.00214 mol; the ratio is 1 : 1, so [HCl] = 0.00214 / 0.0250 = 0.0856 mol dm-3. Marks: equation (1); concentration with units (1). Common error: treating the titre as the volume of the acid being titrated.
6	A 0.100 mol dm-3 ethanoic acid solution has Ka = 1.8 x 10-5 at 25 C. (a) Find [H+] and the pH. (b) Find the degree of dissociation as a percentage.	(a) Let [H+] = x; Ka = x2 / (0.1 – x) is approximately x2 / 0.1, so x = sqrt(1.8 x 10-5 x 0.100) = 1.34e-03 mol dm-3 and pH = 2.87. (b) Degree of dissociation = 1.34e-03 / 0.100 x 100% = 1.34%. Marks: [H+] (1); pH (1). Common error: treating ethanoic acid as a strong acid gives pH 1.00, which is wrong.

Section B (4 marks each, 12 marks)

Show full working. Give answers to three significant figures.

#	Question	Answer and marking notes
7	25.0 cm3 of dilute sulphuric acid is titrated with 0.160 mol dm-3 sodium hydroxide; 22.50 cm3 is required. (a) Write the balanced equation. (b) Find the amount of NaOH used. (c) Find the concentration of the acid.	(a) H2SO4 + 2NaOH -> Na2SO4 + 2H2O. (b) n(NaOH) = 0.160 x 22.50 / 1000 = 0.0036 mol. (c) From the equation n(H2SO4) = 0.0036 / 2 = 0.0018 mol, so [H2SO4] = 0.0018 / 0.0250 = 0.072 mol dm-3. Marks: equation (1); amount of NaOH (1); dividing by 2 and the final concentration (2). Common error: forgetting that sulphuric acid is diprotic.
8	A 0.0500 mol dm-3 solution of a weak monobasic acid HA has pH 3.00 at 25 C. (a) Find [H+]. (b) Find Ka for HA. (c) Find the percentage dissociation of HA.	(a) [H+] = 10-3.00 = 1.00e-03 mol dm-3. (b) [A-] is approximately [H+] = 1.00e-03 and [HA] = 0.05 – 0 = 0.049, so Ka = (1.00e-03)2 / 0.049 = 2.04e-05. (c) Percentage dissociation = 1.00e-03 / 0.05 x 100% = 2%. Marks: Ka expression (2); value (1); percentage (1). Common error: ignoring the acid already dissociated in the denominator.
9	Equal volumes of 0.100 mol dm-3 HA and 0.100 mol dm-3 NaA are mixed to form a buffer (Ka of HA = 1.74 x 10-5 at 25 C). (a) Find the pH of the buffer. (b) 0.0050 mol of HCl is added to 1.00 dm3 of the buffer. Find the new pH, ignoring volume change.	(a) After mixing, [HA] = [A-] = 0.0500 mol dm-3, so pH = pKa + log([A-]/[HA]) = 4.76 + 0 = 4.76. (b) The added acid converts A- into HA: n(HA) = 0.100 + 0.0050 = 0.105 mol, n(A-) = 0.100 – 0.0050 = 0.095 mol, so pH = 4.76 + log(0.095/0.105) = 4.72, a fall of only about 0.04. Marks: buffer equation and pH (2); new pH (2). Common error: taking -log of the added acid concentration and ignoring the buffer action.

Section C (Questions 10 to 12, 26 marks)

Long questions: answer in parts and explain the principles you use.

#	Question	Answer and marking notes
10	A 1.50 g sample of impure sodium carbonate is dissolved and made up to 250.0 cm3. A 25.0 cm3 portion is titrated with 0.100 mol dm-3 hydrochloric acid using methyl orange; the mean titre is 23.50 cm3. (a) Write the balanced equation. (b) Find the amount of HCl used. (c) Find the amount of Na2CO3 in the 25.0 cm3 portion and hence in the whole 250.0 cm3 flask. (d) Find the mass and the percentage purity of Na2CO3 (molar mass of Na2CO3 = 106 g mol-1). (e) Suggest one practical improvement that would increase the accuracy.	(a) Na2CO3 + 2HCl -> 2NaCl + H2O + CO2. (b) n(HCl) = 0.100 x 23.50 / 1000 = 0.00235 mol. (c) The ratio is 1 : 2, so n(Na2CO3) in 25.0 cm3 = 0.00235 / 2 = 0.00118 mol, and the whole flask holds ten times this amount, 0.0118 mol. (d) m(Na2CO3) = 0.0118 x 106 = 1.246 g, so purity = 1.246 / 1.5 x 100% = 83.03%. (e) Improvements: add the acid dropwise near the end point and use half-drops, rinse the flask wall with distilled water, repeat the titration at least three times and average the results. Marks: equation (1); amount of HCl (1); amounts in portion and flask (2); mass and purity (2); improvement (2). Common error: omitting the factor of ten, or assuming a 1 : 1 ratio.
11	25.0 cm3 of 0.100 mol dm-3 ethanoic acid (Ka = 1.74 x 10-5 at 25 C) is titrated with 0.100 mol dm-3 sodium hydroxide. (a) Find the volume of alkali needed to reach the equivalence point. (b) Find the pH at the half-equivalence point and explain your answer. (c) Calculate the pH at the equivalence point. (d) Choose a suitable indicator from phenolphthalein (8.3-10.0) and methyl orange (3.1-4.4), and explain your choice. (e) Explain why the curve is relatively flat before the equivalence point.	(a) n(CH3COOH) = 0.100 x 25.0 / 1000 = 0.0025 mol and neutralisation is 1 : 1, so the volume of alkali needed is 0.0025 / 0.100 x 1000 = 25 cm3. (b) At the half-equivalence point [CH3COOH] equals [CH3COO-], so pH = pKa = 4.76; the mixture is a buffer, which is why the pH is close to pKa. (c) At the equivalence point the total volume is 50.0 cm3, so [CH3COO-] = 0.0025 / 0.0500 = 0.05 mol dm-3. Kb = Kw / Ka = 1.00 x 10-14 / 1.74e-05 = 5.747e-10; [OH-] = sqrt(Kb x 0.05) = 5.36e-06 mol dm-3, pOH = 5.27, so pH = 8.73. (d) Phenolphthalein is suitable because the equivalence pH of 8.73 lies inside its range 8.3-10.0; methyl orange changes near pH 4, which would make the end point appear far too early and give a low result. (e) Before the equivalence point the solution contains both undissociated acid and its conjugate base, forming a buffer that removes most added hydroxide ions, so the pH rises only slowly; the pH shoots up once the buffer capacity is used up near the equivalence point. Marks: (a) 1; (b) 2; (c) 3; (d) 2; (e) 1. Common error: assuming the equivalence pH is 7.00, which is true only for strong acid with strong base.
12	50.0 cm3 of 1.00 mol dm-3 hydrochloric acid and 50.0 cm3 of 1.00 mol dm-3 sodium hydroxide are mixed in a polystyrene cup and the temperature rises by 6.8 C. Assume the solution density is 1.00 g cm-3 and its specific heat capacity is 4.18 J g-1 K-1. (a) Find the amounts of H+ and OH- mixed. (b) Find the heat released in J. (c) Find the enthalpy change of neutralisation in kJ mol-1. (d) The literature value is -57.3 kJ mol-1; give two reasons why the experimental value is smaller. (e) State one assumption made in the calculation.	(a) n(H+) = 1.00 x 50.0 / 1000 = 0.05 mol and n(OH-) = 0.05 mol, so the two are exactly neutralised. (b) Mass of solution = 100.0 g, so Q = mc(dT) = 100.0 x 4.18 x 6.8 = 2842.4 J = 2.84 kJ. (c) n(H2O) formed = 0.05 mol, so dH = -Q / n = -2842.4 / 0.05 J mol-1 = -56848.0 J mol-1, about -56.8 kJ mol-1. (d) Heat is lost to the surroundings and to the cup wall; the thermometer, stirrer and cup absorb heat, so the measured temperature rise is too small; the solutions are not infinitely dilute so the heat of dilution is not accounted for (any two). (e) Assumption: the mixture has the density and specific heat capacity of pure water, the reaction goes to completion and no heat is lost. Marks: (a) 1; (b) 2; (c) 3 including sign and units; (d) 2; (e) 1. Common error: omitting the negative sign, or failing to convert J to kJ.

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