S5 Chemistry Practice Paper: Acids, Bases, pH and Titration

4.0 g of sodium hydroxide (M = 40.0) is dissolved in water to make 250 cm³ of solution.

(a) Find the concentration of the solution in mol/dm³.

(b) 25.0 cm³ of this solution is diluted to 500 cm³. Find the new concentration.

(a) n(NaOH) = 4.0 ÷ 40.0 = 0.100 mol, so c = 0.100 ÷ 0.250 = 0.400 mol/dm³.

(b) The number of moles is unchanged on dilution: 0.400 × 0.0250 = 0.0100 mol, so c = 0.0100 ÷ 0.500 = 0.0200 mol/dm³.

Marking: 2 marks for the moles, 2 for the concentration, 2 for the moles after dilution, 2 for the answer with units.

Calculate the pH of the following solutions at 25 °C, assuming complete dissociation.

(a) 0.010 mol/dm³ hydrochloric acid.

(b) 0.010 mol/dm³ sodium hydroxide solution.

(c) The acid in (a) after dilution by a factor of 1000.

(a) [H⁺] = 0.010 mol/dm³, pH = -log 0.010 = 2.00.

(b) [OH^–] = 0.010 mol/dm³, pOH = 2.00, pH = 14.00 – 2.00 = 12.00.

(c) [H⁺] = 0.010 ÷ 1000 = 1.0 × 10^-5 mol/dm³, pH = 5.00.

Marking: 2 marks for the hydrogen ion concentration and 1 for the pH in each part; part (b) requires the use of Kw.

25.0 cm³ of sodium hydroxide solution of unknown concentration requires 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid to reach the end point.

(a) Write the ionic equation for the neutralisation.

(b) Find the concentration of the sodium hydroxide solution.

(a) H⁺ + OH^– → H₂O.

(b) n(HCl) = 0.100 × 0.0200 = 2.00 × 10^-3 mol; the mole ratio is 1 : 1, so n(NaOH) is the same; c(NaOH) = 2.00 × 10^-3 ÷ 0.0250 = 0.0800 mol/dm³.

Marking: 2 marks for the ionic equation, 2 for the moles of acid, 2 for the mole ratio, 2 for the concentration with units.

The Ka of 0.10 mol/dm³ ethanoic acid is 1.8 × 10^-5 at 25 °C.

(a) Find the concentration of hydrogen ions in the solution.

(b) Find the pH, correct to 2 decimal places.

(c) Find the degree of dissociation as a percentage, correct to 1 decimal place.

(a) [H⁺] = √(Ka × c) = √(1.8 × 10^-5 × 0.10) = √(1.8 × 10^-6) ≈ 1.34 × 10^-3 mol/dm³.

(b) pH = -log(1.34 × 10^-3) ≈ 2.87.

(c) Degree of dissociation = 1.34 × 10^-3 ÷ 0.10 × 100% ≈ 1.3%.

Marking: 3 marks for the Ka expression, 2 for the hydrogen ion concentration, 2 for the pH, 2 for the percentage.

5.0 g of calcium carbonate (M = 100.0) reacts with excess dilute hydrochloric acid to give carbon dioxide. (Molar volume = 24.0 dm³/mol)

(a) Find the number of moles of calcium carbonate.

(b) Find the volume of carbon dioxide formed at room temperature.

(c) Find the mass of carbon dioxide formed.

(a) n(CaCO₃) = 5.0 ÷ 100.0 = 0.050 mol.

(b) CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, ratio 1 : 1, so n(CO₂) = 0.050 mol; V = 0.050 × 24.0 = 1.20 dm³.

(c) m(CO₂) = 0.050 × 44.0 = 2.2 g.

Marking: 2 marks for the moles, 2 for the equation and ratio, 2 for the volume, 2 for the mass.

Each of the following titrations has a different pH at the end point. Choose a suitable indicator for each and explain.

(a) Strong acid titrated with strong base.

(b) Strong base titrated with weak acid such as ethanoic acid.

(c) Strong acid titrated with weak base such as aqueous ammonia.

(a) Strong acid with strong base: the end point is near pH 7 and the sharp change is wide, so either phenolphthalein or methyl orange works.

(b) Strong base with weak acid: the end point is above pH 7 (about 8 to 9), so phenolphthalein is needed because its range lies in the alkaline region.

(c) Strong acid with weak base: the end point is below pH 7 (about 5 to 6), so methyl orange is needed; phenolphthalein would change colour before the end point.

Marking: 1 mark for each indicator, 2 marks each for matching the range to the end point pH.

Are the following salt solutions acidic, alkaline or neutral? Explain each.

(a) Sodium chloride. (b) Sodium carbonate. (c) Ammonium chloride. (d) Sodium ethanoate.

(a) Sodium chloride: neutral, because it is a salt of a strong acid and a strong base and does not undergo hydrolysis.

(b) Sodium carbonate: alkaline, because the carbonate ion hydrolyses to produce hydroxide ions.

(c) Ammonium chloride: acidic, because the ammonium ion hydrolyses to release hydrogen ions.

(d) Sodium ethanoate: alkaline, because the ethanoate ion hydrolyses to produce hydroxide ions.

Marking: 1 mark for each property, 1 for each hydrolysis explanation.

A student needs 250 cm³ of 0.200 mol/dm³ sodium carbonate solution.

(a) Find the number of moles of sodium carbonate required.

(b) If hydrated sodium carbonate (Na₂CO₃·10H₂O, M = 286.0) is used, find the mass needed.

(a) n(Na₂CO₃) = 0.200 × 0.250 = 0.0500 mol.

(b) The hydrated salt supplies the same number of moles, so m = 0.0500 × 286.0 = 14.3 g.

Marking: 4 marks for the moles, 2 for using the correct molar mass, 2 for the mass with units.

In acidic solution, potassium manganate(VII) reacts with iron(II) ions in the mole ratio 1 : 5. 25.0 cm³ of iron(II) sulphate solution needs 24.0 cm³ of 0.0200 mol/dm³ potassium manganate(VII) solution to reach the end point.

(a) Find the number of moles of potassium manganate(VII) used.

(b) Find the moles of iron(II) ions and the concentration of the original solution.

(a) n(KMnO₄) = 0.0200 × 0.0240 = 4.80 × 10^-4 mol.

(b) The mole ratio is 1 : 5, so n(Fe²⁺) = 4.80 × 10^-4 × 5 = 2.40 × 10^-3 mol; c = 2.40 × 10^-3 ÷ 0.0250 = 0.0960 mol/dm³.

Marking: 3 marks for the moles, 2 for the ratio, 1 for the moles of iron(II), 2 for the concentration with units.

Equal moles of ethanoic acid and sodium ethanoate are mixed to form a buffer solution at 25 °C. The pKa of ethanoic acid is 4.74.

(a) Find the pH of the solution.

(b) Explain why the pH changes only slightly when a small amount of strong acid or strong base is added.

(a) With equal moles of the two components, pH = pKa = 4.74.

(b) Added acid is absorbed by the ethanoate ion to form ethanoic acid; added base is absorbed by ethanoic acid to form the ethanoate ion. Both are taken up by the reservoir components, so the pH changes only slightly.

Marking: 4 marks for the pH, 2 for the explanation of acid absorption, 2 for base absorption.

Compare 0.10 mol/dm³ hydrochloric acid with 0.10 mol/dm³ ethanoic acid at the same temperature.

(a) Which has the lower pH? Explain.

(b) Which conducts electricity better? Explain.

(c) Which reacts faster at first with a magnesium ribbon? Is the total number of moles of hydrogen the same?

(a) Hydrochloric acid has the lower pH (pH = 1.00); ethanoic acid is only partly dissociated (pH ≈ 2.87) so its hydrogen ion concentration is lower.

(b) Hydrochloric acid conducts better because it is fully dissociated and the ion concentration is higher.

(c) Hydrochloric acid reacts faster at first because the hydrogen ion concentration is higher; however, both contain the same number of moles of acid, and ethanoic acid continues to dissociate as it is used up, so the total moles of hydrogen are the same.

Marking: 3 marks for (a), 3 for (b), 3 for the conclusion and explanation in (c).

25.0 cm³ of sulphuric acid of unknown concentration is neutralised by 30.0 cm³ of 0.150 mol/dm³ sodium hydroxide solution.

(a) Find the concentration of the sulphuric acid.

(b) Find the mass concentration of the acid in g/dm³ (M(H₂SO₄) = 98.0).

(a) n(NaOH) = 0.150 × 0.0300 = 4.50 × 10^-3 mol; H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, ratio 1 : 2, so n(H₂SO₄) = 2.25 × 10^-3 mol; c = 2.25 × 10^-3 ÷ 0.0250 = 0.0900 mol/dm³.

(b) Mass concentration = 0.0900 × 98.0 = 8.82 g/dm³.

Marking: 2 marks for the moles of alkali, 3 for the ratio, 2 for the concentration, 2 for the mass concentration.