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A Level H1 Chemistry Practice Paper 5

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Questions

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TuitionGoWhere Practice Paper – Chemistry H1 A-Level

TuitionGoWhere Exam Practice (AI)

FieldDetails
Subject:Chemistry H1 (8873)
Level:A-Level
Paper:Practice Paper – Acids, Bases & Salts
Version:5 of 5
Duration:1 hour 15 minutes
Total Marks:50

Name: ___________________________ Class: ___________ Date: _____________

Instructions to Candidates

  1. This paper consists of 20 questions in three sections.
  2. Answer all questions in the spaces provided.
  3. Show all working clearly for calculation questions. Marks are awarded for method.
  4. You may use a calculator and the A-Level Chemistry Data Booklet.
  5. State symbols and units must be included where appropriate.
  6. The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Short Answer & Definitions (Questions 1–6)

Total: 12 marks

1. Define the term Brønsted–Lowry acid. Give one example of a Brønsted–Lowry acid and write an equation to show its behaviour in aqueous solution.

[2 marks]

2. What is meant by the term weak acid? Illustrate your answer with a balanced equation, including state symbols, for the dissociation of ethanoic acid in water.

[2 marks]

3. A student measures the pH of 0.10 mol dm⁻³ hydrochloric acid and finds it to be 1.0. The pH of 0.10 mol dm⁻³ ethanoic acid is measured as 2.9. Explain why these two acids of the same concentration have different pH values.

[2 marks]

4. State the formula of the conjugate base of each of the following acids:

(a) H₂SO₄

(b) H₂PO₄⁻

[2 marks]

5. Calcium oxide (CaO) is classified as a basic oxide. Write a balanced equation, including state symbols, for the reaction of calcium oxide with:

(a) dilute hydrochloric acid

(b) water

[2 marks]

6. Identify the Period 3 element that forms an amphoteric oxide. Write one balanced equation to show this oxide reacting with an acid, and one balanced equation to show it reacting with a base.

[2 marks]

Section B: Calculations (Questions 7–14)

Total: 24 marks

7. Calculate the pH of a 0.0250 mol dm⁻³ solution of nitric acid, HNO₃.

[2 marks]

8. A solution of sodium hydroxide has a pH of 12.30 at 25 °C. Calculate the concentration of hydroxide ions, [OH⁻], in this solution. (Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C)

[2 marks]

9. A 0.150 mol dm⁻³ solution of a weak monoprotic acid, HA, has a pH of 2.85 at 25 °C.

(a) Calculate the concentration of hydrogen ions, [H⁺], in this solution.

[1 mark]

(b) Write the expression for the acid dissociation constant, Ka, of HA.

[1 mark]

(c) Calculate the value of Ka for HA, stating its units.

[2 marks]

10. In a titration, 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution required 20.0 cm³ of sulfuric acid for complete neutralisation.

The balanced equation for the reaction is: 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

Calculate the concentration of the sulfuric acid in mol dm⁻³.

[3 marks]

11. A buffer solution is prepared by mixing 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid (CH₃COOH) with 50.0 cm³ of 0.200 mol dm⁻³ sodium ethanoate (CH₃COONa). The Ka of ethanoic acid is 1.74 × 10⁻⁵ mol dm⁻³ at 25 °C.

(a) Calculate the concentration of ethanoic acid and ethanoate ions in the buffer solution after mixing.

[2 marks]

(b) Calculate the pH of this buffer solution.

[2 marks]

12. A student prepared a standard solution by dissolving 2.65 g of anhydrous sodium carbonate, Na₂CO₃, in distilled water and making the solution up to 250.0 cm³ in a volumetric flask.

(a) Calculate the amount, in moles, of Na₂CO₃ dissolved. (Mr of Na₂CO₃ = 106.0)

[1 mark]

(b) Calculate the concentration of the sodium carbonate solution in mol dm⁻³.

[1 mark]

(c) Calculate the concentration of sodium ions, [Na⁺], in this solution.

[1 mark]

13. The pH of a sample of rainwater was found to be 5.60. Carbon dioxide dissolved in the rainwater forms carbonic acid, H₂CO₃, which dissociates according to the equation:

H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)  Ka = 4.30 × 10⁻⁷ mol dm⁻³

Calculate the concentration of carbonic acid, [H₂CO₃], in the rainwater, assuming that all the acidity is due to the first dissociation of carbonic acid.

[3 marks]

14. A 0.500 g sample of an impure monoprotic weak acid, HX (Mr = 122.0), was dissolved in water and titrated with 0.100 mol dm⁻³ NaOH solution. The titre volume was 28.50 cm³.

(a) Calculate the amount, in moles, of NaOH used in the titration.

[1 mark]

(b) Calculate the mass of pure HX in the sample.

[1 mark]

(c) Calculate the percentage purity of the HX sample.

[1 mark]

Section C: Structured & Data Interpretation (Questions 15–20)

Total: 14 marks

15. The table below shows the pH values of four solutions, W, X, Y, and Z, each of concentration 0.10 mol dm⁻³.

SolutionSolutepH
WHCl1.0
XCH₃COOH2.9
YNaOH13.0
ZNH₃11.1

(a) Explain why solution W has a lower pH than solution X.

[1 mark]

(b) Explain why solution Y has a higher pH than solution Z.

[1 mark]

(c) Solution X and solution Z are mixed in equimolar proportions. State the type of solution formed and explain how it resists changes in pH when small amounts of acid are added.

[2 marks]

16. A student investigates the reaction between marble chips (calcium carbonate) and dilute hydrochloric acid. The equation for the reaction is:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

The student adds excess marble chips to 50.0 cm³ of 0.500 mol dm⁻³ HCl and collects the carbon dioxide gas produced.

(a) Calculate the amount, in moles, of HCl used.

[1 mark]

(b) Calculate the volume of CO₂ gas produced at room temperature and pressure. (Molar volume of gas at r.t.p. = 24.0 dm³ mol⁻¹)

[2 marks]

(c) Suggest why the actual volume of gas collected may be less than the calculated value.

[1 mark]

17. The graph below shows how the pH changes when 0.100 mol dm⁻³ sodium hydroxide solution is added to 25.0 cm³ of a weak acid, HA.

[Assume a typical weak acid–strong base titration curve is provided, showing an initial pH of ~3, a gradual rise, a steep rise around pH 6–10, and an equivalence point at pH ~8.5.]

(a) Use the graph to estimate the pH at the equivalence point.

[1 mark]

(b) Explain why the pH at the equivalence point is greater than 7.

[1 mark]

(c) A suitable indicator for this titration must change colour within the pH range of the vertical portion of the graph. Suggest one suitable indicator for this titration, giving a reason for your choice.

[2 marks]

18. Ammonium chloride, NH₄Cl, is a salt that dissolves in water to form an acidic solution.

(a) Write an equation to show the dissociation of ammonium chloride in water.

[1 mark]

(b) Write an equation to show why the resulting solution is acidic.

[1 mark]

(c) Explain, using your equations, why ammonium chloride solution has a pH less than 7.

[1 mark]

19. A student is asked to prepare a buffer solution of pH 4.50 using ethanoic acid (CH₃COOH, Ka = 1.74 × 10⁻⁵ mol dm⁻³) and sodium ethanoate (CH₃COONa).

(a) Calculate the ratio [CH₃COO⁻] / [CH₃COOH] required to achieve this pH.

[2 marks]

(b) Describe how the student could prepare 100 cm³ of this buffer solution in the laboratory, given 0.10 mol dm⁻³ solutions of both ethanoic acid and sodium ethanoate.

[2 marks]

20. A 0.0100 mol dm⁻³ solution of barium hydroxide, Ba(OH)₂, is prepared.

(a) Write an equation for the complete dissociation of barium hydroxide in water.

[1 mark]

(b) Calculate the pH of this barium hydroxide solution at 25 °C.

[2 marks]

END OF PAPER

Check your work carefully. Ensure all answers are in the spaces provided.

Answers

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TuitionGoWhere Practice Paper – Chemistry H1 A-Level

Answer Key & Marking Scheme

Paper: Practice Paper – Acids, Bases & Salts (Version 5 of 5) Total Marks: 50

Section A: Short Answer & Definitions (12 marks)

1. Define the term Brønsted–Lowry acid. Give one example of a Brønsted–Lowry acid and write an equation to show its behaviour in aqueous solution.

[2 marks]

Answer:

  • A Brønsted–Lowry acid is a proton (H⁺) donor. [1]
  • Example: HCl (or any suitable acid). [½]
  • Equation: HCl(aq) → H⁺(aq) + Cl⁻(aq) OR HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq) [½]

Accept any correct acid and corresponding dissociation equation with state symbols.

2. What is meant by the term weak acid? Illustrate your answer with a balanced equation, including state symbols, for the dissociation of ethanoic acid in water.

[2 marks]

Answer:

  • A weak acid is one that only partially dissociates/ionises in aqueous solution. [1]
  • Equation: CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) [1]

Essential: reversible arrow (⇌) and state symbols. Accept H₃O⁺ instead of H⁺.

3. A student measures the pH of 0.10 mol dm⁻³ hydrochloric acid and finds it to be 1.0. The pH of 0.10 mol dm⁻³ ethanoic acid is measured as 2.9. Explain why these two acids of the same concentration have different pH values.

[2 marks]

Answer:

  • HCl is a strong acid; it dissociates completely in water, so [H⁺] = 0.10 mol dm⁻³, giving pH = 1.0. [1]
  • CH₃COOH is a weak acid; it dissociates only partially, so [H⁺] is much less than 0.10 mol dm⁻³, giving a higher pH of 2.9. [1]

Accept: reference to degree of dissociation/ionisation; equilibrium position for weak acid.

4. State the formula of the conjugate base of each of the following acids:

(a) H₂SO₄ (b) H₂PO₄⁻

[2 marks]

Answer:

  • (a) HSO₄⁻ [1]
  • (b) HPO₄²⁻ [1]

Conjugate base = acid minus one H⁺; charge decreases by 1.

5. Calcium oxide (CaO) is classified as a basic oxide. Write a balanced equation, including state symbols, for the reaction of calcium oxide with:

(a) dilute hydrochloric acid (b) water

[2 marks]

Answer:

  • (a) CaO(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) [1]
  • (b) CaO(s) + H₂O(l) → Ca(OH)₂(aq/s) [1]

Accept Ca(OH)₂ as (aq) or (s); state symbols essential.

6. Identify the Period 3 element that forms an amphoteric oxide. Write one balanced equation to show this oxide reacting with an acid, and one balanced equation to show it reacting with a base.

[2 marks]

Answer:

  • Element: Aluminium / Al [½]
  • With acid: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) [¾]
  • With base: Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq) [¾]

Accept Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄ or Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O. State symbols required for full marks.

Section B: Calculations (24 marks)

7. Calculate the pH of a 0.0250 mol dm⁻³ solution of nitric acid, HNO₃.

[2 marks]

Answer:

  • HNO₃ is a strong monoprotic acid → [H⁺] = 0.0250 mol dm⁻³ [1]
  • pH = −log₁₀[H⁺] = −log₁₀(0.0250) = 1.60 (to 2 d.p.) [1]

Method mark for using pH = −log[H⁺]; accuracy mark for correct answer.

8. A solution of sodium hydroxide has a pH of 12.30 at 25 °C. Calculate the concentration of hydroxide ions, [OH⁻], in this solution. (Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C)

[2 marks]

Answer:

  • pOH = 14.00 − pH = 14.00 − 12.30 = 1.70 [1]
  • [OH⁻] = 10⁻ᵖᴼᴴ = 10⁻¹·⁷⁰ = 0.0200 mol dm⁻³ (or 2.00 × 10⁻² mol dm⁻³) [1]

Alternative: [H⁺] = 10⁻¹²·³⁰ = 5.01 × 10⁻¹³; [OH⁻] = Kw/[H⁺] = 1.00 × 10⁻¹⁴ / 5.01 × 10⁻¹³ = 0.0200 mol dm⁻³.

9. A 0.150 mol dm⁻³ solution of a weak monoprotic acid, HA, has a pH of 2.85 at 25 °C.

(a) Calculate the concentration of hydrogen ions, [H⁺], in this solution. [1 mark]

Answer:

  • [H⁺] = 10⁻²·⁸⁵ = 1.41 × 10⁻³ mol dm⁻³ [1]

(b) Write the expression for the acid dissociation constant, Ka, of HA. [1 mark]

Answer:

  • Ka = [H⁺][A⁻] / [HA] [1]

Must have correct species; accept [H₃O⁺] for [H⁺].

(c) Calculate the value of Ka for HA, stating its units. [2 marks]

Answer:

  • [H⁺] = [A⁻] = 1.41 × 10⁻³ mol dm⁻³; [HA] ≈ 0.150 − 1.41 × 10⁻³ ≈ 0.149 mol dm⁻³ [1]
  • Ka = (1.41 × 10⁻³)² / 0.149 = 1.33 × 10⁻⁵ mol dm⁻³ [1]

Accept 1.33–1.34 × 10⁻⁵; units: mol dm⁻³.

10. In a titration, 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution required 20.0 cm³ of sulfuric acid for complete neutralisation.

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

Calculate the concentration of the sulfuric acid in mol dm⁻³.

[3 marks]

Answer:

  • n(NaOH) = c × V = 0.100 × (25.0/1000) = 0.00250 mol [1]
  • Mole ratio NaOH : H₂SO₄ = 2 : 1 → n(H₂SO₄) = 0.00250 / 2 = 0.00125 mol [1]
  • c(H₂SO₄) = n / V = 0.00125 / (20.0/1000) = 0.0625 mol dm⁻³ [1]

Method marks for correct mole calculation and ratio; accuracy for final answer.

11. A buffer solution is prepared by mixing 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid with 50.0 cm³ of 0.200 mol dm⁻³ sodium ethanoate. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

(a) Calculate the concentration of ethanoic acid and ethanoate ions in the buffer solution after mixing. [2 marks]

Answer:

  • Total volume = 100.0 cm³ (dilution factor ×2 for both) [1]
  • [CH₃COOH] = 0.200 × (50.0/100.0) = 0.100 mol dm⁻³
  • [CH₃COO⁻] = 0.200 × (50.0/100.0) = 0.100 mol dm⁻³ [1]

(b) Calculate the pH of this buffer solution. [2 marks]

Answer:

  • Ka = [H⁺][CH₃COO⁻] / [CH₃COOH] → [H⁺] = Ka × [CH₃COOH] / [CH₃COO⁻] [1]
  • [H⁺] = 1.74 × 10⁻⁵ × (0.100/0.100) = 1.74 × 10⁻⁵ mol dm⁻³
  • pH = −log₁₀(1.74 × 10⁻⁵) = 4.76 [1]

Accept pH = pKa = 4.76 when [acid] = [salt].

12. A student dissolved 2.65 g of anhydrous Na₂CO₃ in distilled water and made the solution up to 250.0 cm³. (Mr = 106.0)

(a) Calculate the amount, in moles, of Na₂CO₃ dissolved. [1 mark]

Answer:

  • n = m / Mr = 2.65 / 106.0 = 0.0250 mol [1]

(b) Calculate the concentration of the sodium carbonate solution in mol dm⁻³. [1 mark]

Answer:

  • c = n / V = 0.0250 / (250.0/1000) = 0.100 mol dm⁻³ [1]

(c) Calculate the concentration of sodium ions, [Na⁺], in this solution. [1 mark]

Answer:

  • Na₂CO₃ → 2Na⁺ + CO₃²⁻ → [Na⁺] = 2 × 0.100 = 0.200 mol dm⁻³ [1]

13. The pH of rainwater is 5.60. H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq); Ka = 4.30 × 10⁻⁷ mol dm⁻³. Calculate [H₂CO₃] in the rainwater.

[3 marks]

Answer:

  • [H⁺] = 10⁻⁵·⁶⁰ = 2.51 × 10⁻⁶ mol dm⁻³ [1]
  • [H⁺] = [HCO₃⁻] = 2.51 × 10⁻⁶ mol dm⁻³ [1]
  • Ka = [H⁺][HCO₃⁻] / [H₂CO₃] → [H₂CO₃] = (2.51 × 10⁻⁶)² / (4.30 × 10⁻⁷) = 1.47 × 10⁻⁵ mol dm⁻³ [1]

Accept 1.46–1.47 × 10⁻⁵ mol dm⁻³.

14. A 0.500 g sample of impure monoprotic weak acid HX (Mr = 122.0) was titrated with 0.100 mol dm⁻³ NaOH. Titre = 28.50 cm³.

(a) Calculate the amount, in moles, of NaOH used. [1 mark]

Answer:

  • n(NaOH) = 0.100 × (28.50/1000) = 0.00285 mol [1]

(b) Calculate the mass of pure HX in the sample. [1 mark]

Answer:

  • n(HX) = n(NaOH) = 0.00285 mol (1:1 ratio)
  • mass = n × Mr = 0.00285 × 122.0 = 0.348 g (or 0.3477 g) [1]

(c) Calculate the percentage purity of the HX sample. [1 mark]

Answer:

  • % purity = (0.348 / 0.500) × 100 = 69.6% (or 69.5%) [1]

Section C: Structured & Data Interpretation (14 marks)

15. pH data for four 0.10 mol dm⁻³ solutions.

(a) Explain why solution W (HCl, pH 1.0) has a lower pH than solution X (CH₃COOH, pH 2.9). [1 mark]

Answer:

  • HCl is a strong acid and dissociates completely, producing a higher [H⁺] than the weak acid CH₃COOH, which only partially dissociates. Lower pH corresponds to higher [H⁺]. [1]

(b) Explain why solution Y (NaOH, pH 13.0) has a higher pH than solution Z (NH₃, pH 11.1). [1 mark]

Answer:

  • NaOH is a strong base and dissociates completely, producing a higher [OH⁻] (and thus lower [H⁺], higher pH) than the weak base NH₃, which only partially ionises in water. [1]

(c) Solution X and solution Z are mixed in equimolar proportions. State the type of solution formed and explain how it resists changes in pH when small amounts of acid are added. [2 marks]

Answer:

  • Type of solution: Buffer solution (specifically, an acidic buffer). [1]
  • Explanation: The buffer contains CH₃COOH (weak acid) and CH₃COO⁻ (conjugate base, from reaction with NH₃). When acid (H⁺) is added, the conjugate base CH₃COO⁻ reacts with it: CH₃COO⁻ + H⁺ → CH₃COOH, removing added H⁺ and resisting pH change. [1]

Accept: reference to equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ shifting left upon addition of H⁺.

16. Marble chips + HCl reaction.

(a) Calculate the amount, in moles, of HCl used. [1 mark]

Answer:

  • n(HCl) = 0.500 × (50.0/1000) = 0.0250 mol [1]

(b) Calculate the volume of CO₂ gas produced at r.t.p. (Molar volume = 24.0 dm³ mol⁻¹). [2 marks]

Answer:

  • Mole ratio: 2HCl : 1CO₂ → n(CO₂) = 0.0250 / 2 = 0.0125 mol [1]
  • Volume = n × 24.0 = 0.0125 × 24.0 = 0.300 dm³ (or 300 cm³) [1]

(c) Suggest why the actual volume of gas collected may be less than the calculated value. [1 mark]

Answer:

  • Some CO₂ may dissolve in the water/aqueous solution. OR gas may escape before collection. OR reaction may not go to completion. [1]

Accept any reasonable suggestion.

17. Titration curve analysis (weak acid vs. strong base).

(a) Use the graph to estimate the pH at the equivalence point. [1 mark]

Answer:

  • pH ≈ 8.5 (accept 8.0–9.0 based on described curve) [1]

(b) Explain why the pH at the equivalence point is greater than 7. [1 mark]

Answer:

  • At the equivalence point, the solution contains the salt of a weak acid and strong base (e.g., NaA). The anion A⁻ undergoes hydrolysis: A⁻ + H₂O ⇌ HA + OH⁻, producing OH⁻ ions and making the solution alkaline (pH > 7). [1]

(c) Suggest one suitable indicator for this titration, giving a reason for your choice. [2 marks]

Answer:

  • Indicator: Phenolphthalein (pH range 8.2–10.0) OR Thymol blue (pH range 8.0–9.6). [1]
  • Reason: The indicator's pH range falls within the steep/vertical portion of the titration curve (pH ~6–10), so the colour change occurs at the equivalence point. [1]

Accept any indicator with range within the vertical section.

18. Ammonium chloride, NH₄Cl, dissolves in water to form an acidic solution.

(a) Write an equation to show the dissociation of ammonium chloride in water. [1 mark]

Answer:

  • NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq) [1]

State symbols required.

(b) Write an equation to show why the resulting solution is acidic. [1 mark]

Answer:

  • NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) [1]

Accept NH₄⁺ ⇌ NH₃ + H⁺. Reversible arrow essential.

(c) Explain, using your equations, why ammonium chloride solution has a pH less than 7. [1 mark]

Answer:

  • The NH₄⁺ ion acts as a Brønsted–Lowry acid, donating a proton to water to form H₃O⁺ (or H⁺). The presence of excess H⁺/H₃O⁺ ions makes the solution acidic (pH < 7). [1]

19. Buffer preparation (pH 4.50) using CH₃COOH and CH₃COONa. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

(a) Calculate the ratio [CH₃COO⁻] / [CH₃COOH] required. [2 marks]

Answer:

  • [H⁺] = 10⁻⁴·⁵⁰ = 3.16 × 10⁻⁵ mol dm⁻³ [1]
  • Ka = [H⁺][CH₃COO⁻] / [CH₃COOH] → [CH₃COO⁻]/[CH₃COOH] = Ka / [H⁺] = 1.74 × 10⁻⁵ / 3.16 × 10⁻⁵ = 0.551 [1]

Accept 0.55 or 0.551.

(b) Describe how the student could prepare 100 cm³ of this buffer solution in the laboratory, given 0.10 mol dm⁻³ solutions of both ethanoic acid and sodium ethanoate. [2 marks]

Answer:

  • Ratio [salt]/[acid] = 0.551. Since both stock solutions have the same concentration (0.10 mol dm⁻³), the volume ratio V(salt) / V(acid) = 0.551. [1]
  • Let V(acid) = x cm³, then V(salt) = 0.551x. Total volume = x + 0.551x = 100 → 1.551x = 100 → x = 64.5 cm³ (acid), V(salt) = 35.5 cm³.
  • Method: Measure 64.5 cm³ of 0.10 mol dm⁻³ CH₃COOH and 35.5 cm³ of 0.10 mol dm⁻³ CH₃COONa using burettes/measuring cylinders, mix in a beaker, and make up to 100 cm³ with distilled water in a volumetric flask if needed. [1]

Accept any clear description with correct volumes; allow minor rounding differences.

20. A 0.0100 mol dm⁻³ solution of Ba(OH)₂ is prepared.

(a) Write an equation for the complete dissociation of barium hydroxide in water. [1 mark]

Answer:

  • Ba(OH)₂(s) → Ba²⁺(aq) + 2OH⁻(aq) [1]

State symbols required.

(b) Calculate the pH of this barium hydroxide solution at 25 °C. [2 marks]

Answer:

  • [OH⁻] = 2 × 0.0100 = 0.0200 mol dm⁻³ [1]
  • pOH = −log₁₀(0.0200) = 1.70 → pH = 14.00 − 1.70 = 12.30 [1]

Alternative: [H⁺] = Kw/[OH⁻] = 1.00 × 10⁻¹⁴ / 0.0200 = 5.00 × 10⁻¹³; pH = −log₁₀(5.00 × 10⁻¹³) = 12.30.

END OF ANSWER KEY

Total: 50 marks