AS Chemistry: Acids and Bases — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Acids and Bases revision notes

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Questions

1. Define a Brønsted–Lowry acid and base. [2]

2. Identify the two conjugate acid–base pairs in: HNO₂ + H₂O ⇌ NO₂⁻ + H₃O⁺ [2]

3. Explain the difference between a strong acid and a concentrated acid. [2]

4. Calculate the pH of:

(a) 0.050 mol dm⁻³ hydrochloric acid [2] (b) 0.100 mol dm⁻³ sodium hydroxide (K_w = 1.0 × 10⁻¹⁴) [3]

5. Ethanoic acid is a weak acid with K_a = 1.75 × 10⁻⁵ mol dm⁻³.

(a) Write the expression for K_a. [1] (b) Calculate the pH of a 0.100 mol dm⁻³ solution. [3] (c) State two assumptions made in the calculation. [2]

6. Explain how a buffer solution made from ethanoic acid and sodium ethanoate resists a change in pH when a small amount of acid is added. [3]

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Answers

1. An acid is a proton (H⁺) donor [1]; a base is a proton acceptor [1].

2. HNO₂ / NO₂⁻ [1]; H₃O⁺ / H₂O [1].

3. Strong refers to the degree of dissociation — a strong acid dissociates completely [1]. Concentrated refers to the amount of acid per unit volume of solution [1]. A concentrated weak acid and a dilute strong acid are both perfectly possible.

4. (a) HCl is strong, so [H⁺] = 0.050 mol dm⁻³ [1] pH = −log(0.050) = 1.30 [1].

(b) [OH⁻] = 0.100 mol dm⁻³ [H⁺] = K_w ÷ [OH⁻] = 1.0 × 10⁻¹⁴ ÷ 0.100 [1] = 1.0 × 10⁻¹³ [1] pH = 13.0 [1].

5. (a) K_a = [H⁺][CH₃COO⁻] ÷ [CH₃COOH] [1].

(b) [H⁺] = √(K_a × c) = √(1.75 × 10⁻⁵ × 0.100) [1] = √(1.75 × 10⁻⁶) = 1.32 × 10⁻³ [1] pH = 2.88 [1].

(c) [H⁺] = [CH₃COO⁻], i.e. the dissociation of water is negligible [1]; the concentration of undissociated acid is approximately equal to the initial concentration, since dissociation is slight [1].

6. The solution contains a large reserve of ethanoate ions from the salt [1]. Added H⁺ ions react with the ethanoate ions to form undissociated ethanoic acid: CH₃COO⁻ + H⁺ → CH₃COOH [1]. The added H⁺ is therefore removed from solution, so the pH changes only slightly [1].

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Additional questions

7. State which indicator (methyl orange or phenolphthalein) is suitable for a titration between a strong acid and a weak base, and explain why. [3]

8. Explain, using an equation, how the same ethanoic acid/sodium ethanoate buffer described in question 6 resists a change in pH when a small amount of alkali (OH⁻) is added instead. [3]

9. A student dilutes a solution of a strong, fully-dissociated acid — starting well above neutral acidity, so that [H⁺] from the acid is much greater than the 10⁻⁷ mol dm⁻³ from water’s own autoionisation — by a factor of 10. State and explain the effect on the pH. [2]

Answers to additional questions

7. Methyl orange (range 3.1–4.4) [1]. A strong acid + weak base titration has its vertical section over a lower, more acidic pH range (roughly pH 3–7), and a suitable indicator must have its full colour-change range falling entirely within that vertical section [1] [1] — phenolphthalein’s range (8.3–10.0) falls outside it entirely, so it would change colour too early or not show a sharp end point.

8. The buffer contains a large reserve of undissociated ethanoic acid [1]. Added OH⁻ ions react with this ethanoic acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O [1]. The added OH⁻ is therefore removed from solution as it is converted to water, so the pH rises only slightly [1].

9. The pH increases by 1 [1]. Diluting a strong, fully-dissociated acid by a factor of 10 reduces [H⁺] by a factor of 10, and since pH = −log[H⁺], each tenfold decrease in [H⁺] increases the pH by exactly 1 [1]. (This exact one-unit rule assumes the acid’s own [H⁺] stays much greater than the ~10⁻⁷ mol dm⁻³ contributed by water itself — for extremely dilute strong acid approaching neutral pH, water’s autoionisation becomes significant and the simple rule breaks down.)

A note on titration curves and indicator choice

Questions 7 and, by extension, the acid-base calculations in questions 4-5 all connect to the same underlying idea: the equivalence point of a titration is not always pH 7 — it depends on the salt formed by the specific acid and base combined. A strong acid with a strong base gives a neutral salt and an equivalence point at pH 7, but a strong acid with a weak base gives a salt that hydrolyses to produce an acidic solution, shifting the equivalence point below 7, while a weak acid with a strong base shifts it above 7. An indicator is only suitable for a given titration if its own colour-change range sits entirely within the steep, vertical section of that specific titration’s curve, which is why the same indicator cannot be used for every acid-base combination.

A note on weak-acid pH calculations

Question 5’s approach relies on two assumptions that are worth understanding rather than just stating: that the dissociation of water itself is negligible compared with the acid’s own dissociation, and that the equilibrium concentration of undissociated acid is approximately equal to its initial concentration, since a weak acid only dissociates slightly. These assumptions are what allow the simplified square-root formula to be used instead of solving the full equilibrium expression exactly, and they are also why this method cannot be applied to a strong acid, which dissociates completely rather than only slightly – using the weak-acid method on a strong acid, or vice versa, is one of the most common and avoidable errors in this topic.

Where marks are usually lost

• Confusing strong with concentrated.
• Forgetting to use K_w when calculating the pH of an alkali.
• Using the strong-acid method for a weak acid.
• Not stating both assumptions in a weak-acid pH calculation.
• Explaining a buffer without naming the reserve species.