A Level H1 Chemistry Acids Bases Salts Quiz

A-Level Chemistry H1 Quiz - Acids Bases Salts

Answer Key

Section A: Multiple Choice

1. Answer: B

A weak acid is defined as an acid that partially dissociates in aqueous solution. This is a fundamental distinction in acid-base chemistry. The term "weak" refers to the degree of dissociation, not the concentration of the acid. A dilute strong acid can have a higher pH than a concentrated weak acid. Option A is incorrect because a weak acid does not necessarily have a very low pH — it depends on both strength and concentration. Option C is incorrect because weak acids do conduct electricity, albeit poorly, due to the presence of some ions. Option D confuses strength with concentration.

[1 mark]

2. Answer: B

In pure water, $[H^+] = [OH^-]$. Since $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$, it follows that $[H^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}$ mol dm⁻³. Therefore, $pH = -\log(1.0 \times 10^{-14}) = 7$. Pure water is neutral at 298 K.

[1 mark]

3. Answer: C

Adding dilute hydrochloric acid introduces $H^+$ ions. In the buffer, the ethanoate ions ($CH_3COO^-$) react with the added $H^+$ to form more ethanoic acid, but the addition of $H^+$ still causes a net increase in $[H^+]$, which decreases the pH. Option A would increase the pH (more conjugate base shifts equilibrium to the left, reducing $[H^+]$). Option B would increase the pH (OH⁻ reacts with the weak acid, producing more conjugate base). Option D, dilution, does not significantly change the pH of a buffer because the ratio $[A^-]/[HA]$ remains constant.

[1 mark]

4. Answer: C

Hydrochloric acid is a strong acid and sodium hydroxide is a strong base. At the equivalence point, the salt formed is sodium chloride ($NaCl$), which is derived from a strong acid and a strong base. The ions $Na^+$ and $Cl^-$ do not hydrolyse in water, so the solution is neutral with pH = 7.

[1 mark]

5. Answer: C

Sodium ethanoate is the salt of a weak acid (ethanoic acid) and a strong base (sodium hydroxide). When dissolved in water, the ethanoate ion ($CH_3COO^-$) undergoes hydrolysis:

$CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)$

This produces $OH^-$ ions, making the solution alkaline (pH > 7). Sodium chloride and potassium nitrate are salts of strong acids and strong bases (pH = 7). Ammonium chloride is a salt of a weak base and a strong acid (pH < 7).

[1 mark]

Section B: Structured Questions

6. (a) A strong acid is an acid that completely dissociates (or ionises) in aqueous solution.

[1 mark]

Common mistake: Students sometimes say "fully ionises in water" — this is acceptable. Saying "dissociates into H⁺ ions" without specifying "completely" is incomplete.

(b) $H_2SO_4(aq) \rightarrow 2H^+(aq) + SO_4^{2-}(aq)$

or equivalently: $H_2SO_4(aq) + 2H_2O(l) \rightarrow 2H_3O^+(aq) + SO_4^{2-}(aq)$

[1 mark] — Award the mark for correct products and state symbols. The equation must show complete dissociation (single arrow, not reversible).

(c) Hydrochloric acid is a strong acid and dissociates completely in water, producing a high concentration of $H^+$ ions. Ethanoic acid is a weak acid and only partially dissociates, so at the same concentration (0.10 mol dm⁻³), it produces a lower concentration of $H^+$ ions. Since $pH = -\log[H^+]$, a lower $[H^+]$ means a higher pH (less acidic). Therefore, the HCl solution has a lower pH.

[2 marks] — 1 mark for identifying HCl as strong and ethanoic acid as weak. 1 mark for linking degree of dissociation to $[H^+]$ and hence pH.

7. Nitric acid ($HNO_3$) is a strong acid, so it dissociates completely:

$HNO_3(aq) \rightarrow H^+(aq) + NO_3^-(aq)$

Therefore, $[H^+] = 0.050$ mol dm⁻³.

$pH = -\log[H^+] = -\log(0.050) = -\log(5.0 \times 10^{-2})$

$= -(\log 5.0 + \log 10^{-2}) = -(0.70 - 2) = 1.30$

Answer: pH = 1.30

[2 marks] — 1 mark for stating that $[H^+] = 0.050$ mol dm⁻³ (complete dissociation). 1 mark for correct pH calculation.

Common mistake: Students who treat $HNO_3$ as a weak acid and try to use $K_a$ will get the wrong answer. Always identify whether the acid is strong or weak first.

8. (a) $K_a = \frac{[C_2H_5COO^-][H^+]}{[C_2H_5COOH]}$

[1 mark] — The expression must have products over reactant. State symbols are not required in $K_a$ expressions.

(b) For a weak acid, $K_a = \frac{[H^+]^2}{[HA]}$ (assuming $[H^+] = [A^-]$ and $[HA] \approx$ initial concentration).

$[H^+]^2 = K_a \times [HA] = 1.3 \times 10^{-5} \times 0.15 = 1.95 \times 10^{-6}$

$[H^+] = \sqrt{1.95 \times 10^{-6}} = 1.396 \times 10^{-3} \text{ mol dm}^{-3}$

$pH = -\log(1.396 \times 10^{-3}) = 2.85$

Answer: pH = 2.85

[3 marks] — 1 mark for correct $K_a$ expression rearrangement. 1 mark for correct substitution. 1 mark for correct final answer (accept 2.85–2.86).

Common mistake: Forgetting to take the square root, or using the wrong concentration. Also, students sometimes write $[H^+] = K_a \times [HA]$ without squaring — this gives an incorrect answer.

9. (a) Neutralisation (or acid-base reaction).

[1 mark]

(b) Moles of $CH_3COOH$ initially = $0.10 \times 1.0 = 0.10$ mol. Moles of $NaOH$ added = 0.020 mol.

The reaction is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

NaOH is the limiting reagent. After reaction:

• Moles of $CH_3COOH$ remaining = $0.10 - 0.020 = 0.080$ mol
• Moles of $CH_3COONa$ (conjugate base) formed = 0.020 mol

The resulting solution contains both a weak acid and its conjugate base, so it is a buffer solution.

[2 marks] — 1 mark for calculating remaining moles of weak acid and moles of conjugate base. 1 mark for concluding it is a buffer because both components are present.

10. (a) Total volume of mixture = $50.0 + 50.0 = 100.0$ cm³ = 0.100 dm³.

$[CH_3COOH] = \frac{0.20 \times 0.050}{0.100} = 0.10 \text{ mol dm}^{-3}$

$[CH_3COO^-] = \frac{0.30 \times 0.050}{0.100} = 0.15 \text{ mol dm}^{-3}$

[2 marks] — 1 mark for each correct concentration.

(b) Using the Henderson-Hasselbalch equation:

$pH = pK_a + \log\frac{[A^-]}{[HA]}$

$pK_a = -\log(1.8 \times 10^{-5}) = 4.74$

$pH = 4.74 + \log\frac{0.15}{0.10} = 4.74 + \log(1.5) = 4.74 + 0.18 = 4.92$

Answer: pH = 4.92

[2 marks] — 1 mark for correct $pK_a$ calculation. 1 mark for correct substitution and final answer.

Alternative method: Using $K_a$ expression directly:

$K_a = \frac{[H^+][A^-]}{[HA]}$, so $[H^+] = K_a \times \frac{[HA]}{[A^-]} = 1.8 \times 10^{-5} \times \frac{0.10}{0.15} = 1.2 \times 10^{-5}$

$pH = -\log(1.2 \times 10^{-5}) = 4.92$

11. Sodium carbonate ($Na_2CO_3$) is a salt of a strong base (NaOH) and a weak acid ($H_2CO_3$). The carbonate ion ($CO_3^{2-}$) undergoes hydrolysis in water:

$CO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HCO_3^-(aq) + OH^-(aq)$

This produces $OH^-$ ions, making the solution alkaline (pH > 7), so universal indicator turns blue-purple.

Ammonium chloride ($NH_4Cl$) is a salt of a weak base ($NH_3$) and a strong acid ($HCl$). The ammonium ion ($NH_4^+$) undergoes hydrolysis:

$NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)$

This produces $H^+$ (or $H_3O^+$) ions, making the solution acidic (pH < 7), so universal indicator turns orange-red.

[3 marks] — 1 mark for identifying sodium carbonate as a salt of strong base/weak acid and explaining alkaline pH. 1 mark for identifying ammonium chloride as a salt of weak base/strong acid and explaining acidic pH. 1 mark for correct hydrolysis equations.

12. Reagents: Hydrochloric acid and sodium hydroxide solution.

Procedure:

1. Using a pipette, transfer 25.0 cm³ of a known concentration of sodium hydroxide solution into a conical flask.
2. Add 2–3 drops of a suitable indicator (e.g., phenolphthalein) to the conical flask.
3. Fill a burette with hydrochloric acid of known concentration.
4. Add the acid from the burette to the sodium hydroxide solution in the conical flask, swirling continuously, until the indicator changes colour at the endpoint (e.g., phenolphthalein changes from pink to colourless).
5. Record the volume of acid used. Repeat the titration until concordant results are obtained.

Obtaining pure crystals: 6. Using the exact volume of acid needed (from the titration), mix the acid and alkali in the correct proportions without adding indicator. 7. Pour the resulting sodium chloride solution into an evaporating basin. 8. Heat the solution gently to evaporate most of the water until the solution is saturated (crystals begin to form at the edges). 9. Allow the saturated solution to cool slowly so that crystals form. 10. Filter the crystals, wash with a small amount of cold distilled water, and dry between filter papers or in a warm oven.

[3 marks] — 1 mark for correct reagents and titration procedure. 1 mark for using results without indicator to prepare the salt. 1 mark for correct crystallisation method (evaporate, cool, filter, dry).

13. Neutralisation is the reaction between $H^+(aq)$ ions (from an acid) and $OH^-(aq)$ ions (from a base) to form water:

$H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

[1 mark] — The answer must refer to $H^+$ and $OH^-$ ions forming water. Simply saying "acid reacts with base" is insufficient at A-Level.

14. (a) $Ca(OH)_2(aq) \rightarrow Ca^{2+}(aq) + 2OH^-(aq)$

[1 mark] — Must show 2 moles of $OH^-$ per mole of $Ca(OH)_2$.

(b) From the equation: $[OH^-] = 2 \times 0.025 = 0.050$ mol dm⁻³.

$pOH = -\log[OH^-] = -\log(0.050) = 1.30$

$pH = 14.00 - 1.30 = 12.70$

Answer: pH = 12.70

[3 marks] — 1 mark for calculating $[OH^-] = 0.050$ mol dm⁻³. 1 mark for calculating $pOH$. 1 mark for correct pH.

Common mistake: Forgetting that $Ca(OH)_2$ produces 2 moles of $OH^-$ per mole of compound. Students who use $[OH^-] = 0.025$ will get pH = 12.40, which is incorrect.

15. In the ammonia–ammonium chloride buffer, the equilibrium is:

$NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$

When a small amount of HCl is added, the $H^+$ ions react with the $NH_3$ (the base component of the buffer):

$NH_3 + H^+ \rightarrow NH_4^+$

The buffer absorbs the added $H^+$ by converting it into $NH_4^+$, so the change in $[H^+]$ (and hence pH) is very small.

In pure water, there is no buffer system. The added $H^+$ ions directly increase $[H^+]$, causing a sharp drop in pH. For example, adding enough HCl to make $[H^+] = 0.001$ mol dm⁻³ to pure water gives pH = 3, a dramatic change from pH 7.

[3 marks] — 1 mark for explaining that added $H^+$ reacts with $NH_3$ in the buffer. 1 mark for stating that the buffer minimises the pH change. 1 mark for contrasting with pure water where added $H^+$ causes a sharp pH drop.

Section C: Application and Data Interpretation

16. (a) The titration curve should show:

• Starting pH around 2.9 (weak acid).
• A gradual rise through the buffer region (approximately 5–20 cm³), with a relatively flat section around pH 4.7 (the $pK_a$ region, at half-equivalence).
• A steep rise near 25.0 cm³ (equivalence point).
• Equivalence point at pH ≈ 11.0 (above 7 because the salt of a weak acid and strong base is alkaline).
• A levelling off at high pH as excess NaOH is added.

The equivalence point should be labelled at approximately 25.0 cm³, pH 11.0. The buffer region should be labelled in the relatively flat portion before the steep rise (approximately 5–20 cm³).

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Generated image for this question.

(b) From the data, the pH at the equivalence point (25.0 cm³) is 11.0.

[1 mark]

(c) At the equivalence point, all the ethanoic acid has been neutralised to form sodium ethanoate ($CH_3COONa$). Sodium ethanoate is the salt of a weak acid and a strong base. The ethanoate ion ($CH_3COO^-$) undergoes hydrolysis:

$CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)$

This produces $OH^-$ ions, making the solution alkaline, so the pH is greater than 7.

[2 marks] — 1 mark for identifying the salt as being from a weak acid/strong base. 1 mark for explaining hydrolysis producing $OH^-$ ions.

(d) Phenolphthalein is a suitable indicator. Its colour change range is approximately pH 8.2–10.0, which falls within the steep portion of the titration curve near the equivalence point. This means the colour change will occur very close to the true equivalence point, giving an accurate result.

[2 marks] — 1 mark for naming phenolphthalein (or any indicator with range within the steep region). 1 mark for explaining that its pH range falls within the steep portion of the curve.

17. (a) Weak acid: $H_2CO_3$ (carbonic acid)

Conjugate base: $HCO_3^-$ (hydrogen carbonate ion)

[1 mark] — Both must be correct.

(b) When excess $H^+$ enters the blood, it reacts with the hydrogen carbonate ion (the conjugate base):

$HCO_3^-(aq) + H^+(aq) \rightarrow H_2CO_3(aq)$

The equilibrium shifts to the right (or the added $H^+$ is removed by reaction with $HCO_3^-$), so the increase in $[H^+]$ is minimised and the pH remains relatively stable.

[2 marks] — 1 mark for the correct equation showing $H^+$ reacting with $HCO_3^-$. 1 mark for explaining that this removes the added $H^+$ and minimises pH change.

(c) When excess $OH^-$ enters the blood, it reacts with the carbonic acid (the weak acid component):

$H_2CO_3(aq) + OH^-(aq) \rightarrow HCO_3^-(aq) + H_2O(l)$

Alternatively, $OH^-$ reacts with $H^+$ from the equilibrium, causing the equilibrium to shift to the left to replace the $H^+$, and the $H_2CO_3$ dissociates to replenish $H^+$. The net effect is that the added $OH^-$ is neutralised and the pH remains relatively stable.

[2 marks] — 1 mark for the correct equation or explanation. 1 mark for explaining that the buffer minimises the pH change.

18. (a) Titration 2 and Titration 3 both give 24.30 cm³ (concordant). Titration 1 (24.30 cm³) is also concordant. Titration 3 (24.40 cm³) is very close. The rough titration (24.80 cm³) is discarded.

Average titre = $\frac{24.30 + 24.30 + 24.40}{3} = \frac{73.00}{3} = 24.33$ cm³

(Alternatively, using only titrations 1 and 2: $\frac{24.30 + 24.30}{2} = 24.30$ cm³ — both approaches are acceptable.)

Answer: 24.33 cm³ (or 24.30 cm³)

[1 mark]

(b) The equation for the reaction is:

$HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$

The mole ratio is 1:1.

$c_{HCl} \times V_{HCl} = c_{NaOH} \times V_{NaOH}$

$c_{HCl} \times 25.0 = 0.100 \times 24.33$

$c_{HCl} = \frac{0.100 \times 24.33}{25.0} = 0.0973 \text{ mol dm}^{-3}$

Answer: 0.0973 mol dm⁻³ (or 0.0972 mol dm⁻³ if using 24.30 cm³)

[3 marks] — 1 mark for correct equation and 1:1 ratio. 1 mark for correct substitution. 1 mark for correct answer with appropriate significant figures.

19. (a) $NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$

[1 mark] — Must show the reversible arrow (equilibrium) since ammonia is a weak base.

(b) Given: pH = 11.1, so $pOH = 14.0 - 11.1 = 2.90$

$[OH^-] = 10^{-2.90} = 1.259 \times 10^{-3} \text{ mol dm}^{-3}$

At equilibrium: $[NH_4^+] = [OH^-] = 1.259 \times 10^{-3}$ mol dm⁻³

$[NH_3] = 0.10 - 1.259 \times 10^{-3} \approx 0.0987$ mol dm⁻³ (or approximately 0.10 mol dm⁻³)

$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = \frac{(1.259 \times 10^{-3})^2}{0.0987}$

$= \frac{1.585 \times 10^{-6}}{0.0987} = 1.61 \times 10^{-5} \text{ mol dm}^{-3}$

Answer: $K_b = 1.6 \times 10^{-5}$ mol dm⁻³ (accept $1.6 \times 10^{-5}$ to $1.8 \times 10^{-5}$ depending on approximation used)

[4 marks] — 1 mark for calculating $[OH^-]$ from pH. 1 mark for identifying $[NH_4^+] = [OH^-]$. 1 mark for correct $K_b$ expression and substitution. 1 mark for correct final answer.

20. (a) Calcium carbonate (limestone) or calcium hydroxide (slaked lime) or calcium oxide (quicklime).

[1 mark] — Any one correct answer.

(b) Calcium carbonate reacts with the acid (e.g., $H^+$) in the soil:

$CaCO_3(s) + 2H^+(aq) \rightarrow Ca^{2+}(aq) + H_2O(l) + CO_2(g)$

This removes $H^+$ ions from the soil, thereby raising the pH (making it less acidic).

[1 mark] — Correct equation showing acid neutralisation.

(c) Sodium hydroxide is a strong base and is highly corrosive. It is difficult to control the amount added, and excess NaOH would make the soil too alkaline, which is also harmful to crops. Additionally, NaOH is expensive for large-scale agricultural use and can damage soil structure and harm microorganisms. Calcium carbonate (limestone) is preferred because it is cheap, safe to handle, and insoluble in water, so it does not easily over-treat the soil.

[2 marks] — 1 mark for stating that NaOH is too corrosive/dangerous or would make soil too alkaline. 1 mark for explaining why an alternative (e.g., limestone) is more practical (cheap, safe, or self-regulating).

END OF ANSWER KEY