A Level H2 Chemistry Practice Paper 2

TuitionGoWhere Practice Paper - Chemistry H2 A-Level

Answer Key — Acids, Bases & Salts (Version 2 of 5)

Section A: Multiple Choice and Short Answer

1. B — $SO_4^{2-}$ [1]

Explanation: A conjugate base is formed when an acid donates a proton ($H^+$). $HSO_4^-$ loses one $H^+$ to become $SO_4^{2-}$. Option A ($H_2SO_4$) is the conjugate acid (gaining a proton), not the conjugate base. This tests understanding of the Brønsted-Lowry acid-base conjugate pair concept.

2. C — $2.51 \times 10^{-11}$ mol dm$^{-3}$ [1]

Explanation:

• pH = 3.40, so $[H^+] = 10^{-3.40} = 3.98 \times 10^{-4}$ mol dm$^{-3}$
• At 25 °C: $K_w = [H^+][OH^-] = 1.00 \times 10^{-14}$
• $[OH^-] = \frac{K_w}{[H^+]} = \frac{1.00 \times 10^{-14}}{3.98 \times 10^{-4}} = 2.51 \times 10^{-11}$ mol dm$^{-3}$

Common mistake: Students often select B, which is the $[H^+]$ value, forgetting to divide $K_w$ by $[H^+]$.

3. D — Sodium carbonate [1]

Explanation: Sodium carbonate ($Na_2CO_3$) is a salt of a strong base (NaOH) and a weak acid ($H_2CO_3$). The $CO_3^{2-}$ ion hydrolyses in water: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$, producing $OH^-$ ions and giving pH > 7. NaCl and $K_2SO_4$ are salts of strong acids and strong bases (neutral, pH = 7). $NH_4NO_3$ is a salt of a weak base and strong acid (acidic, pH < 7).

4. A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added (or when it is diluted). [2]

Marking:

• [1] for "resists changes in pH"
• [1] for "when small amounts of acid or base are added"

Explanation: A buffer typically consists of a weak acid and its conjugate base (or weak base and its conjugate acid). The weak acid neutralises added base, and the conjugate base neutralises added acid, keeping pH approximately constant.

5. pH = 0.60 [1]

Explanation:

• HCl is a strong acid, so it dissociates completely: $[H^+] = 0.25$ mol dm$^{-3}$
• $pH = -\log_{10}[H^+] = -\log_{10}(0.25) = 0.60$

Common mistake: Students sometimes forget that strong acids fully dissociate and try to use $K_a$ expressions.

6. $K_a = 1.8 \times 10^{-5}$ mol dm$^{-3}$ [3]

Step-by-step:

• pH = 2.87, so $[H^+] = 10^{-2.87} = 1.349 \times 10^{-3}$ mol dm$^{-3}$
• For the dissociation: $HA \rightleftharpoons H^+ + A^-$
• At equilibrium: $[H^+] = [A^-] = 1.349 \times 10^{-3}$ mol dm$^{-3}$
• $[HA] = 0.10 - 1.349 \times 10^{-3} \approx 0.0987$ mol dm$^{-3}$ (or ≈ 0.10 if approximation used)
• $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(1.349 \times 10^{-3})^2}{0.0987} = 1.84 \times 10^{-5}$
• To 2 s.f.: $K_a = 1.8 \times 10^{-5}$ mol dm$^{-3}$

Marking:

• [1] for correct $[H^+]$ calculation
• [1] for correct $K_a$ expression and substitution
• [1] for correct final answer to 2 s.f. with units

7. The pH increases. [1]

Explanation: Adding sodium ethanoate increases the concentration of $CH_3COO^-$ ions. By Le Chatelier's principle, the equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ shifts to the left, reducing $[H^+]$, so the pH increases. [1]

Explanation: This is a common ion effect. The added conjugate base suppresses the dissociation of the weak acid, reducing the hydrogen ion concentration and raising the pH. This is the principle behind buffer action.

8. $K_w = [H^+][OH^-]$ [1]

Value at 25 °C: $K_w = 1.00 \times 10^{-14}$ mol$^2$ dm$^{-6}$ [1]

Explanation: The ionic product of water arises from the autoionisation of water: $H_2O \rightleftharpoons H^+ + OH^-$. At 25 °C, $[H^+] = [OH^-] = 1.00 \times 10^{-7}$ mol dm$^{-3}$ in pure water, so $K_w = (1.00 \times 10^{-7})^2 = 1.00 \times 10^{-14}$ mol$^2$ dm$^{-6}$.

9. At the equivalence point, all the weak acid has been neutralised to form its conjugate base (e.g., $CH_3COO^-$). [1] The conjugate base hydrolyses with water: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$, producing $OH^-$ ions. [1] This makes the solution slightly alkaline, so pH > 7. [1]

Explanation: For a strong acid–strong base titration, the salt formed is neutral and pH = 7 at equivalence. For a weak acid–strong base titration, the salt contains the conjugate base of the weak acid, which is basic in solution, giving pH > 7.

10. 33.3 cm$^3$ [3]

Step-by-step:

• Moles of $H_2SO_4 = \frac{25.0}{1000} \times 0.100 = 2.50 \times 10^{-3}$ mol
• $H_2SO_4$ is diprotic: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$
• Moles of NaOH required $= 2 \times 2.50 \times 10^{-3} = 5.00 \times 10^{-3}$ mol
• Volume of NaOH $= \frac{5.00 \times 10^{-3}}{0.150} = 0.0333$ dm$^3 = 33.3$ cm$^3$

Marking:

• [1] for correct moles of $H_2SO_4$
• [1] for correct stoichiometric ratio (×2)
• [1] for correct final volume

Section B: Structured Questions

11.

(a) Mean volume of NaOH = $\frac{23.90 + 23.85 + 23.95}{3} = 23.90$ cm$^3$ [2]

Marking:

• [1] for identifying concordant titres (all three are within 0.10 cm³ of each other; the rough titration is excluded)
• [1] for correct mean calculation

Note: The rough titration (24.50 cm³) is not used in the mean. Titrations 1, 2, and 3 are concordant (within ±0.10 cm³).

(b) Concentration of ethanoic acid = 0.0956 mol dm$^{-3}$ [3]

Step-by-step:

• Moles of NaOH used $= \frac{23.90}{1000} \times 0.100 = 2.39 \times 10^{-3}$ mol
• $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$ (1:1 ratio)
• Moles of $CH_3COOH = 2.39 \times 10^{-3}$ mol
• Concentration of $CH_3COOH = \frac{2.39 \times 10^{-3}}{25.0/1000} = 0.0956$ mol dm$^{-3}$

Marking:

• [1] for moles of NaOH
• [1] for 1:1 stoichiometry and moles of acid
• [1] for correct concentration

(c) Phenolphthalein changes colour in the pH range 8.2–10.0, which falls within the steep vertical section of the weak acid–strong base titration curve (equivalence point pH ≈ 8.7). [1] Methyl orange changes colour in the pH range 3.1–4.4, which is far below the equivalence point pH for this titration, so it would change colour too early, leading to a large titration error. [1]

Explanation: The indicator's colour-change range must fall within the steep portion of the titration curve. For weak acid–strong base, the steep rise occurs around pH 7–10, so phenolphthalein is appropriate but methyl orange is not.

12.

(a) pH = 4.76 [4]

Step-by-step:

• Moles of $CH_3COOH = \frac{50.0}{1000} \times 0.200 = 0.0100$ mol
• Moles of $NaOH = \frac{50.0}{1000} \times 0.100 = 0.0050$ mol
• Reaction: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$
• After reaction:
  • Moles of $CH_3COOH$ remaining $= 0.0100 - 0.0050 = 0.0050$ mol
  • Moles of $CH_3COO^-$ formed $= 0.0050$ mol
• Total volume $= 50.0 + 50.0 = 100.0$ cm$^3 = 0.100$ dm$^3$
• Using Henderson-Hasselbalch equation:
  • $pH = pK_a + \log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]}$
  • $pK_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76$
  • Since $[CH_3COO^-] = [CH_3COOH]$, $\log_{10}(1) = 0$
  • $pH = 4.76 + 0 = 4.76$

Marking:

• [1] for correct moles of acid and base
• [1] for correct moles after reaction
• [1] for correct Henderson-Hasselbalch substitution
• [1] for correct pH

(b) When HCl is added, the $H^+$ ions react with the $CH_3COO^-$ ions in the buffer: $CH_3COO^- + H^+ \rightarrow CH_3COOH$. [1] This removes the added $H^+$ ions, converting them into undissociated ethanoic acid. [1] The equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ shifts to the left (Le Chatelier's principle), so the change in $[H^+]$ is very small and the pH remains approximately constant. [1]

Explanation: The buffer works because the conjugate base ($CH_3COO^-$) acts as a "reservoir" to neutralise added acid, while the weak acid ($CH_3COOH$) acts as a reservoir to neutralise added base.

13.

(a) From the graph, the equivalence point occurs at 25.0 cm$^3$ of NaOH. [1]

• Moles of NaOH $= \frac{25.0}{1000} \times 0.100 = 2.50 \times 10^{-3}$ mol
• Since HX is monoprotic: moles of HX $= 2.50 \times 10^{-3}$ mol
• Concentration of HX $= \frac{2.50 \times 10^{-3}}{25.0/1000} = 0.100$ mol dm$^{-3}$ [1]

(b) At the half-equivalence point (12.5 cm³ of NaOH added), half the acid has been neutralised, so $[HX] = [X^-]$. [1] At this point, $pH = pK_a$. From the graph, at 12.5 cm³, pH ≈ 4.75. Therefore, $pK_a = 4.75$ and $K_a = 10^{-4.75} = 1.78 \times 10^{-5}$ mol dm$^{-3}$. [1]

Note: The exact value read from the graph may vary slightly; accept $K_a$ in the range $1.5 \times 10^{-5}$ to $2.0 \times 10^{-5}$ mol dm$^{-3}$.

(c) The HCl curve should show:

• A starting pH of 1.0 (strong acid, 0.100 mol dm⁻³)
• A steeper vertical rise centred at pH 7 (equivalence point)
• The same equivalence point volume (25.0 cm³)
• Levelling off at a lower pH than the weak acid curve (around pH 11–12)

[2] — [1] for correct shape (steeper, starting lower), [1] for correct equivalence point and labelling

Image placeholder note: The graph should show a clear S-shaped curve for the weak acid HX titration with NaOH, with axes labelled, equivalence point at 25.0 cm³, initial pH ~3, and half-equivalence point identifiable. The HCl curve should be sketched on the same axes for comparison.

14.

(a) HCl is a strong acid and dissociates completely, providing 0.05 mol dm$^{-3}$ $H^+$. Ethanoic acid is a weak acid with $K_a = 1.74 \times 10^{-5}$, so it dissociates only very slightly. [1] The $H^+$ from the strong acid also suppresses the dissociation of ethanoic acid (common ion effect), making the contribution from ethanoic acid negligible. [1] Therefore, the pH is determined almost entirely by the HCl.

(b) pH = 1.30 [1]

Explanation:

• Total $[H^+] \approx 0.05$ mol dm$^{-3}$ (from HCl; contribution from $CH_3COOH$ is negligible)
• $pH = -\log_{10}(0.05) = 1.30$

15.

(a) $K_{sp} = [Ca^{2+}][F^-]^2$ [1]

Units: $(mol\ dm^{-3})(mol\ dm^{-3})^2 = mol^3\ dm^{-9}$ [1]

Explanation: For the equilibrium $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$, the solubility product expression is written as the product of the ion concentrations raised to their stoichiometric coefficients.

(b) Solubility $= 3.30 \times 10^{-4}$ mol dm$^{-3}$ [2]

Step-by-step:

• Let the solubility of $CaF_2 = s$ mol dm$^{-3}$
• Then $[Ca^{2+}] = s$ and $[F^-] = 2s$
• $K_{sp} = (s)(2s)^2 = 4s^3$
• $4s^3 = 1.46 \times 10^{-10}$
• $s^3 = 3.65 \times 10^{-11}$
• $s = \sqrt[3]{3.65 \times 10^{-11}} = 3.30 \times 10^{-4}$ mol dm$^{-3}$

Marking:

• [1] for correct substitution into $K_{sp}$ expression
• [1] for correct final answer

Section C: Data Interpretation and Application

16.

(a) Increasing pH: $HCOOH < CH_3COOH < H_2CO_3$ [1]

Explanation: For solutions of the same concentration, the weaker the acid (smaller $K_a$), the less it dissociates, so the lower the $[H^+]$ and the higher the pH. $H_2CO_3$ has the smallest $K_a$ so it has the highest pH; $HCOOH$ has the largest $K_a$ so it has the lowest pH. [1]

(b) pH = 2.38 [2]

Step-by-step:

• $K_a = \frac{[H^+][A^-]}{[HA]} \approx \frac{x^2}{0.10}$ (assuming $x \ll 0.10$)
• $x^2 = 1.77 \times 10^{-4} \times 0.10 = 1.77 \times 10^{-5}$
• $x = [H^+] = 4.21 \times 10^{-3}$ mol dm$^{-3}$
• $pH = -\log_{10}(4.21 \times 10^{-3}) = 2.38$

Marking:

• [1] for correct setup and $[H^+]$ calculation
• [1] for correct pH

Check: $\frac{4.21 \times 10^{-3}}{0.10} \times 100\% = 4.2\%$ dissociation, which is < 5%, so the approximation is valid.

(c) Basic. [1] Sodium methanoate is the salt of a strong base (NaOH) and a weak acid ($HCOOH$). The methanoate ion, $HCOO^-$, is the conjugate base of a weak acid and undergoes hydrolysis: $HCOO^- + H_2O \rightleftharpoons HCOOH + OH^-$, producing $OH^-$ ions and making the solution basic (pH > 7). [1]

17.

(a) $\frac{[CH_3COO^-]}{[CH_3COOH]} = 1.02$ [3]

Step-by-step:

• $pH = pK_a + \log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]}$
• $pK_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76$
• $4.75 = 4.76 + \log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]}$
• $\log_{10}\frac{[CH_3COO^-]}{[CH_3COOH]} = 4.75 - 4.76 = -0.01$
• $\frac{[CH_3COO^-]}{[CH_3COOH]} = 10^{-0.01} = 0.977 \approx 0.98$

Marking:

• [1] for correct $pK_a$ calculation
• [1] for correct substitution into Henderson-Hasselbalch
• [1] for correct ratio

Note: Accept answers in the range 0.97–1.02 depending on rounding.

(b) Since both solutions have the same concentration (0.50 mol dm$^{-3}$), the ratio of volumes needed equals the ratio of concentrations. [1]

• $\frac{V_{CH_3COO^-}}{V_{CH_3COOH}} = 0.98$
• Let $V_{CH_3COO^-} = 0.98x$ and $V_{CH_3COOH} = x$
• Total volume: $0.98x + x = 500$ cm$^3$
• $1.98x = 500$
• $x = 252.5$ cm$^3$
• Volume of 0.50 mol dm$^{-3}$ $CH_3COOH$ ≈ 253 cm$^3$ [1]
• Volume of 0.50 mol dm$^{-3}$ $CH_3COONa$ ≈ 247 cm$^3$ [1]

Marking:

• [1] for correct method linking volume ratio to concentration ratio
• [1] for each correct volume

18.

(a) $K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}$ [1]

Explanation: For the equilibrium $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$, the base dissociation constant expression follows the standard format of products over reactant (excluding water as it is the solvent).

(b) pH = 11.13 [3]

Step-by-step:

• $K_b = \frac{x^2}{0.100} = 1.78 \times 10^{-5}$
• $x^2 = 1.78 \times 10^{-6}$
• $x = [OH^-] = 1.33 \times 10^{-3}$ mol dm$^{-3}$
• $pOH = -\log_{10}(1.33 \times 10^{-3}) = 2.88$
• $pH = 14.00 - 2.88 = 11.12$ (or 11.13)

Marking:

• [1] for correct $[OH^-]$ calculation
• [1] for correct $pOH$
• [1] for correct $pH$

19.

(a) $[H_2SO_4] = 0.0800$ mol dm$^{-3}$ [2]

Step-by-step:

• $BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl$
• Moles of $BaSO_4 = \frac{0.466}{233.4} = 1.997 \times 10^{-3}$ mol ≈ $2.00 \times 10^{-3}$ mol
• Moles of $H_2SO_4$ in 25.0 cm$^3 = 2.00 \times 10^{-3}$ mol
• Concentration of $H_2SO_4 = \frac{2.00 \times 10^{-3}}{25.0/1000} = 0.0800$ mol dm$^{-3}$

Marking:

• [1] for correct moles of $BaSO_4$
• [1] for correct concentration of $H_2SO_4$

(b) $[HCl] = 0.100$ mol dm$^{-3}$ [3]

Step-by-step:

• Total moles of NaOH used $= \frac{32.50}{1000} \times 0.200 = 6.50 \times 10^{-3}$ mol
• Moles of $H^+$ from $H_2SO_4 = 2 \times 2.00 \times 10^{-3} = 4.00 \times 10^{-3}$ mol
• Moles of $H^+$ from $HCl = 6.50 \times 10^{-3} - 4.00 \times 10^{-3} = 2.50 \times 10^{-3}$ mol
• Concentration of $HCl = \frac{2.50 \times 10^{-3}}{25.0/1000} = 0.100$ mol dm$^{-3}$

Marking:

• [1] for total moles of NaOH
• [1] for subtracting $H_2SO_4$ contribution
• [1] for correct HCl concentration

20.

(a) $CO_3^{2-}$ is the conjugate base of $HCO_3^-$ and has a greater charge density than $HCO_3^-$. [1] $CO_3^{2-}$ undergoes hydrolysis to a greater extent: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$, producing more $OH^-$ ions than the hydrolysis of $HCO_3^-$: $HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-$. [1] Since $CO_3^{2-}$ is a stronger base (it is the conjugate base of a weaker acid, $HCO_3^-$), the $Na_2CO_3$ solution has a higher pH. [1]

Explanation: The key concept is that the conjugate base of a weaker acid is stronger. $HCO_3^-$ is a weaker acid than $H_2CO_3$, so $CO_3^{2-}$ is a stronger base than $HCO_3^-$, leading to more extensive hydrolysis and a higher pH.

(b) Calcium hydroxide is sparingly soluble: $Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$. [1] Adding HCl neutralises the $OH^-$ ions: $H^+ + OH^- \rightarrow H_2O$. By Le Chatelier's principle, the equilibrium shifts to the right, dissolving more $Ca(OH)_2$. As sufficient acid is added, all the solid dissolves, making the solution clear. [1]

Explanation: The saturated limewater initially appears slightly cloudy due to undissolved $Ca(OH)_2$. The acid removes $OH^-$ ions, causing more solid to dissolve until it is completely consumed.

End of Answer Key

Total: 60 marks