A Level H2 Chemistry Practice Paper 2

TuitionGoWhere Practice Paper — Chemistry H2 A-Level

Answer Key: Acids, Bases & Salts (Version 2)

Section A: Multiple Choice

1. B [1]

• Explanation: A conjugate base is formed when a Brønsted–Lowry acid donates a proton ($H^+$). $HSO_4^-$ loses one $H^+$ to become $SO_4^{2-}$. Option A ($H_2SO_4$) is the conjugate acid (gains a proton), not the conjugate base. Option C is the conjugate acid of water. Option D is unrelated.
• Common mistake: Students confuse conjugate acid with conjugate base. Remember: acid loses $H^+$ → conjugate base; base gains $H^+$ → conjugate acid.

2. B [1]

• Explanation: pH = 3.40, so $[H^+] = 10^{-3.40} = 3.98 \times 10^{-4}$ mol dm$^{-3}$. Using $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$: $[OH^-] = \frac{1.0 \times 10^{-14}}{3.98 \times 10^{-4}} = 2.51 \times 10^{-11} \text{ mol dm}^{-3}$
• Common mistake: Students may select A, which is $[H^+]$ not $[OH^-]$, or C, which is a calculation error.

3. C [1]

• Explanation: $K_2CO_3$ is a salt of a strong base ($KOH$) and a weak acid ($H_2CO_3$). The $CO_3^{2-}$ ion hydrolyses in water to produce $OH^-$, making the solution alkaline (pH > 7). $NH_4Cl$ and $AlCl_3$ produce acidic solutions (cations of weak bases). $NaNO_3$ is a salt of a strong acid and strong base, giving pH ≈ 7.
• Teaching note: Salts of strong acid + weak base → acidic; weak acid + strong base → alkaline; strong + strong → neutral.

4. C [1]

• Explanation: For a weak acid, $[H^+] = \sqrt{K_a \times c} = \sqrt{4.0 \times 10^{-6} \times 0.050} = \sqrt{2.0 \times 10^{-7}} = 4.47 \times 10^{-4}$ mol dm$^{-3}$. $pH = -\log(4.47 \times 10^{-4}) = 3.35$
• Wait — recalculating: $\sqrt{4.0 \times 10^{-6} \times 0.050} = \sqrt{2.0 \times 10^{-7}} = 4.472 \times 10^{-4}$. $pH = -\log(4.472 \times 10^{-4}) = 3.35$. Answer is B.
• Correction: The answer is B (3.35).
• Common mistake: Students may forget to take the square root or use the wrong formula.

5. B [1]

• Explanation: Moles of $CH_3COOH$ = $0.050 \times 0.200 = 0.010$ mol. Moles of $NaOH$ = $0.050 \times 0.200 = 0.010$ mol. The acid and base react in a 1:1 ratio, so all the $CH_3COOH$ is completely neutralised to form $CH_3COONa$. The resulting solution contains only the salt $CH_3COONa$ (and water), with no remaining weak acid to form a buffer pair. A buffer requires significant amounts of both a weak acid and its conjugate base.
• Common mistake: Students assume any mixture of acid and base forms a buffer. A buffer requires excess weak acid remaining alongside its conjugate base.

6. A [1]

• Explanation: At the half-equivalence point, exactly half the weak acid has been neutralised, so $[HA] = [A^-]$. Substituting into the Henderson–Hasselbalch equation: $pH = pK_a + log\frac{[A^-]}{[HA]} = pK_a + log(1) = pK_a + 0 = pK_a$
• Teaching note: This is a key concept — at half-equivalence, pH = $pK_a$, which allows experimental determination of $K_a$ from a titration curve.

7. A [1]

• Explanation: Let the solubility of $Mg(OH)_2$ = $s$ mol dm$^{-3}$. $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ $K_{sp} = [Mg^{2+}][OH^-]^2 = s \times (2s)^2 = 4s^3$ $4s^3 = 5.6 \times 10^{-12}$ $s^3 = 1.4 \times 10^{-12}$ $s = \sqrt[3]{1.4 \times 10^{-12}} = 1.12 \times 10^{-4} \approx 1.1 \times 10^{-4} \text{ mol dm}^{-3}$
• Common mistake: Forgetting the coefficient 2 for $[OH^-]$ when setting up the $K_{sp}$ expression, leading to $s^3$ instead of $4s^3$.

8. A [1]

• Explanation: This is a weak base ($NH_3$) titrated with a strong acid ($HCl$). The equivalence point pH is acidic (below 7) because the salt $NH_4Cl$ hydrolyses to produce an acidic solution. The equivalence point pH is approximately: $[NH_4^+] \approx 0.050 \text{ mol dm}^{-3} \quad (\text{after dilution})$ $K_a(NH_4^+) = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$ $[H^+] = \sqrt{K_a \times c} = \sqrt{5.56 \times 10^{-10} \times 0.050} = 5.27 \times 10^{-6}$ $pH \approx 5.28$ Methyl orange (pH range 3.1–4.4) is the closest match. The steep portion of the curve passes through the methyl orange range.
• Teaching note: For weak base–strong acid titrations, the equivalence point is acidic, so an indicator with an acidic pH range is needed.

9. B [1]

• Explanation: Using the Henderson–Hasselbalch equation: $pH = pK_a + log\frac{[A^-]}{[HA]}$ $pK_a = -\log(1.8 \times 10^{-4}) = 3.74$ $pH = 3.74 + log\frac{0.15}{0.10} = 3.74 + log(1.5) = 3.74 + 0.18 = 3.92$
• Common mistake: Swapping $[A^-]$ and $[HA]$ in the log term, giving pH = 3.57 (option A).

10. B [1]

• Explanation: At the equivalence point, all $CH_3COOH$ has been converted to $CH_3COONa$. The $CH_3COO^-$ ion is the conjugate base of a weak acid and undergoes hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$ This produces $OH^-$ ions, making the solution slightly alkaline (pH > 7).
• Common mistake: Choosing A (excess NaOH) — at the equivalence point, there is no excess reagent.

Section B: Structured Questions

11. [4 marks]

(a) [1] A Brønsted–Lowry acid is a proton ($H^+$) donor.

• Marking: Must mention "proton" and "donor" (or "donates").

(b) [2]

• Pair 1: acid = $H_2O$ / conjugate base = $OH^-$ [1]
• Pair 2: acid = $NH_4^+$ / conjugate base = $NH_3$ [1]
• Explanation: $H_2O$ donates $H^+$ to become $OH^-$. $NH_3$ accepts $H^+$ to become $NH_4^+$. The conjugate acid–base pairs are: $H_2O$/$OH^-$ and $NH_4^+$/$NH_3$.

(c) [1] $NH_3$ can act as a Brønsted–Lowry base because it has a lone pair of electrons on the nitrogen atom that can accept a proton ($H^+$).

• Marking: Must mention lone pair and ability to accept $H^+$.

12. [5 marks]

(a) [1] $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$

• Marking: Correct formula of all species and balanced equation. Award 1 mark.

(b) [3]

• Moles of $NaOH$ = $\frac{25.0}{1000} \times 0.120 = 3.00 \times 10^{-3}$ mol [1]
• From the equation, mole ratio $H_2SO_4 : NaOH = 1 : 2$
• Moles of $H_2SO_4$ = $\frac{3.00 \times 10^{-3}}{2} = 1.50 \times 10^{-3}$ mol [1]
• Volume of $H_2SO_4$ = $\frac{1.50 \times 10^{-3}}{0.100} = 1.50 \times 10^{-2}$ dm$^3$ = 15.0 cm$^3$ [1]
• Marking: 1 mark for moles of NaOH, 1 mark for correct mole ratio application, 1 mark for final answer with unit.

(c) [1] Indicator: phenolphthalein [½] Colour change: pink to colourless [½] (or colourless to pink, depending on direction — here acid in burette, alkali in flask, so pink → colourless)

• Note: Since $H_2SO_4$ (strong acid) is added to $NaOH$ (strong base), the equivalence point is at pH 7. Either phenolphthalein or methyl orange is acceptable. However, phenolphthalein is more commonly used for strong acid–strong base titrations in school labs.

13. [6 marks]

(a) [2]

• Titrations 2 and 3 are concordant (within 0.10 cm$^3$ of each other). [1]
• Average titre = $\frac{24.30 + 24.30}{2} = 24.30$ cm$^3$ [1]
• Marking: 1 mark for identifying concordant values, 1 mark for correct average. Titration 1 (24.35) is also close but the two identical values (24.30, 24.30) should be used. If student uses all three concordant values: $\frac{24.35 + 24.30 + 24.30}{3} = 24.32$ cm$^3$, this is also acceptable.

(b) [2]

• Moles of $NaOH$ = $\frac{24.30}{1000} \times 0.100 = 2.43 \times 10^{-3}$ mol [1]
• Since $HA$ is monoprotic, mole ratio $HA : NaOH = 1 : 1$
• Concentration of $HA$ = $\frac{2.43 \times 10^{-3}}{25.0/1000} = \frac{2.43 \times 10^{-3}}{0.0250} = 0.0972$ mol dm$^{-3}$ [1]

(c) [2] The pH at the equivalence point is 8.72 (greater than 7) because the salt formed (sodium salt of the weak acid, $NaA$) undergoes hydrolysis. The conjugate base $A^-$ reacts with water: $A^- + H_2O \rightleftharpoons HA + OH^-$ This produces $OH^-$ ions, making the solution slightly alkaline. [2]

• Marking: 1 mark for identifying hydrolysis of the conjugate base, 1 mark for the explanation/equation showing production of $OH^-$.

14. [5 marks]

(a) [2]

• Initial moles of $HCOOH$ = $\frac{100}{1000} \times 0.300 = 0.0300$ mol
• Moles of $NaOH$ = $\frac{100}{1000} \times 0.150 = 0.0150$ mol [1]
• $NaOH$ reacts with $HCOOH$ in 1:1 ratio:
  • Moles of $HCOOH$ remaining = $0.0300 - 0.0150 = 0.0150$ mol
  • Moles of $HCOO^-$ formed = $0.0150$ mol [1]
• Marking: 1 mark for moles of NaOH, 1 mark for both remaining HCOOH and HCOO⁻ moles.

(b) [2]

• Total volume = $100 + 100 = 200$ cm$^3$ = 0.200 dm$^3$
• $[HCOOH] = \frac{0.0150}{0.200} = 0.0750$ mol dm$^{-3}$
• $[HCOO^-] = \frac{0.0150}{0.200} = 0.0750$ mol dm$^{-3}$
• $pK_a = -\log(1.8 \times 10^{-4}) = 3.74$ [1]
• $pH = 3.74 + log\frac{0.0750}{0.0750} = 3.74 + 0 = 3.74$ [1]
• Alternative (using $K_a$ expression): $pH = pK_a$ when $[HA] = [A^-]$, so $pH = 3.74$.

(c) [1] The pH will decrease only very slightly (remain almost unchanged). The added $H^+$ ions react with the $HCOO^-$ in the buffer: $HCOO^- + H^+ \rightarrow HCOOH$ This removes the added $H^+$ ions, minimising the pH change. The buffer resists changes in pH.

• Marking: Award 1 mark for stating pH remains almost unchanged with correct explanation involving the buffer action.

15. [5 marks]

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

• Marking: Must include the squared term for $F^-$. No mark for $K_{sp}$ without the square.

(b) [3]

• Let solubility of $CaF_2$ = $s$ mol dm$^{-3}$ $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$ $[Ca^{2+}] = s, \quad [F^-] = 2s$ $K_{sp} = s \times (2s)^2 = 4s^3$ [1] $4s^3 = 3.9 \times 10^{-11}$ $s^3 = 9.75 \times 10^{-12}$ [1] $s = \sqrt[3]{9.75 \times 10^{-12}} = 2.14 \times 10^{-4} \text{ mol dm}^{-3}$ [1]
• Answer: $2.1 \times 10^{-4}$ mol dm$^{-3}$ (to 2 s.f.)
• Marking: 1 mark for correct $K_{sp}$ expression in terms of $s$, 1 mark for correct substitution, 1 mark for final answer.

(c) [1] The solubility of $CaF_2$ would decrease. The $NaF$ solution provides $F^-$ ions (common ion effect). According to Le Chatelier's principle, the increased $[F^-]$ shifts the equilibrium to the left (towards the solid), reducing the solubility of $CaF_2$.

• Marking: 1 mark for "decrease" with correct explanation referencing common ion effect or Le Chatelier's principle.

Section C: Free Response

16. [8 marks]

(a) [1] Volume of $NaOH$ at equivalence point = 22.5 cm$^3$ (read from the graph at the steepest point of the curve).

• Marking: Accept 22.4–22.6 cm$^3$ depending on reading precision from the graph.

(b) [2]

• Moles of $NaOH$ at equivalence = $\frac{22.5}{1000} \times 0.100 = 2.25 \times 10^{-3}$ mol [1]
• Mole ratio $CH_3COOH : NaOH = 1 : 1$
• Concentration of $CH_3COOH$ = $\frac{2.25 \times 10^{-3}}{25.0/1000} = \frac{2.25 \times 10^{-3}}{0.0250} = 0.0900$ mol dm$^{-3}$ [1]

(c) [2]

• At the half-equivalence point, volume of $NaOH$ = $\frac{22.5}{2} = 11.25$ cm$^3$
• From the graph, at 11.25 cm$^3$, pH ≈ 4.74 [1]
• At half-equivalence: $pH = pK_a$, so $pK_a = 4.74$
• $K_a = 10^{-4.74} = 1.8 \times 10^{-5}$ mol dm$^{-3}$ [1]
• Marking: 1 mark for identifying half-equivalence point and reading pH, 1 mark for calculating $K_a$.

(d) [2] Phenolphthalein changes colour in the pH range 8.2–10.0. From the graph, the equivalence point occurs at pH ≈ 8.7, which falls within the phenolphthalein colour-change range. The steep portion of the titration curve passes through the phenolphthalein range, so a sharp colour change (colourless to pink) is observed at the equivalence point, giving an accurate endpoint. [2]

• Marking: 1 mark for stating the pH range of phenolphthalein, 1 mark for linking it to the equivalence point pH from the graph.

(e) [1] Methyl orange changes colour in the pH range 3.1–4.4, which is well below the equivalence point pH of 8.7. The colour change would occur before the true equivalence point (in the buffer region), leading to an underestimate of the volume of $NaOH$ required and therefore an underestimate of the concentration of $CH_3COOH$. This would not give an accurate result.

• Marking: 1 mark for explaining that methyl orange changes colour too early, leading to inaccurate results.

17. [7 marks]

(a) [1] $NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)$

• Accept: $NH_4^+ \rightleftharpoons NH_3 + H^+$ (simplified form)
• Marking: 1 mark for correct equation showing $NH_4^+$ acting as an acid.

(b) [2]

• For the conjugate acid–base pair: $K_a(NH_4^+) \times K_b(NH_3) = K_w$ [1]
• Therefore: $K_a(NH_4^+) = \frac{K_w}{K_b(NH_3)}$ [1]
• Explanation: $NH_4^+$ and $NH_3$ are a conjugate acid–base pair. The product of $K_a$ of the conjugate acid and $K_b$ of the base equals $K_w$.
• Marking: 1 mark for the relationship $K_a \times K_b = K_w$, 1 mark for rearranging to $K_a = K_w / K_b$.

(c) [3]

• $K_a(NH_4^+) = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$ mol dm$^{-3}$ [1]
• For the weak acid $NH_4^+$ at concentration 0.200 mol dm$^{-3}$: $[H^+] = \sqrt{K_a \times c} = \sqrt{5.56 \times 10^{-10} \times 0.200} = \sqrt{1.11 \times 10^{-10}} = 1.05 \times 10^{-5} \text{ mol dm}^{-3}$ [1] $pH = -\log(1.05 \times 10^{-5}) = 4.98$ [1]
• Answer: pH = 4.98 (or 5.0 to 2 s.f.)
• Marking: 1 mark for calculating $K_a$, 1 mark for correct $[H^+]$ calculation, 1 mark for pH.

(d) [1] The student's claim is correct. Adding $NH_4Cl$ to $NH_3$ solution increases the concentration of $NH_4^+$ (the conjugate acid). According to Le Chatelier's principle, this shifts the equilibrium: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$ to the left, reducing $[OH^-]$ and therefore decreasing the pH. This is consistent with the buffer equation: increasing $[NH_4^+]$ decreases pH.

• Marking: 1 mark for agreeing with the claim and providing a correct explanation using Le Chatelier's principle or the Henderson–Hasselbalch equation.

Mark Summary