A Level H1 Chemistry Practice Paper 1

TuitionGoWhere Practice Paper — Chemistry H1 A-Level

Answer Key & Marking Scheme

Paper: Practice Paper — Acids, Bases & Salts | Version: 1 of 5 | Total Marks: 60

Section A: Multiple Choice (10 marks)

1. B [1]
$K_w = [\text{H}^+][\text{OH}^-]$ is the definition of the ionic product of water. It is derived from the equilibrium $\text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)}$. At 298 K, $K_w = 1.00 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}$.
Common mistake: Choosing A — students confuse $K_w$ with the concentration of water itself.

2. A [1]
Working:
$\text{pH} = 3.50 \Rightarrow [\text{H}^+] = 10^{-3.50} = 3.16 \times 10^{-4} \text{ mol dm}^{-3}$
$K_w = [\text{H}^+][\text{OH}^-] = 1.00 \times 10^{-14}$
$[\text{OH}^-] = \dfrac{1.00 \times 10^{-14}}{3.16 \times 10^{-4}} = 3.16 \times 10^{-11} \text{ mol dm}^{-3}$
Teaching note: In any aqueous solution at 298 K, $[\text{H}^+][\text{OH}^-] = K_w = 1.00 \times 10^{-14}$. As $[\text{H}^+]$ increases, $[\text{OH}^-]$ must decrease.

3. B [1]
$\text{NH}_4\text{Cl}$ is a salt of a weak base ($\text{NH}_3$) and a strong acid ($\text{HCl}$). The $\text{NH}_4^+$ ion undergoes hydrolysis: $\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+$, producing $\text{H}^+$ ions and making the solution acidic.

• $\text{Na}_2\text{CO}_3$: salt of strong base + weak acid → basic
• $\text{KNO}_3$: salt of strong base + strong acid → neutral
• $\text{CH}_3\text{COONa}$: salt of strong base + weak acid → basic

4. B [1]
Working:
When equal volumes of equal concentrations are mixed, the concentrations of acid and salt are halved, but their ratio remains 1:1.
Using the Henderson–Hasselbalch equation:
$\text{pH} = \text{p}K_a + \log_{10}\dfrac{[\text{salt}]}{[\text{acid}]}$
$\text{p}K_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76$
$\text{pH} = 4.76 + \log_{10}(1) = 4.76 + 0 = 4.76$
Key point: When $[\text{salt}] = [\text{acid}]$, $\text{pH} = \text{p}K_a$.

5. B [1]
At the equivalence point of a strong acid–strong base titration, the salt formed (e.g., NaCl) does not hydrolyse. The solution is neutral with pH = 7.
Common mistake: Choosing D — the indicator choice does not affect the pH at the equivalence point; it only affects how accurately the end-point is detected.

6. B [1]
A weak acid is defined as an acid that partially dissociates in aqueous solution. This is a question of acid strength, not concentration.
Common mistake: Choosing A — "low concentration" describes a dilute acid, not a weak acid. A weak acid can be concentrated but still only partially dissociate.

7. A [1]
Working:
$\text{p}K_a = -\log_{10}(K_a) = -\log_{10}(2.5 \times 10^{-4}) = 3.60$
Teaching note: $\text{p}K_a$ is the negative logarithm of $K_a$. A smaller $K_a$ (weaker acid) gives a larger $\text{p}K_a$.

8. C [1]
A buffer requires a weak acid and its conjugate base (or a weak base and its conjugate acid). $\text{HCl}$ is a strong acid and $\text{NaCl}$ is its salt — there is no equilibrium to resist pH changes.

• A: weak acid + its salt → buffer ✓
• B: weak base + its salt → buffer ✓
• D: weak acid + its salt → buffer ✓

9. B [1]
The buffer region occurs when significant amounts of both the weak acid and its conjugate base are present. This is most effective at the half-equivalence point, where exactly half the acid has been neutralised, giving $[\text{acid}] = [\text{salt}]$.

10. A [1]
Working:
$\text{pH} = 3.00 \Rightarrow [\text{H}^+] = 10^{-3.00} = 1.00 \times 10^{-3} \text{ mol dm}^{-3}$
For a weak acid HA: $K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \approx \dfrac{[\text{H}^+]^2}{[\text{HA}]_{\text{initial}}}$
$K_a = \dfrac{(1.00 \times 10^{-3})^2}{0.050} = \dfrac{1.00 \times 10^{-6}}{0.050} = 2.0 \times 10^{-5} \text{ mol dm}^{-3}$
Approximation: Since the acid is weak, $[\text{H}^+] \ll [\text{HA}]_{\text{initial}}$, so $[\text{HA}]_{\text{equilibrium}} \approx [\text{HA}]_{\text{initial}}$.

Section B: Structured Questions (30 marks)

11.

(a) A Brønsted–Lowry acid is a proton ($\text{H}^+$) donor. [1]
Marking note: Must mention "proton" and "donor" (or "donates"). Simply saying "produces $\text{H}^+$" is not sufficient for full credit at A-Level.

(b) A Brønsted–Lowry base is a proton ($\text{H}^+$) acceptor. [1]
Marking note: Must mention "proton" and "acceptor" (or "accepts").

(c) A conjugate acid–base pair is a pair of species that differ by one proton ($\text{H}^+$). [1]
Example: $\text{CH}_3\text{COOH}$ and $\text{CH}_3\text{COO}^-$ form a conjugate acid–base pair. The acid has one more proton than its conjugate base.

12.

(a) $\text{CH}_3\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$
or equivalently: $\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)}$ [1]
Marking note: The reversible arrow ($\rightleftharpoons$) is essential. State symbols are required.

(b) Ethanoic acid is a weak acid because it only partially dissociates in water. [1] The equilibrium lies far to the left, meaning only a small fraction of ethanoic acid molecules donate protons to water. [1]
Teaching note: A strong acid (e.g., HCl) would have a single arrow ($\rightarrow$) showing complete dissociation. The reversible arrow indicates incomplete dissociation.

(c) Working: [3]
$K_a = \dfrac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 1.74 \times 10^{-5}$

Let $[\text{H}^+] = x$. Then $[\text{CH}_3\text{COO}^-] = x$ and $[\text{CH}_3\text{COOH}] \approx 0.100 - x \approx 0.100$ (since $x$ is small).

$1.74 \times 10^{-5} = \dfrac{x^2}{0.100}$
$x^2 = 1.74 \times 10^{-6}$
$x = \sqrt{1.74 \times 10^{-6}} = 1.32 \times 10^{-3} \text{ mol dm}^{-3}$

$\text{pH} = -\log_{0}(1.32 \times 10^{-3}) = 2.88$

Answer: pH = 2.88 [3]
Marking scheme:

• [1] Correct $K_a$ expression
• [1] Correct substitution and calculation of $[\text{H}^+]$
• [1] Correct pH value (accept 2.88 ± 0.02)

13.

(a) $\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$
or ionic: $\text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{H}_2\text{O(l)}$ [1]

(b) Working: [3]
Moles of NaOH used $= \dfrac{22.40}{1000} \times 0.100 = 2.24 \times 10^{-3} \text{ mol}$

From the equation, mole ratio HCl : NaOH = 1 : 1
Moles of HCl $= 2.24 \times 10^{-3} \text{ mol}$

Concentration of HCl $= \dfrac{2.24 \times 10^{-3}}{25.0/1000} = \dfrac{2.24 \times 10^{-3}}{0.0250} = 0.0896 \text{ mol dm}^{-3}$

Answer: 0.0896 mol dm⁻³ [3]
Marking scheme:

• [1] Correct moles of NaOH
• [1] Correct use of 1:1 stoichiometry
• [1] Correct final concentration (accept 0.0896 or 0.090 to 2 s.f.)

(c) Any of: methyl orange, methyl red, or bromophenol blue. [1]
For a strong acid–strong base titration, the equivalence point is at pH 7, and the pH change is very steep (approximately pH 3–11). Any indicator with a transition range within this steep portion is suitable. [1]
Note: Phenolphthalein is also commonly accepted, though its transition range (pH 8.2–10.0) is slightly above the equivalence point. The key justification is that the vertical portion of the curve spans the indicator's transition range.

14.

(a) Working: [3]
When equal volumes of equal concentrations are mixed, both concentrations are halved:
$[\text{HCOOH}] = 0.100 \text{ mol dm}^{-3}$ and $[\text{HCOONa}] = 0.100 \text{ mol dm}^{-3}$

Using Henderson–Hasselbalch:
$\text{p}K_a = -\log_{10}(1.78 \times 10^{-4}) = 3.75$
$\text{pH} = \text{p}K_a + \log_{10}\dfrac{[\text{HCOO}^-]}{[\text{HCOOH}]} = 3.75 + \log_{10}\dfrac{0.100}{0.100} = 3.75 + 0 = 3.75$

Answer: pH = 3.75 [3]
Marking scheme:

• [1] Correct $\text{p}K_a$ calculation
• [1] Correct use of Henderson–Hasselbalch equation
• [1] Correct final pH

(b) When HCl is added, the $\text{H}^+$ ions react with the methanoate ions ($\text{HCOO}^-$) in the buffer: [1]
$\text{HCOO}^-\text{(aq)} + \text{H}^+\text{(aq)} \rightarrow \text{HCOOH(aq)}$ [1]

The added $\text{H}^+$ is consumed by the conjugate base, converting it to the weak acid. Since both species are present in large amounts, the ratio $[\text{HCOO}^-]/[\text{HCOOH}]$ changes only slightly, so the pH remains nearly constant. [1]
Teaching note: The buffer works because it contains significant amounts of both the weak acid (to react with added base) and its conjugate base (to react with added acid).

15.

(a) pH at equivalence point ≈ 8.7 [1]
Accept any value in the range 8.5–9.0 based on reading from the graph.

(b) At the equivalence point, all the ethanoic acid has been converted to sodium ethanoate ($\text{CH}_3\text{COONa}$). [1] The ethanoate ion ($\text{CH}_3\text{COO}^-$) is the conjugate base of a weak acid and undergoes hydrolysis: $\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$, producing $\text{OH}^-$ ions and making the solution slightly alkaline. [1]

(c) The buffer region should be marked on the graph between approximately 5 cm³ and 20 cm³ of NaOH added — the relatively flat portion of the curve before the steep rise. [1]
Specifically, the region where the curve rises gradually (before the equivalence point) should be indicated.

(d) Working: [3]
At the half-equivalence point (12.5 cm³ of NaOH added), half the ethanoic acid has been neutralised, so $[\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-]$.
At this point, $\text{pH} = \text{p}K_a$ (from Henderson–Hasselbalch: $\text{pH} = \text{p}K_a + \log_{10}(1) = \text{p}K_a$).

From the graph, at 12.5 cm³, pH ≈ 4.76.
Therefore, $\text{p}K_a = 4.76$ and $K_a = 10^{-4.76} = 1.74 \times 10^{-5} \text{ mol dm}^{-3}$.

Answer: $K_a = 1.74 \times 10^{-5} \text{ mol dm}^{-3}$ [3]
Marking scheme:

• [1] Correct identification that at half-equivalence point, $\text{pH} = \text{p}K_a$
• [1] Correct reading of pH from graph at 12.5 cm³
• [1] Correct calculation of $K_a$

Section C: Free Response (20 marks)

16.

(a) $\text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)}$ [1]
Marking note: Reversible arrow and state symbols required. Ammonia accepts a proton from water, forming $\text{NH}_4^+$ and $\text{OH}^-$.

(b) Working: [4]
$K_b = \dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} = 1.78 \times 10^{-5}$

Let $[\text{OH}^-] = x$. Then $[\text{NH}_4^+] = x$ and $[\text{NH}_3] \approx 0.100 - x \approx 0.100$.

$1.78 \times 10^{-5} = \dfrac{x^2}{0.100}$
$x^2 = 1.78 \times 10^{-6}$
$x = \sqrt{1.78 \times 10^{-6}} = 1.33 \times 10^{-3} \text{ mol dm}^{-3}$

$\text{pOH} = -\log_{10}(1.33 \times 10^{-3}) = 2.88$
$\text{pH} = 14.00 - 2.88 = 11.12$

Answer: pH = 11.12 [4]
Marking scheme:

• [1] Correct $K_b$ expression
• [1] Correct calculation of $[\text{OH}^-]$
• [1] Correct calculation of pOH
• [1] Correct conversion to pH (using $\text{pH} + \text{pOH} = 14$)

(c) Ammonium chloride dissociates completely in water to give $\text{NH}_4^+$ and $\text{Cl}^-$ ions. [1] The $\text{NH}_4^+$ ion is the conjugate acid of the weak base $\text{NH}_3$ and undergoes hydrolysis: [1]
$\text{NH}_4^+\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_3\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$

This produces $\text{H}_3\text{O}^+$ (or $\text{H}^+$) ions, making the solution acidic. [1]
Teaching note: Salts of weak bases and strong acids are always acidic because the conjugate acid of the weak base hydrolyses water to produce $\text{H}^+$ ions.

17.

(a) Working: [3]
Using the Henderson–Hasselbalch equation:
$\text{pH} = \text{p}K_a + \log_{10}\dfrac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$

$\text{p}K_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76$

$5.00 = 4.76 + \log_{10}\dfrac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$
$\log_{10}\dfrac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 0.24$
$\dfrac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 10^{0.24} = 1.74$

Answer: $[\text{CH}_3\text{COO}^-] : [\text{CH}_3\text{COOH}] = 1.74 : 1$ [3]
Marking scheme:

• [1] Correct $\text{p}K_a$ value
• [1] Correct substitution into Henderson–Hasselbalch
• [1] Correct ratio (accept 1.74:1 or 1.7:1)

(b) Working: [4]
From (a), $\dfrac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = 1.74$

Moles of ethanoic acid in 100 cm³ of 0.500 mol dm⁻³ $= \dfrac{100}{1000} \times 0.500 = 0.0500 \text{ mol}$

Since the volume is the same for both species, the mole ratio equals the concentration ratio:
$\dfrac{n(\text{CH}_3\text{COO}^-)}{n(\text{CH}_3\text{COOH})} = 1.74$
$n(\text{CH}_3\text{COO}^-) = 1.74 \times 0.0500 = 0.0870 \text{ mol}$

Mass of sodium ethanoate $= 0.0870 \times 82.0 = 7.13 \text{ g}$

Answer: 7.13 g [4]
Marking scheme:

• [1] Correct moles of ethanoic acid
• [1] Correct use of ratio from (a) to find moles of sodium ethanoate
• [1] Correct molar mass used
• [1] Correct final mass (accept 7.1–7.14 g)

18.

(a) $\text{H}_3\text{PO}_4\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_2\text{PO}_4^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$
or: $\text{H}_3\text{PO}_4\text{(aq)} \rightleftharpoons \text{H}_2\text{PO}_4^-\text{(aq)} + \text{H}^+\text{(aq)}$ [1]

(b) $K_{a1} \gg K_{a2} \gg K_{a3}$ because: [3]

• After the first dissociation, the remaining species ($\text{H}_2\text{PO}_4^-$) carries a negative charge. Removing a positively charged proton ($\text{H}^+$) from a negatively charged species is progressively more difficult due to increasing electrostatic attraction. [1]
• Each successive dissociation removes $\text{H}^+$ from an increasingly negative ion, requiring more energy. [1]
• The negative charge that builds up on the phosphate species after each dissociation makes it increasingly difficult to remove the next proton. [1]
  Teaching note: This is a general pattern for polyprotic acids. The first proton is easiest to remove; each subsequent proton is harder to remove because the remaining anion holds onto its protons more tightly.

(c) Working: [3]
$K_{a1} = \dfrac{[\text{H}^+][\text{H}_2\text{PO}_4^-]}{[\text{H}_3\text{PO}_4]} = 7.11 \times 10^{-3}$

Let $[\text{H}^+] = x$. Then:
$7.11 \times 10^{-3} = \dfrac{x^2}{0.100 - x}$

Since $K_{a1}$ is relatively large, we cannot assume $x \ll 0.100$. Solving the quadratic:
$x^2 + 7.11 \times 10^{-3}x - 7.11 \times 10^{-4} = 0$

Using the quadratic formula:
$x = \dfrac{-7.11 \times 10^{-3} + \sqrt{(7.11 \times 10^{-3})^2 + 4(7.11 \times 10^{-4})}}{2}$
$x = \dfrac{-7.11 \times 10^{-3} + \sqrt{5.06 \times 10^{-5} + 2.844 \times 10^{-3}}}{2}$
$x = \dfrac{-7.11 \times 10^{-3} + \sqrt{2.895 \times 10^{-3}}}{2}$
$x = \dfrac{-7.11 \times 10^{-3} + 5.38 \times 10^{-2}}{2} = \dfrac{4.67 \times 10^{-2}}{2} = 2.34 \times 10^{-2}$

$\text{pH} = -\log_{10}(2.34 \times 10^{-2}) = 1.63$

Answer: pH = 1.63 [3]
Marking scheme:

• [1] Correct $K_{a1}$ expression
• [1] Correct solution of quadratic (or valid approximation method)
• [1] Correct pH value (accept 1.60–1.65)
  Note: If the student uses the approximation $x \ll 0.100$, they would get $x = \sqrt{7.11 \times 10^{-4}} = 2.67 \times 10^{-2}$, giving pH = 1.57. This approximation is less accurate here because $K_{a1}$ is relatively large. Award partial credit [2/3] for this approach.

19.

(a) Hydrochloric acid is a strong acid and dissociates completely, contributing $0.100 \text{ mol dm}^{-3}$ of $\text{H}^+$ ions. [1] Ethanoic acid is a weak acid with $K_a = 1.74 \times 10^{-5}$, so it contributes a negligible additional concentration of $\text{H}^+$ ions compared to the strong acid. The presence of the high $[\text{H}^+]$ from HCl also suppresses the dissociation of ethanoic acid (common ion effect). [1]
Teaching note: The common ion effect means that the $\text{H}^+$ from HCl pushes the ethanoic acid equilibrium even further to the left, making its contribution to $[\text{H}^+]$ even smaller.

(b) Add a small amount of solid calcium carbonate (or magnesium ribbon) to the solution. [1] The rate of effervescence (gas evolution) will be rapid initially due to the strong acid (HCl), then continue at a slower rate due to the weak acid (ethanoic acid). Alternatively, measure the pH over time — the pH will remain relatively constant for a period as the weak acid continues to react, whereas a solution of only strong acid would show a different reaction profile. [1]
Alternative acceptable answer: Measure the electrical conductivity — the mixture will have higher conductivity than ethanoic acid alone due to the fully dissociated HCl. Or: titrate with NaOH using a pH meter — the titration curve will show two distinct equivalence points (or a buffered region) indicating the presence of two acids.

20.

(a) $K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2$ [1]
Marking note: The expression must include the correct species with the correct powers (1 for $\text{Mg}^{2+}$, 2 for $\text{OH}^-$).

(b) Working: [3]
Let the solubility of $\text{Mg(OH)}_2 = s \text{ mol dm}^{-3}$.
$\text{Mg(OH)}_2\text{(s)} \rightleftharpoons \text{Mg}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)}$

$[\text{Mg}^{2+}] = s$ and $[\text{OH}^-] = 2s$

$K_{sp} = (s)(2s)^2 = 4s^3 = 5.50 \times 10^{-12}$
$s^3 = \dfrac{5.50 \times 10^{-12}}{4} = 1.375 \times 10^{-12}$
$s = \sqrt[3]{1.375 \times 10^{-12}} = 1.11 \times 10^{-4} \text{ mol dm}^{-3}$

Answer: $1.11 \times 10^{-4} \text{ mol dm}^{-3}$ [3]
Marking scheme:

• [1] Correct relationship between solubility and ion concentrations ($[\text{OH}^-] = 2s$)
• [1] Correct substitution into $K_{sp}$ expression
• [1] Correct final answer (accept $1.1 \times 10^{-4}$ to $1.11 \times 10^{-4}$)

(c) The solubility of magnesium hydroxide will decrease. [1] This is due to the common ion effect: adding $\text{NaOH}$ increases $[\text{OH}^-]$, shifting the equilibrium $\text{Mg(OH)}_2\text{(s)} \rightleftharpoons \text{Mg}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)}$ to the left (Le Chatelier's principle), causing more $\text{Mg(OH)}_2$ to precipitate. [1]
Teaching note: The common ion effect is a direct application of Le Chatelier's principle. Adding a product ion ($\text{OH}^-$) shifts the equilibrium towards the solid, reducing solubility.

END OF ANSWER KEY

Total Marks: 60