A Level H1 Chemistry Practice Paper 4

TuitionGoWhere Practice Paper — Chemistry H1 A-Level

Answer Key: Acids, Bases & Salts

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Section A: Short Answer Questions

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1. [1 mark]

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

Teaching note: The key word is "completely." Every molecule of the acid donates its proton to water. Examples include HCl, HNO₃, and H₂SO₄. This is different from concentration — a strong acid can be dilute or concentrated.

Mark scheme: 1 mark for "completely dissociates/ionises in water."

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2. [2 marks]

The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration:

$pH = -\log_{10}[H^+]$

where $[H^+]$ is the concentration of hydrogen ions in mol dm⁻³. A higher $[H^+]$ gives a lower pH (more acidic), and a lower $[H^+]$ gives a higher pH (more alkaline).

Teaching note: pH is a logarithmic scale, so a change of 1 pH unit represents a tenfold change in $[H^+]$. At 25 °C, pH < 7 is acidic, pH = 7 is neutral, and pH > 7 is alkaline.

Mark scheme: 1 mark for the definition/formula; 1 mark for the reference to $[H^+]$ concentration and its relationship to acidity/alkalinity.

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3. [2 marks]

Hydrochloric acid is a strong acid, so it dissociates completely:

$HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)$

Therefore $[H^+] = 0.025$ mol dm⁻³.

$pH = -\log_{10}(0.025) = -\log_{10}(2.5 \times 10^{-2})$ $= -(\log_{10}2.5 + \log_{10}10^{-2})$ $= -(0.398 - 2) = 1.60$

Teaching note: For strong monoprotic acids, $[H^+]$ equals the acid concentration directly. Students should remember that $\log_{10}(2.5 \times 10^{-2}) = \log_{10}2.5 + \log_{10}10^{-2} = 0.398 - 2 = -1.602$, so $pH = 1.60$.

Mark scheme: 1 mark for correct $[H^+]$; 1 mark for correct pH = 1.60.

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4. [1 mark]

$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

Teaching note: The $K_a$ expression follows the equilibrium expression: products over reactants, each raised to the power of their stoichiometric coefficients. Water is omitted because it is the solvent (its concentration is essentially constant). Pure liquids and solids are also omitted from equilibrium expressions.

Mark scheme: 1 mark for the correct expression. No units required for this question.

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5. [3 marks]

Given: concentration = 0.10 mol dm⁻³, pH = 3.20

Step 1: Calculate $[H^+]$ from pH:

$[H^+] = 10^{-3.20} = 6.31 \times 10^{-4} \text{ mol dm}^{-3}$

Step 2: For the dissociation $HX \rightleftharpoons H^+ + X^-$, at equilibrium:

$[H^+] = [X^-] = 6.31 \times 10^{-4} \text{ mol dm}^{-3}$ $[HX] = 0.10 - 6.31 \times 10^{-4} \approx 0.10 \text{ mol dm}^{-3}$

Step 3: Substitute into the $K_a$ expression:

$K_a = \frac{[H^+][X^-]}{[HX]} = \frac{(6.31 \times 10^{-4})^2}{0.10}$ $= \frac{3.98 \times 10^{-7}}{0.10} = 3.98 \times 10^{-6} \text{ mol dm}^{-3}$

Common mistake: Students often forget to square $[H^+]$ or use the initial concentration of acid instead of the equilibrium concentration. Also, some forget to include units (mol dm⁻³).

Mark scheme: 1 mark for correct $[H^+]$ calculation; 1 mark for correct substitution; 1 mark for correct $K_a$ value with units ($3.98 \times 10^{-6}$ mol dm⁻³, accept $4.0 \times 10^{-6}$ mol dm⁻³).

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6. [2 marks]

Difference 1: A strong acid has a lower pH than a weak acid of the same concentration, because the strong acid dissociates completely, producing a higher $[H^+]$.

Difference 2: A strong acid reacts more vigorously (faster rate of reaction) with reactive metals or carbonates than a weak acid of the same concentration, due to the higher $[H^+]$.

Acceptable alternatives:

• A strong acid solution is a better conductor of electricity (higher ion concentration).
• A strong acid has a higher degree of ionisation/dissociation than a weak acid.

Common mistake: Students often confuse "strong/weak" with "concentrated/dilute." Strength refers to the degree of dissociation, not the amount of acid dissolved.

Mark scheme: 1 mark for each valid difference. Answers must compare the same concentration.

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7. [3 marks]

Sodium carbonate is a salt formed from a strong base (NaOH) and a weak acid ($H_2CO_3$). When dissolved in water, the carbonate ion ($CO_3^{2-}$) undergoes hydrolysis (reaction with water):

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

The production of hydroxide ions ($OH^-$) increases the $[OH^-]$ in solution, making the solution alkaline (pH > 7).

Teaching note: This is a key concept: salts of strong bases and weak acids produce alkaline solutions because the conjugate base of the weak acid reacts with water to produce $OH^-$ ions. The parent acid ($H_2CO_3$) is weak, so its conjugate base ($CO_3^{2-}$) is relatively strong and accepts protons from water.

Mark scheme: 1 mark for identifying that $CO_3^{2-}$ undergoes hydrolysis; 1 mark for the correct equation; 1 mark for stating that $OH^-$ ions are produced, making the solution alkaline.

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8. [3 marks]

Reagent: Dilute nitric acid (or dilute hydrochloric acid).

Observation with NaCl: No visible change (no effervescence/bubbles).

Observation with Na₂CO₃: Effervescence/bubbles of gas produced; the gas evolved turns limewater milky.

Equation:

$Na_2CO_3(aq) + 2HNO_3(aq) \rightarrow 2NaNO_3(aq) + H_2O(l) + CO_2(g)$

or in ionic form:

$CO_3^{2-}(aq) + 2H^+(aq) \rightarrow H_2O(l) + CO_2(g)$

Teaching note: Carbonates react with acids to produce carbon dioxide gas. This is a standard test for the carbonate ion. Sodium chloride does not react with dilute acid because HCl and HNO₃ are both strong acids and no gas, precipitate, or weak electrolyte is formed.

Mark scheme: 1 mark for correct reagent (dilute acid); 1 mark for correct observations (no change with NaCl, effervescence with Na₂CO₃); 1 mark for correct equation.

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9. [2 marks]

Step 1: Calculate $[H^+]$ from pH:

$[H^+] = 10^{-12.8} = 1.58 \times 10^{-13} \text{ mol dm}^{-3}$

Step 2: Use the ionic product of water:

$K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$

$[OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{1.58 \times 10^{-13}}$

$[OH^-] = 6.33 \times 10^{-2} \text{ mol dm}^{-3}$

Alternative method: $pOH = 14.0 - 12.8 = 1.2$, so $[OH^-] = 10^{-1.2} = 0.063$ mol dm⁻³.

Common mistake: Students sometimes confuse $K_w$ with $K_a$ or forget that $K_w$ applies at 25 °C. Also, some students calculate $[H^+]$ instead of $[OH^-]$.

Mark scheme: 1 mark for correct method using $K_w$; 1 mark for correct answer ($6.3 \times 10^{-2}$ mol dm⁻³ or 0.063 mol dm⁻³).

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10. [4 marks]

Step 1: Calculate moles of $H_2SO_4$:

$n(H_2SO_4) = c \times V = 0.100 \times \frac{25.0}{1000} = 2.50 \times 10^{-3} \text{ mol}$

Step 2: Use the stoichiometric ratio from the equation:

$H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O$

The mole ratio is $H_2SO_4 : KOH = 1 : 2$.

$n(KOH) = 2 \times 2.50 \times 10^{-3} = 5.00 \times 10^{-3} \text{ mol}$

Step 3: Calculate the volume of KOH solution:

$V(KOH) = \frac{n}{c} = \frac{5.00 \times 10^{-3}}{0.150} = 0.0333 \text{ dm}^3 = 33.3 \text{ cm}^3$

Common mistake: Students often forget that $H_2SO_4$ is diprotic and use a 1:1 ratio instead of 1:2. Also, unit conversion errors (cm³ to dm³) are common.

Mark scheme: 1 mark for moles of $H_2SO_4$; 1 mark for correct mole ratio application; 1 mark for moles of KOH; 1 mark for correct volume = 33.3 cm³.

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Section B: Structured and Data-Response Questions

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11. [10 marks total]

(a) [2 marks]

$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

Units of $K_a$: mol dm⁻³

Teaching note: The units of $K_a$ are derived from the expression: $(mol \, dm^{-3} \times mol \, dm^{-3}) / (mol \, dm^{-3}) = mol \, dm^{-3}$.

Mark scheme: 1 mark for correct expression; 1 mark for correct units (mol dm⁻³).

(b) [3 marks]

Step 1: Set up the equilibrium expression. Since dissociation is small, $[CH_3COOH]_{eq} \approx 0.200$ mol dm⁻³.

$K_a = \frac{[H^+]^2}{[CH_3COOH]}$

Step 2: Rearrange and solve for $[H^+]$:

$[H^+]^2 = K_a \times [CH_3COOH] = 1.74 \times 10^{-5} \times 0.200 = 3.48 \times 10^{-6}$

$[H^+] = \sqrt{3.48 \times 10^{-6}} = 1.865 \times 10^{-3} \text{ mol dm}^{-3}$

Step 3: Calculate pH:

$pH = -\log_{10}(1.865 \times 10^{-3}) = 2.73$

Common mistake: Students may forget to take the square root or may use the full quadratic formula unnecessarily (the approximation is valid here since dissociation is small).

Mark scheme: 1 mark for correct setup of $K_a$ expression; 1 mark for correct $[H^+]$; 1 mark for correct pH = 2.73.

(c) [3 marks]

When a small amount of dilute HCl is added to the buffer:

1. The added $H^+$ ions react with the conjugate base ($CH_3COO^-$) in the buffer:

$H^+(aq) + CH_3COO^-(aq) \rightarrow CH_3COOH(aq)$

2. The conjugate base "mops up" the added $H^+$, converting it back to the weak acid.

3. Since the $[H^+]$ does not increase significantly, the pH remains almost unchanged.

4. The buffer works because it contains significant amounts of both the weak acid and its conjugate base.

Teaching note: The key concept is that the buffer resists pH change by consuming added acid (via the conjugate base) or added base (via the weak acid). Students should identify which component reacts with the added substance.

Mark scheme: 1 mark for stating that $H^+$ reacts with $CH_3COO^-$; 1 mark for the equation; 1 mark for explaining that $[H^+]$ remains nearly constant so pH is maintained.

(d) [2 marks]

When equal volumes of 0.200 mol dm⁻³ $CH_3COOH$ and 0.200 mol dm⁻³ $CH_3COONa$ are mixed, the concentrations of both are halved (due to doubling of volume):

$[CH_3COOH] = 0.100 \text{ mol dm}^{-3}$ $[CH_3COO^-] = 0.100 \text{ mol dm}^{-3}$

Using the Henderson-Hasselbalch equation:

$pH = pK_a + \log_{10}\left(\frac{[CH_3COO^-]}{[CH_3COOH]}\right)$

$pK_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76$

$pH = 4.76 + \log_{10}\left(\frac{0.100}{0.100}\right) = 4.76 + \log_{10}(1) = 4.76 + 0 = 4.76$

Teaching note: When $[acid] = [salt]$, $pH = pK_a$. This is a useful result to remember. The Henderson-Hasselbalch equation is not always provided in exams, so students should know how to derive it from the $K_a$ expression.

Mark scheme: 1 mark for correct $pK_a$ or correct ratio; 1 mark for correct pH = 4.76.

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12. [10 marks total]

(a) [2 marks]

HA is a weak acid.

Reasoning: The initial pH is approximately 2.8, which is higher than the expected pH of 1.0 for a 0.100 mol dm⁻³ strong monoprotic acid. A strong acid at this concentration would have $[H^+] = 0.100$ mol dm⁻³, giving pH = 1.0. The higher initial pH indicates that HA does not fully dissociate, which is characteristic of a weak acid.

Additionally, the titration curve shows a buffer region (a relatively flat section before the steep rise), which is characteristic of a weak acid–strong base titration.

Mark scheme: 1 mark for identifying weak acid; 1 mark for correct reasoning (higher initial pH than expected for strong acid, or reference to buffer region).

(b) [2 marks]

From the graph:

• pH at equivalence point: approximately 8.7–9.0 (the pH at the midpoint of the steep rise)
• Volume of NaOH at equivalence point: 25.0 cm³

Teaching note: The equivalence point for a weak acid–strong base titration occurs at pH > 7 because the salt formed (NaA) is the conjugate base of a weak acid, which hydrolyses to produce an alkaline solution.

Mark scheme: 1 mark for correct volume (25.0 cm³); 1 mark for correct pH range (8.5–9.5).

(c) [2 marks]

The $pK_a$ can be estimated from the graph at the half-equivalence point (where half the volume of NaOH needed to reach the equivalence point has been added).

At the half-equivalence point, volume of NaOH = 12.5 cm³, and at this point $[HA] = [A^-]$, so $pH = pK_a$.

From the graph, at 12.5 cm³, the pH is approximately 4.8.

Therefore, $pK_a \approx 4.8$ and $K_a \approx 1.6 \times 10^{-5}$ mol dm⁻³.

Mark scheme: 1 mark for identifying the half-equivalence point; 1 mark for reading the pH ≈ 4.8 from the graph.

(d) [2 marks]

Phenolphthalein is the most suitable indicator.

Explanation: The equivalence point occurs at approximately pH 8.7–9.0. Phenolphthalein changes colour in the pH range 8.2–10.0, which encompasses the equivalence point. Methyl orange (3.1–4.4) and bromothymol blue (6.0–7.6) change colour well before the equivalence point would be reached, leading to significant titration error.

Mark scheme: 1 mark for selecting phenolphthalein; 1 mark for correct explanation referencing the pH range of the indicator matching the equivalence point pH.

(e) [2 marks]

The solution at the equivalence point is alkaline (pH > 7).

Explanation: At the equivalence point, all the weak acid HA has been neutralised to form the salt NaA. The anion $A^-$ is the conjugate base of the weak acid HA. It undergoes hydrolysis with water:

$A^-(aq) + H_2O(l) \rightleftharpoons HA(aq) + OH^-(aq)$

The production of $OH^-$ ions makes the solution alkaline.

Mark scheme: 1 mark for predicting alkaline; 1 mark for correct explanation involving hydrolysis of $A^-$ producing $OH^-$.

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13. [9 marks total]

(a) [2 marks]

Equation:

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

$K_b$ expression:

$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}$

Mark scheme: 1 mark for correct equation (with reversible arrow); 1 mark for correct $K_b$ expression.

(b) [3 marks]

Given: concentration = 0.100 mol dm⁻³, pH = 11.12

Step 1: Calculate $[H^+]$ and then $[OH^-]$:

$[H^+] = 10^{-11.12} = 7.59 \times 10^{-12} \text{ mol dm}^{-3}$

$[OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{7.59 \times 10^{-12}} = 1.318 \times 10^{-3} \text{ mol dm}^{-3}$

Step 2: At equilibrium: $[NH_4^+] = [OH^-] = 1.318 \times 10^{-3}$ mol dm⁻³

$[NH_3]_{eq} = 0.100 - 1.318 \times 10^{-3} \approx 0.100 \text{ mol dm}^{-3}$

Step 3: Calculate $K_b$:

$K_b = \frac{(1.318 \times 10^{-3})^2}{0.100} = \frac{1.737 \times 10^{-6}}{0.100} = 1.74 \times 10^{-5} \text{ mol dm}^{-3}$

Mark scheme: 1 mark for correct $[OH^-]$; 1 mark for correct substitution; 1 mark for correct $K_b = 1.74 \times 10^{-5}$ mol dm⁻³.

(c) [3 marks]

Prediction: pH < 7 (acidic)

Explanation: Ammonium chloride is formed from a weak base ($NH_3$) and a strong acid (HCl). The ammonium ion ($NH_4^+$) is the conjugate acid of the weak base. It undergoes hydrolysis with water:

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

The production of $H_3O^+$ (or $H^+$) ions makes the solution acidic (pH < 7).

Mark scheme: 1 mark for correct prediction (pH < 7); 1 mark for identifying $NH_4^+$ as the conjugate acid that hydrolyses; 1 mark for the correct equation.

(d) [1 mark]

Use: Maintaining pH in biological systems (e.g., in buffer solutions for enzyme assays, or in maintaining the pH of blood in laboratory simulations).

Acceptable alternatives: Used in household cleaning products, in textile dyeing processes, or in qualitative analysis (e.g., in the identification of metal ions using aqueous ammonia in the presence of ammonium chloride to prevent precipitation of metal hydroxides).

Mark scheme: 1 mark for any valid biological or industrial use of an ammonia/ammonium chloride buffer.

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14. [9 marks total]

(a) [4 marks]

Step 1: Calculate the mass of KOH needed.

Moles of KOH required = $0.098 \times 0.250 = 0.0245$ mol (or any reasonable concentration — the method is what matters).

Mass = moles × molar mass = $0.0245 \times 56.1 = 1.37$ g

Step 2: Weigh out the calculated mass of KOH using an electronic balance. Place a weighing boat on the balance, tare it, and add KOH pellets until the required mass is obtained.

Step 3: Transfer the KOH into a beaker and add approximately 100 cm³ of distilled water. Stir with a glass stirring rod until the KOH has completely dissolved.

Step 4: Transfer the solution into a 250.0 cm³ volumetric flask using a funnel. Rinse the beaker and stirring rod with distilled water and add the washings to the flask.

Step 5: Add distilled water to the flask until the meniscus reaches the 250.0 cm³ mark on the neck of the flask. Use a dropping pipette for the final additions.

Step 6: Stopper the flask and invert several times to ensure the solution is thoroughly mixed.

Mark scheme: 1 mark for correct mass calculation; 1 mark for correct apparatus (balance, beaker, volumetric flask); 1 mark for correct transfer and rinsing procedure; 1 mark for making up to the mark and mixing.

(b)(i) [1 mark]

Titrations 2 and 3 are concordant (within 0.10 cm³ of each other).

Average titre = $\frac{24.35 + 24.35}{2} = 24.35$ cm³

Mark scheme: 1 mark for correct average = 24.35 cm³.

(b)(ii) [3 marks]

Step 1: Moles of KOH used:

$n(KOH) = 0.098 \times \frac{24.35}{1000} = 2.386 \times 10^{-3} \text{ mol}$

Step 2: From the equation $HNO_3 + KOH \rightarrow KNO_3 + H_2O$, the mole ratio is 1:1.

$n(HNO_3) = n(KOH) = 2.386 \times 10^{-3} \text{ mol}$

Step 3: Concentration of $HNO_3$:

$c(HNO_3) = \frac{n}{V} = \frac{2.386 \times 10^{-3}}{25.0/1000} = \frac{2.386 \times 10^{-3}}{0.0250} = 0.0954 \text{ mol dm}^{-3}$

Mark scheme: 1 mark for moles of KOH; 1 mark for correct mole ratio; 1 mark for correct concentration = 0.0954 mol dm⁻³.

(b)(iii) [1 mark]

Acceptable answers:

• Rinse the burette with the KOH solution (not just distilled water) before filling.
• Swirl the conical flask continuously during the titration.
• Read the burette at eye level to avoid parallax error.
• Add the KOH solution dropwise near the endpoint.
• Ensure the jet of the burette is filled (no air bubbles).

Mark scheme: 1 mark for any valid precaution.

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15. [9 marks total]

(a) [1 mark]

For the equilibrium: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$

$K_{sp} = [Pb^{2+}][I^-]^2$

Mark scheme: 1 mark for correct expression.

(b) [3 marks]

Step 1: Let the solubility of $PbI_2$ = $s$ mol dm⁻³.

From the stoichiometry of dissolution:

• $[Pb^{2+}] = s$
• $[I^-] = 2s$

Step 2: Substitute into the $K_{sp}$ expression:

$K_{sp} = [Pb^{2+}][I^-]^2 = s \times (2s)^2 = 4s^3$

Step 3: Solve for $s$:

$4s^3 = 8.49 \times 10^{-9}$

$s^3 = 2.1225 \times 10^{-9}$

$s = \sqrt[3]{2.1225 \times 10^{-9}} = 1.28 \times 10^{-3} \text{ mol dm}^{-3}$

Common mistake: Students often forget that $[I^-] = 2s$ (not $s$) and therefore write $K_{sp} = s^3$ instead of $K_{sp} = 4s^3$. This is a very common error with salts that produce ions in a 1:2 ratio.

Mark scheme: 1 mark for correct relationship between ion concentrations and $s$; 1 mark for correct substitution into $K_{sp}$ expression; 1 mark for correct answer ($1.28 \times 10^{-3}$ mol dm⁻³).

(c) [2 marks]

The solubility of $PbI_2$ would decrease.

Explanation: Adding KI increases the concentration of $I^-$ ions in solution. According to Le Chatelier's principle, the equilibrium:

$PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$

shifts to the left (towards the solid) to counteract the increase in $[I^-]$. This is the common ion effect. As a result, less $PbI_2$ dissolves, and its solubility decreases.

Mark scheme: 1 mark for predicting decreased solubility; 1 mark for correct explanation using Le Chatelier's principle / common ion effect.

(d) [3 marks]

Although $Ba^{2+}$ ions are toxic, $BaSO_4$ is safe to ingest because it is extremely sparingly soluble in water.

The solubility product $K_{sp}$ of $BaSO_4$ is very small ($1.08 \times 10^{-10}$ mol² dm⁻⁶ at 25 °C). This means that only a negligibly tiny concentration of $Ba^{2+}$ ions dissolves in the gastrointestinal tract:

$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$

The concentration of dissolved $Ba^{2+}$ is so low that it does not reach toxic levels in the body. The solid $BaSO_4$ passes through the digestive system without releasing harmful amounts of barium ions.

Mark scheme: 1 mark for stating that $BaSO_4$ is sparingly soluble / has a very small $K_{sp}$; 1 mark for explaining that the concentration of dissolved $Ba^{2+}$ is negligibly small; 1 mark for concluding that the amount is too small to be toxic.

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End of Answer Key

Section A: 25 marks | Section B: 35 marks | Total: 60 marks