A Level H2 Chemistry Practice Paper 4

TuitionGoWhere Exam Practice (AI) - Chemistry H2 A-Level

Marking Scheme - Practice Paper (Version 4)

Topic: Acids, Bases and Salts

Total Marks: 60

Section A: Structured Questions

1. Titration and pH Curves

(a) Suitable Volume Calculation [2 marks]

• Identify concordant results: Titres 1 ($23.80$), 2 ($23.60$), and 3 ($23.80$).
• Note: Titre 2 differs from 1 and 3 by $0.20 \, \text{cm}^3$. Usually, concordant results are within $0.10 \, \text{cm}^3$. However, in many school contexts, if only 3 accurate titres are given, students might average all or exclude the outlier.
  • Strict Interpretation: 1 and 3 are concordant ($23.80$). Average = $23.80 \, \text{cm}^3$.
  • Alternative Interpretation (if 2 is accepted): Average of 1, 2, 3 = $(23.80+23.60+23.80)/3 = 23.73 \, \text{cm}^3$.
  • Standard Exam Expectation: Usually, exclude rough. Check range. $23.80$ and $23.80$ are identical. $23.60$ is an outlier ($>0.1$ difference). Use 1 and 3.
  • Average = $\frac{23.80 + 23.80}{2} = 23.80 \, \text{cm}^3$.
• Marking:
  • 1 mark for identifying concordant titres (1 and 3) or correct exclusion of outlier.
  • 1 mark for correct calculation of mean ($23.80 \, \text{cm}^3$).
  • Note: If student averages all three ($23.73$), award 1 mark for method but deduct 1 for precision/selection error unless specific instructions allow wider tolerance. Let's stick to the rigorous $23.80$.

(b) Concentration of Ethanoic Acid [2 marks]

• Equation: $\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$
• Mole ratio 1:1.
• Moles of $\text{NaOH} = 0.100 \times \frac{23.80}{1000} = 2.38 \times 10^{-3} \, \text{mol}$.
• Moles of $\text{CH}_3\text{COOH} = 2.38 \times 10^{-3} \, \text{mol}$.
• Concentration = $\frac{2.38 \times 10^{-3}}{25.0/1000} = \frac{2.38 \times 10^{-3}}{0.025} = 0.0952 \, \text{mol dm}^{-3}$.
• Marking:
  • 1 mark for correct moles of NaOH.
  • 1 mark for correct concentration ($0.0952 \, \text{mol dm}^{-3}$).

(c) pH at Equivalence Point > 7 [2 marks]

• At equivalence, the solution contains sodium ethanoate ($\text{CH}_3\text{COONa}$).
• The ethanoate ion hydrolyses: $\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$.
• Production of $\text{OH}^-$ ions makes the solution alkaline (pH > 7).
• Marking:
  • 1 mark for equation showing hydrolysis producing $\text{OH}^-$.
  • 1 mark for stating that $\text{OH}^-$ causes alkalinity/pH > 7.

(d) pH Curve Sketch [3 marks]

• Shape: Starts at pH $\approx 2.9$ (weak acid). Gradual rise (buffer region). Steep vertical section at equivalence point ($\approx 23.8 \, \text{cm}^3$). Ends at pH $\approx 12-13$ (excess strong base).
• Equivalence Point: Located at pH $\approx 8-9$ (basic side).
• Buffer Region: Indicated in the flat region before the vertical rise (around half-equivalence, $11.9 \, \text{cm}^3$).
• Marking:
  • 1 mark for correct initial pH and general shape (S-shape).
  • 1 mark for vertical section centered at correct volume and pH > 7.
  • 1 mark for labeling equivalence point and buffer region.

2. Buffer Solutions

(a) Definition [1 mark]

• A solution that resists changes in pH upon the addition of small amounts of acid or alkali.
• Marking: 1 mark for key concept "resists change in pH".

(b) pH of Buffer [2 marks]

• $[\text{Acid}] = [\text{Salt}] = 0.050 \, \text{mol dm}^{-3}$ (diluted by half, but ratio is 1:1).
• Alternatively, use mole ratio directly since volume is same.
• $\text{pH} = \text{p}K_a + \log_{10} \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$
• $\text{p}K_a = -\log_{10}(1.7 \times 10^{-5}) = 4.77$.
• $\text{pH} = 4.77 + \log_{10}(1) = 4.77$.
• Marking:
  • 1 mark for correct $\text{p}K_a$ or setup.
  • 1 mark for correct pH ($4.77$).

(c) pH Change after adding HCl [3 marks]

• Moles of $\text{H}^+$ added = $1.0 \times 10^{-3} \, \text{dm}^3 \times 1.0 \, \text{mol dm}^{-3} = 0.001 \, \text{mol}$.
• Initial moles in buffer (in $100 \, \text{cm}^3$):
  • $\text{CH}_3\text{COOH} = 0.050 \times 0.1 = 0.005 \, \text{mol}$.
  • $\text{CH}_3\text{COO}^- = 0.050 \times 0.1 = 0.005 \, \text{mol}$.
• Reaction: $\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$.
• New moles:
  • $\text{CH}_3\text{COO}^- = 0.005 - 0.001 = 0.004 \, \text{mol}$.
  • $\text{CH}_3\text{COOH} = 0.005 + 0.001 = 0.006 \, \text{mol}$.
• New pH = $4.77 + \log_{10} \left( \frac{0.004}{0.006} \right) = 4.77 + \log_{10}(0.667) = 4.77 - 0.176 = 4.59$.
• Change in pH = $4.77 - 4.59 = 0.18$ (decrease).
• Marking:
  • 1 mark for correct new mole calculations.
  • 1 mark for correct new pH calculation.
  • 1 mark for correct magnitude of change ($0.18$).

(d) Comparison with Water [2 marks]

• In water, $[\text{H}^+] = \frac{0.001 \, \text{mol}}{0.1 \, \text{dm}^3} = 0.01 \, \text{mol dm}^{-3}$.
• $\text{pH} = -\log_{10}(0.01) = 2.0$.
• Change from pH 7 to 2 is 5 units.
• Buffer contains high concentrations of conjugate base which removes added $\text{H}^+$, minimizing the increase in $[\text{H}^+]$.
• Marking:
  • 1 mark for stating pH of water would drop significantly (to ~2).
  • 1 mark for explaining that buffer components react to remove $\text{H}^+$.

3. Solubility Equilibria

(a) $K_{sp}$ Expression [1 mark]

• $K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2$
• Marking: 1 mark for correct expression.

(b) Molar Solubility in Water [2 marks]

• Let solubility be $s \, \text{mol dm}^{-3}$.
• $[\text{Mg}^{2+}] = s$, $[\text{OH}^-] = 2s$.
• $K_{sp} = (s)(2s)^2 = 4s^3$.
• $1.8 \times 10^{-11} = 4s^3$.
• $s^3 = 4.5 \times 10^{-12}$.
• $s = \sqrt[3]{4.5 \times 10^{-12}} \approx 1.65 \times 10^{-4} \, \text{mol dm}^{-3}$.
• Marking:
  • 1 mark for setup ($4s^3$).
  • 1 mark for correct value ($1.65 \times 10^{-4}$).

(c) Precipitation Check [3 marks]

• Total volume = $200 \, \text{cm}^3$. Concentrations halved.
• $[\text{Mg}^{2+}] = 0.005 \, \text{mol dm}^{-3}$.
• $[\text{OH}^-] = 0.005 \, \text{mol dm}^{-3}$.
• Ionic Product (IP) = $[\text{Mg}^{2+}][\text{OH}^-]^2 = (0.005)(0.005)^2 = 1.25 \times 10^{-7}$.
• Compare IP with $K_{sp}$: $1.25 \times 10^{-7} > 1.8 \times 10^{-11}$.
• Since IP > $K_{sp}$, a precipitate will form.
• Marking:
  • 1 mark for correct concentrations after mixing.
  • 1 mark for correct IP calculation.
  • 1 mark for correct conclusion (Precipitate forms).

(d) Solubility in $\text{NH}_4\text{Cl}$ [2 marks]

• $\text{NH}_4^+$ is acidic: $\text{NH}_4^+ \rightleftharpoons \text{NH}_3 + \text{H}^+$.
• $\text{H}^+$ reacts with $\text{OH}^-$ from $\text{Mg(OH)}_2$ equilibrium: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$.
• This reduces $[\text{OH}^-]$, shifting the solubility equilibrium $\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-$ to the right (Le Chatelier's Principle).
• Marking:
  • 1 mark for identifying reaction between $\text{NH}_4^+$/$\text{H}^+$ and $\text{OH}^-$.
  • 1 mark for explaining shift in equilibrium/increased solubility.

4. Indicators

(a) Derivation [2 marks]

• $K_{\text{In}} = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]}$
• $[\text{H}^+] = K_{\text{In}} \frac{[\text{HIn}]}{[\text{In}^-]}$
• $-\log_{10}[\text{H}^+] = -\log_{10} K_{\text{In}} - \log_{10} \left( \frac{[\text{HIn}]}{[\text{In}^-]} \right)$
• $\text{pH} = \text{p}K_{\text{In}} + \log_{10} \left( \frac{[\text{In}^-]}{[\text{HIn}]} \right)$
• Marking:
  • 1 mark for $K_{\text{In}}$ expression.
  • 1 mark for final logarithmic form.

(b) pH Range [1 mark]

• $\text{pH} = \text{p}K_{\text{In}} \pm 1$.
• Range: $8.3 - 10.3$.
• Marking: 1 mark for correct range.

(c) Suitability [2 marks]

• Yes, it is suitable.
• The equivalence point of weak acid-strong base titration is in the basic range (pH 8-9).
• The indicator's colour change range (8.3-10.3) overlaps with the steep vertical portion of the titration curve near the equivalence point.
• Marking:
  • 1 mark for "Yes".
  • 1 mark for linking pH range to equivalence point pH.

Section B: Data Interpretation and Application

5. Acid Strength and pH

(a) Difference between A and B [2 marks]

• HCl is a strong acid; it dissociates completely in water ($[\text{H}^+] = 0.1 \, \text{M}$, pH 1).
• $\text{CH}_3\text{COOH}$ is a weak acid; it dissociates partially ($[\text{H}^+] < 0.1 \, \text{M}$, pH > 1).
• Marking:
  • 1 mark for complete vs partial dissociation.
  • 1 mark for linking to $[\text{H}^+]$ concentration.

(b) Acid B vs Acid C [3 marks]

• Acid C ($\text{ClCH}_2\text{COOH}$) is stronger than Acid B ($\text{CH}_3\text{COOH}$) (lower pH).
• Chlorine is electronegative and exerts a negative inductive effect (-I effect).
• This withdraws electron density from the carboxylate group, stabilizing the conjugate base ($\text{ClCH}_2\text{COO}^-$) by dispersing the negative charge.
• This makes the O-H bond more polar and easier to break, increasing $K_a$.
• Marking:
  • 1 mark for identifying C as stronger.
  • 1 mark for mentioning electronegativity/inductive effect.
  • 1 mark for explaining stabilization of conjugate base.

(c) $K_a$ of Acid D [2 marks]

• $\text{pH} = 2.4 \Rightarrow [\text{H}^+] = 10^{-2.4} = 3.98 \times 10^{-3} \, \text{mol dm}^{-3}$.
• For weak acid: $K_a \approx \frac{[\text{H}^+]^2}{[\text{HA}]}$.
• $K_a = \frac{(3.98 \times 10^{-3})^2}{0.10} = \frac{1.58 \times 10^{-5}}{0.10} = 1.58 \times 10^{-4} \, \text{mol dm}^{-3}$.
• Marking:
  • 1 mark for correct $[\text{H}^+]$.
  • 1 mark for correct $K_a$ ($1.6 \times 10^{-4}$).

(d)(i) pH of Mixture (Strong Acid + Strong Base) [1 mark]

• Moles $\text{H}^+ = 0.020 \times 0.10 = 0.002$.
• Moles $\text{OH}^- = 0.020 \times 0.10 = 0.002$.
• Exact neutralization. Salt is NaCl (neutral).
• $\text{pH} = 7$.
• Marking: 1 mark for pH 7.

(d)(ii) Weak Acid Comparison [2 marks]

• Lower pH (more acidic) than 7? No, wait.
• Titration of Weak Acid (B) + Strong Base.
• At equivalence, salt is $\text{CH}_3\text{COONa}$.
• Hydrolysis produces $\text{OH}^-$.
• pH will be > 7 (Basic).
• Question asks: "If Acid B were used... would final pH be higher, lower or same?"
• Final pH of Strong Acid+Base = 7.
• Final pH of Weak Acid+Base > 7.
• So, pH would be Higher.
• Marking:
  • 1 mark for "Higher".
  • 1 mark for explanation (formation of basic salt/hydrolysis).

6. Qualitative Analysis

(a) Cation [1 mark]

• $\text{Al}^{3+}$ (Aluminium ion).
• (Zn also fits white ppt soluble in excess NaOH, but Zn ppt is soluble in excess NH3. Al ppt is insoluble in excess NH3. So it must be Al).
• Marking: 1 mark for $\text{Al}^{3+}$.

(b) Anion [1 mark]

• $\text{Cl}^-$ (Chloride ion).
• White ppt with $\text{AgNO}_3$ soluble in dilute $\text{NH}_3$.
• Marking: 1 mark for $\text{Cl}^-$.

(c) Equation with excess NaOH [1 mark]

• $\text{Al}^{3+}\text{(aq)} + 4\text{OH}^-\text{(aq)} \rightarrow [\text{Al(OH)}_4]^-\text{(aq)}$
• Marking: 1 mark for correct complex ion formula and balancing.

(d) Equation with $\text{AgNO}_3$ [1 mark]

• $\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \rightarrow \text{AgCl(s)}$
• Marking: 1 mark for correct equation.

(e) Formula of X [1 mark]

• $\text{AlCl}_3$
• Marking: 1 mark.

(f) Observations for $\text{Cu}^{2+}$ with $\text{NH}_3$ [2 marks]

• Dropwise: Pale blue precipitate ($\text{Cu(OH)}_2$ or basic salt).
• Excess: Precipitate dissolves to form a deep blue solution ($[\text{Cu(NH}_3)_4]^{2+}$).
• Marking:
  • 1 mark for blue ppt.
  • 1 mark for deep blue solution in excess.

(g) Colour of Transition Metals [3 marks]

• Transition metal ions have partially filled d-orbitals.
• Ligands cause d-orbitals to split into different energy levels.
• Electrons absorb visible light energy to transition between these split d-orbitals ($d-d$ transition).
• The colour observed is the complementary colour of the light absorbed.
• $\text{Al}^{3+}$ has an empty d-subshell ($[\text{Ne}]$ configuration), so no $d-d$ transitions are possible.
• Marking:
  • 1 mark for d-orbital splitting.
  • 1 mark for absorption of visible light/$d-d$ transition.
  • 1 mark for Al having no d-electrons/empty d-shell.