A Level Chemistry H3 Practice Paper 1

TuitionGoWhere Practice Paper - Chemistry H3 A-Level: Answer Key and Marking Scheme

Subject: Chemistry H3 Level: A-Level (H3, Syllabus 9813) Paper: Practice Paper (Version 1 of 5) Total Marks: 100

Section A (60 Marks)

Question 1 (Total: 9 marks)

(a) Brønsted–Lowry acid: A species that donates a proton ($H^+$). [1] Brønsted–Lowry base: A species that accepts a proton ($H^+$). [1]

(b) (i) $K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$ [1]

(ii) Step 1: Set up ICE table (or use approximation). $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ Initial: 0.250, 0, 0 Change: -x, +x, +x Equilibrium: 0.250 - x, x, x

Step 2: Substitute into $K_a$ expression. $1.74 \times 10^{-5} = \frac{x^2}{0.250 - x}$

Step 3: Assume x is negligible compared to 0.250 (since $K_a$ is small). $1.74 \times 10^{-5} \approx \frac{x^2}{0.250}$ $x^2 = 4.35 \times 10^{-6}$ $x = 2.09 \times 10^{-3} \text{ mol dm}^{-3}$

Step 4: Calculate pH. $pH = -\log[H^+] = -\log(2.09 \times 10^{-3}) = 2.68$ [2]

Assumption: The extent of dissociation of ethanoic acid is negligible, so $[CH_3COOH]_{equilibrium} \approx 0.250 \text{ mol dm}^{-3}$. [1]

(c) (i) Step 1: Calculate moles of acid and salt. Moles of $CH_3COOH = \frac{50.0}{1000} \times 0.500 = 0.0250 \text{ mol}$ Moles of $CH_3COONa = \frac{25.0}{1000} \times 0.500 = 0.0125 \text{ mol}$

Step 2: Calculate concentrations in the mixture. Total volume = 75.0 cm³ = 0.0750 dm³. $[CH_3COOH] = \frac{0.0250}{0.0750} = 0.333 \text{ mol dm}^{-3}$ $[CH_3COO^-] = \frac{0.0125}{0.0750} = 0.167 \text{ mol dm}^{-3}$

Step 3: Use the Henderson–Hasselbalch equation. $pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$ $pH = -\log(1.74 \times 10^{-5}) + \log\frac{0.167}{0.333}$ $pH = 4.76 + \log(0.500) = 4.76 - 0.301 = 4.46$ [2]

(ii) The added $H^+$ ions react with the ethanoate ions, $CH_3COO^-$, to form undissociated ethanoic acid, $CH_3COOH$. This removes most of the added $H^+$ ions, so the pH change is small. [1]

Question 2 (Total: 10 marks)

(a) (i) $B + H_2O \rightleftharpoons BH^+ + OH^-$ [1]

(ii) Step 1: Calculate $[OH^-]$ from pH. $pOH = 14 - pH = 14 - 11.13 = 2.87$ $[OH^-] = 10^{-2.87} = 1.35 \times 10^{-3} \text{ mol dm}^{-3}$

Step 2: Set up ICE table. $B + H_2O \rightleftharpoons BH^+ + OH^-$ Initial: 0.100, 0, 0 Change: -x, +x, +x Equilibrium: 0.100 - x, x, x where $x = [OH^-] = 1.35 \times 10^{-3} \text{ mol dm}^{-3}$

Step 3: Substitute into $K_b$ expression. $K_b = \frac{[BH^+][OH^-]}{[B]} = \frac{x^2}{0.100 - x} = \frac{(1.35 \times 10^{-3})^2}{0.100 - 1.35 \times 10^{-3}}$ $K_b = \frac{1.82 \times 10^{-6}}{0.09865} = 1.85 \times 10^{-5} \text{ mol dm}^{-3}$ [3]

(b) (i) The ionic product of water, $K_w$, is the product of the concentrations of hydrogen ions and hydroxide ions in water, $K_w = [H^+][OH^-]$. At 298 K, $K_w = 1.00 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}$. [1]

(ii) Step 1: Calculate $[OH^-]$. $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$ $[OH^-] = 2 \times 0.0100 = 0.0200 \text{ mol dm}^{-3}$

Step 2: Calculate $[H^+]$. $[H^+] = \frac{K_w}{[OH^-]} = \frac{1.00 \times 10^{-14}}{0.0200} = 5.00 \times 10^{-13} \text{ mol dm}^{-3}$

Step 3: Calculate pH. $pH = -\log(5.00 \times 10^{-13}) = 12.30$ [2]

(c) (i) Step 1: Calculate $[H^+]$ of the original solution. $[H^+] = 10^{-1.30} = 0.0501 \text{ mol dm}^{-3}$

Step 2: Calculate $[H^+]$ after dilution. Dilution factor = 100, so $[H^+]_{diluted} = \frac{0.0501}{100} = 5.01 \times 10^{-4} \text{ mol dm}^{-3}$

Step 3: Calculate pH. $pH = -\log(5.01 \times 10^{-4}) = 3.30$ [2]

(ii) At very low concentrations, the auto-ionisation of water ($[H^+] = 1.00 \times 10^{-7} \text{ mol dm}^{-3}$) becomes significant and cannot be neglected. The pH will approach 7, not continue to increase indefinitely. [1]

Question 3 (Total: 11 marks)

(a) (i) $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$ [1]

(ii) Step 1: Calculate moles of $Na_2CO_3$. Molar mass of $Na_2CO_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0 \text{ g mol}^{-1}$ Moles of $Na_2CO_3 = \frac{2.65}{106.0} = 0.0250 \text{ mol}$

Step 2: Calculate concentration. Concentration $= \frac{0.0250}{0.250} = 0.100 \text{ mol dm}^{-3}$ [2]

(b) (i) Step 1: Calculate moles of $Na_2CO_3$ in the titre. Moles of $Na_2CO_3 = \frac{25.0}{1000} \times 0.100 = 2.50 \times 10^{-3} \text{ mol}$

Step 2: Calculate moles of HCl. From the equation, mole ratio $Na_2CO_3 : HCl = 1:2$. Moles of HCl $= 2 \times 2.50 \times 10^{-3} = 5.00 \times 10^{-3} \text{ mol}$

Step 3: Calculate concentration of HCl. Concentration $= \frac{5.00 \times 10^{-3}}{23.50/1000} = 0.213 \text{ mol dm}^{-3}$ [2]

(ii) Indicator: Methyl orange. [1] Reason: The titration is between a strong acid (HCl) and a weak base ($Na_2CO_3$). The pH at the equivalence point is acidic (pH < 7) because $CO_2$ is formed and dissolved in the solution. Methyl orange changes colour in the pH range 3.1–4.4, which is within the sharp pH change at the equivalence point for this titration. [1]

(c) (i) Sketch: The graph should show:

• A curve starting at pH ~13 (strong base).
• A gradual decrease in pH as HCl is added.
• A sharp, vertical drop in pH around the equivalence point (at 25 cm³ of HCl added).
• The equivalence point at pH 7.
• The curve levelling off at pH ~1 after the equivalence point.
• The buffer region is before the equivalence point, where the pH changes gradually. [3]

(ii) At the equivalence point, the solution contains only NaCl, which is a salt of a strong acid and a strong base. Neither $Na^+$ nor $Cl^-$ hydrolyses significantly in water, so the solution is neutral, and pH = 7. [1]

Question 4 (Total: 10 marks)

(a) (i) $K_{sp} = [Ag^+][Cl^-]$ [1]

(ii) Step 1: Let solubility = s mol dm⁻³. $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ $[Ag^+] = s$, $[Cl^-] = s$

Step 2: Substitute into $K_{sp}$ expression. $K_{sp} = s \times s = s^2 = 1.77 \times 10^{-10}$

Step 3: Calculate s. $s = \sqrt{1.77 \times 10^{-10}} = 1.33 \times 10^{-5} \text{ mol dm}^{-3}$ [2]

(b) Step 1: Consider the common ion effect. In a $0.100 \text{ mol dm}^{-3}$ NaCl solution, $[Cl^-] = 0.100 \text{ mol dm}^{-3}$.

Step 2: Let solubility in NaCl solution = s' mol dm⁻³. $[Ag^+] = s'$, $[Cl^-] = 0.100 + s' \approx 0.100$ (since s' is very small).

Step 3: Substitute into $K_{sp}$ expression. $K_{sp} = s' \times 0.100 = 1.77 \times 10^{-10}$

Step 4: Calculate s'. $s' = \frac{1.77 \times 10^{-10}}{0.100} = 1.77 \times 10^{-9} \text{ mol dm}^{-3}$ [3]

(c) (i) $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$ [1]

(ii) Step 1: Let solubility = s mol dm⁻³. $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$ $[Ag^+] = 2s$, $[CrO_4^{2-}] = s$

Step 2: Substitute into $K_{sp}$ expression. $K_{sp} = (2s)^2 \times s = 4s^3 = 1.12 \times 10^{-12}$

Step 3: Calculate s. $s^3 = \frac{1.12 \times 10^{-12}}{4} = 2.80 \times 10^{-13}$ $s = \sqrt[3]{2.80 \times 10^{-13}} = 6.54 \times 10^{-5} \text{ mol dm}^{-3}$ [3]

Question 5 (Total: 10 marks)

(a) (i) $K_{sp} = [Mg^{2+}][OH^-]^2$ [1]

(ii) Step 1: Calculate $pOH$ and $[OH^-]$. $pOH = 14 - 10.45 = 3.55$ $[OH^-] = 10^{-3.55} = 2.82 \times 10^{-4} \text{ mol dm}^{-3}$

Step 2: Calculate $[Mg^{2+}]$. From the equation $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$, $[Mg^{2+}] = \frac{1}{2}[OH^-] = 1.41 \times 10^{-4} \text{ mol dm}^{-3}$

Step 3: Substitute into $K_{sp}$ expression. $K_{sp} = (1.41 \times 10^{-4}) \times (2.82 \times 10^{-4})^2$ $K_{sp} = 1.41 \times 10^{-4} \times 7.95 \times 10^{-8} = 1.12 \times 10^{-11} \text{ mol}^3 \text{ dm}^{-9}$ [3]

(b) (i) Colourless. [1]

(ii) The solution is basic (pH 10.45 > 8.2), so phenolphthalein turns pink. [1]

(c) (i) The pink colour fades and the solution becomes colourless. [1] This is because the added $H^+$ ions neutralise the $OH^-$ ions, shifting the equilibrium $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ to the right. As $[OH^-]$ decreases, the pH drops below 8.2, and phenolphthalein becomes colourless. [1]

(ii) $H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$ [1]

(d) Magnesium hydroxide is sparingly soluble, so it dissolves slowly and provides a gradual, prolonged neutralisation of stomach acid without causing a sudden, drastic change in pH. This makes it a mild, safe antacid. [1]

Question 6 (Total: 10 marks)

(a) (i) [4]

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

(b) (i) Step 1: Set up ICE table. $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$ Initial: 0.200, 0, 0 Change: -x, +x, +x Equilibrium: 0.200 - x, x, x

Step 2: Substitute into $K_a$ expression. $K_a = \frac{[NH_3][H_3O^+]}{[NH_4^+]} = \frac{x^2}{0.200 - x} = 5.62 \times 10^{-10}$

Step 3: Assume x is negligible. $x^2 = 5.62 \times 10^{-10} \times 0.200 = 1.124 \times 10^{-10}$ $x = 1.06 \times 10^{-5} \text{ mol dm}^{-3}$

Step 4: Calculate pH. $pH = -\log(1.06 \times 10^{-5}) = 4.97$ [3]

(c) (i) $CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)$ [1]

(ii) Step 1: Calculate $pOH$ and $[OH^-]$. $pOH = 14 - 8.88 = 5.12$ $[OH^-] = 10^{-5.12} = 7.59 \times 10^{-6} \text{ mol dm}^{-3}$

Step 2: Substitute into $K_b$ expression. $K_b = \frac{[CH_3COOH][OH^-]}{[CH_3COO^-]} = \frac{(7.59 \times 10^{-6})^2}{0.100 - 7.59 \times 10^{-6}}$ $K_b \approx \frac{5.76 \times 10^{-11}}{0.100} = 5.76 \times 10^{-10} \text{ mol dm}^{-3}$ [1]

Section B (40 Marks)

Question 7 (Total: 20 marks)

(a) (i) $HCOOH(aq) \rightleftharpoons HCOO^-(aq) + H^+(aq)$ [1]

(ii) Step 1: Calculate $[H^+]$. $[H^+] = 10^{-2.38} = 4.17 \times 10^{-3} \text{ mol dm}^{-3}$

Step 2: Set up ICE table. $HCOOH \rightleftharpoons HCOO^- + H^+$ Initial: 0.100, 0, 0 Change: -x, +x, +x Equilibrium: 0.100 - x, x, x where $x = 4.17 \times 10^{-3} \text{ mol dm}^{-3}$

Step 3: Substitute into $K_a$ expression. $K_a = \frac{[HCOO^-][H^+]}{[HCOOH]} = \frac{x^2}{0.100 - x} = \frac{(4.17 \times 10^{-3})^2}{0.100 - 4.17 \times 10^{-3}}$ $K_a = \frac{1.74 \times 10^{-5}}{0.09583} = 1.82 \times 10^{-4} \text{ mol dm}^{-3}$ [3]

(b) (i) The chlorine atom is more electronegative than the hydrogen atom. It exerts a negative inductive effect (-I effect), withdrawing electron density from the O–H bond through the sigma bond framework. This weakens the O–H bond, making it easier for the proton to be released. Therefore, chloroethanoic acid is a stronger acid than methanoic acid. [3]

(c) (i) Step 1: Calculate moles of acid and salt. Moles of HCOOH $= \frac{25.0}{1000} \times 0.200 = 5.00 \times 10^{-3} \text{ mol}$ Moles of HCOONa $= \frac{20.0}{1000} \times 0.150 = 3.00 \times 10^{-3} \text{ mol}$

Step 2: Calculate concentrations in the mixture. Total volume = 45.0 cm³ = 0.0450 dm³. $[HCOOH] = \frac{5.00 \times 10^{-3}}{0.0450} = 0.111 \text{ mol dm}^{-3}$ $[HCOO^-] = \frac{3.00 \times 10^{-3}}{0.0450} = 0.0667 \text{ mol dm}^{-3}$

Step 3: Use the Henderson–Hasselbalch equation. $pH = pK_a + \log\frac{[HCOO^-]}{[HCOOH]}$ $pH = -\log(1.82 \times 10^{-4}) + \log\frac{0.0667}{0.111}$ $pH = 3.74 + \log(0.600) = 3.74 - 0.222 = 3.52$ [4]

(d) (i) Step 1: Calculate moles of HCl added. Moles of HCl $= \frac{1.0}{1000} \times 1.00 = 1.00 \times 10^{-3} \text{ mol}$

Step 2: Calculate new moles of acid and salt after reaction. The added $H^+$ reacts with $HCOO^-$: $HCOO^- + H^+ \rightarrow HCOOH$. Moles of HCOOH $= 5.00 \times 10^{-3} + 1.00 \times 10^{-3} = 6.00 \times

<stage5_exam_answers_md>