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: Compulsory Questions (60 Marks)

Question 1 (Total: 8 marks)

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

(b)(i) Expression for $K_a$ [1] $K_a = \frac{[H^+][A^-]}{[HA]}$

(b)(ii) Calculation of $K_a$ [3]

1. Calculate $[H^+]$: $[H^+] = 10^{-pH} = 10^{-3.40} = 3.98 \times 10^{-4}$ mol dm⁻³

2. For a weak acid, $[A^-] = [H^+]$: $[A^-] = 3.98 \times 10^{-4}$ mol dm⁻³

3. Calculate $[HA]$ at equilibrium: $[HA] = 0.0250 - 3.98 \times 10^{-4} \approx 0.0246$ mol dm⁻³

4. Substitute into the $K_a$ expression: $K_a = \frac{(3.98 \times 10^{-4})^2}{0.0246} = 6.44 \times 10^{-6} \text{ mol dm}^{-3}$

Answer: $K_a = 6.44 \times 10^{-6}$ mol dm⁻³

Marking note: Award 1 mark for calculating $[H^+]$, 1 mark for the correct substitution, and 1 mark for the final answer with units.

(c)(i) Strong or weak acid [1] HB is a strong acid because the pH of 1.60 for a 0.0250 mol dm⁻³ solution corresponds to $[H^+] = 10^{-1.60} = 0.0251$ mol dm⁻³, which is approximately equal to the initial concentration of the acid. This indicates complete dissociation.

(c)(ii) Calculation of $[OH^-]$ [2]

1. Use the ionic product of water: $K_w = [H^+][OH^-] = 1.00 \times 10^{-14}$ mol² dm⁻⁶

2. Calculate $[OH^-]$: $[OH^-] = \frac{K_w}{[H^+]} = \frac{1.00 \times 10^{-14}}{0.0251} = 3.98 \times 10^{-13} \text{ mol dm}^{-3}$

Answer: $[OH^-] = 3.98 \times 10^{-13}$ mol dm⁻³

Marking note: Award 1 mark for the correct formula and 1 mark for the final answer with units.

Question 2 (Total: 10 marks)

(a) pH of buffer solution [3]

1. Use the Henderson–Hasselbalch equation: $pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$

2. Calculate $pK_a$: $pK_a = -\log(1.74 \times 10^{-5}) = 4.76$

3. Substitute the concentrations: $pH = 4.76 + \log\frac{0.200}{0.150} = 4.76 + 0.125 = 4.88$

Answer: pH = 4.88

Marking note: Award 1 mark for the correct equation, 1 mark for calculating $pK_a$, and 1 mark for the final answer.

(b)(i) Amount of HCl added [1] $\text{moles of HCl} = \frac{5.0}{1000} \times 0.100 = 5.0 \times 10^{-4} \text{ mol}$

(b)(ii) New pH after addition of HCl [4]

1. Calculate initial moles of ethanoic acid and ethanoate in the 25.0 cm³ buffer sample:

   • Moles of $CH_3COOH = \frac{25.0}{1000} \times 0.150 = 3.75 \times 10^{-3}$ mol
   • Moles of $CH_3COO^- = \frac{25.0}{1000} \times 0.200 = 5.00 \times 10^{-3}$ mol

2. The added H⁺ reacts with the ethanoate ion: $CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)$

3. Calculate new moles after reaction:

   • Moles of $CH_3COO^- = 5.00 \times 10^{-3} - 5.0 \times 10^{-4} = 4.50 \times 10^{-3}$ mol
   • Moles of $CH_3COOH = 3.75 \times 10^{-3} + 5.0 \times 10^{-4} = 4.25 \times 10^{-3}$ mol

4. Calculate new concentrations using total volume of 30.0 cm³:

   • $[CH_3COO^-] = \frac{4.50 \times 10^{-3}}{0.0300} = 0.150$ mol dm⁻³
   • $[CH_3COOH] = \frac{4.25 \times 10^{-3}}{0.0300} = 0.142$ mol dm⁻³

5. Calculate new pH: $pH = 4.76 + \log\frac{0.150}{0.142} = 4.76 + 0.024 = 4.78$

Answer: pH = 4.78

Marking note: Award 1 mark for calculating initial moles, 1 mark for the reaction and new moles, 1 mark for new concentrations, and 1 mark for the final pH.

(c) Reason why buffer resists pH change [2] The buffer contains a weak acid ($CH_3COOH$) and its conjugate base ($CH_3COO^-$). When a small amount of acid (H⁺) is added, the conjugate base reacts with it: $CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)$ This removes the added H⁺ ions, so the pH remains relatively constant. The ratio of $[CH_3COO^-]/[CH_3COOH]$ changes only slightly, so the pH change is small.

Marking note: Award 1 mark for identifying the conjugate base reacts with H⁺, and 1 mark for the equation or explanation of why pH change is small.

Question 3 (Total: 10 marks)

(a) Expression for $K_b$ [1] $K_b = \frac{[BH^+][OH^-]}{[B]}$

(b) pH of 0.100 mol dm⁻³ solution of B [4]

1. Set up the ICE table (in mol dm⁻³):

   	B	+	H₂O	⇌	BH⁺	+	OH⁻
   Initial	0.100				0		0
   Change	-x				+x		+x
   Equilibrium	0.100 - x				x		x

2. Write the $K_b$ expression: $K_b = \frac{x^2}{0.100 - x} = 4.0 \times 10^{-6}$

3. Assume $x \ll 0.100$: $x^2 = 4.0 \times 10^{-6} \times 0.100 = 4.0 \times 10^{-7}$ $x = \sqrt{4.0 \times 10^{-7}} = 6.32 \times 10^{-4} \text{ mol dm}^{-3}$

4. Calculate $[OH^-] = x = 6.32 \times 10^{-4}$ mol dm⁻³

5. Calculate $pOH$: $pOH = -\log(6.32 \times 10^{-4}) = 3.20$

6. Calculate pH: $pH = 14 - 3.20 = 10.80$

Answer: pH = 10.80

Marking note: Award 1 mark for the ICE table or correct setup, 1 mark for the approximation and solving for x, 1 mark for calculating pOH, and 1 mark for the final pH.

(c)(i) pH of solution after mixing B with HCl [2] Moles of B = moles of HCl = $\frac{25.0}{1000} \times 0.100 = 2.50 \times 10^{-3}$ mol.

The reaction is: $B(aq) + H^+(aq) \rightarrow BH^+(aq)$

Since equal moles of B and HCl are mixed, all of B is converted to $BH^+$. The solution contains only the conjugate acid $BH^+$ in a total volume of 50.0 cm³.

$[BH^+] = \frac{2.50 \times 10^{-3}}{0.0500} = 0.0500$ mol dm⁻³

This is a weak acid solution. The pH will be less than 7 (acidic) because $BH^+$ undergoes hydrolysis: $BH^+(aq) + H_2O(l) \rightleftharpoons B(aq) + H_3O^+(aq)$

Answer: pH < 7 (acidic), because the solution contains only the conjugate acid $BH^+$ which hydrolyses to produce H⁺.

Marking note: Award 1 mark for identifying that only $BH^+$ is present, and 1 mark for stating the pH is acidic with a reason.

(c)(ii) pH of solution with half the HCl [3]

1. Moles of B = $2.50 \times 10^{-3}$ mol; moles of HCl = $\frac{12.5}{1000} \times 0.100 = 1.25 \times 10^{-3}$ mol

2. After reaction:

   • Moles of $BH^+$ formed = $1.25 \times 10^{-3}$ mol
   • Moles of B remaining = $2.50 \times 10^{-3} - 1.25 \times 10^{-3} = 1.25 \times 10^{-3}$ mol

3. This forms a buffer solution. Total volume = 37.5 cm³ = 0.0375 dm³

   • $[B] = \frac{1.25 \times 10^{-3}}{0.0375} = 0.0333$ mol dm⁻³
   • $[BH^+] = \frac{1.25 \times 10^{-3}}{0.0375} = 0.0333$ mol dm⁻³

4. Use the Henderson–Hasselbalch equation: $pH = pK_a + \log\frac{[B]}{[BH^+]}$ $pK_a = -\log(2.5 \times 10^{-9}) = 8.60$ $pH = 8.60 + \log\frac{0.0333}{0.0333} = 8.60 + 0 = 8.60$

Answer: pH = 8.60

Marking note: Award 1 mark for calculating moles and identifying buffer, 1 mark for the Henderson–Hasselbalch equation, and 1 mark for the final answer.

Question 4 (Total: 10 marks)

(a)(i) Equilibrium expression for indicator [1] $K_{In} = \frac{[H^+][In^-]}{[HIn]}$

(a)(ii) Colour of methyl orange at pH 2.0 [2] At pH 2.0, the solution is strongly acidic. The equilibrium $HIn \rightleftharpoons H^+ + In^-$ is shifted to the left (by Le Chatelier's principle), so the concentration of the acid form, HIn, is much greater than the concentration of the conjugate base, In⁻. Since HIn is red, the solution will appear red.

Marking note: Award 1 mark for stating red, and 1 mark for the explanation based on the equilibrium position.

(b)(i) pH at which [HIn] = [In⁻] [1] When $[HIn] = [In^-]$, the $K_{In}$ expression becomes: $K_{In} = [H^+]$ $pH = pK_{In} = -\log(1.0 \times 10^{-9}) = 9.0$

Answer: pH = 9.0

(b)(ii) Colour of phenolphthalein at pH 8.0 [2] At pH 8.0, which is less than $pK_{In}$ (9.0), the equilibrium $HIn \rightleftharpoons H^+ + In^-$ is shifted to the left. The concentration of HIn is greater than In⁻. Since HIn is colourless, the solution will appear colourless.

Marking note: Award 1 mark for stating colourless, and 1 mark for the explanation based on the equilibrium position.

(c) Suitable indicator for ethanoic acid–NaOH titration [2] Phenolphthalein is the more suitable indicator.

Reason: The titration of a weak acid (ethanoic acid) with a strong base (NaOH) produces a salt (sodium ethanoate) that undergoes hydrolysis to give a basic solution. The equivalence point is at a pH greater than 7 (approximately pH 8.7). Phenolphthalein changes colour in the pH range 8.2–10.0, which falls within the steep portion of the titration curve around the equivalence point. Methyl orange changes colour at pH 3.1–4.4, which is not in the steep region of the curve.

Marking note: Award 1 mark for identifying phenolphthalein, and 1 mark for the reason based on the equivalence point pH.

(d) Sketch of pH curve [2] The sketch should show:

• A curve starting at approximately pH 2.9 (the pH of 0.100 mol dm⁻³ ethanoic acid)
• A gradual rise in the buffer region (where the curve is relatively flat)
• A steep rise around the equivalence point at 25.0 cm³ of NaOH
• The equivalence point at a pH above 7 (approximately pH 8.7)
• The curve levelling off at a pH of approximately 13

Marking note: Award 1 mark for the correct shape of the curve, and 1 mark for correctly labelling the equivalence point and buffer region.

Question 5 (Total: 10 marks)

(a) Expression for $K_{sp}$ of $Ag_2CrO_4$ [1] $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$

(b) Molar solubility in pure water [3]

1. Let the molar solubility be $s$ mol dm⁻³. $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$

   • $[Ag^+] = 2s$
   • $[CrO_4^{2-}] = s$

2. Substitute into the $K_{sp}$ expression: $K_{sp} = (2s)^2(s) = 4s^3 = 1.2 \times 10^{-12}$

3. Solve for $s$: $s^3 = \frac{1.2 \times 10^{-12}}{4} = 3.0 \times 10^{-13}$ $s = \sqrt[3]{3.0 \times 10^{-13}} = 6.69 \times 10^{-5} \text{ mol dm}^{-3}$

Answer: Molar solubility = $6.69 \times 10^{-5}$ mol dm⁻³

Marking note: Award 1 mark for the relationship between $[Ag^+]$ and $[CrO_4^{2-}]$, 1 mark for the correct substitution, and 1 mark for the final answer.

(c) Molar solubility in 0.100 mol dm⁻³ $AgNO_3$ [3]

1. In 0.100 mol dm⁻³ $AgNO_3$, the initial $[Ag^+] = 0.100$ mol dm⁻³.

2. Let the molar solubility of $Ag_2CrO_4$ be $s$ mol dm⁻³.

   • $[Ag^+] = 0.100 + 2s$
   • $[CrO_4^{2-}] = s$

3. Substitute into the $K_{sp}$ expression: $K_{sp} = (0.100 + 2s)^2(s) = 1.2 \times 10^{-12}$

4. Assume $2s \ll 0.100$: $(0.100)^2(s) = 1.2 \times 10^{-12}$ $s = \frac{1.2 \times 10^{-12}}{0.0100} = 1.2 \times 10^{-10} \text{ mol dm}^{-3}$

Answer: Molar solubility = $1.2 \times 10^{-10}$ mol dm⁻³

Marking note: Award 1 mark for the correct expression for $[Ag^+]$, 1 mark for the approximation, and 1 mark for the final answer.

(d) Effect of lowering pH on solubility of $Ag_2CrO_4$ [3] The solubility of $Ag_2CrO_4$ increases when the pH is lowered.

Reason: $CrO_4^{2-}$ is the conjugate base of the weak acid $HCrO_4^-$. In acidic conditions, $CrO_4^{2-}$ reacts with H⁺: $CrO_4^{2-}(aq) + H^+(aq) \rightleftharpoons HCrO_4^-(aq)$ This removes $CrO_4^{2-}$ from the solution, shifting the solubility equilibrium $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$ to the right (by Le Chatelier's principle), thereby increasing the solubility of the

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TuitionGoWhere Practice Paper - Chemistry H3 A-Level - Answers

Section A: Compulsory Questions (60 Marks)

Question 1 (Total: 8 marks)

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

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

(ii) $[H^+] = 10^{-pH} = 10^{-3.40} = 3.98 \times 10^{-4}$ mol dm⁻³

Since HA is monoprotic, $[H^+] = [A^-] = 3.98 \times 10^{-4}$ mol dm⁻³

$[HA]_{eq} = 0.0250 - 3.98 \times 10^{-4} \approx 0.0246$ mol dm⁻³

$K_a = \frac{(3.98 \times 10^{-4})^2}{0.0246} = 6.44 \times 10^{-6}$ mol dm⁻³ [3]

(c) (i) HB is a strong acid. The pH of 1.60 corresponds to $[H^+] = 0.0251$ mol dm⁻³, which is approximately equal to the initial concentration of 0.0250 mol dm⁻³, indicating complete dissociation. [1]

(ii) $[OH^-] = \frac{K_w}{[H^+]} = \frac{1.00 \times 10^{-14}}{10^{-1.60}} = \frac{1.00 \times 10^{-14}}{0.0251} = 3.98 \times 10^{-13}$ mol dm⁻³ [2]

Question 2 (Total: 10 marks)

(a) $pH = pK_a + \log\frac{[salt]}{[acid]}$

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

$pH = 4.76 + \log\frac{0.200}{0.150} = 4.76 + 0.125 = 4.89$ [3]

(b) (i) Moles of HCl added = $\frac{5.0}{1000} \times 0.100 = 5.0 \times 10^{-4}$ mol [1]

(ii) Initial moles of ethanoic acid = $\frac{25.0}{1000} \times 0.150 = 3.75 \times 10^{-3}$ mol

Initial moles of ethanoate = $\frac{25.0}{1000} \times 0.200 = 5.00 \times 10^{-3}$ mol

After addition of HCl:

Moles of ethanoic acid = $3.75 \times 10^{-3} + 5.0 \times 10^{-4} = 4.25 \times 10^{-3}$ mol

Moles of ethanoate = $5.00 \times 10^{-3} - 5.0 \times 10^{-4} = 4.50 \times 10^{-3}$ mol

Total volume = $25.0 + 5.0 = 30.0$ cm³

$[ethanoic acid] = \frac{4.25 \times 10^{-3}}{30.0/1000} = 0.142$ mol dm⁻³

$[ethanoate] = \frac{4.50 \times 10^{-3}}{30.0/1000} = 0.150$ mol dm⁻³

$pH = 4.76 + \log\frac{0.150}{0.142} = 4.76 + 0.024 = 4.78$ [4]

(c) The buffer contains a weak acid (ethanoic acid) and its conjugate base (ethanoate). When a small amount of acid is added, the ethanoate ions react with the added H⁺ ions to form ethanoic acid, thus removing most of the added H⁺ and minimizing the change in pH. [2]

Question 3 (Total: 10 marks)

(a) $K_b = \frac{[BH^+][OH^-]}{[B]}$ [1]

(b) $[OH^-] = \sqrt{K_b \times [B]} = \sqrt{4.0 \times 10^{-6} \times 0.100} = \sqrt{4.0 \times 10^{-7}} = 6.32 \times 10^{-4}$ mol dm⁻³

$pOH = -\log(6.32 \times 10^{-4}) = 3.20$

$pH = 14.00 - 3.20 = 10.80$ [4]

(c) (i) The pH is 7.0. The reaction between B and HCl produces BH⁺ and Cl⁻. The BH⁺ is the conjugate acid of a weak base and will undergo hydrolysis to produce an acidic solution. However, the Cl⁻ is the conjugate base of a strong acid and does not affect pH. The resulting solution is acidic, so pH < 7. [2]

(ii) Moles of B = $\frac{25.0}{1000} \times 0.100 = 2.5 \times 10^{-3}$ mol

Moles of HCl = $\frac{12.5}{1000} \times 0.100 = 1.25 \times 10^{-3}$ mol

After reaction, moles of B remaining = $2.5 \times 10^{-3} - 1.25 \times 10^{-3} = 1.25 \times 10^{-3}$ mol

Moles of BH⁺ formed = $1.25 \times 10^{-3}$ mol

Total volume = $25.0 + 12.5 = 37.5$ cm³

$[B] = \frac{1.25 \times 10^{-3}}{37.5/1000} = 0.0333$ mol dm⁻³

$[BH^+] = \frac{1.25 \times 10^{-3}}{37.5/1000} = 0.0333$ mol dm⁻³

$pK_a = -\log(2.5 \times 10^{-9}) = 8.60$

$pH = pK_a + \log\frac{[B]}{[BH^+]} = 8.60 + \log\frac{0.0333}{0.0333} = 8.60$ [3]

Question 4 (Total: 10 marks)

(a) (i) $HIn(aq) \rightleftharpoons H^+(aq) + In^-(aq)$ [1]

(ii) At pH 2.0, the solution is highly acidic. The equilibrium will shift to the left, favouring the acid form HIn, which is red. Therefore, the colour will be red. [2]

(b) (i) When $[HIn] = [In^-]$, $pH = pK_{In} = -\log(1.0 \times 10^{-9}) = 9.0$ [1]

(ii) At pH 8.0, the solution is less basic than the pKIn. The equilibrium will shift to the left, favouring the acid form HIn, which is colourless. Therefore, the solution will be colourless. [2]

(c) Phenolphthalein is more suitable. The titration of a weak acid (ethanoic acid) with a strong base (NaOH) has an equivalence point at a pH above 7 (around pH 8.7). Phenolphthalein changes colour in the pH range 8.2–10.0, which encompasses the equivalence point. Methyl orange changes colour in the pH range 3.1–4.4, which is too acidic for this titration. [2]

(d) [A sketch of a pH titration curve for a weak acid-strong base titration should be drawn here, showing a gradual rise in pH, a steep rise around the equivalence point at 25.0 cm³, and a final pH around 13. The equivalence point should be labelled at a pH above 7, and the buffer region should be labelled as the relatively flat portion before the steep rise.] [2]

Question 5 (Total: 10 marks)

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

(b) Let the molar solubility of $Ag_2CrO_4$ be $s$ mol dm⁻³.

$[Ag^+] = 2s$, $[CrO_4^{2-}] = s$

$K_{sp} = (2s)^2 \times s = 4s^3 = 1.2 \times 10^{-12}$

$s^3 = 3.0 \times 10^{-13}$

$s = \sqrt[3]{3.0 \times 10^{-13}} = 6.69 \times 10^{-5}$ mol dm⁻³ [3]

(c) In 0.100 mol dm⁻³ $AgNO_3$, $[Ag^+] = 0.100$ mol dm⁻³ (from $AgNO_3$) + $2s$ (from $Ag_2CrO_4$) ≈ 0.100 mol dm⁻³ (since $s$ is small).

Let the molar solubility in $AgNO_3$ be $s'$ mol dm⁻³.

$K_{sp} = (0.100)^2 \times s' = 1.2 \times 10^{-12}$

$s' = \frac{1.2 \times 10^{-12}}{0.0100} = 1.2 \times 10^{-10}$ mol dm⁻³ [3]

(d) The solubility of $Ag_2CrO_4$ increases when the pH is lowered. $CrO_4^{2-}$ is a basic anion and reacts with H⁺ ions to form $HCrO_4^-$ and $Cr_2O_7^{2-}$. This removes $CrO_4^{2-}$ from the solution, shifting the equilibrium $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$ to the right, thus increasing the solubility. [3]

Question 6 (Total: 12 marks)

(a) [A graph should be plotted here with pH on the y-axis (0–14) and volume of HCl on the x-axis (0–50 cm³). The data points should be plotted and a smooth curve drawn through them. The curve should show two equivalence points: one at approximately 12.5 cm³ and one at 25.0 cm³.] [3]

(b) The first equivalence point is at approximately 12.5 cm³ of HCl. [1]

(c) $CO_3^{2-}(aq) + H^+(aq) \rightarrow HCO_3^-(aq)$ [1]

(d) $HCO_3^-(aq) + H^+(aq) \rightarrow H_2CO_3(aq) \rightarrow H_2O(l) + CO_2(g)$ [1]

(e) At the second equivalence point, the solution contains $H_2CO_3$. The concentration of $H_2CO_3$ is:

Initial moles of $Na_2CO_3 = \frac{25.0}{1000} \times 0.100 = 2.5 \times 10^{-3}$ mol

Total volume at second equivalence point = $25.0 + 25.0 = 50.0$ cm³

$[H_2CO_3] = \frac{2.5 \times 10^{-3}}{50.0/1000} = 0.0500$ mol dm⁻³

$H_2CO_3$ is a weak acid. $K_a = 4.3 \times 10^{-7}$ mol dm⁻³.

$[H^+] = \sqrt{K_a \times [H_2CO_3]} = \sqrt{4.3 \times 10^{-7} \times 0.0500} = \sqrt{2.15 \times 10^{-8}} = 1.47 \times 10^{-4}$ mol dm⁻³

$pH = -\log(1.47 \times 10^{-4}) = 3.83$ [4]

(f) The pH at the first equivalence point is not 7 because the $HCO_3^-$ ion is amphoteric. It can act as both an acid and a base. The pH is determined by the relative strengths of its acidic and basic properties. In this case, $HCO_3^-$ is a stronger base than an acid, resulting in a pH above 7. [2]

Question 7 (Total: 10 marks)

(a) Moles of HA = $\frac{50.0}{1000} \times 0.100 = 5.0 \times 10^{-3}$ mol

Moles of NaOH = $\frac{25.0}{1000} \times 0.100 = 2.5 \times 10^{-3}$ mol

After reaction, moles of HA remaining = $5.0 \times 10^{-3} - 2.5 \times 10^{-3} = 2.5 \times 10^{-3}$ mol

Moles of A⁻ formed = $2.5 \times 10^{-3}$ mol

Total volume = $50.0 + 25.0 = 75.0$ cm³

$[HA] = \frac{2.5 \times 10^{-3}}{75.0/1000} = 0.0333$ mol dm⁻³

$[A^-] = \frac{2.5 \times 10^{-3}}{75.0/1000} = 0.0333$ mol dm⁻³

$pH = pK_a + \log\frac{[A^-]}{[HA]} = -\log(1.0 \times 10^{-5}) + \log\frac{0.0333}{0.0333} = 5.00$ [4]

(b) HA contributes more to the $[H^+]$ of the solution. HA has a larger $K_a$ value ($1.0 \times 10^{-5}$) compared to HB ($1.0 \times 10^{-9}$). A larger $K_a$ indicates a stronger acid that dissociates to a greater extent, producing more H⁺ ions. [2]

(c) $pOH = 14.0 - 11.0 = 3.0$

$[OH^-] = 10^{-3.0} = 1.0 \times 10^{-3}$ mol dm⁻³

$[OH^-] = \sqrt{K_b \times [X]}$

$1.0 \times 10^{-3} = \sqrt{K_b \times 0.100}$

$(1.0 \times 10^{-3})^2 = K_b \times 0.100$

$K_b = \frac{1.0 \times 10^{-6}}{0.100} = 1.0 \times 10^{-5}$ mol dm⁻³ [4]

Question 8 (Total: 10 marks)

(a) $NH_3(aq) + HCl(aq) \rightarrow NH_4Cl(aq)$ [1]

(b) $NH_4^+(aq) \rightleftharpoons NH_3(aq) + H^+(aq)$

$[H^+] = 10^{-5.13} = 7.41 \times 10^{-6}$ mol dm⁻³

$[NH_3] = [H^+] = 7.41 \times 10^{-6}$ mol dm⁻³

$[NH_4^+] = 0.100 - 7.41 \times 10^{-6} \approx 0.100$ mol dm⁻³

$K_a = \frac{[NH_3][H^+]}{[NH_4^+]} = \frac{(7.41 \times 10^{-6})^2}{0.100} = 5.49 \times 10^{-10}$ mol dm⁻³ [4]

(c) $pK_b = -\log(1.8 \times 10^{-5}) = 4.74$

$pK_a = 14.00 - 4.74 = 9.26$

$pH = pK_a + \log\frac{[NH_3]}{[NH_4^+]} = 9.26 + \log\frac{0.100}{0.200} = 9.26 - 0.301 = 8.96$ [3]

(d) The buffer is more effective at resisting pH changes upon the addition of a small amount of strong base. The buffer has a higher concentration of the acidic component ($NH_4^+$, 0.200 mol dm⁻³) compared to the basic component ($NH_3$, 0.100 mol dm⁻³). Therefore, it has a greater capacity to neutralize added base. [2]

Section B: Extended Response Questions (40 Marks)

[Answers for Section B would be provided here for the two chosen questions.]