R3.1 Exam Practice

(a) A and B are acidic (pH < 7); C is basic (pH > 7) [1]

(b)

• HCl is a strong acid; fully ionises, so [H⁺] = 0.0100 mol dm⁻³ [1]
• CH₃COOH is a weak acid; partially ionises, producing a lower [H⁺] and therefore a higher pH [1]

(c) [H⁺] = Kw/[OH⁻] = 1.00 × 10⁻¹⁴/0.0100 = 1.00 × 10⁻¹² mol dm⁻³ [1]

(a)

• Pair 1: CH₃COOH (acid) and CH₃COO⁻ (conjugate base) [1]
• Pair 2: H₂O (base) and H₃O⁺ (conjugate acid) [1]

(b)

• Amphiprotic = can both donate and accept a proton (H⁺) [1]
• Water can donate H⁺ to form OH⁻ or accept H⁺ to form H₃O⁺ [1]

(a) Brønsted-Lowry acid = proton (H⁺) donor [1]; Brønsted-Lowry base = proton (H⁺) acceptor [1]

(b) Strong acid = complete (~100%) ionisation in water [1]; Weak acid = partial ionisation (equilibrium lies to the left) [1]

(c) Concentration refers to moles of solute per dm³; strength refers to the proportion of molecules that ionise [1]

(a) pH = -log₁₀[H⁺] [1]

(b) HNO₃ is a strong acid; pH = -log(0.0500) = 1.30 [1]

(c)

• [H⁺] = 10⁻³ = 1.00 × 10⁻³ mol dm⁻³ [1]
• [OH⁻] = Kw/[H⁺] = 1.00 × 10⁻¹⁴/1.00 × 10⁻³ = 1.00 × 10⁻¹¹ mol dm⁻³ [1]

(a) SO₂ and NO_x (nitrogen oxides) [1] (Both required)

(b)

• SO₂ dissolves in rainwater / is oxidised to SO₃ [1]
• SO_2(g) + H_2O(l) → H_2SO_3(aq) OR SO_3(g) + H_2O(l) → H_2SO_4(aq) [1]

(c) Any two of: acidification of lakes kills aquatic life [1]; corrosion of limestone buildings / statues [1]; leaching minerals from soil / releasing toxic Al³⁺ ions; damage to forests / vegetation

(a)

• \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.76 \times 10^{-5} \times 0.100} = 1.33 \times 10^{-3}\text{ mol dm}^{-3}\) [1]
• \(\text{pH} = -\log_{10}(1.33 \times 10^{-3}) = 2.88\) [1]
• Assumption: the dissociation of ethanoic acid is negligible / \([\text{CH}_3\text{COOH}]_{\text{eqm}} \approx [\text{CH}_3\text{COOH}]_{\text{initial}}\) OR the only source of \(\text{H}^+\) is the acid dissociation / water dissociation is negligible [1]

(b)

• After mixing, volume doubles, so concentrations are halved: \([\text{CH}_3\text{COOH}] = 0.0500\text{ mol dm}^{-3}\) and \([\text{CH}_3\text{COO}^-] = 0.100\text{ mol dm}^{-3}\) (or use ratio of moles directly since volumes cancel) [1]
• \([\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{salt}]} = 1.76 \times 10^{-5} \times \frac{0.0500}{0.100} = 8.80 \times 10^{-6}\text{ mol dm}^{-3}\) OR \(\text{pH} = \text{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]} = 4.75 + \log(2) = 5.06\) [1]