AQA A2 Chemistry C21 Acid,Base and Buffer Kerboodle Answers

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C21.1 Defining an acid: Page No AQA A2 Chemistry C21 Acid Base And Buffer Kerboodle Answers.

63 1 a HNO₃ + OH^– → NO₃^– + H₂O Acid: HN0₃ Base: OH^– b CH₃COOH + H₂O → CH₃COO^– + H₃O⁺ Acid: CH₃COOH Base: H₂O 2 1 x 10-¹⁰ mol dm^-3 3 a OH^– H₂O b NH₃ NH₄⁺ c H₂O H₃O⁺ d Cl^– HCl

21.2 The pH scale AQA A2 Chemistry C21 Acid Base And Buffer Kerboodle Answers : Page No. 67

1 2.00 2 1 x 10^-6 moldm-³ 3 1 x 10-⁵ moldm-³ 4 1.70 5 13.30

21.3 Weak acids and bases AQA A2 Chemistry C21 Acid Base And Buffer Kerboodle Answers : Page No. 70

1 Chloroethanoic acid 2 They are the same. 3 a 1.94 b 0.01DD mol dm^-3 benzoic acid 3.10

21.4 Acid-base titrations AQA A2 Chemistry C21 Acid Base And Buffer Kerboodle Answers : Page No. 73

1 a 2NaOH(aq) + H₂S0₄(aq) -+ Na₂S0₄(aq) + 2H₂O(l) b 0.50 c 1.5×10^-3 d 0.120 moldm^-3 2 a Which curve represents i  a strong acid A ii a weak acid? B b Which one could represent: i ethanoicacid.CH-jCOOH B ii hydrochloricacid,HCI? A c Strong, as there is a rapid pH change in the alkaline region at the equivalence point. Banner 2

21.5 Choice of indicators for titrations AQA A2 Chemistry C21 Acid Base And Buffer Kerboodle Answers : Page No. 76

1 b and d 2b and d 21.6 Buffer solutions: Page No. 81 1 a 5.07 b 4.20 Practice questions: Page No. 82-83 1 (a) K_a = [H+]² / [CH₃CH₂COOH] [H+] = √(1.35 × 10^-5 × 0.169) = 1.51 ×10^-3 pH = 2.82 (b) (i) CH₃CH₂COOH + NaOH → CH₃CH₂COONa + H₂O OR CH₃CH₂COOH + OH^– → CH₃CH₂COO^– + H₂O (ii) mol propanoic acid = 0.250 – 0.015 = 0.235 mol propanoate ions = 0.190 + 0.015 = 0.205 (iii) [H+] = Ka × [CH₃CH₂COOH] / [CH₃CH₂COO^–] = (1.35 × 10-5) × 0.235 / 0.205 = 1.548 × 10-5 pH = 4.81 2. (a) The ionic product of water has the symbol K (i) [H⁺][OH^–] OR [H3O⁺][OH^–] (ii) √(3.46 × 10−14) = 1.86 × 10–7 pH = 6.73 (iii) [H+] = 10−11.36 (= 4.365 × 10^-12 OR 4.37 × 10^-12) Kw = [4.365 ×10^-12 OR 4.37 ×10^-12 × 0.047] = 2.05 × 10^-13 (b) (i) HCOOH ⇌ HCOO– + H⁺ OR HCOOH + H2O ⇌ HCOO^– + H3O⁺ (ii) Ka = [H⁺][HCOO^–] / [HCOOH] OR [H3O⁺][HCOO^–] / [HCOOH] (iii) Ka = [H+]2 / [HCOOH] ([H+]2 = 1.78 × 10–4 × 0.056 = 9.97 × 10–6) [H+] = 3.16 × 10–3 pH = 2.50 (iv) Decrease

• Equilibrium shifts/moves to RHS OR more H+ OR Ka increases OR more dissociation
• To reduce temperature or oppose increase/change in temperature

(c) (i) [H+] = Ka × [HX] / [X−] OR pH = pKa – log ([HX] / [X−]) 1.78 × 10−4 × 2.35 × 10−2 / 1.84 × 10−2 OR pH = 3.75 – log (2.35 × 10−2 / (1.84 × 10−2)) = 2.27 × 10−4 pH = 3.64 (ii) Mol H+ added = 5.00 × 10–4 Mol HCOOH = 2.40 × 10–2 AND Mol HCOO– = 1.79 × 10–2 [H+] = Ka × [HX] / [X−] = 1.78 × 10−4 × 2.40 × 10−2 / (1.79 × 10−2) = 2.39 × 10−4 OR pH = 3.75 – log (2.40 × 10−2 / (1.79 × 10−2)) pH = 3.62 3 © (a) (i) Write an expression for the ionic product of water, K Kw = [H+][OH–] (ii) 2.34 ×10^-7 (ili) 2.34 ×10^-7 (iv) 5.48 to 5.50 ×10^-14 (b) [H+] = 10^-14 / 0.136 = 7.35 × 10^-14 OR pOH = 0.87 pH = 13.13 4 All solutions have a concentration of 0.1 mol dm-1. (a) (i) C (ii) A (iii) D (b) (i) Bromocresol green. (ii) Purple to yellow. Banner 3

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