AQA A2 Chemistry C21 Acid,Base and Buffer Kerboodle Answers

This page contains the AQA A2 Chemistry C21 Acid Base and Buffer Questions and kerboodle answers for revision and understanding.This page also contains the link to the notes and video for the revision of this topic.
 
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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) Ka = [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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