RI 2025 Acid-Base Equilibria Tut 15 - Answers to Self-check questions

-6- Suggested Answers to Self–Check Questions 1 (a) (i) A Brønsted–Lowry acid is a proton donor while a Brønsted–Lowry base is a proton acceptor. In this case, an acid–base reaction involves the transfer of a proton from the acid to the base. (ii) Acid 1 and base 1 constitute one conjugate acid–base pair. Acid 2 and base 2 constitute another conjugate acid–base pair. I. CH3COOH + H2SO4 ⇌ CH3COOH2+ + HSO4– base 1 acid 2 acid 1 base 2 II. CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH–(aq) base 1 acid 2 acid 1 base 2 III. HNO2(aq) + CN–(aq) ⇌ HCN(aq) + NO2–(aq) acid 1 base 2 acid 2 base 1 (b) (i) A Lewis acid is an electron–pair acceptor while a Lewis base is an electron–pair donor. In this case, an acid–base reaction involves the formation of a dative covalent bond between the acid and the base. (ii) I. (CH3)3N(g) + BF3(g) ⇌ (CH3)3NBF3(s) base acid II. A l(OH)3(s) + OH–(aq) ⇌ [Al(OH)4]–(aq) acid base (c) (i) HBr(g) + H2O(l) → H3O+(aq) + Br–(aq) Brønsted–Lowry acid base Lewis acid base (ii) HIO + NH2– ⇌ NH3 + IO– Brønsted–Lowry acid base Lewis acid base (iii) Brønsted–Lowry base acid Lewis base acid 2 (a) 0.010 mol dm−3 H2SO4 (strong dibasic acid) H2SO4(aq)  2H+(aq) + SO42–(aq) [H+] = 2  0.010 = 0.0200 mol dm–3 pH = lg (0.0200) = 1.70 (b) 0.40 g dm −3 NaOH (strong base) NaOH(aq)  Na+(aq) + OH–(aq) . ... .    304 0OH 0 0100 mol dm23 0 16 0 1 0 pOH = lg (0.0100) = 2.00 pH = 14  2.00 = 12.0 (c) 14.00 cm 3 of 0.10 mol dm−3 H2SO4 is added to 20.00 cm3 of 0.10 mol dm–3 NaOH (strong acid partially neutralised by strong base) 14.002 0.10 0.0028 mol1000    Hn 20.00 0.10 0.002 mol1000   OHn in resultant solution 0.0028 0.002 0.0008 mol  Hn  3 0.0008H in resultant solution 100014.00 20.00 0.02353 mol dm       pH of resultant solution = lg (0.02353) = 1.63 (d) 0.30 mol dm −3 CH3CH2COOH (weak acid) CH3CH2COOH(aq) ⇌ H+(aq) + CH3CH2COO–(aq) 𝐾௔ ൌ ሾ𝐻ାሿሾ𝐶𝐻ଷ𝐶𝐻ଶ𝐶𝑂𝑂– ሿ ሾ𝐶𝐻ଷ𝐶𝐻ଶ𝐶𝑂𝑂𝐻ሿ 4.89 5 310 10 1.288 10 mol dm     apK aK Since CH3CH2COOH is a weak acid with a small Ka, [CH3CH2COOH]eqm  [CH3CH2COOH]initial = 0.30 mol dm–3  5 32 33 H CH CH COOH 1.288 10 0.30 1.966 10 mol dm        aK pH = –lg (1.966 x 10 –3) = 2.71

-7- (e) 2.00 mol dm −3 CH3CH2NH2 (weak base) CH3CH2NH2(aq) + H2O(l) ⇌ CH3CH2NH3+(aq) + OH– (aq) 𝐾௕ ൌ ሾ𝑂𝐻ିሿሾ𝐶𝐻ଷ𝐶𝐻ଶ𝑁𝐻ଷ ାሿ ሾ𝐶𝐻ଷ𝐶𝐻ଶ𝑁𝐻ଶሿ Since CH3CH2NH2 is a weak base with a small Kb, [CH3CH2NH2]eqm  [CH3CH2NH2]initial = 2.00 mol dm–3  4 322 3 OH CH CH NH 5.1 10 2.00 0.03194 mol dm        bK pOH = –lg (0.03194) = 1.496 pH = 14 – 1.496 = 12.5 (f) 0.015 mol dm−3 C6H5COOK+ (basic salt containing C6H5COO which is the conjugate base of the weak acid C6H5COOH) C6