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) C6H5COO(aq) + H2O(l) ⇌ C6H5COOH(aq) + OH–(aq) 𝐾௕ ൌ ሾOHିሿሾC6H5COOHሿ ሾC6H5COOെሿ Since C6H5COO is a weak base with a small Kb, [C6H5COO]eqm  [C6H5COO]initial = 0.015 mol dm–3 [OH−] = ඥ𝐾௕ሾ𝐶଺𝐻ହ𝐶𝑂𝑂ିሿ 14 65 5 63 1.0 10OH C H COO 0.015 6.5 10 1.519 10 mol dm              w a K K pOH = –lg (1.519 x 10 –6) = 5.818 pH = 14 – 5.818 = 8.18 (g) 0.25 mol dm−3 methylammonium nitrate CH3NH3NO3 (acidic salt containing CH3NH3+ which is the conjugate acid of the weak base CH3NH2) CH3NH3+(aq) + H2O(l) ⇌ CH3NH2(aq) + H3O+(aq) 𝐾௔ ൌ ሾHଷOାሿሾCHଷNHଶሿ ሾCHଷNHଷ ାሿ Since CH 3NH3+ is a weak acid with a small Ka, [CH3NH3+]eqm  [CH3NH3+]initial = 0.25 mol dm–3                 33 3 33 14 4 63 HO C HN H CH NH 1.0 10 0.254.4 10 2.384 10 mol dm a w b K K K pH = –lg (2.384 x 10 –6) = 5.62 (h) A solution containing 0.50 mol dm–3 CH3CH2COOH and 0.35 mol dm–3 CH3CH2COO–K+ (buffer) CH 3CH2COOH(aq) ⇌ CH3CH2COO–(aq) + H+(aq)   32 32 5 CH CH COO lg CH CH COOH 0.35lg 1.3 10 lg 4.73 0.50         apH pK (i) A solution containing 0.060 mol dm –3 of NH3 and 0.080 mol dm–3 of NH4Cl (buffer) NH 3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)   4 3 5 NH lg NH 0.080lg 1.74 10 lg 0.060         bpOH pK 4. 884 pH = 14 – 4.884 = 9.12 3 (a) HX(aq) + H 2O(l) ⇌ H3O+(aq) + X–(aq) At equilibrium, [H 3O+] = [X–] = 10–3.25 = 5.623 x 10-4 mol dm–3 [HX] = 1.00 x 10 –3 – 5.623 x 10-4 = 4.377 x 10–4 mol dm–3    24 3 -4 -3 4 5.623 10HO X 7.23 × 10 mol dmHX 4.377 10           aK

-8- (b) CH3COO–Na+(aq)  CH3COO–(aq) + Na+(aq) CH3COO–(aq) + H2O(l) ⇌ CH3COOH(aq) + OH–(aq) At equilibrium, [CH 3COOH] = [OH–] = 10–(14 – 8.37) = 2.344 x 10–6 mol dm–3 [CH3COO–] = 1.00 x 10–2 – 2.344 x 10–6 = 9.998 x 10–3 mol dm–3    26 3 -10 -3 3 3 2.344 10CH COOH OH 5.497 × 10 mol dm9.998 10CH COO       bK pKb of CH3COO– = –lg (5.497 x 10–10) = 9.260 pKa of CH3COOH = 14 – 9.260 = 4.74 (c) (NH4)2SO4(aq)  2NH4+(aq) + SO42–(aq) NH 4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) Initial [NH4+] = (2)(0.050) = 0.100 mol dm–3 At equilibrium, [NH 3] = [H3O+] = 10–5.12 = 7.586 x 10–6 mol dm–3 [NH 4+] = 0.100 – 7.586 x 10–6 = 0.100 mol dm–3    26 33 10 3 4 7.586 10NH H O 5.755 10 mol dm0.100NH         aK pKa of NH4+ = –lg (5.755 x 10–10) = 9.240 pKb of NH3 = 14 – 9.240 = 4.76 4 (Ans: B) In the absence of water, HCl(g) remains as an undissociated simple molecule, and is not dissociated into H+ and Cl ions, thus it does not have acidic properties (options A, C and D are incorrect) and does not act as mobile charge carriers (option B is correct). 5 (Ans: D) Ka only changes with temperature. Since the temperature is constant, Ka remains constant as volume changes. 6 (Ans: D) Overall Equation: 2MnO4+ 5SO2 + 2H2O  2Mn2+ + 5SO42 + 4H+ Since H+ is produced, [H+] increases, causing pH to decrease gradually. 7 (Ans: B) At 25 oC, pH = – lg [H+] = – lg (10–7) = 7 and pKw = 14 At equilibrium, [H+] = [OH–] = 10-7 mol dm3 [H2O] = 1000/18 = 55.56 mol dm3 (See note below) Ka= ሾH൅ሿሾOH-ሿ ሾH2Oሿ = ൫10-7൯ 2 55.56 =1.800×10-16 mol dm-3 pKa of H2O = –lg (1.800 x 10–16) = 15.7 Hence, pH < pKw < pKa Note: How to find [H2O]? Consider 1 dm3 of water, Assume density of water = 1 g cm–3  mass of 1 dm3 of H2O = 1000 g 2 1000 55.56 mol2 1.0 16.0  HOn [H2O] = 55.56 mol dm–3 8 (Ans: A) Pyruvic acid is a weak acid, hence upon titration with NaOH (strong base), at equivalence point, pH > 7 due to anion hydrolysis. CH3COCOO- CH3COCOOH + H2O + OH-