RI 2025 Tutorial 15 Acid-Base Equilibrium Ans

-1- Answers to Practice Questions 1 (a) pH = −lg [H+] [H+] = 10–3.50 = 3.16 x 10–4 mol dm–3 (b) HA(aq) + NaOH(aq)  NaA(aq) + H2O(l) 3 NaOH HA 27.50 reacted present in apple juice 0.100 2.75 10 mol1000    nn 3 32.75 10[HA] in apple juice 1000 0.110 mol dm25.0    Since [H+] << [HA], the acid HA in apple juice undergoes only partial dissociation and is therefore a weak acid. (c) HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) 3 HA 3 HA 4 dissociated in 1 dm of apple juicedegree of dissociation of HA initial in 1 dm of apple juice 3.16 10 0.110 n n    32.87 10  At equilibrium, [H 3O+] = [A–] = 3.16 x 10–4 mol dm–3 [HA] = 0.110 – 3.16 x 10–4 = 0.1097 mol dm–3   24 3 3.16 10HO A HA 0.1097       aK 739.10 10 mol dm (d) Since both acids have the same pH (3.50), both acids have the same [H +]. However, partial dissociation occurs in apple juice, hence the total acid concentration is higher (than hydrochloric acid). [HCl] = [H +] = 3.16 x 10-4 mol dm-3 From (b): [HA] in apple juice = 0.110 mol dm-3 In 1 dm 3, the sample of apple juice will have a greater amount of acid present. Since the volume of H 2 to be produced is proportional to the amount of acid present and a greater amount of acid is present in the sample of apple juice, the apple juice sample would yield a larger volume of hydrogen gas than the hydrochloric acid sample. 2 (a) (i) CH3COOH(aq) ⇌ CH3COO–(aq) + H+(aq) At equilibrium, [H +] = [CH3COO–] [ C H 3COOH]eqm  [CH3COOH]initial = 0.50 mol dm–3, since CH 3COOH is a weak acid with a small Ka     2 3 33 4.74 3 3 3 3 CH COO H H CH COOH CH COOH H CH COOH 10 0.50 3.016 10 mol dm lg 3.016 10 a a K K pH                      2.52

-2- (ii) Let volume of CH3COOH used be V dm3. 3 Initial 0.50 molCH COOHnV  Initial 0.20 molNaOHnV  Upon mixing, the following reaction takes place: CH3COOH + NaOH  CH3COONa+ + H2O Hence CH3COOH is in excess. 3 formed reacted 0.20 molNaOHCH COOnn V   3 Unreacted 0.50 0.20 0.30 molCH COOHnV V V   The resultant solution is a buffer.  0.20 3 2 0.30 3 2 CH COO of resultant solution lg 4.74 lg CH COOH V V a V V pH pK         4.56 (b) (i) CH3COO(aq) + H2O(l) ⇌ CH3COOH(aq) + OH(aq) CH3COO undergoes hydrolysis in water to form OH , which causes the solution to be alkaline and pH is above 7. (Note: [OH] > [H3O+]) (ii)  3 3 CH COOH OH CH COO bK     Kb = Kw Ka = 1.0×10-14 10-4.74 = 5.50×10-10 mol dm-3 (iii) CH3COO(aq) + H2O(l) ⇌ CH3COOH(aq) + OH(aq) At equilibrium, [OH –] = [CH3COOH] [ C H 3COO–]eqm  [CH3COO–]initial = 0.050 mol dm–3, since CH 3COO– is a weak base with a small Kb   2 3 33 10 6 3 3 6 CH COOH OH OH CH COO CH COO OH CH COO 5.50 10 0.050 5.242 10 mol dm pOH lg 5.242 10 5.28 pH 14 5.28       