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                             b b K K 8.72

-3- (c) Let the volume of CH3COONa+(aq) needed be V dm3. Amount of CH3COO = V x 0.10 = 0.10 V mol. Amount of CH3COOH = 0.025 x 0.12 = 0.0030 mol  3 3 0.10 0.0030 0.26 CH COO pH of resultant solution lg CH COOH 5.00 4.74 lg 0.10lg 0.260.0030 0.003010 0.10             total total a V V V pK V V 330.0546 dm 54.6 cm 3 (a) The resultant solution is a buffer. 4 750 present 0.20 0.15 mol1000NHn    3 500 present 0.10 0.05 mol1000 NHn   At equilibrium in the resultant solution, [NH 4+]eqm  [NH4+]initial and [NH3]eqm  [NH3]initial 14 10 1.0 10lg( ) lg( ) 4.7786.00 10 w b a KpK K        0.15 4 0.05 3 NH 0.15pOH of resultant solution lg 4.778 lg 4.778 lg 5.255NH 0.05           total total V b V pK pH = 14 – 5.255 = 8.745 = 8.75 (b) (i) Upon addition of OH–, the following reaction occurs: NH4+(aq) + OH–(aq) → NH3(aq) + H2O(l) 4 present in resultant solution 0.15 0.002 0.148 molNHn    3 present in resultant solution 0.05 0.002 0.052 molNHn   The resultant solution is still a buffer.           0.148 4 0.052 3 NH 0.148pOH of resultant solution lg 4.778 lg 4.778 lg 5.232NH 0.052 total total V b V pK pH = 14 – 5.232 = 8.768 Change in pH = 8.768 – 8.745 = +0.023  There is an increase in pH of 0.023 units.

-4- (ii) 1.0 added 2.00 0.002 mol1000Hn    Upon addition of H+, the following reaction occurs: NH3(aq) + H+(aq) → NH4+(aq) 4 present in resultant solution 0.15 0.002 0.152 molNHn    3 present in resultant solution 0.05 0.002 0.048 molNHn   The resultant solution is still a buffer.  0.152 4 0.048 3 NH 0.152pOH of resultant solution lg 4.778 lg 4.778 lg 5.279NH 0.048          total total V b V pK pH = 14 – 5.279 = 8.721 Change in pH = 8.721 – 8.745 = –0.024  There is a decrease in pH of 0.024 units. 4 (a) A buffer solution is a solution which is able to resist changes in pH when a small amount of an acid or a base is added. The buffer in blood plasma consists of a mixture of H2CO3(aq) and HCO3–(aq). When [H +] increases, by Le Chatelier’s Prin ciple, the equilibrium position of H2CO3(aq) ⇌ HCO3–(aq) + H+(aq) shift to the left to partially offset the increase in [H +] by favouring the backward reaction of the equi librium, resulting in a small decrease in pH. Equation: HCO3–(aq) + H+(aq)  H2CO3(aq) When [OH−] increases, by Le Chatelier’s Principle, the equilibrium position of HCO3–(aq) + H2O(l) ⇌ H2CO3(aq) + OH−(aq) shift to the left to partially offset the increase in [OH −] by favouring the backward reaction of the equilibrium, resulting in a small increase in pH. Equation: H2CO3(aq) + OH−(aq)  HCO3–(aq) + H2O(l) (b) Method 1: Method 2:    3 1 23 7.40 37 23 7 3 7.40 23 HCO H HC O HCO 10 7.90 10 HC O HCO 7.90 10 HC O 10 aK                    19.8      7 3 23 37 23 7.40 lg 7.90 103 23 HCO lg HC O HCO 7.40 lg 7.90 10 lg HC O HCO 10HC O apH pK             19.8 Since [HCO3] is higher than [H2CO3], the buffer solution is more effective in resisting pH changes when there is a sma ll increase in the amount of acid (H +) than when there is a small increase in the amount of base (OH–).

-5- [Note: This is probably due to the fact that many reactions in the body produce H+ and hence the buffer in the blood is required to be more able to resist increase in the amount of H + rather than OH–.] (c) H2CO3(aq) ⇌ H+(aq) + HCO3(aq) -------- (1) H2CO3(aq) ⇌ CO2(aq) + H2O(l) -------- (2) CO2(aq) ⇌ CO2(g) -------- (3) When lactic acid is released into blood, [H +] in blood increases. In accordance to Le Chatelier’s principle, the equilibrium position of reaction (1) shifts to the left to decrease [H+]. This leads to an increase in the [H2CO3]. This increase in the [H2CO3] causes the equilibrium position of reaction (2) to shift to the right, resulting in an increase in [CO2(aq)]. This in turn causes the equilibrium position of reaction (3) to shift right, causing an increase in [CO2(g)]. When [CO2(g)] increases, this causes the rate of exhaling CO2(g) to increase. Thus, the rate of breathing increases as well. 5 (Ans: D) Option A Incorrect. The correct pH of the buffer solution is: pH = pKa + lg [lactate] [lactic acid] = –lg(1.4  10−4) + lg 0.5 1.5 = 3.38 Option B Incorrect. Upon dilution with some water, the pH of the buffer solution should remain the same as the amount of lactic acid and sodium lactate in the buffer solution remains unchanged (although their concentrations decrease). pH = pK a + lg [lactate] [lactic acid] = pKa + lg n(lactate) / Vbuffer n(lactic acid) / Vbuffer = pKa + lg n(lactate) n(lactic a