RI 15. Acid-Base Equilibria Tutorial (Suggested Answers to Practice Questions)

-1- Raffles Institution Year 6 H2 Chemistry 2026 Tutorial 15 – Acid-Base Equilibria Suggested 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, [H3O+] = [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 10H O 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 dm3, 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–] [CH3COOH]eqm  [CH3COOH]initial = 0.50 mol dm–3, since CH3COOH is a weak acid with a small Ka

-2-       ( ) 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 (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 COOn n V − == 3 Unreacted 0.50 0.20 0.30 molCH COOHn V 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] [CH3COO–]eqm  [CH3COO–]initial = 0.050 mol dm–3, since CH3COO– 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, [NH4+]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 l g 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 l g 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 l g 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) at pH = 8.2, [H+] = 10−8.2 = 6.3096  10−9 At pH = 8.1, [H+] = 10−8.1 = 7.9433  10−9 % change in [H+] = (7.9433 – 6.3096)/6.3096  100% = 25.9% (b) As the amount of CO 2 gas increases in the atmosphere, [CO 2(aq)] increases. This causes the position of equilibrium in reaction 4.1 to shift to the right, increasing [H+]. To counteract the increase in [H+], position of equilibrium in reaction 4.2 shifts left, partially offsetting the increase in H + concentration, thus maintaining the pH in seawater. In other words, with respect to reaction 4.2, the backward reaction takes place to neutralise additional H+, minimising the change in [H+] and resisting pH change. CO32– + H+ ⎯→ HCO3– (c) At pH = 8.1, [H+] = 10−8.1 = 7.9433  10−9 From reaction 6.1, [HCO3−]/[CO2] = K1/[H+] = 182.5 HCO3− is more abundant than CO2 From reaction 6.2, [CO32−]/[HCO3−] = K2/[H+] = 0.1385 HCO3− is more abundant than CO32− Hence, HCO3− is most abundant inorganic carbon species at pH 8.1.

-5- 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 = pKa + lg [lactate] [lactic acid] = pKa + lg n(lactate) / Vbuffer n(lactic acid) / Vbuffer = pKa + lg n(lactate) n(lactic acid) (If the buffer solution is infinitely diluted with water, the [H +] in the buffer solution tends towards that of pure water (i.e. 10 −7 mol dm −3) and pH increases and approaches 7.) Hence, pH of the buffer solution will not decrease upon dilution with water. Option C Incorrect. Buffering capacity depends on the amounts of lactic acid and sodium lactate in the buffer solution. Since the amounts of lactic acid and sodium lactate in the buffer solution remain unchanged upon dilution, the buffering capacity remains the same. Option D Correct. Since the buffer solution has a greater amount of lactic acid (weak acid) than the lactate ions (conjugate base), the buffer solution is more effective in buffering the effect of addition of small amounts of base than acid.

-6- 6 (a) [Note: marking out the equivalence point on the graph is not required] The two major organic species are CH3CH(OH)CO2H and CH3CH(OH)CO2−. In the solution, the following equilibria exist, CH3CH(OH)CO2H(aq) ⇌ CH3CH(OH)CO2–(aq) + H+(aq) ……………….(1) CH3CH(OH)CO2−(aq) + H2O(l) ⇌ CH3CH(OH)CO2H(aq) + OH−(aq) ……………….(2) When a small amount of OH– ions is added to the solution, concentration o