RI Acid-Base Remedial

Self-Check Answers 1 Calculate the pH of the following solutions at 25 °C. NaOH is a strong base that undergoes complete ionisation in water. NaOH (aq) → Na+(aq) + OH–(aq) [OH–] = [NaOH (aq)] = 0.1 mol dm–3 pOH = –lg [OH–] = –lg (0.1) = 1.00 pH = 14 – pOH = 14 – 1.00 = 13.0 NH4+ is the conjugate acid of NH3. NH4+ undergoes hydrolysis in water. Ka of NH4+ = 10–14 1.8×10–5 = 5.555 × 10–10 mol dm–3 [H+] = �5.555 x 10 -10 × 0.1 = 7.453 x 10–6 mol dm–3 pH = –lg(7.453 x 10–6) = 5.13 CH3NH2 and CH3NH3+ are a conjugate acid-base pair. This solution is a buffer. pOH = pKb + lg ( [CH3NH3+] [CH3NH2] ) = –lg(4.4 x 10−4) + lg( 0.2 0.1) = 3.66 pH = 14 – pOH = 14 – 3.66 = 10.3 2 Answer: B A buffer is formed if the resulting solution has a weak acid and its conjugate base, or a weak base and its conjugate acid. After mixing A NH3 (weak base), NH4+ (conjugate acid)  buffer B All K2CO3(aq) and H2SO4(aq) reacted to form K2SO4(aq), CO2 and H2O. Thus, no conjugate acid-base pair present to form a buffer. C (Note: CH3CO2–, being a conjugate base of a weak acid, can react with H + from HCl to form CH3CO2H) Since equal volumes of CH3CO2– and HCl are used, CH3CO2– is in excess. After the reaction, there will be an equal amount of CH 3CO2– and CH 3CO2H present. CH3CO2H (weak acid), CH3CO2– (conjugate base)  buffer D HOOC–COOH is in excess while NaOH is limiting . HOOC–COOH will undergo neutralisation reaction with NaOH to form its conjugate base HOOC–COO–. HOOC–COOH (weak acid), HOOC–COO– (conjugate base)  buffer 0.1 mol dm−3 NaOH 0.1 mol dm−3 NH4+ Kb of NH3 = 1.8 x 10−5 0.1 mol dm−3 CH3NH2 0.2 mol dm−3 CH3NH3+ Kb of CH3NH2 = 4.4 x 10−4

3 (i) FCH2CO2H + H2O ⇌ FCH2COO‾ + H3O+ [1] Acid: FCH2CO2H Conjugate base: FCH2COO– [1] Base: H2O Conjugate acid: H3O+ [1] (ii) [H+] =�Ka1×c0 = �1.4125×10-3×0.1 = 1.1885 x 10–2 mol dm–3 [1] pH = –lg[H+] = 1.93 [1] (iii) (iv) At point A, it is the second equivalence point and the species present is – O2CCH2CO2–Being the conjugate base of a weak acid, –O2CCH2CO2– undergoes hydrolysis to give OH– ions causing [OH–] > [H+]. –O2CCH2CO2– + H2O ⇌ –O2CCH2CO2H + OH– (v) Note: The solution containing HO2CCH2CO2⁻ (weak acid with p Ka2) and ⁻O2CCH2CO2⁻ (conjugate base of HO2CCH2CO2⁻) is a buffer. pH = pKa2 + lg [conjugate base] [weak acid] = 5.70 + lg 0.20 0.50 = 5.30 [1] Answer should include: • Initial pH (ans from (a)(iii)) AND volume of NaOH at both equivalence points [1] • pH and volume of NaOH at both maximum buffer capacities (acidic buffer) [1]

Name: ………………………………………. ( ) Class: ……………. Date: ……………. Raffles Institution Year 6 H2 Chemistry 2025 Set A T1W10 – Acid-Base Equilibria Notes: Thought process Examples Calculate the pH of the following solutions at 25 °C. HCl is a ___