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 ________ acid that undergoes __________ ionisation in water. CH3COOH is a ________ acid that undergoes __________ ionisation in water. CH3COO− is the conjugate _______ of CH3COOH. CH3COO− undergoes __________ in water. 0.1 mol dm−3 HCl 0.1 mol dm−3 CH3COOH Ka of CH3COOH = 1.8 x 10−5 0.1 mol dm−3 CH3COO− Ka of CH3COOH = 1.8 x 10−5 ⎯ Pre-ChemFocus Activity ⎯ • Familiarise yourself with the equations and the use of the equations. • To view infopack at IVY > C2025 – H2 CHEMISTRY > Pages > [ChemFocus] T1W10 Acid-Base Equilibria > FIRST VIDEO ONLY • • •

Self-Check Questions 1 Calculate the pH of the following solutions at 25 °C. NaOH is a ________ base that undergoes __________ ionisation in water. NH4+ is the ____________ acid of NH3. NH4+ undergoes __________ in water. CH3NH2 and CH3NH3+ are a conjugate acid-base pair. This solution is a __________________. 2 Which pair of substances, when mixed in equal volumes, does not produce a buffer? A 0.1 mol dm–3 NH3(aq) and 0.1 mol dm–3 NH4Cl(aq) B 0.1 mol dm–3 K2CO3(aq) and 0.1 mol dm–3 H2SO4(aq) C 0.1 mol dm–3 CH3CO2Na(aq) and 0.05 mol dm–3 HCl(aq) D 0.1 mol dm–3 HO2C–CO2H(aq) and 0.05 mol dm–3 NaOH(aq) 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 Check your answers at IVY > C2025 – H2 CHEMISTRY > Pages > [ChemFocus] T1W10 Acid-Base Equilibria > Remaining videos

3 The South African plant Dichapetalum cymosum contains fluoroethanoic acid, FCH2CO2H, and malonic acid, HO2CCH2CO2H. Data about these acids is given in Table 1.1. Table 1.1 acid formula pKa1 pKa2 malonic HO2CCH2CO2H 2.85 5.70 fluoroethanoic FCH2CO2H 2.57 – (i) Write the equation for the dissociation of FCH 2CO2H in water and identify the conjugate acid-base pairs that are present. ……………………………………………………………………………………………….. ..……………………………………………………………………………………………[2] (ii) Calculate the pH of a 0.10 mol dm–3 solution of malonic acid (ignore the effect of pKa2 on the pH). [2] The pH–volume curve when 30 cm 3 of 0.10 mol dm–3 NaOH is added to 10 cm 3 of 0.10 mol dm–3 malonic acid is shown below. (iii) Using your answer from (a)(ii) and the information provided in Table 1.1, label various key points on the axes above. A 7 pH  volume of NaOH added / cm3 0

(iv) With the aid of an equation, explain why the pH at point A on the graph has a value greater than 7. ..……………………………………………………………………………………………[1] (v) Calculate the pH of a solution that contains 0.50 mol of HOOCCH2COO– and 0.20 mol of –OOCCH2COO– dissolved in 1.0 dm3 of water. [1]

Tutorial Discussion Questions (to complete before Chemfocus session) 1a The pKb value of CH3CH2NH2 and pKw at 60 °C are given below. temperature / °C pKw pKb of CH3CH2NH2 60 13.0 3.0 A buffer mixture at 60 °C was prepared using 10 cm 3 of 0.10 mol dm ─3 HNO3(aq) and 20 cm3 of 0.10 mol dm─3 CH3CH2NH2(aq). Deduce the resultant pH. [1] 2a Paracetamol is a common painkiller used to relief minor aches and pains. Its structure is shown below. Paracetamol Paracetamol is a weak monobasic acid with its –OH group having a pKa value of 9.50 at 25 °C. In an experiment, 25.0 cm 3 of a solution of paracetamol was titrated against 0.100 mol dm–3 of aqueous NaOH. It was found that 13.20 cm3 of NaOH was required for complete neutralisation. (i) Calculate the concentration of the paracetamol solution and hence, its initial pH. [3] HO N H C O CH3

(ii) Identify the species present at equivalence and calculate its concentration. [2] (iii) Hence, calculate the pH at equivalence. [2]

Tutorial Practice Questions (to be attempted IN CLASS) 1b The pKb value of NH3 and pKw at 30 °C are given below. temperature / °C pKw pKb of NH3 30 13.8 4.7 A buffer mixture at 30 °C was prepared using equal volumes of 0.10 mol dm─3 NH4Cl (aq) and 0.020 mol dm–3 NaOH(aq). Calculate the resultant pH. [2] 2b A student titrated 20 cm 3 of the buffer sample containing an acid and its salt, with 0.1 mol dm –3 NaOH(aq) and obtained the titration curve below. The maximum buffer capacity is obtained when 20 cm3 of NaOH is added. The equivalence point of the titration is obtained when 60 cm3 of aqueous sodium hydroxide is added. [You may use HA and A– to represent the species in the buffer sample.] pH Vol of NaOH / cm3 7.2 20 60