Question

1. When a 21.5 mL sample of a 0.329 M aqueous hypochlorous acid solution is titrated with a 0.456 M aqueous potassium hydroxide solution, what is the pH after 23.3 mL of potassium hydroxide have been added?

pH =

2. When a 16.0 mL sample of a 0.404 M aqueous hydrocyanic acid solution is titrated with a 0.418 M aqueous barium hydroxide solution, what is the pH at the midpoint in the titration?

pH =

Answer 1

a) Addition of 23.3 mL of KOH: Number of moles of HClO = M*V = 0.329 M* 21.5 mL = 7.0735 mmol Number of moles of KOH = M*V = 0.456 M* 23.3 mL = 10.6248 mmol This is after the equivalence point. At this point, the pH of the solution is determined by the amount of unreacted KOH in the solution. Net moles of KOH = 10.6248 – 7.0735 = 3.5513 mmol Total volume of the solution = 21.5 + 23.3 = 44.8 mL Conc. of KOH = [OH-] = moles/ volume = 3.5513 mmol / 44.8 mL = 0.07927 M pOH =- log [OH-] = -log[0.07927 ] = 1.10 pH = 14 - pOH = 14 – 1.10 = 12.90 (b) Ka for HCN = 4 x 10-10 pka = -log(4 x 10-10) = 9.40 At mid point, the number of moles of acid (HCN) is equal to its conjugate base (CN). So, pH = pka = 9.40 pH of the solution at mid point = 9.40