Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M HClO(aq) with 0.150 M KOH(aq). The ionization constant for HClO can be found here.

(a) before addition of any KOH

(b) after addition of 25.0 mL of KOH

(c) after addition of 30.0 mL of KOH

(d) after addition of 50.0 mL of KOH

(e) after addition of 60.0 mL of KOH

Answer 1

ionization constant for HClO is 2.9 * 10^-8

PKa = -log (2.9*10^-8) = 7.54

a) HClO is wea acid.

PH = 0.5 * PKa - 0.5 * log C

PH = 0.5 * - log (2.9 * 10^-8) - 0.5 * log (0.150)

PH = 3.77 + 0.412

PH = 4.18

b) 25.0 mL of 0.150 M KOH = 0.025 * 0.150 = 0.00375 mole.

50.0 mL of 0.150 M HClO(aq) = 0.050 * 0.150 = 0.0075 mole.

PH = PKa + log [salt] / [acid] = 7.54 + log (0.00375) / (0.0075 - 0.00375)

PH = 7.54

c) 30.0 mL of 0.150 M KOH = 0.030 * 0.150 = 0.0045 mole.

PH = PKa + log [salt] / [acid] = 7.54 + log (0.00375) / (0.0075 - 0.0045)

PH = 7.64

d) all the acid is neutralized by base.

so salt hydrolysis occurs.

[salt] = 0.150 / 2 = 0.075 M

PH = 0.5 * PKw + 0.5 * Pka + 0.5 * log C = 7 + 3.77 + 0.5 * log (0.075)

PH = 10.2

e) excess KOH = 10.0 ml of 0.150 M KOH = 0.010 * 0.150 = 0.0015 mole.

[KOH] = 0.0015 * 1000 / (50 + 60) = 0.0136 M

POH = - log (0.0136)

POH = 1.87

PH = 14 - 1.87 = 12.13