Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.110 M HClO(aq) with 0.110 M KOH(aq)..

A before additon of any KOH

B After additon of 25.0 mL KOH

C after addition of 35.0 ML of KOH

D after addition of 50.0 mL of KOH

E after addition of 60 mL KOH

Answer 1

Titration

A) before addition of KOH

HClO + H2O <==> H3O+ + ClO-

4 x 10^-8 = x^2/0.110

x = [H3O+] = 6.63 x 10^-5 M

pH = -log[H3O+] = 4.18

B) 25 ml KOH added

This is half-equivalence point

[HClO] left = [ClO-] formed

pH = pKa = 7.54

C) 35 ml KOH added

molar concentration of [HClO] left = 0.11 M x 15 ml/85 ml = 0.0194 M

moles concentration of [ClO-] formed = 0.11 M x 35 ml/85 ml = 0.0453 M

pH = pKa + log(base/acid)

     = 7.54 + log(0.0453/0.0194) = 7.91

D) 50 ml KOH added

Equivalence point

[ClO-] formed = 0.11 M x 50 ml/100 ml = 0.055 M

ClO- + H2O <==> HClO + OH-

Kb = 1 x 10^-14/4.0 x 10^-8 = x^2/0.055

x = [OH-] = 1.17 x 10^-4 M

pOH = -log[OH-] = 3.93

pH = 14 - pOH = 10.07

E) 60 ml KOH added

excess [OH-] = 0.11 M x 10 ml/110 ml = 0.01

pOH = -log[OH-] = 2

pH = 14 - pOH = 12