Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.100 M HClO(aq) with 0.100 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

pKa for HClO = 7.53

=> Ka = 2.95 x 10^-8

Given,

Molarity of HClO = 0.1 M

Volume of HClO = 50 mL

=> Millimoles of HClO = 0.1 x 50 = 5

(a) before addition of any KOH

HClO + H2O -----> H3O+ + ClO-

0.1 - X...................X............X

Ka = [H3O+] [ClO-] / [HClO]

=> 2.95 x 10^-8 = X^2 / (0.1 - X)

=> X = 5.43 x 10^-5 M = [H+]

pH = - log [H+] = - log (5.43 x 10^-5) = 4.265

(b) after addition of 25.0 mL of KOH

Milllimoles of KOH added =25 x 0.1 = 2.5

Millimoles of HClO present = 5

Reaction is:

HClO + KOH -----> KClO + H2O

1 mole of HClO reacts with 1 mole of KOH to procude 1 mole of KClO

=> Millimoles of KClO produced = 2.5

Millimoles of HClO remaining = 5 - 2.5 = 2.5

This solution will act as an acidic butter with HClO as acid and KClO as salt

pH for buffer = pKa + log (Salt / Acid)

=> pH = 7.53 + log (2.5 / 2.5) = 7.53

(c) after addition of 30.0 mL of KOH

Milllimoles of KOH added =30 x 0.1 = 3

Millimoles of HClO present = 5

Reaction is:

HClO + KOH -----> KClO + H2O

1 mole of HClO reacts with 1 mole of KOH to procude 1 mole of KClO

=> Millimoles of KClO produced = 3

Millimoles of HClO remaining = 5 - 3 = 2

This solution will act as an acidic butter with HClO as acid and KClO as salt

pH for buffer = pKa + log (Salt / Acid)

=> pH = 7.53 + log (3 / 2) = 7.71

(d) after addition of 50.0 mL of KOH

Complete neutralization will occur

Millimoles of KClO produced = 5

Total Volume of solution = 100 mL

=> [KClO] = 5 / 100 = 0.05 M

Kb for KClO = 10^-14 / 2.95 x 10^-8 = 3.39 x 10^-7

KClO + H2O -----> HClO + OH- + K+

0.05 -X ....................X.........X

Kb = [HClO] [OH-] / [KClO]

=> 3.39 x 10^-7 = X^2 / (0.05 - X)

=> X = 1.3 x 10^-4 M = [OH-]

pOH = - log [OH-] = - log (1.3 x 10^-4) = 3.89

pH = 14 - pOH = 10.11

(e) after addition of 60.0 mL of KOH

Milllimoles of KOH added =60 x 0.1 = 6

Millimoles of HClO present = 5

Reaction is:

HClO + KOH -----> KClO + H2O

1 mole of HClO reacts with 1 mole of KOH to procude 1 mole of KClO

=> Millimoles of KClO produced = 5

Millimoles of KOH remaining = 6 - 5 = 1

Total Volume = 50 + 60 = 110 mL

[KOH] = 1 / 110 = 9.1 x 10^-3 M = [OH-]

pOH = - log [OH-] = 2.04

pH = 14 - pOH =11.96