Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.190 M HClO(aq) with 0.190 M KOH(aq).

a) before addition of any KOH

b) after addition of 25.0 mL of KOH

c) after addition of 35.0 mL of KOH

d) after addition of 50.0 mL of KOH

e) after addition of 60.0 mL of KOH

Answer 1

(a) Before adding any KOH

HClO <==> H+ + ClO-

let x amount has dissociated,

Ka = 2.9 x 10^-8 = x^2/0.190

x = [H+] = 5.51 x 10^-9 M

pH = -log[H+] = 8.26

(b) after 25 ml of 0.190 M KOH is added

This is half equivalence point,

Moles of acid = moles of salt formed

log([salt]/acid]) = 0

So, pH = pKa = 7.54

(c) after 35 ml of 0.19 M KOH is added

moles of acid = 0.19 M x 0.05 L = 9.5 x 10^-3 mols

moles of KOH added = 0.19 M x 0.035 L = 6.65 x 10^-3 mols

excess moles of acid present = 2.85 x 10^-3 mols

molarity of salt formed = 6.65 x 10^-3/(0.085) = 0.078 M

molarity of acid remaining = 2.85 x 10^-3/0.085 = 0.033 M

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

      = 7.54 + log(0.078/0.033)

      = 7.91

(d) after 50 ml of 0.19 M KOH is added

this is the equivalence point

moles of salt formed = 9.5 x 10^-3 mols

molarity of salt = 9.5 x 10^-3/0.1 = 0.095 M

Salt hydrolyzes in water,

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

let x amount has hydrolyzed,

Kb = Kw/Ka = 1 x 10^-14/2.9 x 10^-8 = x^2/0.095

x = [OH-] = 1.81 x 10^-4 M

pOH = 3.74

pH = 14 - pOH = 10.26

(e) after 60 ml of 0.19 M KOH is added

excess moles of KOH = 0.19 M x 0.010 L = 1.9 x 10^-3 mols

molarity of excess KOH = 1.9 x 10^-3/0.11 = 0.0173 M

pOH = -log(0.0173) = 1.762

pH = 14 - pOH = 12.24