Question

Hclo is a weak acid (ka = 4.0 × 10–8) and so the salt naclo acts as a weak base. what is the ph of a solution that is 0.057 m in naclo at 25 °c?

Answer 1

Chemical reaction 1: NaClO(aq) → Na⁺(aq) + ClO⁻(aq).
Chemical reaction 2: ClO⁻(aq) + H₂O(l) ⇄ HClO(aq)+ OH⁻(aq).
c(NaClO) = 0,057 M.
[ClO⁻] = 0,057 M - x.
Ka(HClO) = 4·10⁻⁸.
Ka · Kb = 10⁻¹⁴.
Kb(ClO⁻) = 2,5·10⁻⁷.
Kb(ClO⁻) = [OH⁻] · [HClO] / [ClO⁻].
[OH⁻] = [HClO] = x.
2,5·10⁻⁷ = x² / (0,057 M -x).
x = [OH⁻] = 0.00012 M.
pOH = -log(0.00012 M) = 3.92.
pH = 14 - 3.92.
pH = 10.08.

Answer 2

The pH of the solution is  10.1.

NaClO(aq) ⇄ Na^+(aq) + ClO^-(aq)

Also;

      ClO^-(aq) + H2O(l) ⇄ HClO^-(aq) + OH^-(aq)

I       0.057                           0                   0

C        -x                               +x                  +x

E      0.057 - x                     x                      x

Kb = 1 × 10^-14/Ka

Kb = 1 × 10^-14/ 4.0 × 10–8

Kb = 2.5  × 10^-7

Recall that [HClO^-] = [OH^-]

Kb =  [HClO^-]  [OH^-]/[ClO^-]

2.5  × 10^-7 = [x] [x]/[0.057 - x]

2.5  × 10^-7 [0.057 - x] = [x] [x]/

1.425 × 10^-8 - 2.5  × 10^-7x = x^2

x^2 + 2.5  × 10^-7x  - 1.425 × 10^-8 = 0

x = 0.00012 M

Hence;

x = [OH^-] = 0.00012 M

pOH = - log[OH^-]

pOH = - log[0.00012 M]

pOH = 3.92

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 3.92

pH = 10.1

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