Question

2- Calculate the pH and fraction of dissociation of a 0.00753 M sodium hypochlorite solution (NaOCl) solution.

Answer 1

OCl-(aq) + H2O(l) <---> HOCl(aq) + OH-(aq)
Base ..........acid ............... acid ......... base

Kb = [HOCl][OH-] / [OCl-]

Dissociation of acid:
HOCl(aq) <-------> H+(aq) + OCl-(aq)

Ka = [H+][OCl-] / [HOCl] = 3.5x10^-8

How can we calculate Kb ?

We also know the self-dissociation of water:
H2O(l) <----> H+(aq) + OH-(aq)
Kw=[H+][OH-] = 1x10^-14

Kw / Ka = Kb

{ [H+][OH-] } / { [H+][OCl-] / [HOCl] } =

[HOCl][OH-] / [OCl-] (gives the hydrolysis expression)

Therefore, Kb = Kw / Ka

Kb = 1x10^-14 / 3.5x10^-8 = 2.85 x 10^-7

OCl-(aq) + H2O(l) <---> HOCl(aq) + OH-(aq)
0.00573 - x M ....................... x M ........... x M

[HOCl][OH-] / [OCl-] = 2.85 x 10^-7

(x)(x) / (0.00573 - x) = 2.85 x 10^-7

Since x will be very small compared to 0.00573, x can be neglected.

(x)(x) / (0.00573) = 2.85 x 10^-7

x^2 = 1.635x10^-9

x = 4.04x10^-5 M = [OH-] <<<<<<<<<<<<answer

pOH = -log[OH-] = -log(4.041x10^-5) = 4.3935

pH = 14.00 - pOH

pH= 14.00 - 4.393

pH= 9.606<<<<<<<<<<<<<<<<Answer