Question

What is the pH of a buffer solution containing 0.410 M hypochlorous acid (HOCl) and 0.050 M (sodium hypochlorite) NaOCl? (Ka(HOCL))=3.2*10^-8

Please show work.

Answer 1

Since NaOCl is the salt of a strong base of NaOH and the weak acid of HOCl, when it dissolves it produces Na+ and OCl- ions. The hypochlorite ion hydrolyses in water.

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

The conjugate acid of hypochlorite ion is hypochlorous acid (HOCl) since it is a proton donor. Since in your question is not given Ka value of HOCl is found from the literature as 3.5x10^-8.

For the above equilibrium reaction; the hydrolysis equilbrium constant (Kh) which is Kb in this case, since the solution is basic, will be;

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

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

Ka = [H+][OCl-] / [HOCl] = 3.2x10^-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 / 3.2x10^-8 = 3.125 x 10^-7

OCl-(aq) + H2O(l) <---> HOCl(aq) + OH-(aq)
0.050 - x M ....................... 0.41M ........... x M

[0.41][OH-] / [OCl-] = 3.125 x 10^-7
(0.41)(x) / (0.050 - x) = 3.125 x 10^-7

Since x will be very small compared to 0.050 , x can be neglected.

(0.41)(x) / (0.05) = 3.125 x 10^-7
x = 3.81x10-7


pOH = -log[OH-] = -log(3.81x10-7) = 6.41

pH = 14.00 - pOH = 14.00 - 6.41 = 7.58