Question

Hypochlorous acid, HOCl, reacts with water as shown below: HOCl(aq) + H2O(l) <---> OCl-(aq) + H3O+(aq) What is the pH of a 0.033 M aqeous solution of acetic acid? Ka=1.1e-2

Answer 1

This is a weak acid at LOW concentration, whose dissociation we probably have to solve analytically...

Explanation:

We address the equation...

`HOCl(aq) +H_2O(l)rightleftharpoonsH_3O^+ +""^(-)OCl`..

...where `K_a=0.011=([H_3O^+][""^(-)OCl])/([HOCl])`

And if `x*mol*L^-1` dissociates...

`0.011=x^2/(0.033-x)`

And so `x^2=0.011xx(0.033-x)`

And thus...`x^2+0.011x-3.63xx10^-4=0`

We get `x=0.014*mol*L^-1` as the positive root...

And so AT EQUILIBIRUM...

`[HOCl]=(0.033-0.014)*mol*L^-1=0.019*mol*L^-1`

`[""^(-)OCl]=0.014*mol*L^-1`

`[H_3O^+]=0.014*mol*L^-1=0.014*mol*L^-1`

And it is typical for a weak acid at LOW concentrations, to dissociate more than it would normally would...

`pH=-log_10(0.014)=1.85`