Question

What is `K_a` for `HOCl`, given that initially `[HOCl]=0.08*mol*dm^3`, and `pH=2.85`?

Answer 1

`K_a=([H_3O^+][""^(-)OCl])/([HOCl])=2.54xx10^-5`.........

Explanation:

The `"acid dissociation constant,"` `K_a`, measures the particular acid dissociation............

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

Where `K_a=([H_3O^+][""^(-)OCl])/([HOCl])`

We are given that the initial concentration was `0.08*mol*dm^-3`, and that `pH=2.85`. But we know that `pH=-log_(10)[H_3O^+]`.

And thus `[H_3O^+]=10^(-2.85)*mol*L^-1`.

But, by stoichiometry `[""^(-)OCl]=10^(-2.85)*mol*L^-1`.

And, thus......................

`[HOCl]=(0.08-10^(-2.85))*mol*L^-1=...............`

`=0.0786*mol*L^-1`.

And thus we can calculate `K_a` sans approximation, i.e.

`K_a=((10^(-2.85))^2)/(0.0786)=2.54xx10^-5`.