Question

A buffer solution of a weak acid and its salt had a pH of 7.85 when the acid was 0.251 M and the anion was 1.10 M. How do you calculate the equilibrium dissociation constant (Ka)?

Answer 1

You can do it like this:

Explanation:

We will call the acid `HA` which dissociates:

`HA_((aq))rightleftharpoonsH_((aq))^(+)+A_((aq))^-`

We know the `pH` is `7.95`.

`:.-log[H^+]=7.85`

From which `[H^+]=1.41xx10^(-8)"mol/l"`

The expression for `K_a` is:

`K_a=([H_((aq))^(+)][A_((aq))^-])/([HA_((aq))])`

These are equilibrium concentrations which the question has already provided us with.

`:.K_a=(1.41xx10^(-8)xx1.1)/(0.251)`

`K_a=6.18xx10^(-8)"mol/l"`