# 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)?

Mar 22, 2016

You can do it like this:

#### Explanation:

We will call the acid $H A$ which dissociates:

$H {A}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + {A}_{\left(a q\right)}^{-}$

We know the $p H$ is $7.95$.

$\therefore - \log \left[{H}^{+}\right] = 7.85$

From which $\left[{H}^{+}\right] = 1.41 \times {10}^{- 8} \text{mol/l}$

The expression for ${K}_{a}$ is:

${K}_{a} = \frac{\left[{H}_{\left(a q\right)}^{+}\right] \left[{A}_{\left(a q\right)}^{-}\right]}{\left[H {A}_{\left(a q\right)}\right]}$

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

$\therefore {K}_{a} = \frac{1.41 \times {10}^{- 8} \times 1.1}{0.251}$

${K}_{a} = 6.18 \times {10}^{- 8} \text{mol/l}$