# If the K_a of a monoprotic weak acid is 3.4 x10^-6, what is the pH of a 0.14 M solution of this acid?

Jan 28, 2017

$\textsf{p H = 3.15}$

#### Explanation:

Let $\textsf{H X}$ be the weak acid:

$\textsf{H X r i g h t \le f t h a r p \infty n s {H}^{+} + {X}^{-}}$

For which:

$\textsf{{K}_{a} = \frac{\left[{H}^{+}\right] \left[{X}^{-}\right]}{\left[H X\right]}}$

If $\textsf{{K}_{a}}$ lies in the range $\textsf{{10}^{- 4}}$ - $\textsf{{10}^{- 10}}$ we can assume that the equilibrium concentrations used in the expression are a good enough approximation to the initial concentrations.

This applies here.

Rearranging and taking negative logs of both sides gives:

$\textsf{p H = \frac{1}{2} \left[p {K}_{a} - \log a\right]}$

$\textsf{a}$ is the concentration of the acid.

$\textsf{p {K}_{a} = - \log {K}_{a} = - \log \left(3.4 \times {10}^{- 6}\right) = 5.46}$

$\therefore$$\textsf{p H = \frac{1}{2} \left[5.46 - \left(- 0.854\right)\right] = 3.15}$