# What is the pH of a 0.500 M solution of acetylsalicylic acid, pK 3.52?

Sep 19, 2016

$\textsf{1.91}$

#### Explanation:

For a weak acid:

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

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

$\therefore$sf(pH=1/2[3.52-log(0.5)]

$\textsf{p H = \frac{1}{2} \left[3.52 - \left(- 0.301\right)\right] = 1.91}$