# Question #ab631

Mar 3, 2017

You can do it like this:

#### Explanation:

If you have a weak acid then you can use this expression:

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

Where a is the concentration of the acid. We assume that it is a close approximation to the equilibrium concentration.

$\textsf{{M}_{r} = 204.22}$

$\therefore$$\textsf{n = \frac{m}{M} _ r = \frac{0.180}{204.22} = 8.814 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol}}$

$\textsf{c = \frac{n}{v} = \frac{0.180 \times {10}^{- 4}}{\frac{50.0}{1000}} = 0.01763 \textcolor{w h i t e}{x} \text{mol/l}}$

$\therefore$$\textsf{p H = \frac{1}{2} \left[5.4 - \log \left(0.01763\right)\right]}$

$\textsf{p H = \frac{1}{2} \times 7.154 = 3.58}$