# A solution of ethanoic acid in water has a pH of 3.2. How do you calculate the concentration of the solution? K_a = 1.7 x 10^-5 mol dm^-3?

Dec 29, 2016

You can do it like this:

#### Explanation:

For a weak acid you can derive this expression from an ICE table:

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

Where a is the concentration of the acid.

$\textsf{p {K}_{a} = - \log {K}_{a} = - \log \left(1.7 \times {10}^{- 5}\right) = 4.77}$

$\therefore$$\textsf{3.2 = \frac{1}{2} \left(4.77 - \log a\right)}$

$\textsf{\log \frac{a}{2} = 2.835 - 3.2 = - 0.815}$

$\textsf{\log a = - 1.63}$

From which $\textsf{a = 0.023 \textcolor{w h i t e}{x} \text{mol/l}}$