# How to calculate ph value of the solution? Two salt buffer.

## solution contains: c(Na3PO4) = 0,10 mol/L and c(Na2HPO4)=0,05 mol/L Answer in my book is ph=12.96

Jan 31, 2017

You can do it like this:

#### Explanation:

Phosphoric acid has 3 ionisations. Here we are dealing with the 3rd:

$\textsf{H P {O}_{4}^{2 -} r i g h t \le f t h a r p \infty n s P {O}_{4}^{3 -} + {H}^{+}}$

For which:

$\textsf{{K}_{a 3} = \frac{\left[P {O}_{4}^{2 -}\right] \left[{H}^{+}\right]}{\left[H P {O}_{4}^{2 -}\right]} = 2.2 \times {10}^{- 13} \textcolor{w h i t e}{x} \text{mol/l}}$

You should be provided with this value in the question.

To find the pH we need the concentration of the $\textsf{{H}^{+}}$ ions.

Rearranging gives:

$\textsf{\left[{H}^{+}\right] = {K}_{a 3} \times \frac{\left[H P {O}_{4}^{2 -}\right]}{\left[P {O}_{4}^{2 -}\right]}}$

$\therefore$$\textsf{\left[{H}^{+}\right] = 2.2 \times {10}^{- 13} \times \frac{0.05}{0.1} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{\left[{H}^{+}\right] = 1.1 \times {10}^{- 13} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left(1.1 \times {10}^{- 13}\right)}$

$\textsf{p H = 12.95}$