# How do you calculate the pH of the solution made by adding 0.50 mol of HOBr and 0.30 mol of KOBr to 1.00 L of water?

## The value of Ka for $H O B r$ is $2.0 \cdot {10}^{- 9}$.

Jun 26, 2016

You can do it like this:

#### Explanation:

$H O B r$ dissociates:

$H O B {r}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + O B {r}_{\left(a q\right)}^{-}$

The expression for ${K}_{a}$ is:

${K}_{a} = \frac{\left[{H}_{\left(a q\right)}^{+}\right] \left[O B {r}_{\left(a q\right)}^{-}\right]}{\left[H O B {r}_{\left(a q\right)}\right]}$

These are equilibrium concentrations.

To find the $p H$ we need to know the ${H}_{\left(a q\right)}^{+}$ concentration so rearranging gives:

$\left[{H}_{\left(a q\right)}^{+}\right] = {K}_{a} \times \frac{\left[H O B {r}_{\left(a q\right)}\right]}{\left[O B {r}_{\left(a q\right)}^{-}\right]}$

Because the value of ${K}_{a}$ is so small we can see that the position of equilibrium lies well to the left.

This means that the initial moles given will be a very close approximation to the equilibrium moles so we can use them in the expression.

There will be a volume change on adding these substances to water so the final volume will not now be 1 litre.

This does not matter as the volume is common to both so cancels out:

$\therefore \left[{H}_{\left(a q\right)}^{+}\right] = 2 \times {10}^{- 9} \times \frac{0.5}{\cancel{V}} / \frac{0.3}{\cancel{V}} = 3.33 \times {10}^{- 9} \text{ ""mol/l}$

$p H = - \log \left[{H}_{\left(a q\right)}^{+}\right]$

$\therefore p H = - \log \left[3.33 \times {10}^{- 9}\right]$

$\textcolor{red}{p H = 8.47}$