# What is the pH of a 1.15 * 10^-12 M solution of KOH?

Aug 22, 2016

$\textsf{p H \cong 7}$

#### Explanation:

This is such a dilute solution of KOH that its concentration is negligible. As such I would expect the pH to be very close to 7.

You need to consider the auto - ionisation of water:

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

For which:

$\textsf{{K}_{w} = \left[{H}^{+}\right] \left[O {H}^{-}\right] = {10}^{- 14} \textcolor{w h i t e}{x} {\text{mol"^2"/"l}}^{- 2}}$ at $\textsf{{25}^{\circ} C}$

This means that $\textsf{\left[{H}^{+}\right]}$ and $\textsf{\left[O {H}^{-}\right]}$ are both equal to $\textsf{{10}^{- 7} \textcolor{w h i t e}{x} \text{mol/l}}$.

You might, therefore, conclude that in this solution the total $\textsf{O {H}^{-}}$ concentration is equal to $\left(\textsf{{10}^{- 7} + 1.15 \times {10}^{- 12}}\right) \textcolor{w h i t e}{x} \text{mol/l}$ but this is not quite right.

You have disturbed a system at equilibrium by adding extra $\textsf{O {H}^{-}}$ ions so the system would shift to the left according to Le Chatelier's Principle.

You could then set up an ICE table to find out the equilibrium concentration of the $\textsf{O {H}^{-}}$ ions and find the pH.

However, because $\textsf{{10}^{- 7}}$ is so much greater than sf(1.15xx10^(-12) the perturbation will be negligible and it is a fair assumption that $\textsf{\left[O {H}^{-}\right] = {10}^{- 7} \text{mol/l}}$.

This gives a $\textsf{p H}$ of $\textsf{7}$.

Thought experiment:

There are sf(10^(12) litres in $\textsf{1 k {m}^{3}}$ of water. It would be the typical volume of a Scottish Loch (a small lake).

Take 1 litre of 1.15 M NaOH and make it up to $\textsf{1 k {m}^{3}}$ of water.

That is the sort of dilution we are talking about. It should be dilute enough for a pH of 7.