Question

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

Answer 1

`sf(pH~=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:

`sf(H_2OrightleftharpoonsH^++OH^-)`

For which:

`sf(K_w=[H^+][OH^-]=10^(-14)color(white)(x)"mol"^2"/"l"^(-2))` at `sf(25^@C)`

This means that `sf([H^+])` and `sf([OH^-])` are both equal to `sf(10^(-7)color(white)(x)"mol/l")`.

You might, therefore, conclude that in this solution the total `sf(OH^-)` concentration is equal to `(sf(10^(-7)+1.15xx10^(-12)))color(white)(x)"mol/l"` but this is not quite right.

You have disturbed a system at equilibrium by adding extra `sf(OH^-)` 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 `sf(OH^-)` ions and find the pH.

However, because `sf(10^(-7))` is so much greater than `sf(1.15xx10^(-12)` the perturbation will be negligible and it is a fair assumption that `sf([OH^-]=10^(-7)"mol/l")`.

This gives a `sf(pH)` of `sf(7)`.

Thought experiment:

There are `sf(10^(12)` litres in `sf(1km^(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 `sf(1km^(3))` of water.

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