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

1 Answer
Aug 22, 2016

Answer:

#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.