Under standard conditions, water is known to undergo autoprotolysis:
#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#
This is simply another equilibrium reaction. It is bond-breaking, so it is likely to lie strongly on the left hand side.
At #298K# and #"1 atmosphere"#, the ion product has been measured (and also under other conditions).
#[H_3O^+][HO^-]=10^-14#
Now we can take #log_10# of both sides, and mulitply by #-1# to give #pH#, #pOH#.
#log_10{[H_3O^+][HO^-]}=10^-14#
#=# #log_10[H_3O^+]+log_10[HO^-]=log_10[10^-14]#
On rearrangement (for reasons that will become apparent!),
#-log_10[10^-14]=-log_10[H_3O^+]-log_10[HO^-]#
But since #-log_10[10^-14]=14#, and #-log_10[H_3O^+]=pH#, and #-log_10[HO^-]=pOH# by definition.
#14=pH+pOH#
So instead of a daunting equilibrium expression, you have got a simple arithmetic relationship, provided that you can take logarithms.
Back to your problem (finally). We know that
#[HO^-]=6.0*mol^-1#.
Thus #pOH=-log_10(6.0) ~= -0.8#
And #pH~=13.2#
If you have just been introduced to it, you might want to review the #log# function (so find a maths text). If I say #log_ab=c#, then #a^c=b#. In other words I am asking to what power I raise the base #a# to get #b#.
Can you tell me what #log_(10)100#, and #log_(10)1000000# are? Use your calculator if you do not see it immediately. In the days before electronic calculators, which is not so long ago, log tables were routinely used in the classroom for multiplication and division.
