But what is pH? Simply, pH=-log_10[H_3O^+], and pOH=-log_10[HO^-].
We know that water undergoes autoprotolysis, which we could represent in the following fashion:
2H_2O rightleftharpoons OH^(-) + H_3O^+.
As with any equilibrium, we can define and measure an equilibrium constant, i.e.
K'_w = ([H_3O^+][HO^-])/([H_2O]^2)
But because [H_2O]^2 is so large, it does not change much effectively, and can be incorporated on the left hand side of the equation to give:
K_w = [H_3O^+][HO^-] = 10^-14 under standard conditions of 1*atm, and at 298*K. K_w is low because the reaction involes the breaking of a strong O-H bonds, and thus the equilibrium lies strongly to the left. Now we can manipulate this expression, provide we do it to BOTH sides of the equation. One thing that we can do is to take log_10 of BOTH SIDES.
And thus log_10K_w = log_10[H_3O^+]+log[HO^-].
And, mulitiplying both sides by -1, we get:
-log_10K_w=-log_10(10^-14)=-log_10[H_3O^+]-log[HO^-]
But -log_10(10^-14)=14 by definition, i.e. log_a(a^x)=x, the power to which you raise the base a to get a^x is clearly x.
And, by definition, pOH=-log[HO^-], and pH=-log[H_3O^+]
So pK_w=14=pH+pOH.
