We can write the equilibrium as follows:
2H_2OrightleftharpoonsH_3O^+ + HO^-
And, K_w = [H_3O^+][HO^-]. This equilibrium has been carefully measured (by conductivity experiments) over a range of temperatures. [H_2O] does not appear in the equilibrium because its concentration is effectively constant.
At 298*K we know that K_w = 10^-14. How do we know this? By careful and repeated measurements of the conductivity of the water solvent.
So K_w = [H_3O^+][HO^-]=10^-14.
Now this is an arithmetic expression, that we can divide, multiply, add to, subtract from, provided that we do it to both sides of the equation. We can take log_10 of both sides to give:
log_10K_w = log_10[H_3O^+]+ log_10[HO^-]=log_10(10^-14)
But, by definition of logarithms, if a^b=c, log_(a)c=b, then log_10(10^-14)=-14, and thus,
log_10K_w = log_10[H_3O^+]+ log_10[HO^-]=-14
But by definition, -log_10[H_3O^+]=pH, and -log_10[HO^-]=pOH
pK_w = pH+ pOH=14 as required.
Given that this is a bond-breaking reaction, how would you expect the equilibrium to evolve at temperatures greater than 298K? Would pH increase or decrease?
