Why does pH + pOH = 14?

1 Answer
May 17, 2016

Answer:

Because water undergoes a measurable equilibrium at #298*K#.

Explanation:

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?