What is the pOH of a 0.001 M HCl solution?

1 Answer
May 15, 2017

Answer:

Well, we know that #pH+pOH=14#, so..............

#pOH=11#

Explanation:

For the autoprotolysis reaction:

#2H_2O(l) rightleftharpoons H_3O^+ +HO^(-)#,

we know that #K_w=[H_3O^+][HO^-]=10^-14# under standard conditions............

And taking #log_10# of both sides we gets............

#log_10K_w=log_(10)10^-14=log_10[H_3O^+]+log_10[HO^-]#

And thus #-log_10[H_3O^+]-log_10[HO^-]=-log_(10)10^(-14)=14#

But by definition, #-log_10[H_3O^+]=pH#, and #-log_10[HO^-]=pOH#

So................

#pH+pOH=14#

You will not be asked to reproduce this as a first year student or as an A-level student. You will be asked (as you have been here!) to use the relationship.

And #[HO^-]=10^(-11)*mol*L^-1#.