What is the pH of a #1.0 * 10^-3# #M# sodium hydroxide solution?

1 Answer
Dec 28, 2016

#pH=11#

Explanation:

We examine the equilibrium:

#2H_2O(l) rightleftharpoons H_3O^+ + ""^(-)OH#

Under standard conditions, #[H_3O^+][HO^-]=10^-14#.

And if we take #log_10# of both sides:

#log_10[H_3O^+]+log_10[HO^-]=-14#

Om rearrangement:

#14=-log_10[H_3O^+]-log_10[HO^-]#

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

And thus our definining relationship: #pH+pOH=14#.

Since (finally!) #pOH=-log_10[HO^-]=-log_10(1xx10^-3)=-(-3)=3#.

And if #pOH=3#, #pH=14-3=??#

In A level, you do have to remember the result:

#pH+pOH=14#. With this is mind, these problems become trivial.