`50` litres of `0.1M` HCl is thoroughly mixed with `50` litres of `0.2M NaOH`. What is the `pOH` of the resulting solution?

Explanation

`[OH]=(V_(1)N_(1)-V_(2)N_(2))/((V_(1)+V_(2)))=(50xx0.2-50xx0.1)/((50+50))`
`=(10-5)/(100)=(5)/(100)=0.05=5xx10^(-2)`
`p^(OH)=-log_(10)[OH^(-)]`
`=-log_(10)5xx10^(-2)=2-log_(10)5`
`=2-0.6990=1.3`