`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)))`
`[OH^(-)] = (50xx 0.2 - 50 xx 0.1)/((50 + 50))`
`= (10 - 5)/(100) = (5)/(100) = 0.05`
`= 5 xx 10^(-2)`
`P^(OH) = - log_(10)[OH^(-)]`
`= -log_(10) xx 10^(-2)`
`= 2 -log_(10)5`
`= 2 - 0.6990 = 1.3`