What will be the resultant pH when 200 ml of an aqueous solution of HCl having `pH=2` is mixed with 300 ml of a aqueous solution of NaOH having `pH=12`?

Explanation

Concentration of HCl `=10^(-2)`
`pOH` of `NaOH=14-12=10^(-2)`
200 ml of `10^(-2)` HCl is mixed with 300 mL of `10^(-2)` NaOH. Since NaOH is more the resultant solution is basic.
Concentration of NaOH left
`=((300xx10^(-2))-(200xx10^(-2)))/500`
`=(100xx10^(-2))/500=0.2xx10^(-2)=2xx10^(-3)`
`pOH=2.6990`
`pH=14-2.699=11.3010`