`500mL` of `10^(-5)MNaOH` is mixed with `500mL` of `2.5xx10^(-5)M` of `Ba(OH_(2))`,To the resulting solution ,`99L` water is added ,calculate `pH` of final solution.Take `log0.303=-0.52`.

Explanation

`[OH^(+)]_(f)=((500xx10^(-5))+(500xx2xx2.5xx10^(-5)))/(1000)=3xx10^(-5)M`
`V_(1)=1L&V_(f)=100L`
no of moles of `[OH^(+)]` in resulting solution =no. of moles of `[OH^(+)]` in final
`3xx10^(-5)=[OH^(-)]_(f)xx100`
`:.[OH^(-)]_(f)=3xx10^(-7)M( lt 10^(-6)M)`
so `OH^(-)`ions coming form `H_(2)O` should also be considered.
`H_(2)OhArrunderset(x) H^(+)+underset(x+3xx10^(-7))(OH^(-))`
`K_(w)=x(x+3xx10^(-7))=10^(-14)`
`,.x=((sqrt(13)-3)/(2))xx10^(-7)M=[H^(+)]`
So `pH=7-log0.303=7.52`