The pH of `10^(-8) M` NaOH will be

Explanation

`10^(-8) M NaOH`
`NaOH to Na^(+)+ OH^(-)`
`10^(-8)" 0 0"`
`0" "10^(-8)" "10^(-8)`
`therefore [OH^(-)]=10^(-8)M`
Now proceed for `OH^(-)` as in problem 50 part (e)