How much KOH must be dissolved in one litre of solution to get a pH of 12 at `25^(@)C`?

Answer

`0.56 g L^(-1)`

Explanation

`pH =- log [H_(3)O^(+)] " or " 12 =- log [H_(3)O^(+)]`
`log[1//H_(3)O^(+)] =12 " or " [1//H_(3)O^(+)] = " Antilog " 12 =10^(12)`
`:. " "[H_(3)O^(+)] =10^(-12) M`
`[OH^(-)] =(K_(w))/[[H_(3)O^(+)]] =((1xx 10^(-14) M^(2)))/((10^(-12)M)) =10^(-2) M`
Amount of KOH`//`litre of the solution `=10^(-2) " mol "L^(-1) xx `(Molar mass of KOH)
`=(10^(-2) " mol " L^(-1)) xx ( 56 g " mol"^(-1)) =0.56 g L^(-1)`