Calculate pH of basic solution having `OH^(-) ` as ( i) `10^(-2)` M and (ii) `10^(-8) M`

Explanation

(i) `10^(-2)` M `OH^(-)`
`OH^(-) = 10^(-2) , POH = - log [ OH ] = 2 `
PH = 14-2 = 12
(ii) `10^(-8) M OH^(-)`
In presence of `10^(-8) M OH^(-)` , water dissociates as
`H_(2) O rarr H^(+) + OH^(-)`
`( c-x ) ` x `( 10^(-8)+ x )`
`K_( w ) = [ H^(+)] [OH^(-) ] = ( 10^(-8) + x ) = 10^(-14)`
Which reduce to quadratic equation
`x = ( - 10^(-8) +- sqrt( 10^(-8) + 4 xx 10^(-14)))/( 2)`
`:. x = 0.95 xx 10^(-7) mol //L`
`= [OH ] = ( 10^(-8) +x ) = 10^(-8) + 0.95 xx 10^(-7)`
`= 1.05 xx 10^(-7)`
POH `= - log [OH] = - log ( 1.05 xx 10^(-7)) = 6.9788 `
`PH = 14 - POH - 6.9788 = 7.0212`