Correct Answer - B
Concentration of `NH_(4)OH` solution = 0.001 M
Degree of dissociation = `20 % = (20)/(100) = 0.2`
Concentration of `OH^(-)` i.e., `[OH^(-)] =`
Concentration of solution `xx` degree of dissociation
`= 0.001 xx 0.02 = 2 xx 10^(-4)`
`because [H^(+)] [OH^(-)] = 1 xx 10^(-14)`
`therefore [H^(+)] = (1 xx 10^(-14))/(2 xx 10^(-4)) = (1)/(2) xx 10^(-10)`
Again `, because p H = - log [H^(+)]`
`therefore pH = - log ((1)/(2) xx 10^(-10))`
= `- log (-10 - 0.3010) = 10.3010`
