Correct Answer - A
`{:("Mg(OH)"_(2(s)),"Mg"^(2+),+,2OH^(-)),(,x,,2x):}`
`K_(sp) = [Mg]^(2+) [ OH^(-)]^(2) = 1.96 xx 10^(-11)`
`x xx (2 x)^(2) = 1.96 xx 10^(-11)` (concentration of solid is unity)
`4x^(3) = 1.96 xx 10^(-11)`
`x = ((1.96 xx 10^(-11))/(4))^(1//3)`
`x = (4.9 xx 10^(-12))^(1//3) = 1.6 xx 10^(-4)`
So, `OH^(-)` concentration `= 2 xx 1.6 xx 10^(-4)`
i.e. `[OH^(-)] = 3.2 xx 10^(-4)`
`= -log [3.2 xx 10^(-4)] = 4- 0.505 = 3.495`
`:. pH = 14 - 3.495 = 10.505`.
