Solubility product of Mg(OH)2 at ordinary temperature is `1.96xx10^(-11)`, pH of a saturated solution of `Mg(OH)_(2)` will be :

Options

A. `10.50`
B. `8.47`
C. `6.94`
D. `3.47`

Answer

A

Explanation

`"Mg"(OH)_(2(s))hArr underset (x)(Mg^(2+)) +underset (2x) (2OH^(-))`
`K_(sp) = [Mg^(2+) ] [OH^(-)]^(2) = 1.96 xx 10^(-11) x xx (2x)^(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 = ` 2xx 1.6 xx 10^(-4)`
i.e `[OH^(-)] = 3.2 xx 10^(-4)`
Now, pOH = `-log[OH^(-)]`
`= - log [ 3.2 xx 10^(-4)] = 4 - 0.505 = 3.495`
` :. pH = 14 - 3 . 495 = 10.505`