The `K_(SP)` for `Cr(OH)_(3)` is `1.6xx10^(-30)`. The molar solubility of this compound in water is

Options

A. `2sqrt(1.6xx10^(-30))`
B. `4sqrt(1.6xx10^(-30))`
C. `4sqrt((1.6xx10^(-30))/(27))`
D. `1.6xx10^(-30)//27`

Answer

C

Explanation

Let molar solubility of `Cr(OH)_(3)=s mol L^(-1)`
`Cr(OH)_(3)(s)hArrunderset(s)(Cr^(3+)(aq)+underset(3s)(3OH^(-)(aq))`
`K_(SP)=1.6xx10^(-30)`
`=[Cr^(3+)][OH^(-)]^(3)=(s)(3s)^(3)=27s^(4)`
`s^(4)=(1.6xx10^(-30))/(27)`
`s=4sqrt((1.6xx10^(-30))/(27))`