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The K(sp) for Cr(OH)(3) is 1.6xx10^(-30...

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

A

`root(3)(1.6xx10^(-30))`

B

`root(4)(1.6xx10^(-30)//27)`

C

`1.6xx10^(-30)//27`

D

`root(2)(1.6xx10^(-30))`

Text Solution

Verified by Experts

The correct Answer is:
B
If s is the solubility of `Cr(OH)_(3)`
`Cr(OH)_(s)hArr underset(s)(Cr^(3+))+underset(3s)(3OH^(-))`
`K_(sp)=(s)(3s)^(3)=27s^(4)`
`therefore 27s^(4)=1.6xx10^(-30)`
`s= root(4)(1.6xx10^(-30)//27)`
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