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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

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

B

`4sqrt(1.6xx10^(-30))`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
Let molar solubility of `Cr(OH)_(2)= S mol L^(-1)`
`Cr(OH)_(3)(s) to Cr_(s)^(3+) (aq) + 3OH_(3s)^(-)(aq) underset(x to infty)(lim)`
`K_(sp)=1.6xx10^(-30)=[Cr^(3+)][OH^(-3)]=(S)(3S)^(3)=27S^(4)`
`therefore s^(4)=(1.6xx10^(-30))/27 = therefore S = root(4)((1.6xx10^(-30))/27)`
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