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The solubility product of Cr(OH)(3) at 2...

The solubility product of `Cr(OH)_(3)` at 298 K is `6.0xx10^(-31)`. The concentration of hydroxide ions in a saturated solution of `Cr(OH)_(3)` will be :

A

`(2.22xx10^(-31))^(1//4)`

B

`(18xx10^(-31))^(1//2)`

C

`(4.86xx10^(-29))^(1//4)`

D

`(18xx10^(-31))^(1//4)`

Text Solution

Verified by Experts

The correct Answer is:
D
`Cr(OH)_(3) to underset(s)(Cr^(+3))+underset(3s)(3OH^(-))`
`K_(sp)=s.(3s)^(3)`
`implies6xx10^(-31)=27.s^(4)`
`impliess=((6)/(27)xx10^(-31))^(1//4)`
`[OH^(-)]=3s`
`=3xx((6)/(27)xx10^(-31))^(1//4)=(18xx10^(-31))^(1//4)M`
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