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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

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

B

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

C

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

D

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

Text Solution

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The correct Answer is:
A
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