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Given K(sp) for Cr(OH)3 is 6 xx 10^(-31)...

Given `K_(sp)` for `Cr(OH)_3` is `6 xx 10^(-31)` then determine `[OH^- ]`. (Neglect the contribution of `OH^-` ions from `H_2O`)

A

`(2.22 xx 10^(-31))^(1//4)`

B

`(4.86 xx 10^(-29))^(1//4)`

C

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

D

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

Text Solution

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The correct Answer is:
C
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