1.

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

Answer»

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

Solution :`Cr(OH)_(s)(s) hArr Cr^(3+)(aq)underset(S)+3OH^(-)underset(3S)(aq)`
`(s) (3s)^(3)=K_(SP)`
`27S^(4)=K_(sp)`
`s=((K_(sp))/(27))^(1//4)=((1.6xx10^(-30))/(27))^(1//4)`


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