1.

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

Answer»

`root(3)(1.6xx10^(-30))`
`root(4)(1.6xx10^(-30)//27)`
`1.6xx10^(-30)//27`
`root(2)(1.6xx10^(-30))`

Solution :If s is the SOLUBILITY of `CR(OH)_(3)`
`Cr(OH)_(s)hArr UNDERSET(s)(Cr^(3+))+underset(3S)(3OH^(-))`
`K_(sp)=(s)(3s)^(3)=27s^(4)`
`therefore 27s^(4)=1.6xx10^(-30)`
`s= root(4)(1.6xx10^(-30)//27)`


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