If x mol `L^(-1)` is the solubility of `Kal(SO_(4))_(2)` , then `K_(sp)` is equal to

Options

A. `x^(3)`
B. `4x^(4)`
C. `x^(4)`
D. `4x^(3)`

Answer

B

Explanation

`KAl(SO_(4))_(2)hArr K^(+)+Al^(3+)+2SO_(4)^(2-)`
If x is solubility in mol `L^(-1)`, then `[K^(+)]=x, [Al^(3+)]=x,[SO_(4)^(2-)]=2x`
`K_(sp)=[K^(+)][Al^(3+)][SO_(4)^(2-)]=x xx x xx (2x)^(2)=4x^(4)`