The solubility of `PbCl_(2)` is

Options

A. `sqrt(K_(sp))`
B. `(K_(sp))^(1//3)`
C. `(K_(sp)//4)^(1//3)`
D. `(8K_(sp))^(1//2)`

Answer

C

Explanation

`PbCl_(2)hArr Pb^(2+)+2Cl^(-)`,
If solubility is x mol `L^(-1)`
`:. K_(sp)=(x)(2x)^(2)=4x^(3)` or `x=(K_(sp)//4)^(1//3)`