The solubility of `PbCl_(2)` in water is S `molL^(-1)` and its solubility product is `K_(sp)`. The relation between `K_(sp)` and S is represented as `S=root(3)((K_(sp))/(x))`. The value of x is ………….

Answer

4

Explanation

`underset((S))(PbCl_(2))hArrunderset((S))(Pb^(2+))+underset((2S))(2Cl)`
`K_(SP)=[Pb^(2+)][Cl^(-)]^(2)`
`=Sxx(2S)^(2)=4S^(3)`
`:.S=root(3)((K_(sp))/(4))`
`:.` Value of x=4.