Calculate the solubility of `PbCI_(2)` in grams `//`lite if the solubility product of

Answer

`5,56 g L^(-1)`

Explanation

Let the solubility of `PbCI_(2) =S " moles " L^(-1)`
The solubility product of `PbCI_(2) = 3.2 xx 10^(-5)`
the solubility equilibrium of `PbCI_(2)` may be represented as :
`PbCI_(2) overset((aq))(hArr) underset(S)(Pb^(2+) (aq)) +underset(2S)(2CI^(-) (aq))`
`K_(sp) =[Pb^(2+)(aq)][CI_(-) (aq)]^(2) =S xx (2s)^(2) =4S^(3)`
`" or " " "S =((K_(sp))/(4))^(1//3) =((3.2 xx 10^(-5))/(4))^(1//3) =(8xx 10^(-6))^(1//3) =2xx 10^(-2) " moles "L^(-1)`
`:.` The solubility of `PbCI_(2) =2 xx 10^(-2) " moles" L^(-1)`
`=2 xx 278 xx 10^(-2) =5.56 g L^(-1)" "(":. Molar mass of " PbCI_(2) = 278)`