Calculate the volume of water required to dissolve`0.1g` lead (II) chloride to get a saturaed solution `(K_(sp)` of `PbCI_(2) = 3.2 xx 10^(-8)`, atomic mass of `Pb = 207 u)`. Multiply your answer with 10 to get answer.

Explanation

The solubility equilibrium is :
`PbCl_2(s) hArr Pb^(2+)(aq) +2Cl^(-)(aq)`
Let the solubility of `PbCl_2` be s mol `L^(-1)`
`[Pb^(2+)]=s, Cl^(-) =2s`
`K_(sp)=[Pb^(2+)][Cl^-]^2`
`=s xx(2s)^2=4s^3`
`4s^3=3.2xx10^(-8)`
`s^3=(3.2xx10^(-8))/4=8xx10^(-9)`
`therefore s=(8xx10^(-9))^(1//3)=2.0xx10^(-3) "mol L"^(-1)`
Solubility of `PbCl_2` in `gL^(-1) =2.0xx10^(-3)xx278`(Molar mass =278)
`=556xx10^(-3) gL^(-1)`
=0.556 g
To get saturated solution, 0.556 g of `PbCl_2` is dissolved in 1L water.
0.1 g of `PbCl_2` is dissolved in `0.1/0.556`L = 0.1798 L water
`therefore` To make saturated solution , 0.1 g of `PbCl_2` be dissolved in 0.1798 L or `approx` 0.2 L of water .