Calculate, volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. [`K_(sp)` of `PbCl_2 = 3.2 xx 10^(-8)`, atomic mass of Pb = 207 u).

Explanation

Suppose , solubility of `PbCl_2` in water is s mol `L^(-1)`
`{:(PbCl_((s)) hArr, Pb_((aq))^(2+) +, 2Cl_((aq))^(-)),((1-s), s,2s):}`
`K_(sp)=[Pb^(2+)]. [Cl^-]^2`
`K_(sp)=[s] [2s]^2=4s^3`
`32xx10^(-8)=4s^3`
`s^3=(3.2+10^(-8))/4=0.8xx10^(-8)`
`s^3=8.0xx10^(-9)`
Solubility of `PbCl_2 , s= 2xx10^(-3) "mol L"^(-1)`
Solubility of `PbCl_2` in g `L^(-1)=278xx2xx10^(-3)`
=0.556 g `L^(-1)`
(` because` Molar mass of `PbCl_2=207+(2xx35.5)`=278)
0.556 g of `PbCl_2` dissolve in 1L of water
`therefore` 0.1 g of `PbCl_2` will dissolve in `=(1xx1.01)/0.556` L of water
= 0.1798 L
To make a saturated solution, dissolution of 0.19 `PbCl_2` in 0.1798 L =0.2 L of water will be required .