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), atomicmass of Pb = 207 u).

Calculate, volume of water required to dissolve 0.1 g lead (II) chlori

1.	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), atomicmass of Pb = 207 u).
Answer» Solution :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 .