Question

The solubility product of PbCl₂ at 298K is 1.7 × 10^-5. Calculate the solubility of PbCl₂ in g/lit. at 298K. 
Atomic Weights: [Pb = 207 and Cl = 35.5]

Answer 1

\[\ce{\underset{\text{(s)}}{PbCl2} ⇌[water] \underset{\text{(aq)(S)}}{Pb^2+} + \underset{\text{(aq)(S)}}{2Cl-}}\]

S = Solubility in mol L^-1

`"K"_"sp" = "S" xx "S"`

`"K"_"sp" = "S"^2`

`1.7 xx 10^-5 = "S"^2`

`"S"^2 = 1.7 xx 10^-5`

= `17 xx 10^-6`

`"S" = sqrt(17) xx 10^-3  "mol"  "L"^-1`

= `4.12 xx 10^-3` mol L^-1

(Mol. Wt. of PbCl₂ = 207 + 71) = 278

∴ S = `278 xx 4.12 xx 10^-3`

S = 1.145 g L^-1