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