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The solubility of $PbF_2$ in water at $25\,^oC$ is $ \sim 10^{-3}\, M$. What is its solubility in $0.05\, M\, NaF$ solution? Assume the latter to be fully ionised.
$1.6\times10^{-6}\, M$
$1.2\times10^{-6}\, M$
$1.2\times10^{-5}\, M$
$1.6\times10^{-4}\, M$
Solution
Solubility of $Pb{F_2} \approx {10^{ – 3}}\,M$
$\therefore \,{K_{sp}} = 4{S^3} = 4 \times {10^{ – 9}}$
In $0.05\,M\,NaF$ we have $0.05\,M$ of $F^-$ ion contributed by $NaF.$ If the solubility of $PbF_2$
in this solution is $S\,M$, then
total $[{F^ – }] = [2S + 0.05]\,M$
$\therefore S{[2S + 0.05]^2} = 4 \times {10^{ – 9}}$
Assuming $2S < < 0.05,$
$S \times 25 \times {10^{ – 4}} = 4 \times {10^{ – 9}}$
$\therefore \,S \times 0.16 \times {10^{ – 5}}\,M \Rightarrow 1.6 \times {10^{ – 6}}\,M$
We observe that our approximation that $2S < < 0.05$ is justified.
