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6-2.Equilibrium-II (Ionic Equilibrium)
medium
$PbCl_2$ has max. concentration of $1.0\times10^{-3}\, M$ in its saturated aq. solution at $25\,^oC$. Its solubility in $0.1\, M\, NaCl$ solution will be
A
$4\times10^{-7}\, M$
B
$4\times10^{-9}\, M$
C
$2\times10^{-7}\, M$
D
$2\times10^{-9}\, M$
Solution
$\mathrm{PbCl}_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})$
$\therefore \quad \mathrm{K}_{\mathrm{sp}}=4 s^{3} \Rightarrow 4 \times\left(10^{-3}\right)^{3}=4 \times 10^{-9}$
Now in $\mathrm{NaCl}$ solution.
$\mathrm{K}_{\mathrm{sp}}\left(\mathrm{PbCl}_{2}\right)=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}$
$\Rightarrow 4 \times 10^{-9}=s \times(0.1)^{2}$
$\Rightarrow \mathrm{s}=4 \times 10^{-7} \,\mathrm{M}$
Standard 11
Chemistry
