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6-2.Equilibrium-II (Ionic Equilibrium)
medium
The solubility product of $PbC{l_2}$ at ${20\,^o}C$ is $1.5 \times {10^{ - 4}}.$ Calculate the solubility
A
$3.75 \times {10^{ - 4}}$
B
$3.34 \times {10^{ - 2}}$
C
$3.34 \times {10^2}$
D
None of these
Solution
(b) $\mathop {PbC{l_2}}\limits_S ⇌\mathop {P{b^{ + 2 }}}\limits_{S\,\,\,} + \mathop {2C{l^ – }}\limits_{{{(2S)}^2}} $
${K_{sp}} = S \times {(2S)^2} = 4{S^3}$
$S = \sqrt[3]{{\frac{{{K_{sp}}}}{4}}} = \sqrt[3]{{\frac{{1.5 \times {{10}^{ – 4}}}}{4}}} = 3.34 \times {10^{ – 2}}$.
Standard 11
Chemistry
