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6-2.Equilibrium-II (Ionic Equilibrium)
hard
The solubility (in $mol\,L^{-1}$ ) of $AgCl\,\,(K_{sp} = 1.0 \times 10^{-10}$ ) in a $0.1\,M\,KCl$ solution will be
A
$1.0\times 10^{-9}$
B
$1.0\times 10^{-10}$
C
$1.0\times 10^{-5}$
D
$1.0\times 10^{-11}$
(AIEEE-2012)
Solution
Let solubility of $\bar AgCl = x\,mole/L$
$AgCl \rightleftharpoons A{g^ + } + C{l^ – }$
i.e., ${K_{sp(AgCl)}} = x \times x$
$KCl \to {K^ + } + \mathop {C{l^ – }}\limits_{0.1} $
$[C{l^ – }]\,$ from $KCl = 0.1\,m$
Total $[C{l^ – }]$ in solution $ = x + 0.1$
${K_{sp}}(AgCl) = [A{g^ + }][C{l^ – }] = x(x + 0.1)$
$1.0 \times {10^{ – 10}} = x(x + 0.1)$
$1.0 \times {10^{ – 10}} = {x^2} + 0.1x$
$1.0 \times {10^{ – 10}} = 0.1x$ (as ${x^2} < < 1$)
$x = 1.0 \times {10^{ – 9}}\,mol/L$
Standard 11
Chemistry
