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The solubility of $AgCl$ in $0.2\,\,M\,\,NaCl$ solution $({K_{sp}}$ for $AgCl = 1.20 \times {10^{ - 10}})$ is
$0.2\,\,M$
$1.2 \times {10^{ - 10}}\,M$
$0.2 \times {10^{ - 10}}\,M$
$6\times {10^{ - 10}}\,M$
Solution
(d) $\mathop {AgCl}\limits_a \rightleftharpoons \mathop {A{g^ + }}\limits_a + \mathop {C{l^ – }}\limits_a $
$\mathop {NaCl}\limits_{0.02} $ $ \rightleftharpoons $ $\mathop {N{a^ + }}\limits_{0.02\,\,\,\,\,} \,\, + \mathop {C{l^ – }}\limits_{0.02\,\,\,\,\,\,} $
${K_{sp}}\,\,AgCl = 1.20 \times {10^{ – 10}}$
${K_{sp}}\,\,AgCl = [A{g^ + }]\,\,[C{l^ – }]$$ = a \times [a + 0.2]$ $ = {a^2} + 0.2a$
${a^2}$ is a very small so it is a neglected.
${K_{sp}}\,\,AgCl = 0.2a$
$1.20 \times {10^{ – 10}} = 0.2a$
$a = \frac{{1.20 \times {{10}^{ – 10}}}}{{0.20}} = 6 \times {10^{ – 10}}$ $mole$
