• # question_answer The solubility of $AgCl$ in $0.2\,\,M\,\,NaCl$ solution $({{K}_{sp}}$for $AgCl=1.20\times {{10}^{-10}})$ is   [MP PET 1996] A)                 $0.2\,\,M$        B)                 $1.2\times {{10}^{-10}}\,M$ C)                 $0.2\times {{10}^{-10}}\,M$     D)                 $0.2\times {{10}^{-10}}\,M$

$\underset{a}{\mathop{AgCl}}\,$⇌$\underset{a}{\mathop{A{{g}^{+}}}}\,+\underset{a}{\mathop{C{{l}^{-}}}}\,$                    $\underset{0.02}{\mathop{NaCl}}\,$⇌ $\underset{0.02\,\,\,\,\,}{\mathop{N{{a}^{+}}}}\,\,\,+\underset{0.02\,\,\,\,\,\,}{\mathop{C{{l}^{-}}}}\,$                    ${{K}_{sp}}\,\,AgCl=1.20\times {{10}^{-10}}$                    ${{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