A) \[8\times {{10}^{-11}}M\]
B) \[8\times {{10}^{-10}}M\]
C) \[8\times {{10}^{-13}}M\]
D) \[8\times {{10}^{-12}}M\]
Correct Answer: B
Solution :
Let the solubility of \[A{{g}_{2}}C{{O}_{3}}\]be s. Then, \[A{{g}_{2}}C{{O}_{3}}\underset{2s}{\mathop{2A{{g}^{+}}}}\,+\underset{s}{\mathop{CO_{3}^{2-}}}\,\] \[AgN{{O}_{3}}\underset{0.1}{\mathop{A{{g}^{{}}}}}\,+\underset{0.1}{\mathop{NO_{3}^{-}}}\,\] \[[A{{g}^{+}}]=(2s+0.1)\] \[[CO_{3}^{2-}]=s\] \[{{K}_{sP}}={{[A{{g}^{+}}]}^{2}}[CO_{3}^{2-}]={{(2s+0.1)}^{2}}(s)\] \[=s\times (4{{s}^{2}}+0.01+0.4s)=4{{s}^{3}}+0.01s+0.4{{s}^{2}}\] Neglecting\[{{s}^{3}}\]and\[{{s}^{2}},\]we get\[8\times {{10}^{-12}}=0.01s\] \[s=\frac{8\times {{10}^{-12}}}{0.01}=8\times {{10}^{-10}}M\]You need to login to perform this action.
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