A) \[1.8\times {{10}^{-11}}M\]
B) \[9\times {{10}^{-10}}M\]
C) \[6.5\times {{10}^{-12}}M\]
D) \[5.6\times {{10}^{-11}}M\]
Correct Answer: B
Solution :
Let s be the solubility of \[AgCl\]. \[AgCl\rightleftharpoons \overset{+}{\mathop{Ag}}\,+\underset{s}{\mathop{C{{l}^{-}}}}\,\] \[[C{{l}^{-}}]\] from \[NaCl=0.2\] \[\therefore \] Concentration of \[C{{l}^{-}}=s+0.2\] \[{{K}_{sp}}=(s)\,(s+0.2)\] \[1.8\times {{10}^{-10}}={{s}^{2}}+0.2s\] (s is very small as \[AgCl\] is sparingly soluble in \[{{H}_{2}}O\]thus i. e., \[{{s}^{2}}<<<1\] So, neglecting \[{{s}^{2}}\]) \[1.8\times {{10}^{-10}}=0.2s\] \[s=\frac{1.8\,\times {{10}^{-10}}}{0.2}\,=9.0\,\times {{10}^{-10}}\,M\]
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