A) \[1\times {{10}^{-10}}\]
B) \[1\times {{10}^{-5}}\]
C) \[1\times {{10}^{-9}}\]
D) \[1\times {{10}^{-4}}\]
Correct Answer: C
Solution :
\[{{K}_{sp}}\] of \[AgCl={{(solubility\text{ }of\text{ }AgCl)}^{2}}\] \[={{(1\times {{10}^{-5}})}^{2}}=1\times {{10}^{-10}}\] Suppose, its solubility in 0.1 M NaCI is x mol/L. \[AgCl\underset{x}{\mathop{\overset{+}{\mathop{Ag}}\,}}\,+\underset{x}{\mathop{C{{l}^{-}}}}\,\] \[NaCl\underset{0.1M}{\mathop{N{{a}^{+}}}}\,+\underset{0.1M}{\mathop{C{{l}^{-}}}}\,\] \[[C{{l}^{-}}]=(x+0.1)M\] \[{{K}_{sp}}\] or \[AgCl=[A{{g}^{+}}][C{{l}^{-}}]\] \[=x\times (x+0.1)\] \[1\times {{10}^{-10}}={{x}^{2}}+0.1x\] Higher power of x can be neglected. \[1\times {{10}^{-10}}=0.1x\] \[x=1\times {{10}^{-9}}M\]
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