A) \[1.0\times {{10}^{-9}}\]
B) \[1.0\times {{10}^{-10}}\]
C) \[1.0\times {{10}^{-5}}\]
D) \[1.0\times {{10}^{-11}}\]
Correct Answer: A
Solution :
Let solubility of\[AgCl=xmole\text{/}L\] \[AgClA{{g}^{+}}+C{{l}^{-}}\]i.e.,\[{{K}_{sp}}_{(AgCl)}=x\times x\] \[KCl\xrightarrow[{}]{{}}{{K}^{+}}+C{{l}^{-}}0.1\] \[[C{{l}^{-}}]\]from \[KCl=0.1m\] Total\[[C{{l}^{-}}]\]in solution\[=x+0.1\] \[{{K}_{sp}}(AgCl)=[A{{g}^{+}}][C{{l}^{-}}]=x(x+0.1)\] \[1.0\times {{10}^{-10}}=x(x+0.1)\] \[1.0\times {{10}^{-10}}={{x}^{2}}+0.1x\] \[1.0\times {{10}^{-10}}=0.1\] \[(as\,{{x}^{2}}<<1)\] \[x=1.0\times {{10}^{-9}}mol/L\]
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