A) 0.11
B) 99.9
C) 0.01
D) 9.99
Correct Answer: B
Solution :
[b] To precipitate the \[AgCl\] \[[A{{g}^{+}}]\] required \[=\frac{{{K}_{sp}}(AgCl)}{[C{{l}^{-}}]}=\frac{1.0\times {{10}^{-10}}}{0.1}=1.0\times {{10}^{-9}}M\] \[\left[ B{{r}^{-}} \right]\] left at this stage \[=\frac{{{K}_{sp}}(AgBr)}{[A{{g}^{+}}]}\] \[=\frac{1.0\times {{10}^{-13}}}{1.0\times {{10}^{-9}}}=1.0\times {{10}^{-4}}M\] % of remaining \[[B{{r}^{-}}]=\frac{1.0\times {{10}^{-4}}}{0.1}\times 100=0.1\] % of \[B{{r}^{-}}\] to be precipitated \[=100-0.1=99.9\]You need to login to perform this action.
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