A) \[1.2\times {{10}^{-10}}g\]
B) \[1.2\times {{10}^{-9}}g\]
C) \[6.2\times {{10}^{-5}}g\]
D) \[5.0\times {{10}^{-8}}g\]
Correct Answer: B
Solution :
[b] \[[AgBr]=[A{{g}^{+}}]=0.05M\] \[{{K}_{sp}}[AgBr]=[A{{g}^{+}}][B{{r}^{-}}]\] \[\Rightarrow [B{{r}^{-}}]=\frac{{{K}_{sp}}[AgBr]}{[A{{g}^{+}}]}\] \[=\frac{5.0\times {{10}^{-13}}}{0.05}={{10}^{-11}}M[mol{{L}^{-1}}]\] Moles of KBr needed to precipitate AgBr \[=[B{{r}^{-}}]\times V={{10}^{-11}}mol\,{{L}^{-1}}\times 1\,L={{10}^{-11}}mol\] Therefore, amount of KBr needed to precipitate AgBr \[={{10}^{-11}}mol\times 120gmo{{l}^{-1}}=1.2\times {{10}^{-9}}g\]
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