A) \[Agl\]
B) \[AgCl\]
C) \[AgBr\]
D) \[A{{g}_{2}}Cr{{O}_{4}}\]
Correct Answer: D
Solution :
\[A{{g}_{2}}Cr{{O}_{4}}2A{{g}^{+}}+CrO_{4}^{2-}\] Solubility product \[{{K}_{sp}}={{(2s)}^{2}}\times s=4{{s}^{3}}\] \[{{K}_{sp}}=(1.1\times {{10}^{-12}})\] \[S=\sqrt[3]{\frac{{{K}_{sp}}}{4}}\,=0.65\times {{10}^{-4}}\] \[AgClA{{g}^{+}}+C{{l}^{-}}\] \[{{K}_{sp}}=S\times S\] \[({{K}_{sp}}=1.8\times {{10}^{-10}})\] \[S=\sqrt{{{K}_{sp}}}\,=1.34\times {{10}^{-5}}\] \[AgBr\,A{{g}^{+}}+B{{r}^{-}}\] \[{{K}_{sp}}=S\times S\] \[({{K}_{sp}}=5\times {{10}^{-13}})\] \[S=\sqrt{{{K}_{sp}}}=0.71\times {{10}^{-6}}\] \[AgIA{{g}^{+}}+{{I}^{-}}\] \[{{K}_{sp}}=S\times S\] \[({{K}_{sp}}=8.3\times {{10}^{-17}})\] \[S=\sqrt{{{K}_{sp}}}=0.9\times {{10}^{-8}}\] \[\because \] Solubility of \[A{{g}_{2}}Cr{{O}_{4}}\] is highest. So, it will precipitate last.
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