A) \[AgCl\]and\[PbCr{{O}_{4}}\]
B) \[AgI\] and\[A{{g}_{2}}C{{O}_{3}}\]
C) \[AgCl\]and\[A{{g}_{2}}C{{O}_{3}}\]
D) \[A{{g}_{2}}C{{O}_{3}}\]and \[AgI\]
E) \[A{{g}_{2}}C{{O}_{3}}\]and \[PbCr{{O}_{4}}\]
Correct Answer: D
Solution :
(i) \[AgCl(s)\underset{s}{\mathop{A{{g}^{+}}}}\,(aq)+\underset{s}{\mathop{C{{l}^{-}}}}\,(aq)\] \[{{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]\] \[=s\times s={{s}^{2}}\] \[{{K}_{sp}}=1.1\times {{10}^{-10}}\](given) \[\therefore \] \[{{s}^{2}}=1.1\times {{10}^{-10}}\] or \[s=1.04\times {{10}^{-5}}mol\,{{L}^{-1}}\] (ii) \[AgI\underset{s}{\mathop{A{{g}^{+}}}}\,+\underset{s}{\mathop{{{I}^{-}}}}\,\] \[{{K}_{sp}}=[A{{g}^{+}}][{{I}^{-}}]\] \[=s\times s={{s}^{2}}\] \[{{K}_{sp}}={{s}^{2}}=1.0\times {{10}^{-16}}\] or \[s=1.0\times {{10}^{-8}}mol\,{{L}^{-1}}\] (iii) \[PbCr{{O}_{4}}\underset{s}{\mathop{P{{b}^{2+}}}}\,+\underset{s}{\mathop{CrO_{4}^{2-}}}\,\] \[{{K}_{sp}}=[P{{b}^{2+}}][CrO_{4}^{2-}]\] \[=s\times s={{s}^{2}}\] \[{{K}_{sp}}={{s}^{2}}=4.0\times {{10}^{-14}}\] or \[s=2.0\times {{10}^{-7}}mol\text{ }{{L}^{-1}}\] (iv) \[A{{g}_{2}}C{{O}_{3}}\underset{s}{\mathop{2A{{g}^{+}}}}\,+\underset{s}{\mathop{CO_{3}^{2-}}}\,\] \[=4{{s}^{3}}\] \[{{K}_{sp}}=[A{{g}^{+}}][CO_{3}^{2-}]\] \[={{(2s)}^{2}}\times s=4{{s}^{3}}\] \[{{K}_{sp}}=4{{s}^{3}}=8.0\times {{10}^{-12}}\] Or \[s=1.26\times {{10}^{-4}}mol\,{{L}^{-1}}\]
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