A) \[AgCl\]
B) \[AgBr\]
C) \[AgCr{{O}_{4}}\]
D) None of these
Correct Answer: C
Solution :
For binary salts (like \[AgCl,\,AgBr\]), \[s=\sqrt{{{K}_{sp}}}\] \[\therefore \] Solubility of \[AgCl=\sqrt{1.8\times {{10}^{-10}}}\] \[=1.35\times {{10}^{-7}}mol/L\] Solubility of \[AgBr=\sqrt{5.0\times {{10}^{-13}}}\] \[=7.1\times {{10}^{-7}}mol/L\] For, \[A{{g}_{2}}Cr{{O}_{4}},\,{{K}_{sp}}=4{{s}^{3}}\] \[\therefore \] Solubility of \[A{{g}_{2}}Cr{{O}_{4}}\] \[=\sqrt[3]{\frac{{{K}_{sp}}}{4}}=\sqrt[3]{\frac{2.4\times {{10}^{-12}}}{4}}\] \[=\sqrt[3]{600\times {{10}^{-15}}}\] \[=8.44\times {{10}^{-15}}mol/L\] As \[A{{g}_{2}}Cr{{O}_{4}}\] has maximum solubility, it will give maximum \[A{{g}^{+}}\] ions in solution. Hence, it will be used.
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