A) AgCl
B) AgBr
C) \[A{{g}_{2}}Cr{{O}_{4}}\]
D) Any of these
Correct Answer: C
Solution :
For binary salts (like AgCl, AgBr) \[AgClA{{g}^{+}}+C{{\underset{S}{\mathop{l}}\,}^{-}}\] \[{{K}_{sp}}=[A{{g}^{+}}+C{{\underset{S}{\mathop{l}}\,}^{-}}\] \[=(s)\,\,(s)\] \[{{K}_{sp}}={{s}^{2}}\] \[s=\sqrt{{{K}_{sp}}}\] \[\therefore \] Solubility of \[AgCl=\sqrt{1.8\times {{10}^{-10}}}\] \[=1.35\times {{10}^{-5}}mol/L\] Solubility of AgBr \[=\sqrt{5.0\times {{10}^{-13}}}\] \[=7.1\times {{10}^{-7}}mol/L\] For ternary salt-like \[A{{g}_{2}}Cr{{O}_{4}}\] \[{{K}_{sp}}4{{s}^{3}}\] \[s={{\left[ \frac{{{K}_{sp}}}{4} \right]}^{1/3}}\] \[\therefore \] Solubility of \[A{{g}_{2}}Cr{{O}_{4}}=\sqrt[3]{\frac{{{K}_{sp}}}{4}}\] \[=\sqrt[3]{\frac{2.4\times {{10}^{-12}}}{4}}\] \[=\sqrt[3]{600\times {{10}^{-15}}}\] \[=8.44\times {{10}^{-5}}mol/L\] The solubility of \[A{{g}_{2}}Cr{{O}_{4}}\] is maximum so, it will give maximum \[A{{g}^{+}}\] ions in solution. Hence, it will be used.
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