A) \[8\times {{10}^{-8}}mol\,{{L}^{-1}}\]
B) \[16\times {{10}^{-4}}mol\,{{L}^{-1}}\]
C) \[2\times {{10}^{-4}}mol\,{{L}^{-1}}\]
D) \[8\times {{10}^{-4}}mol\,{{L}^{-1}}\]
Correct Answer: C
Solution :
Key Idea: First write the dissociation reaction for \[A{{g}_{2}}Cr{{O}_{4}}\]. Then find relationship between concentration of ions and solubility product. Now do the calculation to find the concentration of \[{{[Cr{{O}_{4}}]}^{2-}}\]. Let the solubility of \[A{{g}_{2}}Cr{{O}_{4}}=x\,mol/L\] \[A{{g}_{2}}Cr{{O}_{4}}\to 2A{{g}^{+}}Cr{{O}_{4}}^{2-}\] ions \[2x\] \[x\] \[\therefore \] \[{{K}_{sp}}={{[A{{g}^{+}}]}^{2}}[Cr{{O}_{4}}^{-2}]\] \[={{(2x)}^{2}}(x)\] or \[{{K}_{sp}}=4{{x}^{3}}\] or \[x=\sqrt[3]{\frac{{{K}_{sp}}}{4}}\] Given, \[{{K}_{sp}}=\sqrt[3]{\frac{32\times {{10}^{-2}}}{4}}\] \[=\sqrt[3]{8\times {{10}^{-12}}}\] \[=2\times {{10}^{-4}}mol/L\]
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