A) \[2\times {{10}^{-4}}\]
B) \[8\times {{10}^{-4}}\]
C) \[8\times {{10}^{-8}}\]
D) \[16\times {{10}^{-4}}\]
Correct Answer: A
Solution :
Key Idea: First find relationship between solubility product and solubility after writing dissociation reaction of\[A{{g}_{2}}Cr{{O}_{4}}\]. Then substitute the given values to find the answer. Let the solubility of \[A{{g}_{2}}Cr{{O}_{4}}=x\,g/L\] \[A{{g}_{2}}Cr{{O}_{4}}\to 2A{{g}^{+}}+C{{r}_{2}}O_{4}^{2-}\] Conc. \[x\] \[2x\] \[x\] \[{{K}_{sp}}=[A{{g}^{2+}}][Cr{{O}_{4}}^{2-}]\] \[={{(2x)}^{2}}(x)\] or \[{{K}_{sp}}=4{{x}^{3}}\] \[\therefore \] \[x=\sqrt[3]{\frac{{{K}_{sp}}}{4}}\] Given, \[{{K}_{sp}}=32\times {{10}^{-12}}\] \[\therefore \] \[x=\sqrt[3]{\frac{32\times {{10}^{-12}}}{4}}\]
You need to login to perform this action.
You will be redirected in
3 sec
