A) \[32\times {{10}^{-12}}\]
B) \[8\times {{10}^{-12}}\]
C) \[16\times {{10}^{-12}}\]
D) \[8\times {{10}^{-8}}\]
Correct Answer: A
Solution :
\[A{{g}_{2}}Cr{{O}_{4}}(s)\underset{2s}{\mathop{2A{{g}^{+}}}}\,+CrO_{4}^{2-}\] Given, \[[CrO_{4}^{2-}]=s=2\times {{10}^{-4}}\] \[\therefore \] \[[A{{g}^{+}}]=2s=2\times 2\times {{10}^{-4}}\] \[=4\times {{10}^{-4}}\] \[{{K}_{sp}}={{[A{{g}^{+}}]}^{2}}[CrO_{4}^{2-}]\] \[={{[4\times {{10}^{-4}}]}^{2}}[2\times {{10}^{-4}}]\] \[=32\times {{10}^{-12}}\]
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