A) \[2\times {{10}^{-4}}M\]
B) \[16\times {{10}^{-4}}M\]
C) \[8\times {{10}^{-4}}M\]
D) \[8\times {{10}^{-8}}M\]
Correct Answer: A
Solution :
\[\underset{s}{\mathop{A{{g}_{2}}Cr{{O}_{4}}}}\,\xrightarrow{{}}\underset{2s}{\mathop{2A{{g}^{+}}}}\,+\underset{s}{\mathop{CrO_{4}^{2-}}}\,\] \[{{K}_{sp}}={{(2s)}^{2}}s=4{{s}^{3}}\] \[s={{\left( \frac{{{K}_{sp}}}{4} \right)}^{1/3}}\] \[={{\left( \frac{32\times {{10}^{-12}}}{4} \right)}^{1/3}}\] \[=2\times {{10}^{-4}}M\]
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