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 :
\[\underset{S}{\mathop{A{{g}_{2}}Cu{{O}_{4}}}}\,\xrightarrow{{}}\underset{2S}{\mathop{2A{{g}^{+}}}}\,+\underset{S}{\mathop{CrO_{4}^{2-}}}\,\] \[\therefore \] \[{{K}_{sp}}={{(2S)}^{2}}\times S=4{{S}^{3}}\] or \[S={{\left( \frac{{{K}_{sp}}}{4} \right)}^{1/3}}={{\left( \frac{32\times {{10}^{-12}}}{4} \right)}^{1/3}}\] \[S=2\times {{10}^{-4}}M\]
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