A) \[3.3750\times {{10}^{-12}}\]
B) \[1.6875\times {{10}^{-10}}\]
C) \[1.6875\times {{10}^{-12}}\]
D) \[1.6875\times {{10}^{-11}}\]
Correct Answer: C
Solution :
\[AgCr{{O}_{4}}\]⇌ \[\underset{{{(2S)}^{2}}}{\mathop{2A{{g}^{+}}}}\,+\underset{S\,\,\,}{\mathop{CrO_{4}^{-}}}\,\] \[{{K}_{sp}}=4{{S}^{3}}\] given \[2S=1.5\times {{10}^{-4}}\] \[\therefore \,{{K}_{sp}}={{(2S)}^{2}}\times S\] \[={{(1.5\times {{10}^{-4}})}^{2}}\times \left( \frac{1.5\times {{10}^{-4}}}{2} \right)\] \[=1.6875\times {{10}^{-12}}\]You need to login to perform this action.
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