A) \[6.23\times {{10}^{-11}}M\]
B) \[1.84\times {{10}^{-9}}M\]
C) \[\frac{2.49}{1.84}\times {{10}^{-9}}M\]
D) \[2.49\times {{10}^{-10}}M\]
Correct Answer: D
Solution :
\[{{K}_{sp}}=4{{(s)}^{3}}\] \[=4\times {{(1.84\times {{10}^{-5}})}^{3}}\] \[Cd{{(OH)}_{2}}\rightleftharpoons C{{d}^{+2}}+2O{{H}^{-}}\] \[S'\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,S'\,\,\,\,\,\,\,({{10}^{-2}}+S')\simeq {{10}^{-2}}\] \[S'\times {{({{10}^{-2}})}^{2}}=4\times {{(1.84\times {{10}^{-5}})}^{3}}\] \[S'=4\times {{(1.84)}^{3}}\times {{10}^{-11}}\] \[(S')=2.491\times {{10}^{-10}}M\]
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