A) \[4\times {{10}^{-15}}\]
B) \[4\times {{10}^{-10}}\]
C) \[1\times {{10}^{-15}}\]
D) \[1\times {{10}^{-10}}\]
Correct Answer: A
Solution :
Let the solubility\[=x\] \[x=1.0\times {{10}^{-5}}mol\text{ }{{L}^{-1}}\] (given) \[A{{B}_{2}}\underset{x}{\mathop{{{A}^{2+}}}}\,+\underset{2x}{\mathop{2{{B}^{-}}}}\,\] \[{{K}_{sp}}=[{{A}^{2+}}]{{[{{B}^{-}}]}^{2}}\] \[{{K}_{sp}}=x\times {{(2x)}^{2}}\] \[=4{{x}^{3}}\] \[{{K}_{sp}}=4\times {{(1\times {{10}^{-5}})}^{3}}\] \[=4\times {{10}^{-15}}\]
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