A) \[5.6\times {{10}^{-6}}\]
B) \[3.1\times {{10}^{-4}}\]
C) \[2\times {{10}^{-4}}\]
D) \[4\times {{10}^{-4}}\]
Correct Answer: C
Solution :
\[A{{X}_{2}}\]is ionised as follows \[\underset{S\,mol\,{{L}^{-1}}}{\mathop{A{{X}_{2}}}}\,\underset{S}{\mathop{{{A}^{2+}}}}\,+\underset{2S}{\mathop{2{{X}^{-}}}}\,\] Solubility product of\[A{{X}_{2}}\] \[({{K}_{sp}})=[{{A}^{2+}}]{{[{{X}^{-}}]}^{2}}=S\times {{(2S)}^{2}}=4{{S}^{3}}\] \[\because \] \[{{K}_{sp}}\]of\[A{{X}_{2}}=3.2\times {{10}^{-11}}\] \[\therefore \] \[3.2\times {{10}^{-11}}=4{{S}^{3}}\] \[{{S}^{3}}=0.8\times {{10}^{-11}}=8\times {{10}^{-12}}\] Solubility\[=2\times {{10}^{-4}}mol/L\]
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