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: \[\underset{S\,mol\,{{L}^{-1}}}{\mathop{A{{X}_{2}}}}\,\rightleftharpoons \underset{S}{\mathop{{{A}^{2+}}}}\,+\underset{2S}{\mathop{2{{X}^{-}}}}\,\] Solubility product of \[A{{X}_{2}},\] \[{{K}_{sp}}=[{{A}^{2+}}]{{[{{X}^{-}}]}^{2}}=S\times (2S)=4{{S}^{3}}\] \[\because \]\[{{K}_{sp}}\]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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