A) \[\text{5}.\text{6}\times \text{l}{{0}^{-\text{6}}}\]
B) \[\text{3}.\text{1}\times \text{l}{{0}^{-\text{4}}}\]
C) \[\text{2}\times \text{l}{{0}^{-\text{4}}}\]
D) \[\text{4}\times \text{l}{{0}^{-\text{4}}}\]
Correct Answer: C
Solution :
\[\text{A}{{\text{X}}_{\text{2}}}\] is ionized as follows: \[\begin{align} & A{{X}_{2}}{{A}^{2+}}+2{{X}^{-}} \\ & S\,mol\,{{L}^{-1}}\,\,\,\,\,S\,\,\,\,\,\,\,\,\,\,\,2S \\ \end{align}\] Solubility product of \[A{{X}_{2}}\] \[({{K}_{sp}})=[{{A}^{2+}}]{{[{{X}^{-}}]}^{2}}=S\times {{(25)}^{2}}=4{{S}^{3}}\] \[\because {{K}_{sp}}\]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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