A) \[1\times {{10}^{-2}}M\,C{{a}^{2+}}\]and\[1\times {{10}^{-3}}M\,{{F}^{-}}\]
B) \[1\times {{10}^{-4}}M\,C{{a}^{2+}}\]and\[1\times {{10}^{-4}}M\,{{F}^{-}}\]
C) \[1\times {{10}^{-2}}M\,C{{a}^{2+}}\]and\[1\times {{10}^{-5}}M\,{{F}^{-}}\]
D) \[1\times {{10}^{-3}}M\,C{{a}^{2+}}\]and\[1\times {{10}^{-5}}M\,{{F}^{-}}\]
Correct Answer: A
Solution :
When ionic product i.e. the product of the concentration of ions in the solution exceeds the value of solubility, product, formation of precitpiate occurs. \[Ca{{F}_{2}}C{{a}^{2+}}+2{{F}^{-}}\] Ionic product \[=[C{{a}^{2+}}]{{[{{F}^{-}}]}^{2}}\] when,\[[C{{a}^{2+}}]=1\times {{10}^{-2}}M\] \[{{[{{F}^{-}}]}^{2}}={{(1\times {{10}^{-3}})}^{2}}M\] \[=1\times {{10}^{-6}}M\] \[\therefore \]\[[C{{a}^{2+}}]{{[{{F}^{-}}]}^{2}}=(1\times {{10}^{-2}})(1\times {{10}^{-6}})=1\times {{10}^{-8}}\] In this case, Ionic product \[(1\times {{10}^{-8}})>\]solubility product \[(1.7\times {{10}^{-10}})\] \[\therefore \]Hence (a) is correct option.
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