A) \[1\times {{10}^{-2}}MC{{a}^{2+}}\] and \[1\times {{10}^{-3}}M{{F}^{-}}\]
B) \[1\times {{10}^{-4}}MC{{a}^{2+}}\] and \[1\times {{10}^{-4}}M{{F}^{-}}\]
C) \[1\times {{10}^{-2}}MC{{a}^{2+}}\] and \[1\times {{10}^{-5}}M{{F}^{-}}\]
D) \[1\times {{10}^{-3}}MC{{a}^{2+}}\] and \[1\times {{10}^{-5}}M{{F}^{-}}\]
Correct Answer: A
Solution :
[a] 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. \[Ca{{F}_{2}}\rightleftharpoons C{{a}^{2+}}+2{{F}^{-}}\] Ionic product \[=\left[ C{{a}^{2+}} \right]\text{ }{{\left[ {{F}^{-}} \right]}^{2}}\] When, \[[C{{a}^{2+}}]=1\times {{10}^{-2}}M\] \[{{[{{F}^{-}}]}^{2}}={{(1\times {{10}^{-3}})}^{2}}M\] \[=1\times {{10}^{-6}}M\] \[\therefore \left[ C{{a}^{2+}} \right]{{\left[ {{F}^{-}} \right]}^{2}}=(1\times {{10}^{-2}})(1\times {{10}^{-6}})=1\times {{10}^{-8}}\] In this case, Ionic product \[\left( 1\times {{10}^{-8}} \right)>\] solubility product \[\left( 1.7\times {{10}^{-10}} \right)\]
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