Equilibrium Question 788
Question: lf $ K _{sp} $ of $ CaF_2 $ at $ 25{}^\circ C $ is $ 1.7\times {10^{-10}} $ , the combination amongst the following which gives a precipitate of $ CaF_2 $ is
Options:
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^{-}} $
Show Answer
Answer:
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. $ CaF_2\rightarrow 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}} ) $
