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}} ) $



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