Question

If the $ {{\text{K}}_{\text{sp}}} $ of $ \text{Ca}{{\text{F}}_{\text{2}}} $ at $ \text{2}{{\text{5}}^{\text{0}}}\text{C} $ is $ 1.7\times {{10}^{-10}} $ , the combination amongst the following which gives a precipitate of $ \text{Ca}{{\text{F}}_{\text{2}}} $ is:
(A) $ 1\times {{10}^{-3}}\left( \text{M} \right) $ $ \text{C}{{\text{a}}^{\text{+2}}} $ and $ 1\times {{10}^{-5}}\left( \text{M} \right) $ $ {{\text{F}}^{\text{-}}} $
(B) $ 1\times {{10}^{-4}}\left( \text{M} \right) $ $ \text{C}{{\text{a}}^{\text{+2}}} $ and $ 1\times {{10}^{-4}}\left( \text{M} \right) $ $ {{\text{F}}^{\text{-}}} $
(C) $ 1\times {{10}^{-2}}\left( \text{M} \right) $ $ \text{C}{{\text{a}}^{\text{+2}}} $ and $ 1\times {{10}^{-3}}\left( \text{M} \right) $ $ {{\text{F}}^{\text{-}}} $
(D) $ 1\times {{10}^{-2}}\left( \text{M} \right) $ $ \text{C}{{\text{a}}^{\text{+2}}} $ and $ 1\times {{10}^{-5}}\left( \text{M} \right) $ $ {{\text{F}}^{\text{-}}} $

Answer 1

Hint: The solubility product of a salt is defined the equilibrium constant for the dissociation of the salt when the concentration of the undissociated salt is much higher than the equilibrium concentrations of the cation and the anion of the salt and the concentration of the undissociated salt is considered to be constant.

Complete step by step solution:
The dissociation of calcium fluoride results in the liberation of the calcium $ {\text{C}}{{\text{a}}^{{\text{ + 2}}}} $ cations and $ {{\text{F}}^{\text{ - }}} $ anions according to the following reaction,
 $ \text{Ca}{{\text{F}}_{\text{2}}}\rightleftarrows \text{C}{{\text{a}}^{\text{+2}}}\text{+ }{{\text{F}}^{\text{-}}} $ , the solubility product for the reaction is:
  $ {{\text{K}}_{\text{sp}}} $ = $ \left[ \text{C}{{\text{a}}^{\text{+2}}} \right]{{\left[ {{\text{F}}^{\text{-}}} \right]}^{\text{2}}} $
A salt precipitates out when the product of concentration of the anion and the cation exceeds the solubility product of the salt. So to find out mixing which concentrations of calcium and fluoride will precipitate out the salt, let us multiply the concentration of the cation and the anion given.
 $ \dfrac{\left[ 1\times {{10}^{-3}} \right]}{2}{{\left[ \dfrac{1\times {{10}^{-5}}}{2} \right]}^{2}}=\dfrac{{{10}^{-13}}}{8}=1.25\times {{10}^{-14}} $
 $ \dfrac{\left[ 1\times {{10}^{-4}} \right]}{2}{{\left[ \dfrac{1\times {{10}^{-4}}}{2} \right]}^{2}}=\dfrac{{{10}^{-12}}}{8}=1.25\times {{10}^{-13}} $
 $ \dfrac{\left[ 1\times {{10}^{-2}} \right]}{2}{{\left[ \dfrac{1\times {{10}^{-3}}}{2} \right]}^{2}}=\dfrac{{{10}^{-8}}}{8}=1.25\times {{10}^{-9}} $
 $ \dfrac{\left[ 1\times {{10}^{-2}} \right]}{2}{{\left[ \dfrac{1\times {{10}^{-5}}}{2} \right]}^{2}}=\dfrac{{{10}^{-12}}}{8}=1.25\times {{10}^{-13}} $
From the above calculations it can be seen that the ionic product for the concentrations of the ions in option C exceeds the solubility product and hence it will precipitate in that case.
So, the correct option is C.

Note:
The solubility product of any salt or any substance depends on the temperature of the medium. The higher the temperature of the medium, the greater will be the solubility of the salt and this is due to the increased kinetic energy of the molecules. But this too has some exceptions and there are a lot of substances whose solubility decreases with the increase in the temperature of the medium.