Question
If $${K_{sp}}$$ of $$Ca{F_2}$$ at $${25^ \circ }C$$ is $$1.7 \times {10^{ - 10}},$$ the combination amongst the following which gives a precipitate of $$Ca{F_2}$$ is
A.
$$1 \times {10^{ - 2}}M\,C{a^{2 + }}\,{\text{and}}\,1 \times {10^{ - 3}}M\,{F^ - }$$
B.
$$1 \times {10^{ - 4}}M\,C{a^{2 + }}\,{\text{and}}\,1 \times {10^{ - 4}}M\,{F^ - }$$
C.
$$1 \times {10^{ - 2}}M\,C{a^{2 + }}\,{\text{and}}\,1 \times {10^{ - 5}}M\,{F^ - }$$
D.
$$1 \times {10^{ - 3}}M\,C{a^{2 + }}\,{\text{and}}\,1 \times {10^{ - 5}}M\,{F^ - }$$
Answer :
$$1 \times {10^{ - 2}}M\,C{a^{2 + }}\,{\text{and}}\,1 \times {10^{ - 3}}M\,{F^ - }$$
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.
$$\eqalign{
& Ca{F_2} \rightleftharpoons C{a^{2 + }} + 2{F^ - } \cr
& {\text{Ionic product}} = \left[ {C{a^{2 + }}} \right]{\left[ {{F^ - }} \right]^2} \cr
& {\text{when,}}\,\left[ {C{a^{2 + }}} \right] = 1 \times {10^{ - 2}}M \cr
& {\left[ {{F^ - }} \right]^2} = {\left( {1 \times {{10}^{ - 3}}} \right)^2}M \cr
& = 1 \times {10^{ - 6}}M \cr
& \therefore \left[ {C{a^{2 + }}} \right]{\left[ {{F^ - }} \right]^2} \cr
& = \left( {1 \times {{10}^{ - 2}}} \right)\left( {1 \times {{10}^{ - 6}}} \right) \cr
& = 1 \times {10^{ - 8}} \cr
& {\text{In this case,}} \cr
& {\text{Ionic product}}\left( {1 \times {{10}^{ - 8}}} \right) > {\text{solubility product}}\left( {1.7 \times {{10}^{ - 10}}} \right) \cr} $$