Question
The precipitate of $$Ca{F_2}\left( {{K_{sp}} = 1.7 \times {{10}^{ - 10}}} \right)$$ is obtained when equal volumes of the following are mixed
A.
$${10^{ - 4}}M\,C{a^{2 + }} + {10^{ - 4}}M\,{F^ - }$$
B.
$${10^{ - 2}}M\,C{a^{2 + }} + {10^{ - 3}}M\,{F^ - }$$
C.
$${10^{ - 5}}M\,C{a^{2 + }} + {10^{ - 3}}M\,{F^ - }$$
D.
$${10^{ - 3}}M\,C{a^{2 + }} + {10^{ - 5}}M\,{F^ - }$$
Answer :
$${10^{ - 2}}M\,C{a^{2 + }} + {10^{ - 3}}M\,{F^ - }$$
Solution :
TIPS/Formulae :
For precipitation to occur ionic product > solubility products
$$\eqalign{
& {\text{Given,}}\,\,{K_{sp}}Ca{F_2} = 1.7 \times {10^{ - 10}} \cr
& Ca{F_2} \rightleftharpoons C{a^{2 + }} + 2{F^ - } \cr
& {\text{Ionic product of}}\,\,{\text{Ca}}{{\text{F}}_2} = \left[ {C{a^{2 + }}} \right]{\left[ {{F^ - }} \right]^2} \cr
& {\text{Calculate LP}}{\text{. in each case}} \cr
& {\text{(a) I}}{\text{.P}}{\text{. of Ca}}{{\text{F}}_2} = \left( {{{10}^{ - 4}}} \right) \times {\left( {{{10}^{ - 4}}} \right)^2} = {10^{ - 12}} \cr
& {\text{(b) I}}{\text{.P}}{\text{. of Ca}}{{\text{F}}_2} = \left( {{{10}^{ - 2}}} \right) \times {\left( {{{10}^{ - 3}}} \right)^2} = {10^{ - 8}} \cr
& {\text{(c) I}}{\text{.P}}{\text{. of Ca}}{{\text{F}}_2} = \left( {{{10}^{ - 5}}} \right) \times {\left( {{{10}^{ - 3}}} \right)^2} = {10^{ - 11}} \cr
& {\text{(d) I}}{\text{.P}}{\text{. of Ca}}{{\text{F}}_2} = \left( {{{10}^{ - 3}}} \right) \times {\left( {{{10}^{ - 5}}} \right)^2} = {10^{ - 13}} \cr
& \because \,\,{\text{I}}{\text{.P}} > {\text{solubility in choice (b) only}}{\text{.}} \cr
& \therefore \,\,{\text{ppt of}}\,Ca{F_2}\,{\text{is obtained in case of choice (b) only}} \cr} $$