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If $K_{sp}$ of $CaF_2$ at $25\,^oC$ is $1 .7 \times 10^{-10},$ the combination amongst the following which gives a precipitate of $CaF_2$ is
$1 \times {10^{ - 2}}{\kern 1pt} \,M\,\,C{a^{2 + }}$ and $1 \times {10^{ - 3}}\,\,M\,{F^ - }$
$1 \times {10^{ - 4}}{\kern 1pt} \,M\,\,C{a^{2 + }}$ and $1 \times {10^{ - 4}}\,\,M\,{F^ - }$
$1 \times {10^{ - 2}}{\kern 1pt} \,M\,\,C{a^{2 + }}$ and $1 \times {10^{ - 5}}\,\,M\,{F^ - }$
$1 \times {10^{ - 3}}{\kern 1pt} \,M\,\,C{a^{2 + }}$ and $1 \times {10^{ - 5}}\,\,M\,{F^ - }$
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.
$Ca{F_2} \rightleftharpoons C{a^{2 + }} + 2{F^ – }$
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$
$\therefore \,[C{a^{2 + }}]{[{F^ – }]^2} = (1 \times {10^{ – 2}})(1 \times {10^{ – 6}}) = 1 \times {10^{ – 5}}$
In this case,
Ionic product $(1 \times {10^{ – 8}}) > $ solubility product $(1.7 \times {10^{ – 10}})$
$\therefore $ Hence $(a)$ is correct option.
