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A precipitate of $CaF_2$ $(K_{SP} = 1.7 \times 10^{-10})$ will be obtained when equal volume of the following are mixed
$10^{-4}\,M\,Ca^{2+}$ and $10^{-4}\,M\,F^ -$
$10^{-2}\,M\,Ca^{2+}$ and $10^{-3}\,M\,F^ -$
$10^{-5}\,M\,Ca^{2+}$ and $10^{-3}\,M\,F^ -$
$10^{-3}\,M\,Ca^{2+}$ and $10^{-5}\,M\,F^ -$
Solution
For precipitation, $Q_{ sp }$ (product of molar concentration) $\,>\, K _{ sp }$ (solubility product).
In water, $CaF _2$ is in equilibrium with its ions as follows:
$CaF _2( s ) \rightleftharpoons Ca ^{2+}( aq )+2 F ^{-}( aq )$
Solubility product of $CaF _2$
$K _{ sp }=\left[ Ca ^{2+}\right]\left[ F ^{-2}\right]=1.7 \times 10^{-10}$
In $a), Q_{ sp }=\left[\frac{10^{-4}}{2}\right]\left[\frac{10^{-4}}{2}\right]^2=1.25 \times 10^{-13}$
In $b),Q_{s p}=\left[\frac{10^{-2}}{2}\right]\left[\frac{10^{-3}}{2}\right]^2=1.25 \times 10^{-9}$
In $c), Q_{ sp }=\left[\frac{10^{-5}}{2}\right]\left[\frac{10^{-3}}{2}\right]^2=1.25 \times 10^{-12}$
In $d),Q_{ sp }=\left[\frac{10^{-3}}{2}\right]\left[\frac{10^{-5}}{2}\right]^2=1.25 \times 10^{-14}$
The option $b)$ only has the ionic product greater than the solubility product.
