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If $K_{s p}$ of $\mathrm{CaF}_2$ at $25^{\circ} \mathrm{C}$ is $1.7 \times 10^{-10}$, the combination amongst the following which gives a precipitate of $\mathrm{CaF}_2$ is
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$1 \times 10^{-2} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-3} \mathrm{M} \mathrm{F}^{-}$
$1 \times 10^{-2} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-3} \mathrm{M} \mathrm{F}^{-}$
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.
$\begin{aligned} & \mathrm{CaF}_2 \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{~F}^{-} \\ & \text {Ionic product }=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^2 \\ & \text { when, }\left[\mathrm{Ca}^{2+}\right]=1 \times 10^{-2} \mathrm{M} \\ & {\left[\mathrm{F}^{-}\right]^2=\left(1 \times 10^{-3}\right)^2 \mathrm{M} } \\ & =1 \times 10^{-6} \mathrm{M} \\ \therefore & {\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^2=\left(1 \times 10^{-2}\right)\left(1 \times 10^{-6}\right)=1 \times 10^{-8} } \\ & \text { In this case, } \\ & \text { Ionic product }\left(1 \times 10^{-8}\right)> \\ & \text { solubility product }\left(1.7 \times 10^{-10}\right)\end{aligned}$
$\therefore$ Hence (a) is correct option.
$\begin{aligned} & \mathrm{CaF}_2 \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{~F}^{-} \\ & \text {Ionic product }=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^2 \\ & \text { when, }\left[\mathrm{Ca}^{2+}\right]=1 \times 10^{-2} \mathrm{M} \\ & {\left[\mathrm{F}^{-}\right]^2=\left(1 \times 10^{-3}\right)^2 \mathrm{M} } \\ & =1 \times 10^{-6} \mathrm{M} \\ \therefore & {\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^2=\left(1 \times 10^{-2}\right)\left(1 \times 10^{-6}\right)=1 \times 10^{-8} } \\ & \text { In this case, } \\ & \text { Ionic product }\left(1 \times 10^{-8}\right)> \\ & \text { solubility product }\left(1.7 \times 10^{-10}\right)\end{aligned}$
$\therefore$ Hence (a) is correct option.
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