If K_{sp} of CaF₂ at 25,^oC is 1.7 × 10^{-10} , combination amongst following which gives a preci

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6-2.Equilibrium-II (Ionic Equilibrium)

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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

A

$1 \times 10^{-2}\, M\,\, Ca^{2+}$ and $1 \times 10^{-5}\, M\,\, F^-$

B

$1 \times 10^{-4}\, M\,\, Ca^{2+}$ and $1 \times 10^{-4}\, M\,\, F^-$

C

$1 \times 10^{-3}\, M\,\, Ca^{2+}$ and $1 \times 10^{-5}\, M\,\, F^-$

D

$1 \times 10^{-2}\, M\,\, Ca^{2+}$ and $1 \times 10^{-3}\, M\,\, F^-$

Solution

The precipitate will form only when

Ionic product $>$ Solubility product

CaF $_{2}$ dissociates as follows

Thus, ionic product $=\left[10^{-2}\right]\left[10^{-3}\right]^{2}$

$=10^{-2} \times 10^{-6}=10^{-8}$

This value is greater than $K_{\text {sp }},$ ie,

$1.7 \times 10^{-10},$ so ppt will form.

Standard 11

Chemistry

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