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# Equal volume of the follwing $Ca^{2+}$ and $F^-$ solutions are mixed. In which of the follwing solutions will precipitation occur?

[$K_{sp}$ of $CaF_2$ = $1.7 \times 10^{-10}$]

1. $10^{-2} of Ca^{2+} + 10^{-5} MF^-$

2. $10^{-3} of Ca^{2+} + 10 ^{-3} MF^-$

3. $10^{-4} of Ca^{2+} + 10 ^{-2} MF^-$

4. $10^{-2} of Ca^{2+} + 10 ^{-3} MF^-$

Select the correct answer using codes given below.

(a) In 4 only

(b) In 1 and 2

(c) In 3 and 4

(d) In 2, 3 and 4 Comment
A)

In case 1

$[Ca^{2+}]$ in reaction mixture = $0.5 \times 10^{-2}$ M
$[F^-]$ in reaction mixture = $0.5 \times 10^{-5}$ M
$\therefore$ Ionic product = $[Ca^{2+}][F^-] = (0.5 \times 10^{-2}) \times (0.5 \times 10^{-5})^2 = 125 \times 10^{-15} = 1.25 \times 10^{-13} < 1.7 \times 10^{-10}$
$\therefore$ No. ppt is formed in this case.

In case 2

$[Ca^{2+}]$ in reaction mixture = $0.5 \times 10^{-3}$ M
$[F^-]$ in reaction mixture = $0.5 \times 10^{-3}$ M
$\therefore$ Ionic product = $[Ca^{2+}][F^-] = (0.5 \times 10^{-3}) \times (0.5 \times 10^{-3})^2 = 125 \times 10^{-12} = 1.25 \times 10^{-10} < 1.7 \times 10^{-10}$
$\therefore$ No. ppt is formed in this case.

In case 3

$[Ca^{2+}]$ in reaction mixture = $0.5 \times 10^{-4}$ M
$[F^-]$ in reaction mixture = $0.5 \times 10^{-2}$ M
$\therefore$ Ionic product = $[Ca^{2+}][F^-] = (0.5 \times 10^{-4}) \times (0.5 \times 10^{-2})^2 = 125 \times 10^{-11} = 1.25 \times 10^{-9} > 1.7 \times 10^{-10}$
Thus, a ppt. of $CaF_2$ is formed in this case.

In case 4

$[Ca^{2+}]$ in reaction mixture = $0.5 \times 10^{-2}$ M
$[F^-]$ in reaction mixture = $0.5 \times 10^{-3}$ M
$\therefore$ Ionic product = $[Ca^{2+}][F^-] = (0.5 \times 10^{-2}) \times (0.5 \times 10^{-3})^2 = 125 \times 10^{-11} = 1.25 \times 10^{-9} > 1.7 \times 10^{-10}$
Thus, a ppt. of $CaF_2$ is formed in this case.