If $$K_{sp}$$ of CaF$$_2$$ at $$25^\circ C$$ is $$1.7\times 10^{-10}$$, the combination amongst the following which gives a precipitate of CaF$$_2$$ is

For calcium fluoride, the equilibrium in a saturated solution is

$$CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2\,F^{-}(aq)$$

Its solubility product is defined as

$$K_{sp}= [Ca^{2+}][F^-]^2$$

Whenever a solution actually contains $$Ca^{2+}$$ and $$F^-$$ ions, we calculate the ionic product

$$IP = [Ca^{2+}]_{\text{mixed}}\,[F^-]_{\text{mixed}}^{\,2}$$

Comparison rule:
• If $$IP \gt K_{sp}$$ → the solution is supersaturated and $$CaF_2$$ precipitates.
• If $$IP \le K_{sp}$$ → no precipitation.

Case 1: Option A

$$[Ca^{2+}] = 1 \times 10^{-2}\;M,\quad [F^-] = 1 \times 10^{-3}\;M$$

$$IP = (1\times10^{-2})\,(1\times10^{-3})^{2} = 1\times10^{-2}\times1\times10^{-6} = 1\times10^{-8}$$

$$1\times10^{-8} \gt 1.7\times10^{-10}$$, hence precipitation occurs.

Case 2: Option B

$$IP = (1\times10^{-4})\,(1\times10^{-4})^{2} = 1\times10^{-4}\times1\times10^{-8} = 1\times10^{-12}$$

$$1\times10^{-12} \lt 1.7\times10^{-10}$$, so no precipitate.

Case 3: Option C

$$IP = (1\times10^{-2})\,(1\times10^{-5})^{2} = 1\times10^{-2}\times1\times10^{-10} = 1\times10^{-12}$$

Again $$1\times10^{-12} \lt 1.7\times10^{-10}$$ → no precipitate.

Case 4: Option D

$$IP = (1\times10^{-3})\,(1\times10^{-5})^{2} = 1\times10^{-3}\times1\times10^{-10} = 1\times10^{-13}$$

$$1\times10^{-13} \lt 1.7\times10^{-10}$$ → no precipitate.

Only Case 1 satisfies $$IP \gt K_{sp}$$. Therefore the mixture that produces a $$CaF_2$$ precipitate is:

Option A which is: $$1\times10^{-2}$$ M Ca$$^{2+}$$ and $$1\times10^{-3}$$ M F$$^-$$