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The precipitate of \(CaF_{2}(K_{sp} =1.7\times 10^{-10})\) is obtained when equal volumes of the following are mixed
(A) \(10^{-4}M Ca^{2+}+10^{-4}M F^{-}\)
(B) \(10^{-2}M Ca^{2+}+10^{-3}M F^{-}\)
(C) Both A and B
(D) None of these
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ChemistryClass 11

The precipitate of \(CaF_{2}(K_{sp} =1.7\times 10^{-10})\) is obtained when equal volumes of the following are mixed
(A) \(10^{-4}M Ca^{2+}+10^{-4}M F^{-}\)
(B) \(10^{-2}M Ca^{2+}+10^{-3}M F^{-}\)
(C) Both A and B
(D) None of these

Answer

Answer: B
The expression for the solubility product of \(CaF_2\) is \(K_{sp}=[Ca^{2+}][F^-]^2=1.7\times 10^{-10}\).

When equal volumes of \(10^{-4}M Ca^{2+}\)and\(10^{-4}M F^{-}\) are mixed.

\([Ca^{2+}][F^-]^2=(10^{-4})(10^{-4})^2=10^{-12}<1.7\times 10^{-10}\)

Thus the ionic product is less than the solubility product. Hence precipitation will not occur.

When equal volumes of \(10^{-2}M Ca^{2+}\)and\(10^{-3}M F^{-}\) are mixed.

\([Ca^{2+}][F^-]^2=(10^{-2})(10^{-3})^2=10^{-8}>1.7\times 10^{-10}\)

Thus the ionic product is more than the solubility product. Hence precipitation will occur.

Option B is correct.
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Solution

When ionic product is greater than \(K_{sp}\) then precipitation occur
\(K_{sp} < 10^{-2} M\ Ca^{2+} +10^{-3}M \ F^-\) 
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