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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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## QuestionChemistryClass 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

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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4.6

## Solution

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