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One mixing 500 mL of 10^-6 M Ca^(2+) ion...

One mixing `500 mL` of `10^-6 M Ca^(2+)` ion and `500 mL` of `10^-6 MF^-` ion, no precipitate of `CaF_2` will be obtained. `K_(sp) (CaF_2 = 10^-18)`.
If `K_(sp)` is greater than ionic product, a percipitate will develop.

A

If both assertion and reason are true and the reason is the correct explanation of the assertion.

B

If both assertion and reason are true but reason is not the correct explanation of the assertion.

C

If assertion is true but reason is false.

D

If assertion is false but reason is true.

Text Solution

Verified by Experts

The correct Answer is:
C

`[Ca^(2+)][F^-]^2 = [(500 xx 10^-6)/(100)] xx [(500 xx 10^-6)/(1000)]^2`
=`(1)/(8) xx 10^-18 = 0.12 xx 10^-18 lt K_(sp)`
Thus, precipitation will not take place.
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