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The precipitate of CaF(2) is obtained wh...

The precipitate of `CaF_(2)` is obtained when equal volumes of the following are mixed.
`[K_(sp) (CaF_(2)) = 1.7 xx 10^(-10)]`

A

`10^(-4) M Ca^(2+) + 10^(-4) M F^(Theta)`

B

`10^(-2) M Ca^(2+) + 10^(-3) M F^(Theta)`

C

`10^(-5) M Ca^(2+) + 10^(-3) M F^(Theta)`

D

`10^(-3) M Ca^(2+) + 10^(-5) M F^(Theta)`

Text Solution

Verified by Experts

The correct Answer is:
B
`K_(sp) = 1.7 xx 10^(-10)`
After mixing equal volumes, we get
`[Ca^(2+)] = (10^(-12))/(2) = 0.005 = 5 xx 10^(-3)`
`[F^(Theta)] = (10^(-3))/(2) = 0.005 = 5 xx 10^(-4)`
For `CaF_(2)`,
Ionic product `= [Ca^(2+)] xx [F^(Theta)]^(2)`
`[5 xx 10^(-3)] xx [5 xx 10^(-4)]^(2)`
`= 125 xx 10^(-11)`
` = 1.25 xx 10^(-9)`
SInce ionic product is greater than solubility product, precipitations occurs.
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