The precipitate of `CaF_(2) (K_(sp)=1.7xx10^(-10))` is obtained when equal volumes of the following are mixed

Options

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)`

Answer

B

Explanation

`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.