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The precipitate of CaF(2) (K(sp)=1.7xx10...

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

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C

Precipitation takes place only when ionic product (Q) exceeds the value of solubility product `(K_(sp))`. Calculate the ionic product in each case and compare it with the given `K_(sp)`. Note that when equal volume of solutions containing `Ca^(2+)` and `F^(-)` are mixed, the concentrations of ions will be halved as the volume of solution is doubled:
`CaF_(2)(s)hArrCa^(2+)(aq.)+2F^(-)(aq.), K_(sp)= 1.7xx10^(-10)`
(1) Ionic product `(Q) =C_(ca^(2+)).C_(F^(-))^(2)`
`= ((1)/(2)xx10^(-3))((1)/(2)xx10^(-5))^(2)`
`= (1)/(8)xx10^(-13) lt K_(sp)`, no precipitation
(2) Ionic product `(Q) = ((1)/(2)xx10^(-5))((1)/(2)xx10^(-3))^(2)`
`= (1)/(8)xx10^(-11) lt K_(sp)`, no precipitation
(3) Ionic product `(Q) =((1)/(2)xx10^(-2))((1)/(2)xx10^(-3))^(2)`
`= (1)/(8)xx10^(-8)`
` gt K_(sp)`, precipitation takes place
(4) Ionic product `(Q) = ((1)/(2)xx10^(-4))((1)/(2)xx10^(-4))^(2)`
`= (1)/(8)xx10^(-12) lt K_(sp)`, no precipitation
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