If equal volumes of `BaCI_(2)` and `NaF` solutions are mixed, which of these combination will not give a precipitate? `(K_(sp) of BaF_(2) =1.7 xx 10^(-7))`.

Options

A. `10^(-3)BaCI_(2)` and `2xx10^(-2)M NaF`
B. `10^(-3)M BaCI` and `1.5 xx 10^(-2)M NaF`
C. `1.5 xx 10^(-2)M BaCI_(2)` and `10^(-2)M NaF`
D. `2xx10^(-2)M BaCI_(2)` and `2xx10^(-2)M NaF`

Answer

C

Explanation

When equal volume of `BaCI_(2)` and `NaF` solutions are mixed. (volume becomes double and concentration is halved).
a. `[Ba^(2+)] = (10^(-3))/(2), [F^(Theta)] = (2xx10^(-2))/(2) = 10^(-2)`
`Q_(sp) of BaF_(2) = [Ba^(2+)] [F^(Theta)]^(2)`
`=(0.5 xx 10^(-3)) (10^(-2))^(2) = 5 xx 10^(-6)`
`Q_(sp) gt K_(sp) (6 xx 10^(-6) gt 1.7 xx 10^(-7))` will be precipitated.
b. `[Ba^(2+)] = (10^(-3))/(2), [F^(Theta)] = (1.5 xx 10^(-2))/(2)`
`Q_(sp) of BaF_(2) = [Ba^(2+)] [F^(Theta)]^(2) = (0.5 xx 10^(-2)) (0.75 xx 10^(-2))^(2) = 0.28 xx 10^(-6)`
`Q_(sp) gt K_(sp)`. Hence precipitation occurs.
c. `[Ba^(2+)] = (1.5 xx 10^(-3))/(2), [F^(Theta)] = 10^(-2))/(2)`.
`Q_(sp) pf BaF_(2) = [Ba^(2+)] [F^(Theta)]^(2) = ((1.5 xx 10^(-3))/(2)) ((10^(-2))/(2))^(2)`
`= 0.187 xx 10^(-7)`
`Q_(sp) lt K_(sp) (0.187 xx 10^(-7) lt 1.7 xx 10^(-7))`
So solution in (c) will not precipitate out.
d. `[Ba^(2+)] = (2xx10^(-2))/(2) = 10^(-2)M`,
`[F^(Theta)] = (2xx10^(-2))/(2) = 10^(-2)M`
`Q_(sp) BaF_(2) = (10^(-2)) (10^(-2))^(2) = 10^(-6)`
`:. Q_(sp) gt K_(sp)`. Hence precipitation occurs.