An aqueous solution of sodium fluoride is slowly added to a water sample that co

Question

An aqueous solution of sodium fluoride is slowly added to a water sample that contains barium ion (0.0360 M) and calcium ion (0.0480 M). The K_sp of barium fluoride is 1 .0x10^-6. The K_sp of calcium fluoride is 3.9x10^-11. What is the remaining concentration of the first ion to precipitate when the second ion begins to precipitate? 1.90x10^-4 Since K_sp of calcium fluoride is smaller than that of barium fluoride, the calcium fluoride will precipitate first. The K reaction for calcium fluoride is: CaF_2(s) reversible reaction Ca^2+ (aq) + 2 F^-(aq) How would you determine when the second ion begins to precipitate?

Explanation / Answer

BaF2 <-> Ba+2 + 2F- Ksp = 1*10^-6

CaF2 <-> Ca+2 + 2F- Ksp = 3.9*10^-11

[Ba+2] = 0.036

[Ca+2] = 0.04

when the second ion precipitates...

that is B+2 equilibrium

so

BaF2 <-> Ba+2 + 2F- Ksp = 1*10^-6

Ksp = [Ba+2][F-]^2

1*10^-6 = 0.036 * (F-)^2

[F-] = sqrt((1*10^-6 )/(0.036)) = 0.005270

so..

assume then:

CaF2 <-> Ca+2 + 2F- Ksp = 3.9*10^-11

Ksp = [Ca+2][F-]^2

3.9*10^-11 = [Ca+2](0.005270)^2

[Ca+2] = (3.9*10^-11) / ((0.005270)^2) =1.4042 *10^-6

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