Question

# Selective Precipitation Given: Ksp(Ag2CrO4) = 1.2×10-12 Ksp(BaCrO4) = 2.1×10-10 For a solution which is initially 0.003...

Selective Precipitation

Given: Ksp(Ag2CrO4) = 1.2×10-12 Ksp(BaCrO4) = 2.1×10-10

For a solution which is initially 0.003 M in Ba2+ and 0.003 M in Ag+, it is desired to precipitate one of these ions as its chromate (CrO42−) salt to the maximum possible degree without precipitating any of the other cation. From this solution, which ion can be selectively first-precipitated by controlled addition of CrO42− and, ideally, what is the maximum possible percentage of that ion which can be exclusively so precipitated? (Assume that there is no change in the total volume of the solution.)

Ag+

Ba2+

= ?%

Calculate the CrO42− ion concentration when BaCrO4 first begins to precipitate

Ksp = [Ba2+] [CrO42¯]

2.1×10^-10 = (0.003) [CrO42¯]

[CrO42¯] = 2.1×10^-10 / (0.003)

[CrO42¯] = 7 x 10^-8

Calculate the CrO42− ion concentration when Ag2CrO4 first begins to precipitate

Ksp = [Ag+]2 [CrO42¯]

1.2 ×10^-12 = (0.003)^2 [CrO42¯]

[CrO42¯] = 1.2 ×10^-12 / (0.003)^2

[CrO42¯] = 1.33 x 10^-7

BaCrO4 precipitates first, because CrO42¯ concentration necessary to form BaCrO4 is smaller.

Calculate the [Ba2+] ion when Ag2CrO4 first begins to precipitate

Ksp = [Ba2+] [CrO42¯]

2.1×10^-10 = [Ba2+][1.33 x 10^-7]

[Ba2+] = 2.1×10^-10 /1.33 x 10^-7

[Ba2+] = 1.58 x 10^-3

Calculate the percent Ba2+ remaining in solution:

(1.58 x 10^-3)/ (0.003) x 100 = 52.67%

The maximum possible percentage of Ba2+ ion which can be exclusively precipitated

= 100 – 52.67 = 47.33%