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

Question

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+

= ?%

Explanation / Answer

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%

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