Question

Given: Ksp(Ag2CrO4) = 1.2×10^-12 Ksp(BaCrO4) = 2.1×10^-10 For a solution which is initially 0.004 M in...

Given: Ksp(Ag2CrO4) = 1.2×10^-12 Ksp(BaCrO4) = 2.1×10^-10 For a solution which is initially 0.004 M in Ba2+ and 0.004 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?

Homework Answers

Answer #1

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

Ksp = [Ba2+] [CrO42¯]

2.1×10^-10 = (0.004) [CrO42¯]

[CrO42¯] = 2.1×10^-10 / (0.004)

[CrO42¯] = 5.25 x 10^-8

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

Ksp = [Ag+]2 [CrO42¯]

1.2 ×10^-12 = (0.004)^2 [CrO42¯]

[CrO42¯] = 1.2 ×10^-12 / (0.004)^2

[CrO42¯] = 7.5 x 10^-8

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+][7.5 x 10^-8]

[Ba2+] = 2.1×10^-10 /7.5 x 10^-8

[Ba2+] = 2.8 x 10^-3

Calculate the percent Ba2+ remaining in solution:

(2.8 x 10^-3)/ (0.004) x 100 = 70%

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

= 100 – 70 = 30%

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