Question

# A solution contains 0.010 M Ba2+ and 0.010 M Ag+. Can 99.91% of either ion be...

A solution contains 0.010 M Ba2+ and 0.010 M Ag+. Can 99.91% of either ion be precipitated by

chromate (CrO42-) without precipitating the other metal ion? The Ksp for BaCrO4 is 2.1 x 10-10 and

the Ksp for Ag2CrO4 is 1.2 x 10-12

Since the Ksp for silver chromate is smaller, we know that Ag2CrO4 will form a precipitate first as [CrO42-] increases so that Qsp for Ag2CrO4 also increase from zero to Ksp of Ag2CrO4, at which point, BaCrO4 precipitates. As [CrO42-] increases, [Ag+] decreases. Further increase of [CrO42-] till Qsp for BaCrO4 increases to Ksp of BaCrO4, it then precipitates.

[CrO42-] at the point when precipitation of Ag2CrO4 starts = 1.2 x 10-12 / (0.01 )2 = 1.2 x 10-8

[CrO42-] at the point when precipitation of BaCrO4 starts = 2.1 x 10-10 / (0.01 ) = 2.1 x 10-8

The two values are very close to each other, so separation of then by precipitation is not easy.

[CrO42-] is increased to precipitate Ag2CrO4. When [CrO42-] is increased till Qsp for BaCrO4 increases to Ksp of BaCrO4, it then precipitates.

[Ag] when the first trace of BaCrO4 precipitate starts to form = √(1.2 x 10-12 / 2.1 x 10-8) = 0.00756 M

% precipitation of Ag+ = (0.01 - 0.00756/0.01)*100 = 24.4%

So 99.91% precipitation of Ag is not possible from the solution without precipitation of other ion.

#### Earn Coins

Coins can be redeemed for fabulous gifts.