Question

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

Answer 1

Since the K_sp for silver chromate is smaller, we know that Ag₂CrO₄ will form a precipitate first as [CrO₄^2-] increases so that Q_sp for Ag₂CrO₄ also increase from zero to K_sp of Ag₂CrO₄, at which point, BaCrO₄ precipitates. As [CrO₄^2-] increases, [Ag⁺] decreases. Further increase of [CrO₄^2-] till Q_sp for BaCrO₄ increases to K_sp of BaCrO₄, it then precipitates.

[CrO₄^2-] at the point when precipitation of Ag₂CrO₄ starts = 1.2 x 10-12 / (0.01 )² = 1.2 x 10-8

[CrO₄^2-] 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.

[CrO₄^2-] is increased to precipitate Ag₂CrO₄. When [CrO₄^2-] is increased till Q_sp for BaCrO₄ increases to K_sp 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.