Qualitative Analysis of Cations: Selective Precipitation amount of each cation p

Question

Qualitative Analysis of Cations: Selective Precipitation

amount of each cation present, it is usually necessary to separate the different metal cations. Sometimes all we need to know is which specific cations are present in detectible amounts. By the correct choice of reagents and order of mixing and filtering, we can separate cations from multi-ion mixtures.

Consider an aqueous mixture that contains only the following metal ions:

Ag+, Ba2+, Ca2+, Cd2+, Cu2+, K+, Fe2+, Pb2+, and Sn2+.

Consider an aqueous solution that contains many common metals cations. Before we can determine the

1.         From either Ksp values or your knowledge of which salts are soluble, predict what will happen when concentrated (6M) HCl is added to the mixture (the lack of any Ksp for common metal chloride is an indicator that the chloride is somewhat soluble).

Explanation / Answer

First, we need to write every reaction:

Ag + HCl ----------> AgCl ----> Ksp = 1.8x10-10

Ba + 2Cl ----------> BaCl2; Ksp = not reported

Ca + 2Cl -----------> CaCl2; Ksp =not reported

Cd + 2Cl ------------> CdCl2; Ksp = nnot reported

Cu + 2Cl ------------> CuCl2; Ksp = not reported

K + Cl -----------> KCl; Ksp = not reported

Fe + 2Cl ----------> FeCl2; Ksp = not reported

Pb +´2Cl -----------> PbCl2; Ksp = 1.7x10-5

Sn + 2Cl -----------> SnCl2; Ksp = not repiorted

This means that when a mixture of all this common metals, we will have in solution all of these metals that the ksp are not reported. and the precipitate will contain AgCl and PbCl2.

Do you need any other explanation for this?

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