Question

Please explain how: A sample contains MgNH4(PO4)2, Sb2S3, KBr, Cr(OH)3, and PbCl2. Identify the soluble ions after the addition of 6 M HCl; then H2S and 0.2 M HCl; then OH- to a pH of 8; and then (NH4)2HPO4 with NH3. Explain. a.) Sb2S3 b.) KBr c.) PbCl2 d.) Cr(OH)3 e.) MgNH4(PO4)2

Answer 1

Locate the cations in the group analysis table for cations, and find the corresponding reagent for their precipitation.

Here, the cations are : Mg2+ , NH4+ , Sb3+ , K+ , Cr3+, Pb2+ .

Pb2+ is in group I. It gives a chloride precipitate in acidic solution. It will also give a precipitate with each of the other reagents. Hence, it is placed in group 1, so it can be detected and removed first, and will not continue to give further positives for the rest of the groups, and hinder the detection of other cations.

Sb3+ is in group II. It's sulphide will precipitate in acidic medium. (Similarly, only it's chloride is soluble in acidic medium (I.e , on adding HCl) . It will form a precipitate with all the next reagents).

Cr3+ is in group IIIB. It's hydroxide precipitates in basic solution.

K+ , NH4+ and Mg2+ are in the last group.

However, adding of NH3 + (NH4)2HPO4 will only precipitate Mg(NH4)(PO4)2 (white precipitate. This is the confirmation test for Mg2+). K+ (that is, KBr) will remain in solution for the entirety.

So, when 6M HCl is added, only PbCl2 precipitates. The rest of the Salts are present as their constituent soluble ions : Sb3+, S2- , Mg2+ , NH4+ , PO43-, K+ , Br-, Cr3+ , OH- .

When H2S and 0.2 M HCl is added , Sb2S3 precipitates. Hence , Sb3+ and S2- is removed from the list of soluble ions.

When OH- is added till pH= 8 (making the solution basic, since adding HCl before caused an acidic solution) : Cr(OH)3 will precipitate. So, remove Cr3+ and OH- from the list.

When (NH4)2HPO4 + NH3 is added , Mg(NH4)(PO4)2 precipitates out. Thus, remove Mg2+, NH4+ and PO43- from the list.

At the end , only KBr ( K+ , Br-) is left. So, answer is (b)