Carol has an aqueous solution containing Pb2, Cd2 Fe2+, and Mg ions. These ions
ID: 1025603 • Letter: C
Question
Carol has an aqueous solution containing Pb2, Cd2 Fe2+, and Mg ions. These ions are to be separated, one by one, as precipitates from the aqueous solution. Carol adds 6 M HCI. Next, she changes the supernatant to 0.2 M HCl and then adds H2S. She makes the supernatant more basic, until pH = 8. In a final step, Carol adds (NH2HPO, and NH3. In what sequence does Carol remove the ions as precipitates, one by one, based on the addition scheme of reagents described above? Cd2+ Fe2+ Mg2+ Mg2+ Fe2+ Mg2+ Fe2+ Pb2+ Mg2+ Pb2+ Cd2+ Fe2 + Pb2+ Cd2+ Fe2+ Cd2+Explanation / Answer
Hello dear ,
For solving the above question , the following chart should be just mugged up . Inorganic chemistry is full of such types of questions .
Group seperations fo cations
Hence , the ions will be precipitated in following order :
1) Pb2+ : On adding 6M HCl the solution , Pb2+ ions will be precipitated as follows :
2Pb2+ + 2HCl -> PbCl2 (ppt) + 2H+
2) Cd2+ : On adding 0.2M HCl + H2S to the solution , Cd2+ ions will be precipitated as follows :
Cd2+ + H2S --> CdS(yellow ppt) + 2H+
3) Fe2+ : On making solution of basic (pH=8) , Fe2+ ions will be precipitated with the remaining H2S as follows :
Fe2+ + H2S --> FeS(ppt) + 2H+
4) Mg2+ : On adding (NH4)2PO4 to the solution , Mg2+ ions will be precipitated as follows :
Mg2+ + (NH4)2PO4 --> (Mg)2PO4 + 2NH4+
Hence , the order is 1) Pb2+ 2) Cd2+ 3) Fe2+ 4) Mg2+
Hence , option C is correct .
Thank you !
Group 1 : Ag+ , Hg22+ , Pb2+ Precipitated in 1M HCl Group 2 : Bi3+ , Cd2+ , Cu2+ , Hg2+ , (Pb2+) , Sb3+ and Sb5+ , Sn2+ and Sn4+ Precipitated in 0.1M H2S solution at pH 0.5 Group 3 : Al3+ , (Cd2+), Co2+ , Cr3+ , Fe2+ and Fe3+ , Mn2+ , Ni2+ , Zn2+ Precipitated in 0.1M H2S solution at pH 9 Group 4 : Ba2+ , Ca2+ , Mg2+ Precipitated in 0.2M (NH4)2CO3 solution at pH 10 ; other ions are soluble Group 5 : K+ , Na+ , NH4+ The remaining soluble ions