112 CHM-153 Advance Study Assignment GENERAL UNKNOWN (pages 112- 115) NAME DATE
ID: 556514 • Letter: 1
Question
112 CHM-153 Advance Study Assignment GENERAL UNKNOWN (pages 112- 115) NAME DATE Refer to the procedures on pages 110 and 111 to answer the following questions... e first thing you need to do with the general unknown is to precipitate the chlorides of Ag, Hg2 and Pb2. W drops be enough to precipitate all the Group I cations? Explain. e do this to a mixture of all the possible cations by adding 3 drops of 6M HCI solution. Will: 2) If you did not get even a tiny bit of cloudiness with 3 drops of 6M HCI, what would you conclude? a) You do not have any siver or mercury, but might have lead. b) You do not have any Group I cations. c) You did not add enough 6M HCI. Explain your answer. 3) Give the steps in the procedure for the separation of the Group I cations from the Group II and Ill cations.Explanation / Answer
1- is the solution of HCL can precipitate the Ions because their solubility product are very low. And the concentration of HCL is very high.
2 - on addition of 3 drop of 6 molar HCL did not give any cloudiness then it may show the absence of all the three Ions. 6M HCL have sufficient Chloride ions to precipitate the first group basic radical because there solubility product is low.
3 - add small amount of HCL of strength 1M or 2M. solution gives white precipitate. Now dissolve the white precipitate in hot water and if white precipitate is soluble in water it may be Lead. And if not dissolve in hot water then solution may contain silver and mercury Ions.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.