HW3_26 - 4 to each solution The Pb 2 would form a white...

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Homework due 3/26 Indicate how you would distinguish between aqueous solutions of each pair of ions below with ONE simple test . Indicate what you would observe if each ion was present. In these problems one wants to run a test that will yield different results with each of the possible ions. Ideally, one ion will precipitate and one ion will not, or else there will be a distinctive color difference between the products of the reaction. In most cases there is more than one correct answer to the question, however, only one answer needs to be given for complete credit. Ag 1+ , Al 3+ The simplest test is to add HCl the Ag 1+ will give a white precipitate while the Al 3+ will have no reaction. Pb 2+ , Fe 3+ The simplest test would be to simply observe the color of the solution . Pb 2+ is colorless in solution while Fe 3+ is yellow in solution. Another test would be to add either HCl or H 2 SO
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Unformatted text preview: 4 to each solution. The Pb 2+ would form a white precipitate, while there would not be a precipitate with the Fe 3+. Cr 3+ , Hg 2+ Again the simplest test would be the color of the solution . Cr 3+ solutions are blue-black while those of Hg 2+ are colorless. Addition of excess NaOH causes Cr 3+ solutions to precipitate and the precipitate redissolve while Hg 2+ would lead to a yellow-orange precipitate. Cl 1-, NO 3 1-The easiest test here would be to add AgNO 3 . This would lead to the formation of a white precipitate by the chloride ion and no reaction form the NO31-. Add FeSO 4 and H 2 SO 4 and look for a brown ring at the interface with NO 3 1-and no ring with Cl 1-. Note: since these are ions in aqueous solution, treatment with concentrated sulfuric acid will not work, as that test needs to be run on solid samples of the unknowns....
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This note was uploaded on 04/03/2008 for the course CHEM 114 taught by Professor Martin during the Spring '08 term at Moravian.

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