You first treated your unknown solution with 1 M K 2 C 2 O 4 and 1 M CH 3 COOH a

Question

You first treated your unknown solution with 1 M K2C2O4and 1 M CH3COOH and a yellow precipitate formed. The liquid remaining atop this precipitate was removed after centrifugation, and to it was added 1 M H2SO4. No new precipitate formed. Next, a solution of 0.25 M (NH4)2C2O4was added to the unknown solution which had been treated in the fashion described. A white precipitate formed. This mixture was centrifuged and separated, and the clear solution was then treated with 1 M Na2CO3. No precipitate formed. Which metal ions were present in your unknown? Which metal ions were absent?

Explanation / Answer

Treatement with K2CrO4 and CH3COOH formed a yellow precipitate : Ba2+ present [BaCrO4 formed]

A white precipitate formed upon H2SO4 and (NH4)2C2O4 addition : Sr2+ present (forms SrC2O4)

So,

metal ions present = Ba2+, Sr2+

Metal ions absent = Mg2+, Ca2+

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