You have a solution that contains 0.1 M Ag + , 0.1 M Cu 2+ and 0.1 M Ba 2+ . You

Question

You have a solution that contains 0.1 M Ag + , 0.1 M Cu 2+ and 0.1 M Ba 2+ . You would like to separate these ions by precipitation. For quality p urposes, the relative error must be below 0.1%, i.e., the precipitation of each ion must be complet e (>99.9%) and there must not be more than 0.1% (by weight) of one of the other ions in the pr ecipitate. Explain your answer: nature of precipitate, pH, concentrations etc. Be quantitativ e in your answer. Note: at the end, you want one vial with >99.9% of the silver, one with >99.9% of the copper, and one with >99.9% of the barium

Explanation / Answer

ANSWER:

Case 1: First add 0.1 M HCl solution to precipitate only all Ag+ ions as AgCl. Filter it to seperate from the solution. Here Cu2+ and Ba2+ ions are not precipitated as their clorides.

Case 2: Now add 0.1 M H2SO4 solution to precipitate all Ba2+ ions as BaSO4. Filter it to seperate from the solution. Here Cu2+ is not precipitated as its sulphate.

Case3: Now the solution contains only Cu2+. To precipitate these ions, add 0.1 M NaOH solution which precipitates Cu2+ ions as Cu(OH)2.

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