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A detective was called to investigate a suspected poisoning case. Beside the victim’s...
A detective was called to investigate a suspected poisoning case. Beside the victim’s bed were four bottles containing colorless solutions. The detective labeled the bottles P, Q, R and S. In the rubbish bin were found the labels 'lead nitrate', 'sodium chloride’, ‘barium nitrate' and 'water'.
Can you please help me identify which solution is which and explain what chemicals you would use and how you would do it?
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Unambiguous characterization of the chemicals found in the victim's bedside can be done with few simple chemicals like:
1. NH4HS solution,
2. AgNO3 solution,
3. dilute sulfuric acid solution
with a little knowledge about solubility of sulfides, chlorides and sulfates of the corresponding metal ions (i.e. cations).
From the solubility chart of salts, it is evident that common insoluble salts of the given cations, or anions are: barium sulfate, lead sulfate, silver chloride and lead sulfide.
So first of all, treat portions of all the solutions, A through D with dilute H2SO4. Two of them should produce a white precipitate. Identify these two bottles. These two must be lead nitrate and barium nitrate solutions. Still we do not know which one is which. Now treat a little of these two solutions separately with a little solution of NH4HS (or Na2S). The solution producing a black precipitate of PbS must be lead nitrate, rendering charaterization of the other bottle of this pair as barium nitrate. Barium nitrate should ideally produce a yellowish green solution with NH4HS solution.
The remaining two bottles must be sodium chloride solution and water. Treat little portions of solutions from each bottle with a little silver nitrate solution. The one giving off a curdy white precipitate must be sodium chloride solution.
The solution which produces no precipitate in the above sequence of tests is thus, water.
Posted by llltkl on November 1, 2013 at 2:49 PM (Answer #1)
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