Chemistry 128

November 1, 2002

The is a scheme to determine the presence or absence of specific inorganic chemical species. No effort is made to determine amount or concentration. (This is quantitative analysis.) We will deliberately limit the search to about a half dozen metallic elements. We could look for more elements, including nonmetals, but the scheme would become more complex. It would then take more than the two lab periods we have devoted to the problem.

The detailed, step-by-step procedures will be available in lab on Friday.

You should read this set of notes before class

This is a wet chemical process, conducted in aqueous solutions using relatively simple glassware and fairly ordinary chemical reagents. We will rely on three main procedures

• 1. The addition of a suitable anion will cause some cations to form an insoluble salt and those cations will be precipitated from the solution.
  • Generally some cations will form precipitates and others will not, so this will allow us to begin the separation a mixture.
  • We will often rely on another anion (or a change in conditions) to form other precipitates. This will allow us to further separate the mixture.
  • In the best circumstances, we end up with a set of test tubes, each of which can only contain one element and its easy to come up with a yes or no answer.
• 2. We will rely heavily on the control of pH to determine what will and won't precipitate.
• 3. We will also rely on the formation of complex ions as a way to selectively dissolve precipitates or to prevent the precipitation of a species.

• Once we have separated the mixture we will then rely on specific tests to confirm the presence or absence of a specific element. Most of these tests will not give definitive answers when many other cations might be present.

So why do we follow this historical, but obsolete approach? Basically we use the analytical scheme as a way to illustrate and reinforce many important concepts, generally involving chemical and physical equilibrium. We will examine issues of solubility equilibrium, acid-base equilibrium and equilibrium involving complex ion formation. This is also a good illustration of a logical approach to a problem, dissecting a complex problem into a series of simpler problems.

• You will work with known samples so you can see what to expect.
• You will then be given an unknown that you are to analyze and report,
• Each of you will work on your own unknown sample.

• 1. qualitative analysis only asks if the species is present, not how much of it. Clearly there must be some cutoff value where we say there is so little present we report the answer as "no" even if there are tiny traces of the species. That cutoff value varies greatly with the type and sample and the reason for doing the analysis. We will not try to trick you with tiny concentrations of an element. All samples and unknowns will be about 0.1 Molar.
• 2. several elements are so distinctive that color of the sample can immediately tell you
• if that species is present. (blue Copper or green Nickel.) There are a occasional samples that give false hints so you shouldn't leap at conclusions. We might well add some dyes to your sample to prevent color being a simple answer.
• 3. If we randomly assign unknown samples you could end up with a sample with none of the tested elements or with many of them. Negative results are as important as positive results.
• 4. The sample will contain elements from this list
  • Al, Cu, Cd, Cr, Fe, Ni, Pb, Ag
• 5. Your sample will be a solution and you are to analyze it for all the species in our list. Your solution has a unique ID number and you should perform all your tests on the same sample. You are not sharing this sample with others in the lab.

• We will divide the elements into three groups
  • Group I -- Ag and Pb (Species that form insoluble chlorides)
  • Group II -- Cu and Cd (Species that form insoluble sulfides in acidic solutions)
  • Group III-- Al, Cr, Fe, Ni (Species that form insoluble sulfides or hydroxides

Separation based on simple solubility

Group I consists of two elements that have chlorides of very low solubility-- AgCl and PbCl₂

• The addition of HCl to a solution containing either Ag⁺ or Pb²⁺ will produce a white precipitate. A precipitate tells you that one or both of these elements is present. The absence of a precipitate tells you that both are absent.
  • None of the other elements will precipitate under these conditions.

• There are a few "ifs" that we must watch out for
  • 1. We use HCl because we want to avoid precipitates that form in basic solutions.
  • 2. We test the pH since we could have started with a basic or NH₃ rich solution.
  • 3. We want to avoid extremely high Cl- concentrations since we could redissolve the silver by forming a complex ion :
    • AgCl (s) + Cl- ---> AgCl₂- (in solution)
• The primary reason for precipitating chlorides to simplify the mixture--
  • Group I elements (Pb and Ag) are separated from the mixture
    • (as a solid at the bottom of a test tube)
  • Group II and III elements remain in solution for later analysis

• Rather than waiting for gravity, we use a centrifuge to quickly settle out the solids
  • we can then carefully remove the liquid, transferring it to a second tube.
    • we also add another drop of HCl just to make sure we do not still have some Ag⁺ or Pb²⁺ that didn't react.
  • we can add a small amount of water to rinse the solid
• We now do a series of tests on the solid to determine what we have
  • Silver chloride dissolves in strong ammonia solutions
    • AgCl (s) + 2NH₃ --> Ag(NH₃)²⁺ (aq.) +Cl-.
  • Lead chloride (dissolved) reacts with chromate ion to form a yellow solid

2. Group II-- Solubilities based on the Formation of Sulfides

• We will use H₂S (hydrogen sulfide gas) as our source of sulfide ion.
  • In basic solutions, almost all H₂S is converted to S^2- ion
    • H₂S + 2OH- ---> 2H₂O + S^2-
    • we could get [S^2-] = 0.05 M and almost all cations form insoluble sulfides under those conditions.
    • that's not useful if we want to separate the cations
  • as the pH of the solution gets closer to neutral we get much more HS- ion
    • H₂S + H₂O --> HS^- + H₃O⁺
    • but only a trace concentration of S^2- is produced
    • under these conditions we form precipitates with Cu, Ag, Pb, Cd but we do not form precipitates with Al, Ni, Cr or Fe

• So treating a mixture of ions might produce
  • a precipitate that contains all the Cu and Cd (a little Pb might get through)
  • a solution with the Al, Ni, Cr and Fe
    • Since solid and liquid are still mixtures we will need additional steps.

Group II Separation (Cu and Cd)

• We might be lucky-- CdS is bright yellow and CuS is black.
  • If the solid is yellow, we have Cd and we do not have Cu
  • If we get no precipitate neither Cu or Cd is present
  • If we get a dark precipitate we have more work
    • (mixing black CuS and yellow CdS will probably conceal any color)

• note added to Web page) We can reduce Copper ion to copper metal, removing it from solution. This is done in a way that keeps the Cd ion in solution. Now a sulfide precipitate is easy to recognize.
  • We can then run specific tests on each of those species.

Group III --

• The elements will all react with NaOH to produce insoluble hydroxides.
• Two elements -- Cr and Al will become soluble in a excess of Al
  • Al(OH)₄- and CrO₄2- ions form
    • we can confirm Al with a color test
  • CrO₄2- is easy to spot--
    • it is bright yellow and reacts with Pb²⁺
• The remaining elements (Fe, Ni) remain as precipitates
• Treating the precipitate with NH₃ causes the Ni(OH)₂ to dissolve
  • the new precipitate is Fe only
  • the solution if Ni(NH₃)₄2+
    • We can confirm Fe by reaction with SCN- (bright red color)
    • We can confirm Ni by reaction with dimethylglyoxime (bright red color)

We will add details on the analysis-- additional pages are coming

We hope to add some other links to qualitative analysis schemes soon