1. Separation of the Chloride ($\ce{Cl-}$); Confirmation of Chloride.

    Chloride ion forms an insoluble silver compound. Silver chloride is a white solid.
    $$\ce{Cl^-(aq) + Ag+(aq) -> AgCl(s)}$$ Silver chloride dissolves in $\pu{6 M}$ ammonia, $\ce{NH3}$, forming the colorless ion $\ce{Ag(NH3)2+}$. If nitric acid, $\ce{HNO3}$, is added to a solution containing this ion, the ammonia in the complex reacts with hydrogen ions to form ammonium ions, and the silver recombines with the chloride ions that are still present in solution.
    \begin{align} \ce{AgCl(s) + 2NH3(aq) &-> Ag(NH3)2+(aq) + Cl^-(aq)} \\ \ce{Ag(NH3)2+(aq) + Cl^-(aq) + 2H+(aq) &-> AgCl(s) + 2NH4+(aq)} \end{align}

    a. Place 10 drops of the original test solution (or unknown solution) in a test tube. Test to see if the solution is acidic. If it is not, add $\pu{6 M}$ acetic acid, $\ce{CH3COOH}$, dropwise with stirring until the solution is acidic.

    b. Add 10 drops of $\pu{0.1 M}$ silver nitrate, $\ce{AgNO3}$. A precipitate of $\ce{AgCl}$ will form.

    c. Centrifuge and pour off the supernatant liquid.

    d. Wash the solid with $\pu{0.5 mL}$ distilled water, centrifuge and discard the wash water.

    e. Add $\pu{0.5 mL}$ $\pu{6 M}$ ammonia, $\ce{NH3}$, to the precipitate. Stir to dissolve any $\ce{AgCl}$.

    f. Centrifuge, and pour the supernatant liquid into another test tube to test for chloride ion.

    g. Add $\pu{1 mL}$ $\pu{6 M}$ nitric acid, $\ce{HNO3}$, to the solution containing the dissolved silver chloride. The solution will get hot and smoke from the reaction with the excess ammonia whether or not silver chloride is present.

    h. Test with litmus or pH paper to see if the solution is acidic. If it is not, add more $\ce{HNO3}$ until the solution is acidic. The appearance of the white precipitate of $\ce{AgCl}$ in the acidic solution confirms the presence of chloride.

From the procedure above, if you all could take a loot at step 1a. It makes no sense to me. I do not understand why adding $\ce{CH3COOH}$ has any effect on the solution whatsoever. What is the point of adding it when we later add $\ce{HNO3}$?

  • 4
    $\begingroup$ I think the acetic acid in the first step, serves to ensure that the first precipitation of AgCl occurs in an acidic environment (which is why, from what I infer, the step can be omitted if your starting solution is acidic). The $\ce{HNO_3}$ serves to destroy the ammine complex that you form (since it protonates the ammonia), and thus you obtain re-precipitation. $\endgroup$
    – getafix
    May 31, 2015 at 5:51

1 Answer 1


Playing around with pH values, going from acidic to basic and back to acidic is a major element of the separation methods used in traditional analytic inorganic chemistry. It takes advantage of the facts that both acids and bases were easy to acquire even in the early days of chemistry and that the pH can be monitered well by indicators which had also been known for a long time (even though in the past they might have used different indicators than nowadays). Making sure the pH is at a correct value at a certain time is also important in this cycle.

(Because the Markdown syntax does not allow calling enumerated lists with lowercase letters, I’ll use numbers instead to refer to the different steps.)

  1. You have a solution of potentially many different ions. For the sake of this answer, let’s assume chloride, bromide, nitrate and carbonate. If you measure the pH of this solution, you will notice it is slightly basic due to the carbonate. If you were to add silver nitrate immediately, silver carbonate $\ce{Ag2CO3}$ would also precipitate, complicating later analyses.

    Therefore, you add an acid (whether it is acetic or nitric acid does not matter. It should not be hydrochloric acid for obvious reasons and it should not be too strong.) until the pH of the solution is acidic (i.e. at least 5).

  2. You can now add your silver nitrate. Only chloride and bromide will precipitate as $\ce{AgCl}$ and $\ce{AgBr}$, respectively.

  3. By centrifuging, you now have only the silver chloride and bromide precipitates remaining since you are discarding the supernatant. Due to residual amounts of whichever acid you used (acetic acid in the protocol), the damp precipitate is still acidic. However, most of the original acid will have remained in the supernatant and has been poured away.

  4. Acidic residues remain even after washing albeit to a lesser extent. We can basically consider the precipitate neutral now. This is important for step 5.

  5. Adding diluted ammonia (not $25~\%$ concentrated) will dissolve $\ce{AgCl}$ as $\ce{[Ag(NH3)2]Cl}$ while leaving the $\ce{AgBr}$ untouched. Due to the ammonia you are adding, the pH of the resulting solution is basic. Any remaining residues of acid from step 1 have been neutralised now at the latest.

  6. Centrifuging makes sure to remove any remaining dispersed silver bromide. The supernatant should contain $\ce{[Ag(NH3)2]Cl}$ in solution.

  7. Now we want to reprecipitate the silver chloride to confirm we actually had it in the precipitate. (Note we cannot be certain that silver chloride was present if silver bromide has precipitated out.) We do this by reacidifying the solution until all ammonia is protonated to give ammonium and the ammine ligands therefore dissociate from the $\ce{[Ag(NH3)2]+}$ complex. This causes silver chloride to reprecipitate.

  8. In case you did not add enough acid you are here reminded to do so. Failing to add sufficient amount of acid would lead to the silver ions remaning complexed in solution and you incorrectly assuming there be no chloride in your unknown sample.


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