I am teaching a section of qualitative analysis and I am trying to get a better grasp on the assignments the students are doing. These are the cations they may be presented with and must identify:

  • $\ce{Na+}$
  • $\ce{K+ }$
  • $\ce{NH4+}$
  • $\ce{Ca^2+ }$
  • $\ce{Al^3+ }$
  • $\ce{Mg^2+}$

Now, this lab hasn't changed for the last 20 years, and there is only one way the students perform the separations, which happens to be the same method found in an old lab report that is easily found with Google. The method is:

  1. Add excess 3 M ammonia to the sample.

  2. If a gel-like precipitate forms, then $\ce{Mg^2+ }$ and/or $\ce{Al^3+ }$ ion is present. They have been converted into their respective hydrated hydroxides. Separate the precipitate from the supernatant liquid.

  3. Since hydrated $\ce{Mg^2+ }$ hydroxide is not readily amphoteric, while hydrated $\ce{Al^3+ }$ hydroxide is, addition of excess $\ce{NaOH}$ should cause the an $\ce{Al^3+ }$ precipitate to dissolve. If the precipitate from the previous step does dissolve with excess $\ce{NaOH}$, then $\ce{Al^3+ }$ ion is present in the original sample. If the precipitate does not dissolve, then $\ce{Mg^2+ }$ ion is present.

  4. Time for the supernatant analysis (if there is a supernatant). If there is no ppt, then we can immediately rule out the presence of $\ce{Mg^2+}$ and $\ce{Al^3+ }$. Either way, the solution, non-precipitate portion can only contain $\ce{Na+}$, $\ce{K+ }$, $\ce{NH4+}$, and/or $\ce{Ca^2+ }$ since none of these form insoluble hydroxides at the concentration of ammonia used. Yes, calcium hydroxide is insoluble, but it doesn't form with 3 M $\ce{H3N}$; I've done the $Q_\mathrm{sp}$ vs. $K_\mathrm{sp}$ calculation.

  5. Add sodium carbonate to the liquid portion. If a solid forms, the solid must be calcium carbonate, and therefore $\ce{Ca^2+ }$ was present in the original sample.

  6. If no solid forms, or even if a solid forms, separate the solid from the liquid and perform further tests on the liquid to determine the presence of $\ce{Na+}$, $\ce{K+ }$, and $\ce{H4N+}$. A yellow flame tests tells us $\ce{Na+}$ is present and $\ce{K+}$ may or may not be present. A purple flame test tells us $\ce{Na+}$ is absent.

  7. Heat the remaining solid sample (not any precipitate, but the original sample itself) to ascertain the presence of $\ce{NH4+}$. Heating a sample that contains $\ce{NH4+}$ will result in the release of white smoke. After heating the solid sample until smoking ceases or until one is certain the sample will not start generating white smoke, perform a sodium coboltinitrite test if necessary to determine whether $\ce{K+ }$ is present.

What other valid separations are there? Is this the easiest method, or are there easier ones?

What about initially adding $\ce{NaOH}$?

  1. Add $\ce{NaOH}$. If a white precipitate forms, the sample contains $\ce{Ca^2+}$ (the white ppt is calcium hydroxide). If a gel-like ppt forms, then the sample contains either $\ce{Al^3+}$ or $\ce{Mg^2+}$. Perform the necessary tests described above to determine the nature of the gel-like ppt.

  2. The liquid part of the solution must now only contain $\ce{Na+ }$ or $\ce{K+ }$. Ammonium ion has been converted to ammonia thanks to the hydroxide ion. $\ce{Na+}$ or $\ce{K+ }$ can easily be tested for using a flame test. The sodium coboltinitrite test can also be employed to detect $\ce{K+ }$ if the color of $\ce{K+ }$ has been masked by the color of Na+ in the flame test.

This method seems a bit faster. But are there any pitfalls?

  • 3
    $\begingroup$ Dude, you gotta get with the program in re: typesetting those formulae... $\endgroup$ Commented May 24, 2016 at 2:32
  • $\begingroup$ related chemistry.stackexchange.com/questions/31431/… $\endgroup$
    – Mithoron
    Commented May 24, 2016 at 19:35
  • $\begingroup$ You want more than Mart's answer? o.O $\endgroup$
    – Mithoron
    Commented May 27, 2016 at 0:06
  • $\begingroup$ @Mithoron well I wanted to reward his answer and to see if anyone else wanted to chime in :). $\endgroup$
    – Dissenter
    Commented May 27, 2016 at 0:08

4 Answers 4


In a qualitative analysis, the fast way is always the wrong way. There are already so many issues with the "long" version that I am surprised that this is remotely successful.

The cations of groups 1-3 are always the most difficult to separate and during my time as a teaching assistant were by far the biggest source of error.

The proper course of analysis is

  1. dissolve sample (completely)
  2. precipitate ion
  3. dissolve precipitate
  4. specific reaction
  5. repeat

The reason for separating the cations in the first place is, that they often act in similar ways and the reactions are not very specific. Certain cations should be tested multiple times, too.

All the reactions you are performing are equilibria and very dependent on the overall concentration. None of them form a precipitate quantitatively. Sometimes you might have problems recognising precipitates in the first place.

You need to perform double blind tests for all used chemicals, since already small contaminations may lead to false positives.

The main issue with your proposed strategy is that you will not precipitate aluminium with this strategy, since this is extremely pH dependent. $$\ce{Al^3+ + 4OH- <=> Al(OH)3 + OH- <=> [Al(OH)4]- }$$ Without a buffered solution at around pH 9 and ageing the solution, there should not form a precipitate. I learned this by adding ammonium sulfide first and raising the pH slowly with ammonia. (This will co-precipitate cobalt, nickel, manganese, zinc, and iron [as sulfides], and chromium [as hydroxide]. This might be useful if you have to analyse for more ions later on.)
If a precipitate forms you can dissolve it with sodium hydroxide and use cobalt(II) nitrate to form Cobalt blue.
You can also test the presence of aluminium with morin as it forms a fluorescent complex with it. Note that $\ce{Mg(OH)2}$ is slightly fluorescent, too.

I cannot remember precipitating magnesium with that strategy in any sufficient amount. I would prefer precipitating it as an ammonium phosphate, which you can try to (re-)crystallise as struvite and look at them with a microscope. $$\ce{Mg^2+ + NH4+ + PO4^3- <=> MgNH4PO4 v ->C[H2O] MgNH4PO4.6H2O}$$

If you want to precipitate out aluminium and magnesium (and almost every other heavier element) you can use oxine, $\ce{C9H7NO}$ (abbr. $\ce{Hox}$), solution. $$\ce{ M^{$x+$} + $x$\,Hox <=> [M(ox)_{$x$}] + $x$\,H+ }$$

Then precipitating calcium is straight forward with carbonate solution. Other earth alkaline metals will precipitate, too. You can dissolve this precipitate with acids or you can use the solid to perform flame tests. Since there can only be calcium, strontium, or barium, this should be conclusive. (Of course you could precipitate barium sulfate.)

The main pitfall with your proposed strategy is, that you will not quantitatively precipitate calcium hydroxide and it is as mentioned above quite concentration dependent. The solubility constant is not too small, i.e. $K_\mathrm{sp}=5.5\times10^{-6}$.

If you have precipitated everything so far, then the last thing left over are ammonium, potassium, sodium(, and lithium). The biggest problem is, you cannot mask them properly and you cannot precipitate them. You can use flame tests to determine their presence, but as long as you have sodium in it you might not see anything else. You should use a prism, or a spectrometer to determine the characteristic spectral lines. Apparently phones can do it now, too. Also see publiclab.org... makes you think why separate at all.

You will need to analyse ammonium from the sample. You can add sodium hydroxide, which releases ammonia. You can detect the presence of ammonia in the sample by holding a wetted universal indicator paper over the sample. Additionally, you should be able to smell it.

In other words ...
... be patient with your analysis as there are more pitfalls, false negatives, false positives, contaminations. Don't use short cuts, except atomic emission spectroscopy.


In Qualitative analysis, cations are arranged in groups. For each group, there are a specific set of reactions that determines the cation. If result are negative, we proceed to next group.

\begin{array}{c|c} \mathbf{Group} & \mathbf{Cation} \\\hline \ce{Zero} & \mathrm{\ce{NH4^+}}\\ \ce{I} & \mathrm{\ce{Pb^2+}} \\ \ce{IIA} & \mathrm{\ce{Pb^2+{,Cu^2+}}}\\ \ce{IIB} & \mathrm{\ce{As^3+}}\\ \ce{III} & \mathrm{\ce{Fe^3+ {,Al^3+}}}\\ \ce{IV} & \mathrm{\ce{Co^2+ {,Ni^2+} {,Mn^2+} {,Zn^2+}}}\\ \ce{V} & \mathrm{\ce{Ba^2+ {,Sr^2+}, {,Ca^2+}}}\\ \ce{VI} & \mathrm{\ce{Na+ {,K+} {,Mg^2+}}}\\ \end{array}

Let's start with $\ce{NH4+}$. A simple test is Nessler's reagent test ($\ce{K2[HgI4]}$) which gives brown ppt.

$$\ce{2K2[HgI4] + NH4OH -> NH2.HgO.HgI↓+ 7I- + 2H2O}$$

Then we proceed to $\ce{Al^3+}$. A test called Lake test is performed which gives blue ppt. on a colorless solution.

$$\ce{Al(OH)3 + 3HCl -> AlCl3 + 3H2O}$$ $$\ce{AlCl3 + 3NH4OH -> 3NH4Cl + Al(OH)3↓}$$

Otherwise perform the cobalt nitrate test as described in Martin's answer.

Then proceed to $\ce{Ca^2+}$. For this, add ammonium oxalate. Explanation for this test is given in aventurin's answer.

For the alkali metal ion, there are indeed specific chemical test but the best test for determining them is the Flame test as described in Jan's answer.

  • Magnesium

Ammonium phosphate test (described in Martin's answer).

  • Sodium

Potassium pyroantimonate test (Milky ppt.)

$$\ce{2NaCl + K2H2Sb2O7 -> 2KCl + Na2H2Sb2O7↓}$$

  • Potassium

Sodium cobaltinitrate test (yellow ppt.)

$$\ce{3KCl + Na3[Co(NO2)6] -> 3NaCl + K3[Co(NO2)6]↓}$$

Or Picric acid test (yellow ppt.)

$$\ce{KCl + C6H6(NO2)3OH -> HCl + C6H2(NO2)3OK↓}$$

  • $\begingroup$ Here is the updated table. $\endgroup$ Commented May 26, 2022 at 3:51

Flame test, flame test, flame test!

I assume that the lab has access to a simple optical spectroscope that will actually help you discern the lines different cation give in the flame test. If it doesn’t have one … Seriously how, because the technique is over a hundred years old.

  • Sodium, of course, gives a very recogniseable yellow double line.

  • Lithium (which is not present — why?) would give red lines where nothing else would.

  • Potassium gives a faint red line at the very edge of the spectrum. It can also be detected by looking at the flame through a blue-coloured glass (it was called Kobaltglas, cobalt glass, in our lab) and noting a different flame colour.

  • Calcium gives ‘traffic lights’, a red, yellow and green line approximately at equal distances of each other.


This is easily detected by just adding 6 M $\ce{NaOH}$ (or whatever the highest non-concentrated solution your lab may have) to the sample, putting a watchglas above with a bit of wet indicator paper stuck to the watchglass’ bottom side. If ammonium is present, the indicator paper will quickly get coloured blue (or whatever the colour of ‘alkaline’ is for your indicator). Make sure that no students ‘accidentally’ added ammonia to the sodium hydroxide (i.e. perform a blind experiment without sample).

The remaining

Which leaves magnesium and aluminium. I personally didn’t have magnesium in my ion lottery class, so I don’t really know how to cope with it. However, your method seems reasonable.

A secondary test for working with aluminium may be either the morin fluorescence (which is disturbed by sodium) or alizarin S precipitation (red). The latter worked like a charm in my ion lottery.

I’ll add a link to this beautiful graph from Professor Klüfers of the LMU Munich which shows the prevalent aluminium species at different pH values. The notation is as follows: $\ce{Al}$ is $\ce{[Al(H2O)6]^3+}$, and an added $\ce{H_{$-n$}}$ shows that this number of protons is missing. Hence, $\ce{AlH_{-3}}$ is equal to $\ce{Al(OH)3}$. Since I do not own the graph and am unsure about its copyright status, I prefer to link rather than upload here.


One additional suggestion for the identification of calcium is the precipitation as calcium oxalate (weddellite).

$$\ce{Ca^{2+} + C2O4^{2-} + 2 H2O-> CaC2O4 \cdot 2 H2O}$$

Adding a solution of ammonium oxalate to the weakly ammoniacal or weakly acidic (buffered acetic adid) solution will form a white crystalline precipitate. The reaction is very sensitive and specific if only $\ce{NH4+}$, $\ce{Na+}$, $\ce{K+}$, $\ce{Mg^{2+}}$, $\ce{Ca^{2+}}$, and $\ce{Al^{3+}}$ can be present.


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