There are ten test tubes in the rack at your disposal (1 – 10) and each test tube contains one of the aqueous solutions of the following salts: $\ce{Na2SO4, AgNO3, KI, Ba(OH)2, NH4Cl, Ag2SO4, Pb(NO3)2, NaOH, NH4I, KCl}$. For identification of the particular test tubes, you can use mutual reactions of the solutions in the test tubes only. Determine in which order the solutions of the salts in your rack are and write chemical equations of the reactions you used for the identification of the salts.

I got the above problem from the very first International Chemistry Olympiad. This intrigued me as I have done qualitative analysis, but not like this. Here's what I thought:

The halides can be identified by $\ce{AgNO3}$ solution, the sulfates can be used to precipitate out sulfates of both barium and lead. Ammonium can be identified by the pungent smell that is evolved after adding NaOH, but I have no idea as to which test tube contains what.

Looking at the possibilities I can choose two test - tubes in $10C_2$ ways which is 45 ways, of which the probability of getting my favoured reaction is fairly low.

Can anyone please explain to me how I am to proceed with this?

  • $\begingroup$ Are you allowed to heat the mixtures, are you allowed to smell them, are you allowed to separate precipitate and supernatant, are you allowed to combine three solution, do you have a lot of empty test tubes at your disposal? $\endgroup$
    – Karsten
    Commented Apr 23 at 16:24

3 Answers 3


I understand the task as there cannot be used external means, just reactions between contents of different tubes. I can be wrong, of course. In such a case, the optimal solution would be rather in the Maurice's answer.

The chemical part of the task is simple. The combinatoric part - not so much. The obstacle is that the isolated reactions will not tell you what is in the particular tube and there could be even more than two candidates.

E.g. yellow precipitate of $\ce{PbI2}$, forming "golden rain" would say one of the tubes contains $\ce{Pb(NO3)2}$ lead and the other either $\ce{KI}$ or $\ce{NH4I}$ . In this case it can be distinguished by ammonia test with the tube with $\ce{NaOH}$ or $\ce{Ba(OH)2}$.

Or a white precipitate means one of the two tubes contains barium, lead or silver and the other tube contains sulfate or chloride.

This answer is not the solution, but the methodical suggestion.

  • The experiment can be simulated by a deliberate choice of what is in what tube.
  • Write down list of all reactions among the listed chemicals, together with expected observations and mutual distinguishing, if applicable.
  • Assign a number to each reaction.
  • Assign a latter to each possible tube content.
  • Create a table with cells for each tube combination. With a big advantage, one could use a system of duplicated Excel tables, with keeping tract of choices, trials and retreats.
  • Test each tube combination.
  • If any reaction is detected, write the reaction number to both tubes. If there are more possibilities, write all of them.
  • When done, replace the reaction numbers by both reactants. Remove duplicates.
  • For each tube, combine candidates from all tests it was involved in. Remove duplicates.
  • In the end, you have for each tube all possible candidates.
  • Now it is kind of like a Sudoku puzzle, making recursive choices, trials and retreats.
  • Make the ordered list of tubes by increasing number of candidates.
  • Start with the one with the least number.
  • As a trial hypothesis, assign one of its candidates to it like if it was the one.
  • Remove this candidate for all other tubes.
  • Repeat it for the rest of tubes.
  • When you reach a collision(e.g. some tube has no candidate left), return to the last choice, choose other candidate and continue elimination.
  • If you spent all candidates for some tube, return to the previous choice, choose other candidate and continue.
  • After some number of choices, trials and retreats, you should succeed.
  • With ambiguous reaction results and not enough other results, there could be more than just one possible problem solution.

  • There is possible also the opposite, perhaps even better method. Use the eliminating, "negative selection" approach, considering initially all candidates for every tube and removing not suitable.
  • E.g. if lead iodide formation is detected for two tubes,
    • remove all candidates for them that contain neither lead nor iodide.
    • And as lead can be only in one of these two tubes, remove the lead nitrate candidate from all other eight tubes.
  • This candidate preparation would finally end at the similar starting "Sudoku" one dimension table with the list of final candidates for each tube.
  • Apply on it the same recursive schema of choices, trials and retreats as above.

I have successfully performed the simulation and verified, the "Sudoku solving" approach of choices, trials and retreat is not needed at all. I have chosen the negative selection - elimination approach. having decreasing amount of candidates for each of tubes. Along eliminations due reaction results, there were parallel eliminations of combinations that could not logically coexist.

E.g. If two tubes gave a white(yellow,brown) precipitate, I can eliminate for these two tube all candidates that do not form a white(yellow, brown) precipitate.

Or if two tubes give a white precipitate and I already know one of them contains silver, I can eliminate for the other one all candidates that do not make white precipitate with silver.

Or if I know one compound must be in one of these two tubes, it cannot be in the other eight tubes.

At some points, even massive eliminations occurred, as they were not very obvious in prior steps.

Here is OneDrive link to the excel sheet for viewing. Just the proof of concept made in hurry, without any fancy formatting nor commentary and does not finished evaluation of all reactions yet. but just 2 tubes with 2 options are left, and the final test and choice is trivial.


I will try to follow Poutnik's suggestion. And try to recognize the solutions without using external reagents. The only allowed mixtures are made by mixing two unknown solutions. Let's consider for the present article that the solutions are numbered $(1)$ to $(10)$, by alphabetic order. So the ten solutions are : $(1)$ = $\ce{AgNO3}$, $(2)$ = $\ce{Ag2SO4}$, $(3)$ = $\ce{Ba(OH)2}$, $(4)$ = $\ce{KCl}$, $(5)$ = $\ce{KI}$, $(6)$ = $\ce{NaOH}$, $(7)$ = $\ce{Na2SO4}$, $(8)$ = $\ce{NH4Cl}$ $(9)$ = $\ce{NH4I}$, $(10)$ = $\ce{Pb(NO3)2}$. But this correspondance is not known to the student. It has to be established by the student, who just knows the different substances, but not their order.

It should be mentioned that $\ce{Ag2SO4}$ is only poorly soluble in water : $0.57$ g in $100$ mL at $15°$C. This is about $0.0018$ M. The silver concentration in saturated $\ce{Ag2SO4}$ solutions (and in $(2)$) is then [$\ce{Ag+}$] = $0.0036$ M. Hopefully the other unknown solutions are more concentrated.

$45$ mixtures hcve to be done with these $10$ unknown solutions. A lot of these mixtures produce a precipitate. Let's look first at the mixtures producing a deep yellow precipitate (typical to $\ce{PbI2}$), which is produced twice : in the mixtures $(5)+(10)$, and $(9)+(10)$. As $(10)$ is used in both mixtures, $(10)$ must contain $\ce{Pb(NO3)2}$. Then $(5)$ and $(9)$ must contain $\ce{KI}$ or $\ce{NH4I}$. The only way of separating both iodides without adding chemicals is to heat and evaporate the pure solutions. If a solid remains in the test tube, it is ($5$), or $\ce{KI}$. If the final dry substance slowly sublimates, it is ($9$), or $\ce{NH4I}$. Now we know three substances : $\ce{Pb(NO3)2}$, $\ce{KI}$ and $\ce{NH4I}$.

Then, some mixtures produce a pale yellow precipitate ($\ce{AgI}$), which is observed in four mixtures : $(1)$ + $(5)$, $(1$) + $(9)$, $(2)$ + $(5)$ and $(2)$ + $(9)$. As we know that $(5)$ and $(9)$ contains the same anion iodide, $(1)$ and $(2)$ must contain $\ce{Ag+}$ as cation. But what is the anion in $(1)$ and $(2)$ ? Impossible to know presently and to separate ($1$) and $(2)$. Which one is $\ce{AgNO3}$ and which one is $\ce{Ag2SO4}$? We'll discover it later on.

Now we will heat all mixtures and try to discover which ones smell ammonia when hot. This will happen in mixtures containing $\ce{NH4+ + OH-}$. This odor is observed in mixtures $(3)$ + $(8)$, $(3)$ + $(9)$, $(6)$ + $(8)$ and $(6)$ + $(9$). As we know that $(9)$ is $\ce{NH4I}$, we know that $(3)$ and $(6)$ are hydroxides. Impossible to separate $\ce{NaOH}$ and $\ce{Ba(OH)2}$ presently. But we see that $(8)$ and $(9)$ contain a cation ammonium $\ce{NH4+}$. As $(9)$ is $\ce{NH4I}$, $(8)$ must be $\ce{NH4Cl}$

Now we will look at mixtures using (7). It is worth noticing that $(7)$ produce only two (white) precipitates with unknown substance : $(3)+(7)$, and $(10)$ + $(7)$. As $(10$) is $\ce{Pb(NO3)2}$ and $(3)$ is either $\ce{NaOH}$ or $\ce{Ba(OH)2}$, we might assume that $(3)$ is $\ce{Ba(OH)2}$ to produce $\ce{BaSO4}$ when mixed with $(7)$. As a conséquence, $(6)$ is $\ce{NaOH}$.

Now, $7$ solutions are known : $(3)$ = $\ce{Ba(OH)2}$, ($5$) = $\ce{KI}$, $(6) = \ce{NaOH}$, $(7) = \ce{Na2SO4}$, $(8) = \ce{NH4Cl}$, $(9) = \ce{NH4I}$. $(10)$ = $\ce{Pb(NO3)2}$

To sépare $(1)$ from $(2)$, we mix both with $(3)$. Only $(2)$ makes a precipitate with $(7)$. So $(2)$ is $\ce{Ag2SO4}$ and $(1)$ is $\ce{AgNO3}$.

The last solution to be analyzed is $(4)$ which, by elimination, must be $\ce{KCl}$. It would make precipitates when mixed with silver salts $(1)$ and $(2)$, and maybe with $(10)$, if the concentration of $\ce{PbCl2}$ is sufficient in the mixture (higher than $0.67$ g in $100$ mL = $0.0024$ M.


This task is a typical sort of problem used in the International Chemistry Olympiads. It allows you to quickly judge if the candidate knows the main reactions of usual ions. It looks like you have ten tubes filled with ten different solutions. You also get ten labels, and you have to find which label goes on which tubes. There are many possibilities to solve this problem. Here is one.

I would first define the pH range by using some blue-red indicator paper.

If the solution is blue (basic), it contains $\ce{NaOH}$ or $\ce{Ba(OH)2}$. To distinguish between both, you remove some drops of each solution, and add some drops of any sulfate solution. If it produces a precipitate, the solution contains $\ce{Ba(OH)2}$. If not, the solution contains $\ce{NaOH}$. Put the labels $\ce{NaOH}$ or $\ce{Ba(OH)2}$ on the corresponding tubes. Remove these tubes.

If the solution is not blue, look for halides. Remove some drops of each remaining solution and add some drops of $\ce{NaCl}$ solution. It should produce a white precipitate with only three solutions : $\ce{AgNO3, Ag2SO4, Pb(NO3)2}$. Decide which is which by adding a big amount of concentrated $\ce{HCl}$. If the solution contained $\ce{Pb}$, the precipitate $\ce{PbCl2}$ would get redissolved into $\ce{H2PbCl4}$. Nothing new will happen with silver solutions. This test tells you where is the $\ce{Pb^{2+}}$ solution. To distinguish between $\ce{Ag2SO4}$ and $\ce{AgNO3}$, you remove some new drops of both initial silver solutions, and you add some drops of $\ce{Ba(NO3)2}$ to both solutions. A precipitate should appear only in the $\ce{Ag2SO4}$ solution. Put the labels on newly discovered solutions.

Now you have discovered where are $\ce{NaOH, Ba(OH)2, Pb(NO3)2, AgNO3, Ag2SO4}$ Five solutions remain to be identified : $\ce{Na2SO4, KI,NH4Cl, NH4I and KCl}$. You go on with these five unknown solutions.

The $\ce{Na2SO4}$ will be discovered by picking up a few drops of all five tubes and adding a $\ce{Ba(OH)2}$ solution to all tubes. Only one tube should contain a precipitate : $\ce{Na2SO4}$

The four remaining solutions are difficult to separate : $\ce{KCl, KI, NH4Cl, NH4I}$. A possible approach is to evaporate completely and overheat a few drops of all solutions. Water must be entirely removed. The ammonium compounds will then be vaporized and sublimated in the upper part of the glass tube. Not the potassium salts. When this calcination is done, let cool the tubes, add some drops of an acidified bleach solution ($\ce{NaClO}$ solution). Iodides produce a brown solution or deposit of $\ce{I2}$. Chlorides do not react.

These tests should allow to put the good label to all tubes.

  • $\begingroup$ I understand the task as you cannot use external means, just reactions between content of different tubes. I can be wrong, of course. $\endgroup$
    – Poutnik
    Commented Apr 23 at 10:38

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