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I'm having trouble interpreting the results I got for my experiment.

As you should be well aware of, lithium and strontium both burn with crimson red when they are placed in a flame. Now, I can't decide which one it is.

Then I was told to do an experiment with the same salt where I had to add dilute hydrochloric acid and barium chloride. I had no visible change.

Does this last test help me with identifying between strontium and lithium? What element is it?

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    $\begingroup$ What are the salts you used? Generally in differentiating lithium and strontium you can have another reagent such as hydrochloric acid or barium chloride and create a solid precipitate. $\endgroup$
    – user3735
    Commented May 6, 2014 at 22:53
  • $\begingroup$ @Jun-GooKwak Hi sir, We were given an unknown salt and have to work out the what cation and anion are available. We were only allowed to do two test which I have describe above in the question. Yes, I added hydrochloride acid and barium chloride and I had no change. What does that mean? Is it Lithium that is present or Strontium? $\endgroup$
    – The DON
    Commented May 6, 2014 at 22:58
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    $\begingroup$ I really do not see, that $\ce{LiCl2}$ or $\ce{SrCl2}$ are insoluble. They should be separable with $\ce{(NH4)2CO3}$. $\endgroup$
    – Martin - マーチン ♦
    Commented May 7, 2014 at 1:48
  • $\begingroup$ I'd concur with Martin's last comment: differentiating them by addition of carbonate seems logical, as the strontium salt should precipitate out. $\endgroup$
    – Greg E.
    Commented May 7, 2014 at 4:59

2 Answers 2

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If your solution contains only $\ce{Li+}$ or $\ce{Sr^2+}$, you can differentiate them by adding a sulfate solution (for example, $\ce{Na2SO4}$ or diluted sulfuric acid). If strontium is present, a white, fine-crystalline precipitate of $\ce{SrSO4}$ will form. In the case of lithium, the solution will remain clear. This is because $\ce{SrSO4}$ is much less soluble in water than $\ce{Li2SO4}$ ($\pu{0.0132 g}$ versus $\pu{34.8 g}$ in $\pu{100 g}$ of $\ce{H2O}$ at $\pu{20 ^\circ C}$ (reference)).

Adding diluted $\ce{HCl_{aq}}$ and $\ce{BaCl2}$ to a solution is commonly used as a test for $\ce{SO4^2-}$ (if it is present, $\ce{BaSO4}$ precipitates). Maybe this could help you to distinguish between $\ce{Li+}$ and $\ce{Sr^2+}$ given the different solubilities of their sulfates. If your unknown salt readily dissolves in water and $\ce{SO4^2-}$ is found to be present in the solution, then it is likely that you have lithium sulfate. Strontium sulfate as unknown salt would not readily dissolve and therefore not give a positive test.

However, if your solution can contain more ions besides $\ce{Li+}$ or $\ce{Sr^2+}$, I would suggest to try the separation with ammonium carbonate which is already mentioned in the comments to your question.

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http://en.wikipedia.org/wiki/Solubility_table
http://www.chemicalbook.com/ChemicalProductProperty_EN_CB6723042.htm
http://www.csudh.edu/oliver/chemdata/data-ksp.htm
http://www.chemicalbook.com/ProductChemicalPropertiesCB1184992_EN.htm

Lithium chromate is listed as being exceptionally soluble in water. Strontium chromate is listed as being insoluble.

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    $\begingroup$ The OP clearly states that there is no chromate involved. $\endgroup$
    – Martin - マーチン ♦
    Commented May 7, 2014 at 1:49
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    $\begingroup$ Add a drop of potassium chromate solution on a white drop plate. Precipitate means strontium; no precipitate means lithium. If two horses differ in height by only a quarter hand, how do you tell the black horse from the white horse? $\endgroup$
    – Uncle Al
    Commented May 8, 2014 at 14:52

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