Differentiating Ca2+ and Sr2+ ion

$\ce{Ca^{+2}}$ and $\ce{Sr^{+2}}$ both give a bit similar red color in flame test. What is the easiest way (chemical test) to distinguish $\ce{Ca^{+2}}$ and $\ce{Sr^{+2}}$?




The video document given by AChem is not really clear. It should be replaced by the following method, used in my university.


Let's start from $5$ mL of an initial solution containing possibly any one or all of the three ions $\ce{Ca^{2+}, Sr^{2+}, Ba^{2+}}$.


§$1$. Check if this initial solution contains the ion $\ce{Ba^{2+}}$ by taking $3$ drops of the initial solution, adding $3$ drops of $0.1$ M $\ce{K2Cr2O7}$ plus $1$ mL $0.1$ M $\ce{CH3COONa}$. If a yellow precipitate appears, it means that **Barium** is present in solution : Go to $§2$.  

If the solution remains clear, the solution does not contain $\ce{Ba^{2+}}$ : skip §$2$, and go to §$3$ with the rest of the initial solution.


§$2$. Add the following solutions to the rest of the initial solution : first $3$ mL $0.1$ M $\ce{K2Cr2O7}$ solution, plus $5$ mL $0.1$ M $\ce{CH3COONa}$. Filtrate the yellow $\ce{BaCrO4}$ precipitate. The filtrate must be yellow, and does not contain $\ce{Ba^{2+}}$ any more.


§$3$. Add enough $\ce{Na2CO3}$ to make the solution basic (pH > $8$). Heat to boiling temperature, cool, filter and wash the precipitate containing $\ce{CaCO3 + SrCO3}$. Throw away the yellow filtrate.


$§4$. Add enough $\ce{HNO3}$ to dissolve the precipitate. Usually $1$ mL $1$ M $\ce{HNO3}$ is enough. Drop the liquid in a ceramic capsule. Heat it so as to evaporate it totally, and to eliminate all crystallization waters. Then cool it down and add $1$ mL $100$% pure ethanol. Grind the stuff with a glass stick. Filtrate on a small filter paper. Wash twice with $0.5$ mL pure ethanol ($100$%). The residual insoluble part is made of $\ce{Sr(NO3)2}$. See how to use the filtrate in $§6$.


§$5$. The obtained residue of $\ce{Sr(NO3)2}$ is washed several times to eliminate traces of Calcium nitrate. Dissolve it in $3$ mL water. Add $1$ mL 1 M $\ce{H2SO4}$. A white precipitate is produced if the substance contains **Strontium**. Go to §6 with the filtrate.


§$6$. The filtrate contains calcium nitrate. It can be proven by first evaporating to dryness. Then the residue is dissolved in $1$ mL water. $1$ drop $1$ M $\ce{CH3COOH}$ is added to the obtained solution. The $1$ mL $0.1$ M ammonium oxalate is added. A white $\ce{CaC2O4}$ precipitate shows the presence of **Calcium**.


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A good test for distinguishing calcium ions and strontium ions is **potassium ferrocyanide test.**


$$\ce{Ca^2+ + 2K4[Fe(CN)6] -> \underset{white}{K2Ca[Fe(CN)6]} \downarrow}$$


In presence of ammonium chloride, test is more sensitive. In this case, potassium is replaced by ammonium ions in the precipitate. However, the downside of this test is barium and magnesium interfere. You can separate them as discussed [in my other answer.](https://chemistry.stackexchange.com/a/167916/17368). You can find more details on this test on a paper [linked in a previous comment of mine](https://chemistry.stackexchange.com/questions/170053/can-i-use-potassium-ferricyanide-instead-of-potassium-ferrocyanide-for-the-ident#comment357032_170053). However, the best test is checking the solubility of the corresponding sulfates (explained in my other answer).


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