# finding cation of unknown salt in solution from precipitate

I should probably add that I was originally given an unknown solid mixture of 2 salts. I separated the insoluble salt out by centrifuging and then decanting. I found that the insoluble salt is $\ce{MgCO_3}$ by adding $\ce{HCl}$ (bubbles) and $\ce{NaOH}$ into the solution (jelly like white ppt). The decanted liquid was then what I observed as having 1 soluble salt. The soluble salt I know by testing for $\ce{SO4^{2-}}$ with $\ce{BaCl_2}$ and observing a violet flame in a flame test. Then I had a second bottle with 3 unknown, 2 of which I discovered above. I know that the 3rd one has 2 soluble and then my original question follows.

I have 2 soluble salts in aqueous distilled water solution. I know that one of the salts is potassium sulfate ($\ce{K_2SO_4}$) and I know that the anion of the second salt is chloride ($\ce{Cl^{-}}$). I know that the second salt is either magnesium, sodium, calcium or potassium chloride ($\ce{MgCl_2}$, $\ce{NaCl}$, $\ce{CaCl_2}$, or $\ce{KCl}$).

I tested for the anion for the second salt by adding silver nitrate ($\ce{AgNO_3}$) into the solution and observing a white precipitate. When I added sodium hydroxide ($\ce{NaOH}$) to it, I observed that the cloudy white precipitate disappeared and turned into small black precipitate which then settled to the bottom of the tube. Is there any way to get the cation of the second salt with this information, and are there any other test(s) I can do to find what is the cation for the second salt?

Among the options for the second salt in solution, one is less likely to be found in the presence of $\ce{K_2SO_4}$.

These solutions for analysis are prepared by adding decent amounts of two salts with good solubility to water. Under these conditions, a double displacement reaction may occur and in one case, a precipitate will be formed:

$\ce{K_2SO_4 + CaCl_2 -> 2KCl + CaSO_4}$

I doubt that the person preparing the solutions for analysis used calcium chloride, since

this would have required to remove the precipitate by filtration or decanting off the supernatant solution the concentration of $\ce{Ca^{2+}}$ in the solution would be rather low. Remember that these qualitative tests are not necessary suitable for trace analysis.

The second salt therefore is most likely sodium chloride ($\ce{NaCl}$), potassium chloride ($\ce{KCl}$) or magnesium chloride ($\ce{MgCl_2}$).

I would do a flame test first. If you see a yellow coloured flame, the second cation is the sodium cation ($\ce{Na+}$). Similarly if you see a lilac coloured flame, the second cation is the potassium cation ($\ce{K+})$ and if you initially see a colourless flame and finally see an intense white coloured flame,the second cation is the magnesium cation ($\ce{Mg^{2+}}$). If you want to test for the magnesium cation ($\ce{Mg^{2+}}$), place a drop of the solution on a spot plate, make sure it's alkaline by adding aqueous $\ce{NaOH}$ if necessary, and add a bit of a titan yellow solution. This azo dye with its misleading name (there is no titanium in it) will form a red precipitate with the magnesium cation ($\ce{Mg^{2+}}$).

• Note: yellow flame is not necessarily sodium! Sodium flame coloring is so intense it is known to overwhelm all other colors. So, if you get violet (potassium) or red (calcium), it is valid, but yellow may arise from trace amounts of sodium! May 8, 2015 at 11:55
• Thanks for the suggestions. I should probably add that I was originally given an unknown solid mixture of 2 salts.I separated the insoluble by centrifuge and decanting. I found that the insoluble salt is MgCO3 (by adding HCl (bubbles) and NaOh into that(pcp jelly like white)). The decanted liquid was then what I observed as having 1 soluble salt. The soluble salt I know by testing for SO4 with BaCl2 and observing a violet flame in a flame test. Then I had a second bottle with 3 unknown, 2 of which I discovered above. I know that the 3rd one has 2 soluble and then my original question follows. May 9, 2015 at 7:20
• UPDATE: Looks like I don't have access to Titan Yellow, I will try to acquire either Azo Violet or 4-(p-nitrophenylazo)resorcinol May 13, 2015 at 6:19