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:
Add excess 3 M ammonia to the sample.
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.
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.
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.
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.
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.
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 $\ce{NaOH}$?
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.
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?
