If I have three aqueous ionic solutions in which I know that the cation is $\ce{Al^3+}$, $\ce{Mg^2+}$, or $\ce{Zn^2+}$, how do I find out which is which?
I was thinking to add $\ce{OH-}$ in the form of $\ce{NaOH}$, or something else to produce a precipitate, and testing these precipitates in other liquids, but I'm not sure what other liquids to try.
Answer
First, you add sodium hydroxide to the three different solution. (Remember to safe some of the solution, because we need it for the next test.)
The zinc ion will react with the hydroxide ion in the sodium hydroxide solution to form zinc hydroxide which is white. So, you see white precipitate.
$$\ce{Zn^2+ +2 OH^- -> Zn(OH)2}$$
Adding more sodium hydroxide, the zinc hydroxide will form the zincate ion which is colourless. So, you will see a clear solution.
$$\ce{Zn(OH)2 + 2 OH^- -> Zn(OH)4^2-}$$
The aluminium ion will react with the hydroxide ion in the sodium hydroxide solution to form aluminium hydroxide which is white. So, you see white precipitate.
$$\ce{Al^3+ +3 OH^- -> Al(OH)3}$$
Adding more sodium hydroxide, the aluminium hydroxide will form aluminate ion which is colourless. So, you will see a clear solution.
$$\ce{Al(OH)3 + OH^- -> Al(OH)4^-}$$
The magnesium ion will react with the hydroxide ion in the sodium hydroxide solution to form magnesium hydroxide which is white. So, you see white precipitate.
$$\ce{Mg^2+ +2 OH^- -> Mg(OH)2}$$
However, in this case, it won't form any else by adding more sodium hydroxide. So, you will still see the white precipitate even though you have continued to add sodium hydroxide.
We can differentiate $\ce{Mg^2+}$ solution form the three solutions and now we are left with $\ce{Zn^2+}$ and $\ce{Al^3+}$. Take another sample from the remaining two solutions and add aqueous ammonia solution ($\ce{NH3}$) in the samples.
The zinc ion will react with the ammonia solution to form zinc hydroxide which is white. So, you see white precipitate.
$$\ce{Zn^2+ + 2NH3 + 2H2O <=> Zn(OH)2 + 2NH4^+}$$
Adding more ammonia solution, the zinc hydroxide will form tetraamminezinc(II) ion which is colourless. So, you will see a clear solution.
$$\ce{Zn(OH)2 + 4NH3 -> [Zn(NH3)4]^2+ + 2OH^-}$$
The aluminium ion will react with the ammonia solution to form aluminium hydroxide which is white. So, you see white precipitate.
$$\ce{Al^3+ + 3NH3 + 3H2O <=> Al(OH)3 + 3NH4^+}$$
However, the aluminium hydroxide will not further react with ammonia. So, you will still see the white precipitate even though you have continued to add ammonia.
In conclusion, first put sodium hydroxide in the three solutions (sample in test tube) and keep adding it until only one of the solutions still has white precipitate. That solution is $\ce{Mg^2+}$. Next, take the remaining two solutions (sample in test tube) and keep adding it until one of the solution still has white precipitate. That solution is $\ce{Al^3+}$. And finally the last solution is $\ce{Zn^2+}$.
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