$\ce{NH4+}$: $K_\mathrm{a} = 5.6 \times 10^{-10}$;
$\ce{HPO3^2-}$: $K_\mathrm{a} = 4.5 \times 10^{-13}$

My understanding is that $K_\mathrm{b}(\ce{A-}) = K_\mathrm{w} / K_\mathrm{a}(\ce{HA})$. Thus:

$\ce{NH3}$: $K_\mathrm{b} = 1.8 \times 10^{-5}$; $\ce{PO3^2-}$: $K_\mathrm{b} = 2.2 \times 10^{-2}$

Since the $K_\mathrm{b}$ for phosphate is greater than the $K_\mathrm{b}$ for ammonia, phosphite is a stronger base. However, my professor's solution lists ammonia as the stronger base. Is his solution incorrect?

  • 4
    $\begingroup$ That's phosphite, not phosphate. $\endgroup$ – Ivan Neretin Sep 12 '15 at 23:58
  • $\begingroup$ Also, in my book the pKa of ammonium is about 9. Not that it would change much, though. $\endgroup$ – Ivan Neretin Sep 13 '15 at 0:49
  • $\begingroup$ A bit late but welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. I improved the formatting of your post by using MathJax; for more information on how to do so yourself, check out the help center, this meta post and this one. I also corrected the $\mathrm{p}K_\mathrm{a}$ of ammonia to what it says in my and Ivan’s books and Wikipedia. Ivan was right, it doesn’t change much. $\endgroup$ – Jan Nov 12 '15 at 9:35

Your professor could be wrong, or rather could mean it in some subtly different context that would change everything. The way you put it here, you are right. Normally the constants would tell you the truth. I see no reason to believe this case is an exception.

Besides constants, there is actually a simple way of comparing two acids: let them fight each other to the death. Let us consider $\ce{NH_4^+}$ vs. $\ce{PO_4^{3-}}$ (I know $\ce{PO_4^{3-}}$ is not what you asked; this is just an illustration). Who would end up having that proton? It turns out that ammonium phosphate is unstable, which implies that the proton would go from $\ce{NH_4^+}$ to $\ce{PO_4^{3-}}$, meaning that $\ce{NH_4^+}$ is stronger acid than $\ce{HPO_4^{2-}}$ (in accordance with what the constants would predict), and hence ammonia is a weaker base than phosphate. Now, that was phosphate, the salt of phosphoric acid, while you ask about phosphite, the salt of phosphorous acid, which by all accounts should be weaker. Hence its anion must be an even stronger base.

Simple as it may seem, there is one more thing to mention. Phosphorous acid is actually diprotic, that is, capable of ionizing only two of its hydrogens. The third hydrogen is connected directly to the phosphorus atom. Organic esters $\ce{(RO)_3P}$ are known, but that's another story. The very existence of ion $\ce{PO}_3^{3-}$ seems somewhat dubious. By any means, even if you manage to get one in water solution, it will be instantly hydrolized. This puts it in the ranks of bases too strong to exist in water, which is of course way stronger than our lowly ammonia.


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