Compare the acidity $(K_\mathrm{a1})$ of the following three acids: $\ce{H3PO4},$ $\ce{Ph-PO(OH)2}$ and $\ce{H3C-PO(OH)2}.$


These three acids are of type $\ce{R-PO3H2},$ where $\ce{R} = \ce{OH}, \ce{Ph}, \ce{CH3}.$ The ionization equations of them are

$$\ce{R-PO3H2 <=> R-P(=O)(OH)O- + H+}$$

Comparing with hydrogen, benzene ring can reduce the charge of the anion because of the delocalised $\pi$ bond. Also, methyl group is an electron donating group, which will destabilize the anion. Hence, the $\mathrm{p}K_\mathrm{a1}$ I suggest is $\ce{CH3} > \ce{OH} > \ce{Ph}.$ This answer is analogue to $\ce{R-SO3H}.$

One thing I'm not sure is that this logic does not apply to $\ce{R-CO2H}.$ Clearly, the acidity of carbonic acid is weaker than methanoate acid. I don't know which explanation should be applied in the situation of phosphorus.

  • 2
    $\begingroup$ "Clearly, the acidity of carbonic acid is weaker than methanoate acid" - not really, it's slightly stronger, actually, it's just that it's equilibrium with CO2 messes things up. en.wikipedia.org/wiki/Carbonic_acid $\endgroup$ – Mithoron Dec 15 '19 at 16:29

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