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I was asked to compare the acidic strength of the following phosphorus oxoacids,

enter image description here

I thought the order (of strength) would be $\ce{H3PO4 > H3PO3 > H3PO2}$, because they have three, two and one ionizable $\ce{O-H}$ bonds (that furnish protons / $\ce{H^+}$ ions).

However, looking up the $\pu{pK_a}$ values on the respective Wikipedia pages of each of those acids, suggests the exact opposite order of acidic strength: $\ce{H3PO2 > H3PO3 > H3PO4}$. Apparently the acid with one $\ce{O-H}$ bond is a far stronger than the acid with three $\ce{O-H}$ bonds!

This is horribly counter-intuitive :-(

So, why is the correct order of acidic strength $\ce{H3PO2 > H3PO3 > H3PO4}$, and not the other way around as I believed? What was wrong with my analysisi ( acidity in oxoacids $\propto$ number of $\ce{O-H}$ bonds)?

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marked as duplicate by Mithoron, Community Jan 14 '18 at 20:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ (+1) It's possible that this question should be separated into two parts. The $\pu{p}K_\pu{a}$ values are measured in water where tautomerisation is significant. To directly compare the acidities of the depicted compounds we should move into the gas phase, and compare the $\Delta G_\pu{r}^\circ$ values of deprotonation. Sadly, I only found data for $\ce{H3PO4}$ as shown in OP, written as $\Delta_\pu{r} G^\circ = 1351\ \pu{kJ mol^{-1}}$, doi: 10.1063/1.473465. Further information might be available from computational studies. $\endgroup$ – Linear Christmas Jan 14 '18 at 16:30
  • $\begingroup$ It was asked maybe half a dozen times already, check out the site before asking. $\endgroup$ – Mithoron Jan 14 '18 at 19:40
  • $\begingroup$ There was also chemistry.stackexchange.com/questions/38034/… and terrible chemistry.stackexchange.com/questions/65146/… $\endgroup$ – Mithoron Jan 14 '18 at 20:10
  • $\begingroup$ @Mith Ah, I see :-( $\endgroup$ – paracetamol Jan 14 '18 at 20:10
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    $\begingroup$ There was a better duplicate I think. If I find it mod can add second one. It's recurring question, because it's defying common "rules". $\endgroup$ – Mithoron Jan 14 '18 at 20:13
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Acidic strength of the acids is determined on the relative tendency to donate one proton.

The source where you have seen then $\mathrm{p}K_\mathrm{a}$ value has considered the tendency of first deprotonation of the acids ($\ce{H3PO2}$ deprotonates once, $\ce{H3PO3}$ can deprotonate twice and $\ce{H3PO4}$ can deprotonate thrice).

The double bonded oxygen has negative inductive effect (-I). And this electron withdrawing effect is experienced by one $\ce{H}$ of the one $\ce{-OH}$ group in $\ce{H3PO2}$. Hence it has greatest acidic strength.

Similarly, the -I effect of $\ce{O}$ is experienced by two $\ce{H}$ of the two $\ce{-OH}$ groups in $\ce{H3PO3}$, hence electron withdrawing effect is reduced because it is getting distributed between two groups. Hence, it has lesser acidic strength. Similarly, in $\ce{H3PO4}$, it is distributed between 3 groups and so the acidic strength is least.

Thus the acidic strength, is: $$\ce{H3PO2 > H3PO3 > H3PO4}$$ (I hope it's clear.)

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    $\begingroup$ I'm not convinced by this explanation. Do you have a source? $\endgroup$ – orthocresol Jan 14 '18 at 15:14
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    $\begingroup$ I second what @ortho said. It would be helpful if you could list a source that mentions this. Thanks! :-) $\endgroup$ – paracetamol Jan 14 '18 at 16:14
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    $\begingroup$ @paracetamol Actually, I studied this in my textbooks and also my teacher told this while studying various applications of inductive effect, Mesomeric effect, hyperconjugation, etc. This is a basic concept. But, I think I have the name of a popular book in which I read it.... I'm checking out in google books. I'll upload a screenshot, if possible, shortly. $\endgroup$ – DJ Koustav Jan 14 '18 at 16:36
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    $\begingroup$ Why can't this explanation be applied to the oxyacids of sulfur, namely sulfuric acid and sulfurous acid? $\endgroup$ – Tan Yong Boon Jan 14 '18 at 16:39
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    $\begingroup$ This answer makes no sense, it's just not working like this and this question can't be answered on basis of inductive/mesomeric effects. $\endgroup$ – Mithoron Jan 14 '18 at 20:08

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