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Why is boiling point of sulfuric acid much higher than that of phosphoric acid?

According to the data book,

• The boiling point of sulfuric acid is 337 °C (639 °F; 610 K) When sulfuric acid is above 300 °C (572 °F), it will decompose slowly.

• The boiling point of phosphoric acid is 158 °C (316 °F; 431 K) When phosphoric acid is above 213 °C (415 °F; 486 K), it will decompose slowly.

However, sulfuric acid can only form 2 hydrogen bonds per molecule but phosphoric acid can form 3 hydrogen bonds per molecule. While both of them are strong acids which can undergoes complete ionization, what other factors that I've ignored may lead to the fact that boiling point of sulfuric acid is much higher than that of phosphoric acid?

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Phosphoric acid is not a strong acid and it does not undergo complete deprotonization.

Both acids have a similar molar mass of around 98 grams per mole.

Another property that should be considered is the formation of a net dipole moment on the molecule. With the phosphoric acid 3D structure attached, each of the three oxygen atoms in the $\ce{-OH}$ group pull with equal strength, cancelling each other out, leaving an net dipole moment going upwards coming from the double bonded oxygen.

With sulfuric acid however, both of the hydrogen bonding groups pull in relatively the same direction, while the double bonded oxygen atoms pull in the opposite direction, forming a stronger net dipole.

This is analogous to the net dipole properties of chlorinated methanes. See the attached image and note the similarities in their structures and the structures of the acids we are discussing.

enter image description here

Another key reason for the difference in boiling point is the autoprotolysis of sulfuric acid. Even without the presence of water, sulfuric acid can lose a proton as shown in this equation:

$$\ce{2H2SO4 <=> H3SO4+ + HSO4-}$$

Phosphoric acid is a weak acid and does not have such interactions and only loses protons (in equilibrium) in the presence of water. The autoprotolysis is important because through autoprotolysis, sulfuric acid now forms ion-ion interactions, which are much stronger interactions than hydrogen bonding.

Therefore, as a combination of a slightly stronger dipole moment and more so due to autoprotolysis, sulfuric acid has stronger intermolecular forces, and thus a higher boiling point.

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    $\begingroup$ Pure H3PO4 also undergoes autoprotolysis and to even greater extent then H2SO4. Behavior of pure substances has little in common with their interactions with water, when they are highly diluted. $\endgroup$ – Mithoron Mar 10 '16 at 17:47
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    $\begingroup$ @Mithoron So, does this mean the pKa of $\ce{H3PO4}$ in $\ce{H3PO4}$ is higher than the pKa of $\ce{H2SO4}$ in $\ce{H2SO4}$? I've looked for a source for that but haven't been able to find any, do you have one, by chance? I trust that what you're saying is true, but if there is a consolidated source for this, I'd find it handy. $\endgroup$ – SendersReagent Mar 11 '16 at 0:50
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    $\begingroup$ @DGS books.google.pl/… It's autoprotolysis is probably strongest of all known acids, but rather because of higher basicity not acidity (gets protonated easier). $\endgroup$ – Mithoron Mar 11 '16 at 1:08
  • $\begingroup$ what does the molar mass have to do with the BP? $\endgroup$ – colnegn Mar 13 '16 at 18:23
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    $\begingroup$ @Howsikan Mass turns out not to be the best indicator for dispersion forces because they are exclusively an electronic effect (think about isotopes--albeit molar mass takes care of that issue). Look up a table of atomic polarizabilities, and you will find that the trend is down-left on the PT, and is reflected in the atomic radius and ionization energy. Regarding molecules: effects from things like the molecule's surface area must be considered: two isomers of C16H32 boil at 513 and 554 K; 9 & 16 C main chains, respectively. You might imagine polarizability as how 'floppy' the species is $\endgroup$ – colnegn Mar 17 '16 at 10:14
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In addition to the factors Howsikan mentioned, my intuition leads me to believe there's a parallel with the boiling points of $\ce{NH3}$ and $\ce{H2O}$. I think the oxos ($\ce{O=P}$ and $\ce{O=S}$) are better acceptors than the hydroxyls, so the acids would prefer $\ce{OH...O=}$ over $\ce{OH...OH}$. $\ce{H3PO3}$ has three oxos to one hydroxyl whereas $\ce{H2SO4}$ has two of each, so $\ce{H2SO4}$ can create a favourable network more easily.

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