0
$\begingroup$

Suppose two compounds with, say, $\mathrm{p}K_{\mathrm{a}1}=2.8$ and $\mathrm{p}K_{\mathrm{a}2}=3.2$, and formulae $\ce{C_{$a$}H_{$b$}O_{$c$}}=\ce{C_{$p$}O_{$q$}(OH)_{$r$}}$ and $\ce{C_{$d$}H_{$e$}O_{$f$}}=\ce{C_{$u$}O_{$v$}(OH)_{$w$}}$, respectively, with given $abcdef$, $pqruvw$. Can we explain the above $\mathrm{p}K_\mathrm{a}$ with (ONLY) the knowledge of the given numbers of atoms of $\ce{C}$, $\ce{H}$ and $\ce{O}$ using the theory of Usanovich, and no knowledge of the internal chemical structure of the chemical bonds inside the two molecules?

$\endgroup$
5
$\begingroup$

Can we explain the above pka with (ONLY) the knowledge of the given numbers of atoms of C, H and O using the theory of Usanovich

No, nobody's theory (including Usanovich's) can explain $\ce{pK_{a}}$ differences based on molecular formula coefficients. This is best illustrated by the observation that even compounds with the same coefficients (same values for "abcdef") can have different $\ce{pK_{a}}$ values. For example, $\ce{C3H6O}$

\begin{aligned} \ce{&acetone &&pK_{a}=20}\\ \ce{&allyl~ alcohol &&pK_{a}=$15.5$}\\ \end{aligned}

$\endgroup$
  • $\begingroup$ Even if the two molecular compounds have DIFFERENT C,H and O contents? Your example is a same molecular formula with 2 differents pka, I was arguing about two different compounds (note the numbers/letters were deliberately different). $\endgroup$ – riemannium Jul 3 '14 at 21:37
  • 3
    $\begingroup$ Yes, I understand. My point is, if it fails - as it must - for the same empirical formula, then it must also fail for different empirical formulas. $\endgroup$ – ron Jul 3 '14 at 21:40
  • $\begingroup$ And my point is, that, according to Usanovich theory, acid or base character is defined by the ability of donating/recieving positive/negative species. If you have a compound with differente number of atoms, in principle, there is no match between the possible number of positive/negative ions and possible charges. I can not see your point, ron... $\endgroup$ – riemannium Jul 4 '14 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.