The $K_\mathrm{a}$ or $K_\mathrm{b}$ values provide an idea of how likely a specific step is to happen.

However, how do you specify one acid just being stronger than another, as in the case of $\ce{H2CO3}$ $(\mathrm{p}K_\mathrm{a1} = 3.6,$ but $\mathrm{p}K_\mathrm{a2} = 10.32)$ and phenol $(\mathrm{p}K_\mathrm{a} = 9.95)?$ How do you compare them?

  • 2
    $\begingroup$ Compare pKa1 for both compounds $\endgroup$ Commented Jul 12, 2020 at 10:31
  • $\begingroup$ Isn't it used to measure the likelihood of just the first deprotonation? $\endgroup$
    – harry
    Commented Jul 12, 2020 at 10:54
  • 1
    $\begingroup$ Isn't that what acidity is a measure of? $\endgroup$ Commented Jul 12, 2020 at 10:58
  • $\begingroup$ Acidity is, by definition, the tendency of a compound to act as an H+ donor for Bronsted acids. For Lewis acids, acidity relates to the compound's ability to accept an electron pair. That doesn't necessarily mean the first deprotonation every time, does it? $\endgroup$
    – harry
    Commented Jul 12, 2020 at 11:32
  • 1
    $\begingroup$ It generally is.. atleast to the best of my knowledge $\endgroup$ Commented Jul 12, 2020 at 12:00

2 Answers 2


According to the Brønsted-Lowry acid base theory, an acid is a compound that releases a $\ce{H+}$ ion to give a conjugate base. For most acids, this reaction exists in a dynamic equilibrium. The equilibrium constant of this reaction is what defines $K_\mathrm{a}$. Generally more stable the product, the more of the forward reaction takes place.

Using this last statement, when comparing two compounds, we can say that if the conjugate base of one compound is more stable, it would mean that the value of its $K_\mathrm{a}$ would be greater. This implies its $\mathrm{p}K_\mathrm{a}$ would be less, meaning that the compound would be more acidic than the other.

For example, taking the case of two compounds, $\ce{HCOOH}$ and $\ce{C6H5-OH}$: Here formate ion would have resonance via oxygen whereas phenoxide ion would have resonance via carbon. This makes phenol less acidic than formic acid.

$\mathrm{p}K_\mathrm{a} \ {\text{ of phenol} = 10.0}$

$\mathrm{p}K_\mathrm{a} \ {\text{ of formic acid} = 3.75}$

Now, if we consider polyprotic acids, we can only consider one hydrogen at a time. Successive $K_\mathrm{a}$s decrease in magnitudes of very large orders.

For example, $\ce{H3PO4}$ has $\mathrm{p}K_\mathrm{a1} = 2.12$ whereas $\mathrm{p}K_\mathrm{a2} = 7.21$ and $\mathrm{p}K_\mathrm{a3} = 12.68$.

Hope this answered your question.

  • $\begingroup$ @RahulVerma just trying to follow andelisk's formatting.. $\endgroup$ Commented Jul 12, 2020 at 13:38
  • $\begingroup$ Ah, I didn't noticed it. But, in general, I use $\ce{}$ in most of my equations. I was also suggested in the beginning by some user to use $\ce{} instead of ${}$. Anyways, if @andselisk is using, then it'll be fine. btw, I edited just to make numerals beaultiful. Feel free to make further edits :) $\endgroup$ Commented Jul 12, 2020 at 13:45
  • $\begingroup$ What is the difference between using "dollar\ce{...}dollar", with \ce, and using "dollar{...}dollar" without \ce ? $\endgroup$
    – Maurice
    Commented Jul 12, 2020 at 20:38
  • $\begingroup$ @Maurice, see this answer for mhchem extension, i.e., $\ce{}$ $\endgroup$ Commented Jul 13, 2020 at 8:58

$K_\mathrm{a}$ is the measurement of equilibrium constant when an acid dissociates as: $$\ce{HA <=> H+ + A-}$$ $$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$ The term $\mathrm{p}K_\mathrm{a}$ is defined as $\mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})$. So lower the $\mathrm{p}K_\mathrm{a}$, the better is the acidity.

In the example given, if you wish to compare the acidity, you have to mention which proton of $\ce{H2CO3}$ you are interested to abstract. In other words, $\ce{H2CO3}$ is more acidic then phenol but $\ce{HCO3-}$ is less acidic then phenol.


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