From Wade's Organic Chemistry [1, p. 71]:

In most cases, the $\mathrm{p}K_\mathrm{a}$ of an acid corresponds to the $\mathrm{pH}$ where the acid is about half dissociated. At a lower (more acidic) $\mathrm{pH}$, the acid is mostly undissociated; at a higher (more basic) $\mathrm{pH}$, the acid is mostly dissociated.

I'm having a hard time intuiting this concept. Consider placing $\pu{1 mol}$ of $\ce{HF}$ in $\pu{1 L}$ of water. As the $\ce{HF}$ dissociation reaches equilibrium, I'm assuming the concentration of $\ce{H3O+}$ ions in solution increases, until the maximum concentration of $\ce{H3O+}$ ions are reached at equilibrium. If the concentration of $\ce{H3O+}$ ions in water increases as the dissociation of $\ce{HF}$ proceeds, then wouldn't the $\mathrm{pH}$ of the solution decrease (become more acidic) as the acid dissociates over time?

If any one has any thoughts on how to best interpret the quoted statement, that would be very much appreciated! Any and all new perspectives are welcomed. Thanks in advance for your generous help.


  1. Wade, L. G.; Simek, J. W. Organic Chemistry, 9th ed.; Pearson: Glenview, IL, 2017. ISBN 978-0-321-97137-1.
  • $\begingroup$ In your quote it should have mentioned that this is true only of a weak acid, although this is implied by the mention of pK. Recall that a weak acid is only partly dissociated and so the equilibrium can be pushed 'left and right' as it were by added acid or base. $\endgroup$
    – porphyrin
    May 29, 2018 at 14:01

2 Answers 2


A weak acid (HA) in water: $\ce{HA + H2O <=> H3O+ + A-}$, and thus,

$$K_\mathrm{a} = \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}$$

By Le Chatelier's principle, if you add some $\ce{H3O+}$ from an external source (see Mithoron's comment above) to increase $\mathrm{pH}$, the equilibrium favors the backward reaction in order to reduce some $\ce{A-}$. Consequently, the amount of the dissociation of acid decreases and $K_\mathrm{a}$ will remain constant.

In similar manor, if you decrease $\mathrm{pH}$ of the solution by adding some $\ce{OH-}$, that will react with equal amount of $\ce{H3O+}$. Therefore, the equilibrium favors the forward reaction in order to produce some $\ce{H3O+}$ to keep $K_\mathrm{a}$ constant. Accordingly, $\ce{[A-]}$ in the solution increases while $\ce{[HA]}$ decreases, meaning the amount of the dissociation of acid increases. At one point, $\ce{[A-]} = \ce{[HA]}$, and hence, $K_\mathrm{a} = \ce{[H3O+]}$. At that point, therefore, $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$.

Applying this to your example ($\ce{HA} = \ce{HF}$), $\mathrm{p}K_\mathrm{a}$ of $\ce{HF}$ is $3.14$ ($K_\mathrm{a} = 7.2 \times 10^{-4}$).

When $\pu{1.0 mol}$ of $\ce{HF}$ is dissolve in $\pu{1.0 L}$ of deionized water (no added external acid to adjust $\mathrm{pH}$), $\ce{[HF]} = \pu{1.0 mol\cdot L^{-1}}$ and $\ce{[F-]} = \ce{[H3O+]} = \alpha \gg 1.0$. Thus, $$\ce{[H3O+][A-]} = K_\mathrm{a} \cdot \ce{[HA]}$$ $$\alpha^2 = K_\mathrm{a} \cdot 1.0~~~~\text{or}~~~~\alpha = \sqrt{K_\mathrm{a}}= \sqrt{7.2 \times 10^{-4}}= 2.7 \times 10^{-2}$$

Therefore $\mathrm{pH}$ of the solution is as follows:

$$\mathrm{pH} = -\log \ce{[H3O+]} = -\log \alpha = -\log 2.7 \times 10^{-2}= 1.57$$

Thus, at $\mathrm{pH} = 1.57$, the ionization of the solution is: $\ce{[F-]} = \ce{[H3O+]}= \pu{2.7 \times 10^{-2} mol \cdot L^{-1}}$.

Now, if you added an external base to the solution to increase its $\mathrm{pH}$ until it becomes $\mathrm{pH} = 3.14$, then, at $\mathrm{pH} = 3.14$, $\ce{[F-]} = \ce{[HA]}= \pu{0.5 mol \cdot L^{-1}}$.

  • $\begingroup$ Aaah, I understand now! The quote was referring to an how an external or added acid / base would influence the dissociation of the weak acid. I thought it was referring to how the pH evolves as HF dissociates (ex: how pH changes as a function of time when 1 mol of HF dissolves in 1 L of deionized water). Thanks so much for clearing this up and for going into great detail. $\endgroup$
    – avp29
    May 31, 2018 at 3:33
  • $\begingroup$ @Mathew Mahindaratne> if you decrease or "increase" $\pu{pH}$ of the solution by adding some $\ce{OH−}$ $\endgroup$ Feb 11, 2019 at 6:17
  • 1
    $\begingroup$ @Adnan AL-Amleh: When you add a strong base such as $\ce{OH-}$, it reacts completely with $\ce{H3O+}$ in the solution. As a consequence, the equilibrium is disturbed and shifted right to get back to equilibrium. It means acid dissociate more, but increased $\ce{[H3O+]}$ is much lower than lost of $\ce{[H3O+]}$ by reacting with $\ce{[OH-]}$ so pH increases. $\endgroup$ Feb 12, 2019 at 19:33
  • $\begingroup$ Maybe need to correct:">In similar manor, if you decrease $\pu{pH}$ of the solution by adding some $\ce{OH−}$" $\endgroup$ Feb 13, 2019 at 0:10

RE: "In most cases, the pKa of an acid corresponds to the pH where the acid is about half dissociated. ..."

In general the definition of pKa is the pH at which $\ce{[HA] = [A^-]}$. The only reasons that this would not be true is if:

(1) The pKa is less than 0 since you can't have an aqueous solution with a pH < 0. Thus if the pKa < 0, then the acid is considered to be completely dissociated in aqueous solution. But remember that one drop of water into a liter of concentrated sulfuric acid isn't an aqueous solution.

(2) You allow for the fact that it is really activities that are at equilibrium not concentrations. So for solutions that are more than about 0.1 molar activity and concentration for a species are not the same.


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