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Consider this graph of the titration of a weak acid and a strong base:

Graph of the titration of a weak acid and a strong base

The buffering region is strongest at the midpoint of titration because there is equal concentrations of acetic acid and its conjugate base.

My question is why at the midpoint of titration does the concentration of the acid equal the concentration of its conjugate base? Also, why does the $\mathrm{p}K_{a}$ of the acid = the $\mathrm{pH}$ at this point?

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    $\begingroup$ Using as an example the monoprotic acid HA, by definition the pKa of HA is the pH at which $\ce{[HA] = [A-]}$. $\endgroup$
    – MaxW
    Feb 9, 2020 at 9:45
  • $\begingroup$ Can you explain it please, I don't know what happens. @Karl $\endgroup$
    – Positron12
    Feb 9, 2020 at 13:46
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    $\begingroup$ A + B -> C, OK? How much B do you have to add so that A is eqal to C. $\endgroup$
    – Karl
    Feb 9, 2020 at 14:16

1 Answer 1

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At the midpoint of titration, the concentration of acid is just approximately equal to its conjugated base, as there must be considered the acid dissociation and ( in less extend ) salt hydrolysis.

If we take the definition(simplified using concentrations):

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

we can see that by definition:

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

$\mathrm{p}K_\mathrm{a}$ is $\mathrm{pH}$ when both items of the conjugate pair have the same concentration.

$\mathrm{pH}$ near titration midpoint is approximately equal

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log { \frac {n_\mathrm{base}}{ n_\mathrm{acid, init} - n_\mathrm{base} }}$$

This function has minimal slope = maximal buffer capacity at $$n_\mathrm{acid, init} = 2 \cdot n_\mathrm{base}$$

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  • $\begingroup$ But still, why is just in the midpoint that the acid is equal to the base? $\endgroup$
    – Positron12
    Feb 9, 2020 at 13:44
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    $\begingroup$ @Positron12 If you pass a half of your journey, does the passed segment have the same length as the remaining one, or different ? At the midpoint, the half of acid is neutralized, forming the salt, which is its conjugated base, being in ratio 1:1 to the acid. $\endgroup$
    – Poutnik
    Feb 9, 2020 at 13:49
  • $\begingroup$ Then the same should apply at the equivalence point, why doesn't pKa = pH at it then? $\endgroup$
    – Positron12
    Feb 9, 2020 at 14:00
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    $\begingroup$ Why should it ? Is at the equivalence the acid/base conjugated pair at 1:1 ratio ? There is almost no acid there. $\endgroup$
    – Poutnik
    Feb 9, 2020 at 14:14

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