Below is an image from my textbook representing the titration curve for the titration of $5.00\ \mathrm{mL}$ of $0.010\ \mathrm M$ $\ce{H2S2O3}$ with $0.010\ \mathrm M$ $\ce{KOH}$, which has $\mathrm pK_\mathrm{a1} = 0.6$ and $\mathrm pK_\mathrm{a2} = 1.74$.

The Henderson–Hasselbalch equation implies that at the first half equivalence point, $\mathrm{pH} = \mathrm pK_\mathrm{a1}$, and at the second half equivalence point $\mathrm{pH} = \mathrm pK_\mathrm{a2}$, because there are equal changes of the acid and conjugate base.

This means that at the first buffer region b, I should expect the $\mathrm{pH} = \mathrm pK_\mathrm{a1}=0.6$ and at the second buffer region d, I should expect the $\mathrm{pH} = \mathrm pK_\mathrm{a2} = 1.74$. However, this is clearly not the case with the diagram.

What am I missing here? Is it because I have to account for the autoionization of water in the dilute solution? How would I do so? enter image description here

  • 1
    $\begingroup$ How could you have pH=0.6 if concentration was only 0.01 to begin with? $\endgroup$ – Mithoron Aug 3 '18 at 22:14
  • $\begingroup$ @Mithoron So pKa doesn't always equal pH at the half equivalence point? $\endgroup$ – DrPepper Aug 3 '18 at 23:03
  • $\begingroup$ @KaienYang - Even if both protons ionized you could only get a acid concentration of 0.02 molar which is only a pH of 1.69. In other words the Henderson–Hasselbalch equation doesn't apply in this case. $\endgroup$ – MaxW Aug 3 '18 at 23:27
  • 1
    $\begingroup$ @Zhe - $\ce{H2S2O3}$ is thiosulfuric acid not sulforous acid which is $\ce{H2SO3}$. $\endgroup$ – MaxW Aug 4 '18 at 0:10
  • 1
    $\begingroup$ @MaxW Thanks for the correction, but a pKa value of 0.6 is still a fairly strong acid. $\endgroup$ – Zhe Aug 4 '18 at 3:23

Your acid dissociation constants are $10^{-0.6}$ and $10^{-1.74}$ whereas your thiosulfate species have only a total molar concentration of $10^{-2.00}$. When the dissociation constants of an acidic solute are greater than the solute concentration the solute is rendered like a strong acid by dilution. If you have 0.01 M of an acid with a $pK_a$ of 1.74, and you use the equilibrium relation to determine how much of the acid is dissociated, you find that in fact most of the acid is dissociated.

The proposed use of the H-H equation then does not work because this equation assumes that aqueous hydrogen or hydroxide ions are only a small part of the total solute concentration. When the solute becomes like a strong acid (or strong base) as above, it floods the solution with aqueous hydrogen or hydroxide ions and the assumption needed for the H-H equation fails.

|improve this answer|||||

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.