# Why doesn't pH = pKa1 in the buffer zone for this titration?

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? • How could you have pH=0.6 if concentration was only 0.01 to begin with? – Mithoron Aug 3 '18 at 22:14
• @Mithoron So pKa doesn't always equal pH at the half equivalence point? – DrPepper Aug 3 '18 at 23:03
• @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. – MaxW Aug 3 '18 at 23:27
• @Zhe - $\ce{H2S2O3}$ is thiosulfuric acid not sulforous acid which is $\ce{H2SO3}$. – MaxW Aug 4 '18 at 0:10
• @MaxW Thanks for the correction, but a pKa value of 0.6 is still a fairly strong acid. – 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.