# What are the limitations of the Hendersson-Hasselbalch equation?

Our instructor told us that prior to the equivalence point, the $$\mathrm{pH}$$ of a solution is dependent on the HH equation. However, when I tried practicing for polyprotic acid titration, I came upon this problem that does not make any sense at all. When I'm using the HH equation, the $$\mathrm{pH}$$ I get is lower than the initial $$\mathrm{pH},$$ which is illogical.

The problem goes likes this:

What is the $$\mathrm{pH}$$ of the solution of $$\pu{25 mL}$$ $$\pu{0.100 M}$$ of $$\ce{H2SO3}$$ $$(K_\mathrm{a1} = \pu{1.23e-2},$$ $$K_\mathrm{a2} = \pu{6.6e-8})$$ when $$\pu{5 mL}$$ of standardized $$\pu{0.1000 M}$$ of $$\ce{NaOH}$$ is added?

The initial $$\mathrm{pH}$$ that I've calculated is $$1.53,$$ but every time I try to use HH for the computation of the first $$\pu{5 mL}$$ increment of titration, the $$\mathrm{pH}$$ is lower than the $$\mathrm{pH}$$ of the initial conditions.

## 2 Answers

The key approximation made in deriving the Henderson-Hasselbalch equation is that the equilibrium constant can be written as $$K=\frac{c_{H^+}c_{A^-}}{c_{HA}}$$ that is, we assume activity coefficients are unity. If you take the base-10 logarithm of this equation and rearrange terms you arrive at the HH equation:$$pH=pK_a+\log_{10}\left(\frac{c_{A^-}}{c_{HA}}\right)$$ Problems usually arise with approximations made beyond the HH equation, in the context of a polyprotic compound particularly by ignoring one of the equilibria, including perhaps water autodissociation, when this shouldn't be done.

In the case of the problem you outline, you might

• ignore the second dissociation equilibrium and
• assume that added base neutralizes an equimolar amount of remaining acid $$\ce{H2SO4}$$ to form additional $$\ce{HSO4-}$$.

Then

\begin{align}c_{HA} = c_{HA,\,initial}-c_{added\,OH^-}&=\pu{0.0705 M} - \pu{0.100 M} \frac{\pu{5 mL}}{\pu{30 mL}}\\&=\pu{0.0538 M}\end{align} \begin{align}c_{A^-} = c_{A^-,\,initial}+c_{added\,OH^-}&=\pu{0.0295 M} + \pu{0.100 M} \frac{\pu{5 mL}}{\pu{30 mL}}\\&=\pu{0.0462 M}\end{align}

Applying the HH equation we then have, as the approximate pH \begin{align}pH &=1.91 + \log_{10}\left(\frac{0.0462}{0.0538}\right)\\&=1.84\end{align} which is above the starting pH of 1.53.

A more accurate value (although still approximate) can be obtained starting from the condition of charge balance for the system, and by assuming that the $$\ce{OH^-}$$ concentration is negligible, from which one obtains pH=1.75.

• In connection to your solution, the pH midway from the first equivalence point (12.5 mL of titrant) is defined as -log(ka1). However, utilizing your equation, when 10 mL of titrant is added on the solution, the pH exceeds the pH for the midway point. How can I resolve that? – Kent de los Reyes Aug 5 '19 at 7:14
• I'm afraid this answer is incorrect. The key approximation of HH isn't that the activity coefficients are unity. It's that the equilibrium concentrations of HA and A- are nearly identical to their initial concentrations. Thus one can use an "x is small" approximation (say, when using an ICE table) to calculate the pH. This is a much stronger approximation than merely assuming the activity coefficients are unity, and is what gives the HH equation its functional form. – theorist Aug 5 '19 at 16:10
• @theorist it all depends on how you use the equation. Obviously during a titration (not a buffer solution) the concentrations will change from their initial values. To the extent that the equilibrium expression defines that which it is stated to express, it is correct. Please write an alternate answer or provide a link if you disagree. – Buck Thorn Aug 5 '19 at 16:56
• @a-cyclohexane-molecule You are welcome to downvote. I understand that my interpretation of the question was not in line with your understanding of how the HH equation is often used. I did not associate the question with the use of the HH equation as an approximate way to compute pH values during buffer preparation. I interpreted the OPs question as "what approximations are intrinsic to the HH equation", not to its application in approximating pHs during a buffer prep. The OPs posted interest is in following the effect of adding a certain amount of strong base to an acid. – Buck Thorn Aug 6 '19 at 19:48
• @a-cyclohexane-molecule After re-reading the other answer I notice that it isn't correct, since it states something about what is assumed during the derivation of the HH equation which isn't correct. The HH equation is simply a re-expression of the concentration equilibrium constant in log10 terms. How you use it might involve some approximations which additionally impact the accuracy of the predicted pH (particularly in the case of buffer calcs). – Buck Thorn Aug 6 '19 at 19:53

For the derivation of the Hendersson-Hasselbalch equation, it is assumed that both hydronium and hydroxide ions are minor species, i.e. there concentration is low compared to that of the buffer species.

In your exercise, the hydronium ion concentration (according to your calculation) is about 0.03 M or higher. With a initial concentration of $$\ce{H2SO3}$$ of 0.1 M, this is not negligible. One easy way to see that is to use the calculated pH to determine the concentration of all the other species, determine the reaction quotient Q and compare it to the equilibrium constant.