For example, the $\mathrm{p}K_\mathrm{a}$ values of ethanoic, propanoic, and butanoic acid are related in the following way: $$\text{ethanoic} < \text{butanoic} < \text{propanoic}$$

And their buffering capacities are related in exactly the same way. What causes this?


2 Answers 2


Buffer capacity depends on the $K_\mathrm{a}$ of an acid. I will use the explanation that can be found in David Sheehan: Physical biochemistry: principles and applications. John Wiley & Sons, 2013.

Any aqueous solution containing both $\ce{A-}$ and $\ce{AH}$ is, in principle, capable of resisting change in $\mathrm{pH}$ according to $$\ce {A^- + H^+ <=>> AH}.$$ If alkali is generated in the solution (that would tend to remove protons) we have $$\ce{AH + OH- <=> A- + H_2O}.$$

Most buffers consist of mixtures either of a weak acid and its salt or of a weak base and its salt. The ability to resist change in $\mathrm{pH}$ is finite, especially if the number of protons involve is especially large. This limit is represented by the buffering capacity of the buffer $\beta$. This is defined as the amount of substance in moles of $\ce{[H^+]}$ which must be added to a liter of the buffer to decrease the $\mathrm{pH}$ by one unit. It can be mathematically calculated by $$\beta = \frac{2.3 K_\mathrm{a}[\ce{H+}][C]}{(K_\mathrm{a}+[\ce{H+}])^2},$$ where $[C]$ is the sum of conentrations of $\ce{A-}$ amd $\ce{AH}$. This relationship means that buffering capacity increases with the buffer concentration. Buffers work best at $\mathrm{pH}$ values around their $\mathrm{p}K_\mathrm{a}$ most pH are effective one $\mathrm{pH}$ unit above an one below their $\mathrm{p}K_\mathrm{a}$.
I hope that the formula previously reported help in clarifying how the buffer capacity is related to $\mathrm{p}K_\mathrm{a}$ (since they are indeed related).

Example (based on ZUMDAHL, Steven. World of chemistry. Cengage Learning, 2012.)

A chemist needs a solution buffered at $\mathrm{pH}~4.30$. choosing from ethanoic, propanoic, butanoic acid. We can calculate the ratio $\ce{[HA]/[A^-]}$ required for each system. Considering that a $\mathrm{pH}~4.30$ corresponds to $\ce 5\times 10^{-5} M$ using the equation $$[\ce{H+}]=K_a \frac{[\ce{HA}]}{[\ce{A^-}]}$$ we substitute the required $[\ce{H+}]$ and $\ce{K_a}$ for each acid to calculate the ration $[\ce{H+}]/[\ce{A-}]$ needed in each case is:

\begin{array}{llr} \text{Acid} & [\ce{H+}] = K_\mathrm{a} \frac{[\ce{HA}]}{[\ce{A^-}]} & \frac{[\ce{HA}]}{[\ce{A^-}]} \\ \hline \text{Ethanoic} & 5.0 \times 10^{-5} = 10^{-4.76} \frac{[\ce{HA}]}{[\ce{A^-}]} & 2.88 \\[1ex] \text{Propanoic} & 5.0 \times 10^{-5} = 10^{-4.87} \frac{[\ce{HA}]}{[\ce{A^-}]} & 3.749\\[1ex] \text{Butanoic} & 5.0 \times 10^{-5} = 10^{-4.9} \frac{[\ce{HA}]}{[\ce{A^-}]} & 3.98 \\ \hline \end{array}

In this case the choice would be ethanoic acid since it has the ratio closes to 1.

If we need a buffer near $\mathrm{pH}~5.5$ the best choice is butanoic acid:

\begin{array}{llr} \text{Acid} & [\ce{H+}] = K_\mathrm{a} \frac{[\ce{HA}]}{[\ce{A^-}]} & \frac{[\ce{HA}]}{[\ce{A^-}]} \\ \hline \text{Ethanoic} & 3.2 \times 10^{-6} = 10^{-4.76} \frac{[\ce{HA}]}{[\ce{A^-}]} & 0.18 \\[1ex] \text{Propanoic} & 3.2 \times 10^{-6} = 10^{-4.87} \frac{[\ce{HA}]}{[\ce{A^-}]} & 0.23 \\[1ex] \text{Butanoic} & 3.2 \times 10^{-6} = 10^{-4.9} \frac{[\ce{HA}]}{[\ce{A^-}]} & 0.25 \\ \hline \end{array}

  • 1
    $\begingroup$ Thank you for your effort in continuously improving the quality of the answer. $\endgroup$ Commented Jan 29, 2018 at 17:17
  • $\begingroup$ I tried using this formula for an ethanoic, sodium ethanoate buffer, made by reacting 200 mL of 1M acetic acid and 100 mL of 1M NaOH solution. Am I correct in making the following assumption: $[H^+] = K_a \implies \beta = \frac{2.3 K_\mathrm{a}[\ce{H+}][C]}{(K_\mathrm{a}+[\ce{H+}])^2} = 2.3[C]/4$? If so, wouldn't this mean that this buffer, and a buffer prepared identically but with propanoic acid, would have the same buffering capacity? $\endgroup$
    – Airdish
    Commented Jan 31, 2018 at 17:11
  • $\begingroup$ @Jojostack My question isn't about which is the best buffer for a specific pH. I am asking about how to calculate the buffering capacity of a buffer with equal concentrations of acid and conjugate base. $\endgroup$
    – Airdish
    Commented Feb 1, 2018 at 5:12
  • $\begingroup$ I hope I've understood well. So if you use same amount of acid and conjugated salt you would obtain three solution with different pH. Each of them would be a buffer that will work better in proximity of the pKa of the acid of each buffer solution $\endgroup$
    – Jojostack
    Commented Feb 1, 2018 at 7:23
  • $\begingroup$ That formula, which is demonstrably not correct when pH = pK because the constant is wrong but otherwise is OK (at pH = pK, I haven't checked other values) just shows what should be obvious: for buffers which are strong enough that the charge is dominated by the acid (i.e. not water) and in which if there are multiple protons these have pKs separated by a few pH units buffering capacity does not depend on pK. So no surprise that you found all the acids when used in equal strength produced buffers of the same capacity. $\endgroup$ Commented Feb 1, 2018 at 14:35

Taking "buffering capacity" to mean the amount of acid or base required to change $\mathrm{pH}$ by a fixed small amount at $\mathrm{pH}$ values near one of the $\mathrm pK$'s of the acid on which the buffer is based buffering capacity is not $\mathrm pK$ dependent. To see this write Henderson–Hasselbalch for each proton in the form $\mathrm pr_1=\mathrm pK_1-\mathrm{pH}$, $\mathrm pr_2=\mathrm pK_2-\mathrm{pH}\ldots$ where $\mathrm p(\cdot)=-\log(\cdot)$. Taking the antilog we see $r_j = 10^{(\mathrm{pH}-\mathrm pK_j)}$ Note that $r$ is dependent on the difference between $\mathrm{pH}$ and $\mathrm pK$. Thus we find that $r$ is the same one $\mathrm{pH}$ unit either side of $\mathrm pK$ irrespective of whether $\mathrm pK_\mathrm a$ be 1 or 7.3. This is your first clue that buffering is $\mathrm pK$-independent.

Assuming you have started with $C$ moles of acid the fraction of those that remain undissociated and uncharged at $\mathrm{pH}$ is $f_0(\mathrm{pH})=1/(1+r_1+r_1r_2+r_1r_2r_3\ldots)$, the fraction that has lost one proton and thus carries a single negative charge is $f_1=f_0r_1$, the number that is doubly charged is $f_2=f_1r_2$ and so on so that the total charge on acid anions is $$Q(\mathrm{pH})=C(-f_1-2f_2-3f_3...)$$ (moles of charges or Eq if you prefer). This is clearly a function of $\mathrm{pH}$ (shown) as well as each of the $\mathrm pK$'s (not shown). The number of protons required to shift the buffer's $\mathrm{pH}$ from $\mathrm{pH}$ to $\mathrm{pH}+\delta$ is $\Delta Q=Q(\mathrm{pH}+\delta)-Q(\mathrm{pH})$. This is the buffering (alkalinity, acidity) between $pH$ and $pH+\delta$. The buffering capacity is the slope of the buffering curve. At $pH$ it is: $$\partial Q(\mathrm{pH})/\partial\mathrm{pH}=\lim_{\delta\to0} (Q(\mathrm{pH}+\delta))-Q(\mathrm{pH})/\delta\approx(Q(\mathrm{pH}+\delta)-Q(\mathrm{pH}))/\delta$$ in units of $\mathrm{mol}\cdot\mathrm{pH}^{-1}$. It is the number of moles of protons (1 M strong acid delivers 1 mole of protons per liter) which must be added or absorbed per unit $\mathrm{pH}$ shift.

Now these formulas aren't very complex and if you will take the trouble to put them into a spreadsheet or visualization program and make some plots you will see that the buffering capacity of a buffer near a $\mathrm pK$ depends only on $C$ as long as the $\mathrm pK$'s are separated.

I have been, for simplicity, ignoring the buffering capacity of water here. When $3\lt\mathrm{pH}\lt10$ water does exhibit appreciable buffering and needs to be considered. The net charge on a liter of water ions is $Q_\mathrm w=10^{-\mathrm{pH}}-10^{\mathrm{pH}-\mathrm pK_\mathrm w}$ and the buffering of water, between $\mathrm{pH}$ and $\mathrm{pH}+\delta$, is $$\Delta Q_\mathrm w=10^{-(\mathrm{pH}+\delta)}-10^{-\mathrm{pH}}+(10^{\mathrm{pH}-\mathrm pK_\mathrm w}-10^{\mathrm{pH}+\delta-\mathrm pK_\mathrm w})$$ $Q_\mathrm w$ is added to $Q$ as given above when differentiating to find buffer capacity. Keep in mind that $Q_\mathrm w$ is per liter and $Q$ per mole.

  • $\begingroup$ So if pKa is unrelated to the differing buffering capacities of these acids (assuming that they are all made with the same concentration of acid, and same of concentration of NaOH, at the same temperature), then could you point me to some resources that account for why the difference arises? $\endgroup$
    – Airdish
    Commented Jan 28, 2018 at 12:06
  • $\begingroup$ Based on what we have discussed here there is no difference so I can hardly point you to literature explaining why there is. Perhaps it is in the way you are defining buffering. If you had 0.1 mole of each of these acids and added 0.05 mole of strong base you would wind up at pH 4.76 for acetic acid, 4.88 for propionic and 4.82 for butyric. If you wanted to get acetic acid to pH 4.88 it would take more than 0.05 mole. You might conclude from this that acetic acid has more buffering than butyric acid and it does - to pH 4.88. But the slope of the buffering near the pKs is the same for all 3. $\endgroup$ Commented Jan 28, 2018 at 15:03
  • $\begingroup$ This should really be the accepted answer. Don't see why it is downvoted. $\endgroup$ Commented Jan 28, 2018 at 16:48
  • $\begingroup$ You'll note that I have added water to the answer. In cases where the pK of the acid is small ( < 3 ) or large (> 10) then you are not titrating just your acid but a mix of acids with the other being water. In such cases the buffering capacity will depend on the pK. This would not the case with any of the three acids mentioned in the question unless they were extremely dilute.. $\endgroup$ Commented Jan 28, 2018 at 16:52
  • $\begingroup$ I don't understand that either. If someone sees a problem please let me know what it is so I can fix it. $\endgroup$ Commented Jan 28, 2018 at 16:53

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