Why does buffering capacity increase with the pKₐ of the acid component?

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?




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}


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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.


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