Why does buffering capacity increase with the pKₐ of the acid component?
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 &amp; Sons, 2013.

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

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} &amp; 
[\ce{H+}] = K_\mathrm{a} \frac{[\ce{HA}]}{[\ce{A^-}]} &amp; 
\frac{[\ce{HA}]}{[\ce{A^-}]} \\  \hline
\text{Ethanoic} &amp; 
5.0 \times 10^{-5} = 10^{-4.76} \frac{[\ce{HA}]}{[\ce{A^-}]} &amp; 
2.88 \\[1ex]
\text{Propanoic} &amp;
5.0 \times 10^{-5} = 10^{-4.87} \frac{[\ce{HA}]}{[\ce{A^-}]}  &amp;
3.749\\[1ex]
\text{Butanoic} &amp;
5.0 \times 10^{-5} = 10^{-4.9} \frac{[\ce{HA}]}{[\ce{A^-}]}  &amp;
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} &amp; 
[\ce{H+}] = K_\mathrm{a} \frac{[\ce{HA}]}{[\ce{A^-}]} &amp; 
\frac{[\ce{HA}]}{[\ce{A^-}]} \\  \hline
\text{Ethanoic} &amp; 
3.2 \times 10^{-6} = 10^{-4.76} \frac{[\ce{HA}]}{[\ce{A^-}]} &amp;
0.18 \\[1ex]
\text{Propanoic} &amp;
3.2 \times 10^{-6} = 10^{-4.87} \frac{[\ce{HA}]}{[\ce{A^-}]}  &amp;
0.23 \\[1ex]
\text{Butanoic} &amp;
3.2 \times 10^{-6} = 10^{-4.9} \frac{[\ce{HA}]}{[\ce{A^-}]}  &amp;
0.25 \\ \hline
\end{array}