Am I understanding buffering capacity against strong acid/base correctly? I just want to make sure I'm understanding buffering capacity correctly.

I've been working on this problem:


  Suppose you have an acetic acid buffer, $\mathrm{p}K_\mathrm{a} = 4.74$, at the following $\mathrm{pH}$:
  
  1) $\mathrm{pH} = 4.00$
  
  2) $\mathrm{pH} = 4.35$
  
  3) $\mathrm{pH} = 4.70$
  
  4) $\mathrm{pH} = 5.00$
  
  5) $\mathrm{pH} = 5.40$
  
  6) $\mathrm{pH} = 5.60$
  
  Which buffer solution will have the highest buffering capacity against $\ce{HCl}$? Against $\ce{NaOH}$ ? Which is the best optimal buffer?


From what I understand, optimal buffers are where $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, since that is when the ratio between conjugate base and acid is equal to $1$. This means that there is just enough acid/base relative to each other to minimize $\mathrm{pH}$ changes whether a strong acid or base is added.

As for buffering against $\ce{HCl}$, we would want our $\mathrm{pH}$ to be as high as possible, right?

We want as much conjugate base as possible in our buffer to react with the $\ce{HCl}$, while minimizing the changes to the acid, which means we want $\frac{[\ce{A-}]}{[\ce{HA}]}$ to be as high as possible. Maximizing $\frac{[\ce{A-}]}{[\ce{HA}]}$ means $\log \left(\frac{[\ce{A-}]}{[\ce{HA}]}\right) \rightarrow \infty$, which means $\mathrm{pH} \rightarrow 14$, since $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)$.

The explanation for $\ce{NaOH}$ would be the exact opposite. You want as much acid compared to conjugate base as possible so that the acid reacts with $\ce{NaOH}$ while minimizing changes to the conjugate base, which means $\frac{[\ce{A-}]}{[\ce{HA}]} \rightarrow 0$, $\log \left(\frac{[\ce{A-}]}{[\ce{HA}]}\right) \rightarrow -\infty$, and $\mathrm{pH} \rightarrow 0$. 

Is my understanding correct? The thing that's tripping me up is that I was told the exact amount of conjugate base and acid is more important than the ratio (e.g., $0.5 / 0.9$ is better than $0.05 / 0.09$), but since there's no mention of exact concentrations in the problem, I'm guessing we just assume that we have sufficient amounts of $\ce{A-}$ and $\ce{HA}$.

Is this generalization correct: if my buffer $\mathrm{pH} &lt;&lt;&lt; \mathrm{p}K_\mathrm{a}$, then does that mean it is optimal against strong bases (and vice versa)?

Thank you for your help.
Further to the answer by Poutnik the buffer capacity $\beta$ of a weak acid - conjugate base buffer is defined as the number of moles of strong acid or base $C_B$ needed to change the $p$H by $\pm1$ unit, where

$$\displaystyle \beta=\frac{d[B]}{d\text{pH}}$$

and the equilibrium concentration base present is  $\displaystyle \mathrm{[B]}=\frac{k_w}{\mathrm{[H^+]}} - \mathrm{[H^+]}+\frac{C_BK_a}{\mathrm{[H^+]}+K_a} $. 

(See answer How to set up equation for buffer reaction?)  where $K_w$ is the water ionization equilibrium constant $K_w =\mathrm{ [H^+][OH^-]} = 10^{-14}$, $K_a$ is the acid dissociation constant, and $C_B$ the total concentration of buffer.

After performing the calculation $\displaystyle \beta=2.303\left[\frac{K_w}{[\text{H}^+]}+[\text{H}^+]+\frac{C_BK_a[\text{H}^+]}{ ([\text{H}^+]+K_a)^2 }  \right]$

This can be analysed to find its maximum, but can be simplified first by ignoring the first two terms, because $K_w$ is tiny as is $[H^+]$ compared to the last term close to $K_a$. Plotting this last term with $C_B=0.02$ produces the following curves



where the max buffering is seen to be very close to the $pK_A$. You can see that buffering will be ok only when the $p$H changes by $\approx \pm 1$ and when the $pK_A$ is in the range from about $pK_A $ from $\approx 4 \to 9$. The red curve is the last term in the equation, the blue curve the full equation.

To answer your first question the $p$H should be close to the $pK_A$ and as the total concentration of base ($C_B$) increases the maximum just increases but does not shift.

To find the maximum differentiating the last term only with $[H^+]\equiv x$ produces, after a little algebra, $\displaystyle \frac{d\beta}{dx}=C_BK_a\frac{K_a-x}{(K_a+x)^3}=0$ where the maximum is found when $[H^+]=K_a$