Are the best buffers those that are in a 1:1 ratio with its conjugate acid/base?

I wanted to comment this, but I don't have enough reputation, so I just made this a question. I know that buffers and their conjugate base/acid must exist in similar amounts for the buffer to be effective, so would the buffer be most effective when the ratio is close to 1:1 (of the buffer and its conjugate acid/base)?


Also I'm confused about how much of a weak base/acid buffers should be. If you have a very weak acid, then wouldn't you have a very strong base. In this case will the buffer be ineffective? So the upshot of this paragraph is asking will buffers still work if you have a very weak acid?




Yes, the best buffers are those that are in a 1:1 ratio with its conjugate acid/base, but in a somewhat indirect way; it's because buffer capacity is maximum when $\mathrm{pH}=\mathrm{p}K\_\mathrm{a}$, and for that value of $\mathrm{pH}$, the proportion of the acid/base conjugates is 1:1.


The contribution of an acid/base conjugate pair to buffer capacity is


$\beta\_{\ce{HA/A-}} = C\_\mathrm{A} \cfrac{10^{-\mathrm{pH}-\mathrm{p}K\_\mathrm{a}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K\_\mathrm{a}} \right)^2} \ln{10}$


(For derivation, check [this response](https://chemistry.stackexchange.com/questions/73256/how-does-the-hcl-kcl-buffer-work/94122#94122) to another question)


The maximum of buffer capacity is achieved when $\mathrm{pH} = \mathrm{p}K\_\mathrm{a}$, and is


$\beta\_\mathrm{max}= \cfrac{1}{4} C\_\mathrm{A} \ln{10}$


As you deviate from $\mathrm{pH}=\mathrm{p}K\_\mathrm{a}$ in either direction, concentrations also deviate from 1:1 (as per Henderson-Hasselbalch) *and* buffer capacity will fall.


Note that the maximum buffer capacity *does not depend on the strength of the acid or base*, only on the total concentration of the conjugate system $C\_\mathrm{A}=\ce{[HA]}+\ce{[A-]}$.


What $\mathrm{p}K\_\mathrm{a}$ *will* determine is at which $\mathrm{pH}$ value will this maximum buffer capacity be achieved, so you will choose an acid/base pair with a $\mathrm{p}K\_\mathrm{a}$ that will allow the system to act as a buffer around the $\mathrm{pH}$ that you want to preserve; ideally by choosing a pair with a $\mathrm{p}K\_\mathrm{a}$ equal to the target $\mathrm{pH}$, which is when you will obtain the maximum buffer capacity.


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The buffer capacity is usually used as indication of the 'strength' of a buffer. It is defined as the gradient of total concentration with pH, and is in effect the number of moles of acid/ base needed to change the pH by $\pm 1$. i.e. $\displaystyle \beta = \frac{dC}{dpH}$ and has the form (using concentration instead of number of moles just for simplicity);


$$\displaystyle \beta=2.303\left[\frac{K\_w}{[\text{H}^+]}+[\text{H}^+]+\frac{CK\_a[\text{H}^+]}{ ([\text{H}^+]+K\_a)^2 } \right]$$


where $K\_w=\ce{[H^+][OH^-]}=10^{-14}$ and the equilibrium constant is $K\_a$. 


Usually the first two terms in $\beta$ are small compared to the last one so the buffer capacity is proportional to the total concentration of the species present which is what is expected as the acid and base act as reservoirs as the pH is changed.


The buffer capacity has the form of a peak near to the $\mathrm{p}K\_a$.
[![buffer capacity](https://i.stack.imgur.com/xS5HXm.png)](https://i.stack.imgur.com/xS5HXm.png) 


The figure is that for benzoic acid/sodium benzoate (each at 0.01 M) with a $\mathrm{p}K\_a=4.2$. The maximum is virtually at the $\mathrm{p}K\_a$.


The Henderson - Hasselbalch equation, although it is an approximation, can be used to work out the pH of a buffer solution and in its general form is 


$$\mathrm{[H^+]} = K\_Ac\_a/c\_b, \qquad \mathrm{pH}=\mathrm{p}K\_A - \log\_{10}(c\_a/c\_b) $$


so that when the concentrations are not equal the $\mathrm{p}K\_a$ and pH differ slightly.


The answer in this post [How to set up equation for buffer reaction?](https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566) gives the derivation of the last equation.


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