# When is a buffer solution considered efficient?

By this I mean what is the rigorous definition of a buffer solution?

I know that a buffer is a solution able to resist changes of $$\mathrm{pH}$$, but how much resistance to $$\mathrm{pH}$$ does a solution need to be considered a buffer?

For example I was told that one cannot make an acetic acid buffer $$\left(K_\mathrm{a} = 1.8 \times 10^{-5}\right)$$ at a desired $$\mathrm{pH}$$ of 6.5. Is this true?

• en.wikipedia.org/wiki/Buffer_solution#Buffer_capacity – Mithoron Jun 17 '20 at 0:31
• @Blade: See how I manipulate $K_\mathrm{a}$ using $K_\mathrm{a}$ formatting. Rate constants and equilibrium constants should be itallics such as other variables (e.g., $V, T, R, P, x, y, z,$ etc.). – Mathew Mahindaratne Jun 17 '20 at 1:28
• Who "told you", instead of teaching you how buffers work? – Karl Jun 17 '20 at 7:33
• Buffer solution Is considered efficient, if it meets your criteria for pH stabilization. Note the buffers are defined qualitatively, not quantitatively. 1 M acetic acid buffer at pH 6.5 is roughly comparable in buffer capacity to 0.01 M acetic acid buffer at ideal pH 4.75. So it is inefficient buffer, but still a buffer. – Poutnik Jun 17 '20 at 10:12
• Well, I´m not blaming you, but he seems to have left out a few important points in his teaching. A buffer is obviously most efficient when it is not too dilute and has not already used up part of its buffer capacity in one direction, i.e. it´s fresh. A fresh buffer solution is an equimolar mixture of an acid and its conjugate base. That means you have no choice in the pH. It´s a property of the compound pair. – Karl Jun 17 '20 at 18:52

If buffer differencial capacity ( $$\frac{\mathrm{d}n}{\mathrm{dpH}}\cdot \frac{1}{V}$$, where $$n$$ is molar amount of an added base/acid and $$V$$ is buffer volume) or integral capacity ($$\frac{\Delta n}{\Delta \mathrm{pH}}\cdot \frac{1}{V}$$ per $$\Delta pH = \pm 1$$) meet criteria required by the solution of interest, the buffer is efficient and vice versa. The same buffer may be efficient for one application, while not for another.
As $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[A-]}}{c-\ce{[A-]}}=\mathrm{p}K_\mathrm{a} + \log {\ce{[A-]}} - \log{(c-\ce{[A-]})}$$
then relative differential capacity wrt the maximal capacity at $$\mathrm{pH}=\mathrm{p}K_\mathrm{a}$$ is $$\frac{4}{ \frac{c}{\ce{[A-]}} + \frac{c}{c-\ce{[A-]}}}$$, where $$c$$ is the total molar concentration of the buffer conjugate acid + base pair.