Can I make a buffer with a weak acid without adding it's conjugated base, when the desired pH of the solution is exactly the pKa of the weak acid?

An acid will be 50% dissociated when the solution it's in have a pH equal to the acid pKa ('afirmation 1', for later reference).

We also know that a good buffer solution have equal concentrations of an acid and it's conjugated base.

That's why to make a good buffer we look for an acid that have a pKa close to the pH we want to maintain.

So, suppose I have a solution with a pH 6.2, and I want to add a buffer to maintain this pH at 6.2.

Suppose also, that I have a ideal weak acid for this situation: with a pKa of exactly 6.2.

Using the Henderson–Hasselbalch equation, we know that we need to have the same concentration of the acid and it's conjugated base in the buffer. To add the conjugated base, we use a salt of the refereed acid.

So my question is: If we just throw the acid at the solution without the salt, since the solution pH is exactly the pKa of the acid, the acid will dissociate 50% (afirmation 1), and we will end up with what we wanted: The same concentration of the weak acid and it's conjugated base.

So, in principle, it's possible to make a buffer without adding the salt?

Is this logic correct? If not, what error I'm making?

Thanks

• For a monoprotic acid, HA, there are two variables. First the pKa and second the concentration. There must be a function of the two which would yield a solution with 50% HA and 50% A-. – MaxW Nov 23 '19 at 6:04
• The acid alone solution will have pH approximately 3.1 - 0.5.log(c). Why do you think adding it to solution with pH 6.2 would help you to keep pH 6.2 ? – Poutnik Nov 23 '19 at 7:07

So your solution is already at a pH equal to the pKa of the weak acid you are about to add. Your hunch is that the pH remains the same after adding acid to a solution.

[...] we will end up with what we wanted: The same concentration of the weak acid and it's conjugated base.

It depends on the concentration of what is already in solution and of what you are adding. If you add a weak acid at very low concentration (e.g. an indicator) to a buffered solution, the indicator's protonation state will be governed by the pH of the solution. On the other hand, if you add a weak acid at very high concentration (e.g. a substance that is supposed to act as buffer) to an unbuffered or hardly buffered solution, the buffer will govern the pH.

In your case, the pH will be off (too acidic), and most of the buffer substance will be in the acid (protonated) state. You would have to adjust the pH by adding a strong base to the amount that deprotonates half of the just added acid to get the pH to match the pKa again.

So, in principle, it's possible to make a buffer without adding the salt?

Yes, but not in the scenario you suggested. Most buffers are made by starting with a weak base (pH is much higher than pKa of conjugate weak acid) and adjusting the pH with strong acid, or by starting with a weak acid (pH is much lower than pKa of the weak acid) and adjusting the pH with a strong base. In both cases, you effectively get a mixture with the "salt", i.e. you introduce some spectator ions (often chloride or sodium ions).

So, in principle, it's possible to make a buffer without adding the salt?

Yes and no....

The math is pretty simple. Let's say that there is a monoprotic acid $$\ce{HA}$$.

$$\ce{HA <=> H^+ + A-}$$ $${\mathrm{K}_\mathrm{a}} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}$$ let $$c$$ represent the overall concentration of HA species. If $$\ce{[A-] = [HA] }$$ then $$\ce{[A-] = [HA] = \frac{1}{2}}c$$ and if we neglect the self-hydrolysis of water then $$\ce{[H+] = \frac{1}{2}}c$$ So $$\mathrm{K}_\mathrm{a} = \frac{1}{2}c$$ or $$c = 2\mathrm{K}_\mathrm{a}$$.

However to be a buffer there must be some buffer capacity. For an acid with a pKa = 6.2 we couldn't ignore the autoionization of water, so the calculation would be a bit more complicated. So although there would be a concentration of acid which would yield a solution with a pH of 6.2, the buffer capacity would be so low that the solution couldn't really be considered to be a buffer.