# Henderson-Hasselbalch equation and degree of dissociation

Is it possible to evaluate the degree of dissociation in unbuffered solutions, through the Henderson-Hasselbalch equation? For example: place acetic acid in a sodium hydroxide solution. After the reaction, the pH of the solution is lowered. Is it possible to exploit the Henderson-Hasselbalch to evaluate the degree of dissociation of acetic acid, using this final pH, even if we are not in the presence of a buffer solution? Generally, Henderson-Hasselbalch is introduced in the presence of buffers, which is where my doubt arises. Infact, by placing the same acetic acid in a buffered solution at a given pH, it is very simple to obtain the ratio [A-] / [HA] using the H-H. Is this reasoning also applied to the previous example? Thanks in advance.

• Yes, of course. NaOH is a strong base. pH tells you starting concentration of NaOH. Final pH tells you how much NaOH has reacted. This should agree with the formula pH = pKa + log salt/acid. There will be no buffering effect until pH gets close to pKa. Otherwise it's just an acid-base reaction. Mar 24 at 23:25
• H-H equation is nothing else but equation for the acid dissociation constant, rewritten in the logarithmic form. Any mathematical pH evaluation must respect this equation in either of its forms. Mar 25 at 5:03
• What do you mean with: even if we are not in the presence of a buffer solution? What you have described is the way to create a buffer solution. Mar 25 at 9:44
• @Luckenberg. Right. The "equilibrium" between NaOH and acetic acid is 100% conversion to acetate. Acetic acid will be consumed as any other acid will be until you get close to the PKa of acetic acid. This why other buffers, such as phosphate, are made with KH2PO4 and K2HPO4 in a known ratio (conforming with H-H ) and are only adjusted slightly with H3PO4 or KOH using a pH meter. It might be fun to just try acetic acid/NaOH titration in the lab and experimentally measure pH as more titrant is added. Mar 25 at 14:40
• A higher [A-]/[A] ratio will calculate a higher pH. But an [A-]/[A] ratio of infinity means ... you're not in the buffering range. Mar 25 at 14:47

The equilibrium constant expression for a weak acid is:

$$K_\mathrm{a} = \frac{\ce{[H3O+]}_\mathrm{eq} \ce{[A-]}_\mathrm{eq}}{\ce{[AH]}_\mathrm{eq}}\tag{1}$$

for the reaction

$$\ce{AH(aq) + H2O(l)<=> H3O+(aq) + A-(aq)}$$

In equation (1), the brackets either stand for activity, or for concentration divided by $$\pu{1 mol L-1}$$. If it is the latter, it is an approximation that works best at constant ionic strength and lower concentration of the weak acid/base conjugate pair. The subscripts "eq" specify that these are to be measured at equilibrium.

If you solve (1) for the hydronium ion activity and take the logarithm, you get:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\left( \frac{\ce{[A-]}_\mathrm{eq}}{\ce{[AH]_\mathrm{eq}}}\right)\tag{2}$$

I would argue that this is not the Henderson-Hasselbalch equation, rather it is:

$$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a} + \log\left( \frac{\ce{[A-]}_\mathrm{initial}}{\ce{[AH]_\mathrm{initial}}}\right)\tag{3}$$

Equation (2) is universal and exact. If you know the pH of the solution and the pKa of the weak acid (this could be an indicator, a substance intended as buffer, any weak acid you add to the solution or the conjugate weak acid of a weak base you add to the solution), you can figure out the ratio of protonated to deprotonated species. Granted, the pH might change when you add your substance of interest to the solution, so you should measure it in the final solution.

Equation (3), on the other hand, is an approximation which obviously does not work for a weak acid in water (with nothing else added) because the initial concentration of conjugate base would be zero. It does work fairly well when you add similar concentrations of weak acid and conjugate base to water (one way of making a buffer). In this case, the equilibrium concentrations are not far off from the initial concentrations, so it is a useful approximation.

Is it possible to exploit the Henderson-Hasselbalch to evaluate the degree of dissociation of acetic acid, using this final pH, even if we are not in the presence of a buffer solution?

First, when you mix a sodium hydroxide solution with an excess of acetic acid, you do have a buffer solution (mixing weak acid with strong base while monitoring the pH is a common way of making a buffer). If you measure the pH, you can estimate the ratio of acetic acid to acetate equilibrium concentration using equation (2).

Technically, if you calculate concentrations by assuming that the reaction of sodium hydroxide with acetic acid goes to completion, and call these initial concentrations, you can also apply equation (3).

So it is a matter of opinion or perspective whether the answer to your question is yes or no.

• "Technically, if you calculate concentrations by assuming that the reaction of sodium hydroxide with acetic acid goes to completion, and call these initial concentrations, you can also apply equation (3)". With this, do you mean that the concentrations we find after the reaction are the 'initial' concentrations of the buffer, so we can use (3), that gives us (about) the same result as (2)? Mar 25 at 16:22
• And using the same reasoning, if we have an excess of sodium hydroxide, than we can't find the final pH using H-H, because we must consider the effect of remaining NaOH, which isn't considered by H-H. However, we can still find the degree of dissociation of acetic acid using the final pH, determined by the remaining NaOH.. Mar 25 at 16:41
• @Luckenberg Yes, exactly. With an excess of NaOH (e.g. pH higher than10), less than 1 in 100,000 of the acetate ions will be protonated at any time.
– Karsten
Mar 25 at 18:43