# Can I use the Henderson–Hasselbalch for calculating the pH of non-buffer systems?

Can I use the Henderson–Hasselbalch equation on reactions that are not buffers? When I read up on the derivation of the equation, I noticed that there is absolutely no assumption on the relative concentrations of the acid and conjugate base at least for a monoprotic acid.

Here's a quick derivation of the Henderson Hasselbalch equation: The reaction:

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

$$K_\mathrm{a} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$

Then I take the logarithm of each side, resulting in

$$\log K_\mathrm{a} = \log\frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]} = \log\frac{[\ce{A-}]}{[\ce{HA}]} + \log [\ce{H3O+}]$$

Consequently,

$$-\mathrm{p}K_\mathrm{a} = -\mathrm{pH} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

Can I use the Henderson–Hasselbalch equation on reactions that are not buffers?

Yes, but you should still keep in mind some limitations. The equation you provide

$$K_\mathrm{a} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}$$

should more generally be written

$$K_\mathrm{a} = \frac{a_\ce{A-}a_\ce{H3O+}}{a_\ce{HA}a_\ce{H_2O}}$$

It reduces to your equation$$^\dagger$$ when the concentrations can be used to represent activities $$a_i (=[i]/C^\circ$$), and the activity of water is 1, which generally holds in the limit of very dilute solutions. Therefore it should not be assumed to be accurate (only approximate) at higher concentrations.

$$\dagger$$ Ignoring one other "minor" issue, namely that $$K_a$$ should be unitless, but that is a common fault in the way the equation is written.

As helpfully explained in a comment, this is solved by writing

$$K_a = \frac{[\ce{A^-}][\ce{H_3O^+}]}{[\ce{HA}]C°}$$

where $$C°$$ is the reference standard concentration (usually $$\pu{1 M}$$).

Can I use the Henderson-Hasselbalch equation on reactions that are not buffers?

If you know the equilibrium concentrations of weak acid and conjugate base, you can use the equation. You can also use it when you know the $$\mathrm{pH}$$ and the $$\mathrm pK_\mathrm a$$ and want to know the ratio of protonated and deprotonated species (for example for relating color and $$\mathrm{pH}$$ when working with pH indicators). Written as a formula, the following is always correct for a weak acid and conjugate base at equilibrium:

$$\mathrm{pH} = \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}}$$

or

$$\log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}} = \mathrm{pH} - \mathrm pK_\mathrm a$$

The equation is especially convenient for a buffer because when you mix high concentrations of weak acid and its conjugate base, the concentrations don't change much when the system reaches equilibrium, so you can approximate the $$\mathrm{pH}$$ by plugging in the initial concentrations. So for buffers,

$$\mathrm{pH} = \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}} \approx \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{initial}}}{[\text{weak acid}]_{\text{initial}}}$$

For a solution of a weak acid only or weak base only, that does not work (one of the initial concentrations is zero, and you get a nonsense result). The approximation is best for high and similar concentrations of acid and base when the $$\mathrm pK_\mathrm a$$ is close to $$7$$.