The Henderson-Hasselbalch equation is an essential tool for understanding and calculating the pH of solutions containing weak acids and bases, particularly in the context of buffers in biochemistry and physiology. The equation takes its name for Lawrence Joseph Henderson, who derived the equation for calculating hydrogen ion concentration of a bicarbonate buffer solution in 1908, and Karl Albert Hasselbalch, who expressed Henderson’s expression in logarithmic terms in 1909.

Here is the equation, its derivation, when to use it, when to avoid it, and examples using the Henderson-Hasselbalch equation for both weak acids and weak bases.

### Henderson Hasselbalch Equation for Weak Acids and Weak Bases

The Henderson-Hasselbalch equation is:

- For weak acids:
**pH = pKa + log ([A**^{–}]/[HA]) - For weak bases:
**pH = pKa + log ([B]/[BH**^{+}])

The equation relates the pH of the solution to the pKa (the negative logarithm of the acid dissociation constant, Ka) and the ratio of the molar concentrations of the conjugate base (A^{–} or B) to the undissociated acid (HA or BH^{+}).

Sometimes for weak bases, you have the pKb rather than pKa value. The Henderson-Hasselbalch equation also works for pOH:

**pOH = pKb + log ([B]/[HB ^{+}])**

### Derivation of the Henderson Hasselbalch Equation

The derivation of the Henderson-Hasselbalch equation relies on the relationship between pH, pKa, and the equilibrium constant, Ka.

First, the Ka for a weak acid (HA) is:

Ka = [H+][A-]/[HA]

Taking the negative logarithm of both sides gives the following equation:

-log(Ka) = -log([H+][A-]/[HA])

By definition:

pKa = -log(Ka) and pH = -log([H+])

Substitute these expressions into the equation:

pKa = pH + log([HA]/[A-])

Rearranging the equation gives the Henderson-Hasselbalch equation for weak acids:

pH = pKa + log ([A-]/[HA])

A similar derivation gives the relation for weak bases.

### When to Use the Henderson-Hasselbalch Equation (and Limitations)

The Henderson-Hasselbalch equation is useful in calculating the pH of buffer solutions, determining the isoelectric point of amino acids, and understanding titration curves. It is most accurate when the concentrations of the weak acid and its conjugate base (or weak base and its conjugate acid) are within one order of magnitude of each other and when the pKa of the acid/base is within one pH unit of the desired pH. However, the equation may not be applicable under the following conditions:

- When dealing with strong acids or bases, as their dissociation is nearly complete.
- When the concentrations of the acid/base and its conjugate species are very different, as the equation’s accuracy decreases.
- At extremely low or high pH values, where the activity coefficients of the ions differ significantly from their concentrations.

### pH vs PKa

pH and pKa both appear in the Henderson-Hasselbalch equation. When the concentration of weak acid and its conjugate base are the same, they have the same value:

In this situation:

[HA] = [A^{–}]

pH = pKa + log(1)

pH = pKa

Note that pH is a measure of the acidity or alkalinity of a solution and is the negative logarithm of the hydrogen ion concentration ([H^{+}]). On the other hand, pKa is a measure of the strength of an acid and is the negative logarithm of the acid dissociation constant (Ka). pKa is the pH value where a chemical species donates or accepts a proton (H^{+}). A lower pKa value indicates a stronger acid, whereas a low pH value indicates a more acidic solution.

### Example Problems

#### Weak Acid

Calculate the pH of a solution containing 0.15 M formic acid (HCOOH) and 0.10 M sodium formate (HCOONa). The pKa of formic acid is 3.75.

This is a buffer solution containing a weak acid, formic acid (HCOOH), and its conjugate base, sodium formate (HCOONa). Solve it by applying the Henderson-Hasselbalch equation for weak acids:

pH = pKa + log ([A^{–}]/[HA])

[A^{–}] is the concentration of the conjugate base (formate ion, HCOO-) and [HA] is the concentration of the weak acid (formic acid, HCOOH).

Since sodium formate is a soluble salt, it completely dissociates in water, providing the same concentration of formate ions as the initial concentration of the salt:

[A-] = [HCOO-] = 0.10 M

The concentration of formic acid, the weak acid, is:

[HA] = [HCOOH] = 0.15 M

Now, plug these values into the Henderson-Hasselbalch equation, along with the pKa value of formic acid:

pH = 3.75 + log (0.10/0.15)

Calculating the logarithm and adding it to the pKa:

pH = 3.75 – 0.18 pH ≈ 3.57

Thus, the pH of the solution containing 0.15 M formic acid and 0.10 M sodium formate is approximately 3.57.

#### Weak Base

Calculate the pH of a solution containing 0.25 M ammonia (NH_{3}) and 0.10 M ammonium chloride (NH_{4}Cl). The pKb of ammonia is 4.75.

This is a buffer solution containing a weak base, ammonia (NH_{3}), and its conjugate acid, ammonium chloride (NH_{4}Cl). To find the pH of this solution, apply the Henderson-Hasselbalch equation for weak bases:

pOH = pKb + log ([B]/[HB+])

[B] is the concentration of the weak base (ammonia, NH_{3}) and [HB^{+}] is the concentration of the conjugate acid (ammonium ion, NH_{4}^{+}).

Ammonium chloride is a salt completely dissociates in water, providing the same concentration of ammonium ions as the initial concentration of the salt:

[HB^{+}] = [NH_{4}^{+}] = 0.10 M

The concentration of ammonia, the weak base, is:

[B] = [NH_{3}] = 0.25 M

Now, plug these values into the Henderson-Hasselbalch equation for weak bases, along with the pKb value of ammonia:

pOH = 4.75 + log (0.25/0.10)

Calculate the logarithm and add it to the pKb:

pOH = 4.75 + 0.70 pOH ≈ 5.45

Now, convert pOH to pH. The sum of pH and pOH equals 14:

pH + pOH = 14

Therefore, the pH of the solution is:

pH = 14 – pOH pH = 14 – 5.45 pH ≈ 8.55

Thus, the pH of the solution containing 0.25 M ammonia and 0.10 M ammonium chloride is approximately 8.55.

### References

- Hasselbalch, K. A. (1917). “Die Berechnung der Wasserstoffzahl des Blutes aus der freien und gebundenen Kohlensäure desselben, und die Sauerstoffbindung des Blutes als Funktion der Wasserstoffzahl”. Biochemische Zeitschrift. 78: 112–144.
- Henderson, Lawrence J. (1908). “Concerning the relationship between the strength of acids and their capacity to preserve neutrality”.
*Am. J. Physiol*. 21 (2): 173–179. doi:10.1152/ajplegacy.1908.21.2.173 - Po, Henry N.; Senozan, N. M. (2001). “Henderson–Hasselbalch Equation: Its History and Limitations”.
*J. Chem. Educ*. 78 (11): 1499–1503. doi:10.1021/ed078p1499 - Skoog, Douglas A.; West, Donald M.; Holler, F. James; Crouch, Stanley R. (2004).
*Fundamentals of Analytical Chemistry*(8th ed.). Belmont, Ca (USA): Brooks/ColeISBN 0-03035523-0. - Voet, Donald; Voet, Judith G. (2010).
*Biochemistry*(4th ed.). John Wiley & Sons, Inc. ISBN: 978-0470570951.