**What is the Henderson-Hasselbalch Equation?**

The Henderson-Hasselbalch equation establishes a connection between an acid’s pKa and pH in aqueous solutions (acid dissociation constant). When the concentration of the acid and its conjugate base, or the base and the corresponding conjugate acid, are known, the pH of a buffer solution can be determined with the use of this equation.

**Equation of Henderson-Hasselbalch**

The Henderson-Hasselbalch equation looks like this:

**pH = pK _{a} + log_{10} ([A^{–}]/[HA])**

In this equation, [A-] stands for the molar concentration of the conjugate base (of the acid), and [HA] for the weak acid. The Henderson-Hasselbalch equation can be expressed as follows:

The American scientist Lawrence Joseph Henderson is credited with developing the first equation that could determine the pH level of a specific buffer solution. The Danish scientist Karl Albert Hasselbalch later reformulated this equation in logarithmic terms. The Henderson-Hasselbalch Equation was the name given to the resulting equation.

**Derivation of the Henderson-Hasselbalch Equation**

With the aid of direct methods, the ionisation constants of strong acids and bases can be quickly computed. The extent of ionisation of weak acids and bases, however, precludes the use of the same techniques due to the low degree of ionisation (weak acids and bases hardly ionize). Therefore, the Henderson-Hasselbalch Equation is employed to roughly estimate the pH of these kinds of solutions.

Read Also: Periodic Table

**Take this weak acid HA as an example of ionization:**

Henderson-Hasselbalch equation, also referred to as Henderson equation, is the name given to the aforementioned equation. It is highly helpful for determining the equilibrium pH in acid-base processes as well as for evaluating the pH of a buffer solution. We can determine when from the equation when

both species have the same concentration, or alternatively, the acid will be 50% dissociated.

The same is true for a weak base B:

**Important Points to Remember**

- When exactly half of the acid undergoes dissociation, the value of [A]/[HA] becomes 1, implying that the pK
_{a}of the acid is equal to the pH of the solution at this point. (pH = pK_{a}+ log_{10}(1) = pK_{a}). - For every unit change in the pH to pK
_{a}ratio, a tenfold change occurs in the ratio of the associated acid to the dissociated acid. For example, when the pK_{a}of the acid is 7 and the pH of the solution is 6, the value of [A^{–}]/[HA] is 0.1 but when the pH of the solution becomes 5, the value of [A^{–}]/[HA] becomes 0.01. - The value of [A
^{–}]/[HA] is dependent on the value of the pH and pK_{a}. When pH < pK_{a}; [A^{–}]/[HA] < 1. When pH > pK_{a}; [A^{–}]/[HA] > 1.

Because it assumes that the concentration of the acid and its conjugate base at chemical equilibrium will remain the same as the formal concentration, the Henderson-Hasselbalch equation is unable to predict accurate values for the strong acids and strong bases (the binding of protons to the base is neglected).

The Henderson-Hasselbalch equation cannot provide accurate pH readings for excessively diluted buffer solutions because it ignores the self-dissociation that occurs with water.

**Solved Example**

**A buffer solution is made from 0.4M CH _{3}COOH and 0.6M CH_{3}COO^{–}. If the acid dissociation constant of CH_{3}COOH is 1.8*10^{-5}, what is the pH of the buffer solution?**

As per the Henderson-Hasselbalch equation, pH = pK_{a} + log([CH_{3}COO^{–}]/[CH_{3}COOH])

Here, K_{a} = 1.8*10^{-5} ⇒ pK_{a}= -log(1.8*10^{-5}) = 4.7 (approx.).

Substituting the values, we get:

pH = 4.7 + log(0.6M /0.4M) = 4.7 + log(1.5) = 4.7 + 0.17 = 4.87

Therefore, the pH of the solution is 4.87.

**Frequently Asked Questions – FAQs**

**Are bases compatible with the Henderson-Hasselbalch equation?**The Henderson-Hasselbalch equation holds true when equilibrium concentrations of an acid and conjugate base are included. When solutions containing relatively strong acids are present, equilibrium concentrations can differ greatly from those predicted by neutralization stoichiometry (or not-so-weak bases).

**Is NaOH an effective base?**A fully ionic base, such as sodium hydroxide or potassium hydroxide, is referred to as a solid base. You should think of the chemical as being completely split into metal ions and hydroxide ions in solution. To add one mole of hydroxide ions to the solution, one mole of sodium hydroxide must dissolve.

**What distinguishes KA and pKa from one another?**Ka denotes the acid dissociation constant. pKa is effectively the log of this constant. Similar to this, Kb is the base dissociation constant, and pKb is the -log of Kb. The moles per liter (mol/L) unit of measurement is commonly used to express the constants for acid and base dissociation.

**Is bleach a base or an acid?**It’s quite easy to use chlorine bleach. Actually, we create it by combining chlorine gas with a condensed sodium hydroxide solution to create sodium hypochlorite and sodium chloride.

**Does pKa correspond to pH?**The hydrogen ion concentration of an aqueous solution is determined by the pH scale. It is comparable to pH and pKa (acid dissociation constant), but pKa is more accurate in that it tells you what a molecule would do at a specific pH. The Henderson-Hasselbalch equation explains how pH and pKa relate to one another.