# What is Henderson-Hasselbalch Equation? (Basic Concept)

The **Henderson-Hasselbalch equation** provides a general solution to the quantitative treatment of acid-base equilibrium in biological system. This article explains how to derive Henderson-Hasselbalch Equation.

Buffers are the mixture of **weak acids** and their s**alts of strong bases** (or) the mixture of **weak bases** and their **salts of strong acids**. Simply, Buffers are important concept of Acid-Base Chemistry. Buffers help to maintain a normal pH of the biological systems. When acid (or) alkali is added the pH of the solution changes in the absence of buffers.

- Acids – Bases Concept
- What is pH? What is the importance in living things

**H****ow buffers act:(Henderson-Hasselbalch Equation)**

Buffers act as ‘**Shock absorbers**” against sudden changes of pH by converting injurious strong acids and bases into harmless weak acid salts.

**HA –> H ^{+} + A^{–}**

**BA –> B+ + A ^{–}**

If a buffer solution is composed of weak acid HA and its salt BA, they ionize as follows:

On addition of an Alkali, We shall have,

**Na ^{+} + OH^{–} + H^{+} + A^{– }–> NaA + H_{2}O**

On addition of an acid,

**H ^{+}Cl^{–} + B^{+} + A^{–} —> HA + BCl**

**Example:** Sodium acetate (CH_{3}COO^{–}Na^{+}) + Acetic acid (CH_{3}COOH)

Adding strong acid HCl,

**C H _{3 }COOH + C H_{3 }COO^{–}Na^{+ }+ HCl **

** –> C H _{3 }COOH + C H_{3 }COOH + NaCl**

(Weak acid)

Adding strong alkali NaOH,

**C H _{3 }COO^{–} Na^{+} + C H_{3 }COOH + NaOH **

** –> C H _{3 }COO^{–} Na^{+} + C H_{3 }COO^{–} Na^{+} + H_{2}O**

**He****nderson-Hasselbalch Equation****:**

The pH of a solution containing a weak acid is related to its acid dissociation constant. The relationship can be stated in the convenient form of the “Henderson-Hasselbalch equation”, derived below:

A weak acid, HA, ionizes as follows:

**HA –> H ^{+} + A^{–}**

The equilibrium constant for this dissociation is written as follows:

**[H ^{+}] [A^{–}]**

**K=——————**

**[HA]**

Cross multiply:

**[H ^{+}][A^{–}] = K[HA]**

Divide both sides by [A^{–}]

** [HA]**

**[H ^{+}] = K————–**

** [A ^{–}]**

Take the log of both sides:

** [HA]**

**log [H ^{+}] = log K —————**

** [A ^{–}]**

** [HA] **

**= logK + log ———**

** [A ^{–}]**

Multiply through by –1:

** [HA]**

**-log [H ^{+}] = -log K – log ———-**

** [A ^{–}]**

Substitute pH and for –log [H+] and –logK, respective then:

** [HA]**

**P ^{H}= P^{K} – log ———**

** [A ^{–}]**

Then, to remove the minus sign, invert the last term:

The Handerson-Hasselbalch equation is an expression of great predictive value in protonic equilibria.

**For example,**

**1. When an acid is exactly half-neutralized, [A ^{–}] = [HA] under these conditions,**

** [A ^{–}] 1**

**P ^{H}= P^{K} – log ————- = P^{K} + log —– = P^{K}+ 0**

** [HA] 1**

Therefore, at half neutralization, P^{H} = P^{K}_{.}

**2. When the ratio [A-] /[HA] = 100: 1**

** [A ^{–}]**

**P ^{H}= P^{K} – log ————-**

** [HA]**

**P ^{H }= P^{K} – log 100/1 = P^{K} + 2**

**3. When the ratio [A-] / [HA] = 1:10,**

**P ^{H}= P^{K} – log 1/10 = P^{K} + (-1)**

If the equation is evaluate at several ratio of** [A-] / [HA]** between the limits 10^{3} and 10^{-3} and the calculated pH values plotted, the result obtained describes the titration curves for a weak acid.

**See this Video Tutorial Henderson-Hasselbalch Equation:**