The Henderson-Hasselbalch equation provides a general solution to the quantitative treatment of acid-base equilibrium in a biological system. This article explains how to derive the Henderson-Hasselbalch Equation.
Buffers are the mixture of weak acids and their salts of strong bases (or) the mixture of weak bases and their salts of strong acids. Simply, Buffers are an important concept of Acid-Base Chemistry. Buffers help to maintain a normal pH of the biological systems. When acid (or) alkali has added the pH of the solution changes in the absence of buffers.
How buffers act?
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 + H2O
On addition of an acid,
H+Cl– + B+ + A– —> HA + BCl
Example: Sodium acetate (CH3COO–Na+) + Acetic acid (CH3COOH)
Adding strong acid HCl,
C H3 COOH + C H3 COO–Na+ + HCl
–> C H3 COOH + C H3 COOH + NaCl
Adding strong alkali NaOH,
C H3 COO– Na+ + C H3 COOH + NaOH
–> C H3 COO– Na+ + C H3 COO– Na+ + H2O
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]
Divide both sides by [A–]
[H+] = K————–
Take the log of both sides:
log [H+] = log K —————
= logK + log ———
Multiply through by –1:
-log [H+] = -log K – log ———-
Substitute pH and for –log [H+] and –logK, respective then:
PH= PK – log ———
Then, to remove the minus sign, invert the last term:
The Handerson-Hasselbalch equation is an expression of great predictive value in protonic equilibria.
1. When an acid is exactly half-neutralized, [A–] = [HA] under these conditions,
PH= PK – log ————- = PK + log —– = PK+ 0
Therefore, at half neutralization, PH = PK.
2. When the ratio [A-] /[HA] = 100: 1
PH= PK – log ————-
PH = PK – log 100/1 = PK + 2
3. When the ratio [A-] / [HA] = 1:10,
PH= PK – log 1/10 = PK + (-1)
If the equation is evaluate at several ratio of [A-] / [HA] between the limits 103 and 10-3 and the calculated pH values plotted, the result obtained describes the titration curves for a weak acid.