Henderson's Equation (pH of a buffer):

(i)  Acidic buffer:        

It consists of a mixture of weak acid and its salt (strong electrolyte). The ionisation of the weak acid, HA, can be shown by the equation

HA ↔  H⁺ + A^-

 Applying law of mass action,

K_a =   H⁺A^- /[HA] 

It can be assumed that concentration of A^- ions from complete ionisation of the saltBA is too large to be compared with concentration of A^- ions from the acid HA.

BA ↔  B⁺ + A^-

Thus, [HA] = Initial concentration of the acid as it is feebly ionised in presence of common ion

and [A^-] = Initial concentration of the salt as it is completely ionised.

So [H⁺] = Ka . [Acid]/[Salt]                               ...... (iii)

Taking logarithm and reversing sign,

-log [H⁺] = -log K_a - log[Acid]/[Salt]

or     pH = log[Salt]/[Acid] - log K_a

or     pH = pK_a + log[Salt]/[Acid]                        ...... (iv)

This is known as Henderson's equation.

    When [Salt]/[Acid] = 10 , then

            pH = 1 + pK_a

    and when [Salt]/[Acid] , then

            pH = pK_a -1

So weak acid may be used for preparing buffer solutions having pH values lying within the ranges pK_a + 1 and pK_a -1. The acetic acid gas a pK_a of about 4.8; it may, therefore, be used for making buffer solutions with pH values lying roughly within the ranges 3.8 to 5.8.

(ii)   Basic offer:

It consists of a weak base and its salt with strong acid. Ionization of a weak base, BOH, can be represented by the equation.

BOH ↔  B⁺ + OH^-

Applying law of mass action,

K_b =  [B⁺][OH^-]/[BOH]                                   ....... (i)

or      [OH^-] = K_b [BOH]/[B⁺]                           ...... (ii)

As the salt is completely ionized, it can be assumed that whole of B⁺ ion concentration comes from the salt and contribution of weak base to B⁺ ions can be ignored.

BA ↔ B⁺ + A^-              (Completely ionised)

So   [OH^-]= K_b[Base]/[Salt]                             .... (iii)

or    pOH = log[Salt]/[Base]  log K_b

or    pOH = pKb + log[Salt]/[Base]                   ...... (iv)

Knowing pOH, pH can be calculated by the application of the formula.

pH + pOH = 14