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,

Ka =   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 Ka - log[Acid]/[Salt]

or     pH = log[Salt]/[Acid] - log Ka

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

This is known as Henderson's equation.

    When [Salt]/[Acid] = 10 , then

            pH = 1 + pKa

    and when [Salt]/[Acid] , then

            pH = pKa -1

So weak acid may be used for preparing buffer solutions having pH values lying within the ranges pKa + 1 and pKa -1. The acetic acid gas a pKa 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,

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

or      [OH-] = Kb [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-]= Kb[Base]/[Salt]                             .... (iii)

or    pOH = log[Salt]/[Base]  log Kb

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

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

pH + pOH = 14

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