When does pKa = pH ?

The determination of the K_a, of a weak acid can be difficult. However, its pK_a can be easily estimated by analysis of its titration curve. The pK,, is the pH value at the half equivalence point, that is, the point at which only half of the volume of alkali needed to reach the equivalence point has been added.

‘

The reason for this is that, at the half-quivalence point’, the concentration of the conjugate base, A^-, and that of the undissociated acid, HA, are almost equal. Therefore, they cancel out in the expression for K_a.       

[H⁺_(aq)) ] [A^- _(aq)
        ________________

                                                                                                           K_a =           [HA_(aq)l

                                                                                        So: K_a = [H_{+ (aq)}] and pK_a = pH.

Measuring the pKa of a weak acid from its titration curve: (Graph Missing here!)

Let us consider why this is the case i.e

                                                          that [A^-] = [HA] at ‘half the equivalence point’

.Taking an actual example of a titration involving a weak acid and an alkali may help.

Consider :-

·        CH₃COOH_(aq) with NaOH_(aq)

      This neutralisation reaction will produce the salt CH₃COO Na. Unlike CH₃COOH, this

      product fully dissociates to release all of its ethanoate ions (CH₃COO^-) into solution.

      The relevant equations in this example are : -

·        HA                    ===          A^-             +         H⁺

·        CH₃COOH_(aq)    ===      CH₃COO_(aq)  +         H⁺_(aq)

The equilibrium lies very far to the LHS here so [CH₃COO_(aq)^-]  ==  [CH₃COOH_(aq)]

·        CH₃COOH_(aq)   +   NaOH_(aq)    =   CH₃COONa_(aq)    +   H₂O_(l)

·        CH₃COO Na_(aq)                          CH₃COO^-_(aq)   +   Na⁺_(aq)

           During the titration all of the CH₃COO^- present in the CH₃COOH get converted to

          CH₃COO Na. Half way through the process (at ‘half the equivalence point’) we can      

          see, regarding ‘CH3COO’, that : -

·        half exists as CH₃COOH, and

·        half exists as CH₃COO Na

         Given that very little of the CH₃COOH dissociates and all of the CH₃COO Na

         dissociates, it is fair to say that : -

·        [CH₃COO^-]   =   [CH₃COOH]

So, in general                       [A^-]    =    [HA] at ‘half the equivalence point.

 Chemistry Department: Loreto College, Coleraine.