Calculation of the pH of drug solutions

We have considered above the effect on the ionisation of a drug of buffering the solution at given pH values. When these weakly acidic or basic drugs are dissolved in water they will, of course, develop a pH value in their own right. In this section we examine how the pH of drug solutions of known concentration can be simply calculated from a knowledge of the pKa of the drug. We will consider the way in which one of these expressions may be derived from the expression for the ionisation in solution; the derivation of the other expressions follows a similar route and you may wish to derive these for yourselves.

Weakly acidic drugs

We saw above that the dissociation of these types of drugs may be represented by equation (3.63). We can now express the concentrations of each of the species in terms of the degree of dissociation, a, which is a number with a value between 0 (no dissociation) and 1 (complete dissociation):

Therefore,

Weakly basic drugs

We can show by a similar derivation to that above that the pH of aqueous solutions of weakly basic drugs will be given by pH _ 2pkW +èpKa + 2log c

where c is the initial concentration of the weakly acidic drug in mol dm-3.

Because the drugs are weak acids, a will be very small and hence the term (1 - a) can be approximated to 1. We may therefore write

To introduce pH into the discussion we note that ac = [H3O+] = [H+]

•'• -log[H+] = -log Ka - 2log c pH = 1pKa - ¿log c (3.78)

We now have an expression which enables us to calculate the pH of any concentration of the weakly acidic drug provided that its pKa value is known.

EXAMPLE 3.6 Calculation of the pH of a weak acid

Calculate the pH of a 50 mg cm-3 solution of ascorbic acid (mol. wt = 176.1; pKa = 4.17).

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