Question

Acetate buffer-1 and buffer-2 were developed by your colleague. Acetate buffer 1 is composed of 0.50 M of conjugate base and 0.50 M of acid. Acetate buffer 2 is composed of 0.05 M of conjugate base and 0.05 M of acid. These buffers are required to test the dissolution performance of a strongly basic drug X. The concentration of Drug X added for preliminary screening of Buffer 1 and 2 was 0.06M. Which buffer will you choose for analyzing dissolution performance of DrugX? Explain your answers with desired calculations, [pKa= 4.74].

Answer 1

Buffer -1 will have strong action and better P^H resisting capacity. If constituents of buffer are in concentrated form it will have strong buffering action.it can be shown through calculation also using Henderson equation

P^H = P^ka + log [ CH3COO^- ] / [ CH3COOH]

Let we have taken 1 litre of both the buffer solution.

FOR BUFFER SOLUTION-1

initial P^H = P^ka = 4.74

when 0.06 mol of strong base is added it will neutralise 0.06 mol CH3COOH and form extra 0.06 mol acetate ion according to reaction.

CH3COOH + OH^- $\rightarrow$ CH3COO^- + H2O

so moles of CH3COOH = 0.50 - 0.06 = 0.44 mol

moles of acetate ion = 0.50 + 0.06 = 0.56 mol

P^H = 4.74 + log ( 0.56 / 0.44 ) = 4.74 + 0.1 = 0.75

change in P^H = 0.1

For BUFFER -2

initial P^{H =} 4.74

when 0.06 mol of base is added it will completely neutralise 0.05 mol acetic acid and form extra 0.05 mol acetate ion . So final solution will have 0.1 mol acetate ion and remaining 0.01 mol strong base.which is definitely not a buffer solution.