Question

Pyridine is a conjugate base which reacts with H+ (such as HCl) to form pyridine hydrochloride. The pyridine hydrochloride dissociates to yield H+ with a pKa of 5.36. Describe the preparation of 1.0 liter of a 0.2 M pyridine buffer at pH 5.2 starting with 1.0 M pyridine and 0.5 M HCl.

Answer 1

pH of pyridine buffer = 14 - (pkb+log(salt(or)acid/base))

    pkb of pyridine = 8.75

no of mole of pyridine buffer = 1*0.2 = 0.2mol

pyridine buffer = pyridine + HCl

let us assume , pyridine = x

                   HCl = 0.2-x

   5.2 = 14 - (8.75+log((0.2-x)/x))

       x = 0.0942

no of mol of pyridine = x = 0.0942 mol

no of mol of HCl = 0.2-x = 0.2-0.0942 = 0.1058 mol

no of mole of pyridine converted into pyridinium chloride = 0.1058 mol

total no of mol of pyridine must take = 0.0942+0.1058 = 0.2 mol

volume of pyridine must take = 0.2/1 = 0.2 L = 200 ml

volume of HCl must add = 0.1058/0.5 = 0.2116 L = 211.6 ml

so that, take 200 ml of pyridine add 211.6 ml of HCl and then add water upto 1 L.