Question

A buffer with a pH of 4.18 contains 0.21 M of sodium benzoate and 0.22 M of benzoic acid. What is the concentration of [H'] in the solution after the addition of 0.056 mol of HCI to a final volume of 1.3L? Assume that any contribution of HCl to the volume is negligible. Number al ?»

Answer 1

pH = pka + log [ sodium benzoate ] /[benzoic acid]

4.18 = pka + log (0.21/0.22)

pka = 4.18 +0.02020

pka = 4.2

initial moles of benzoic acid = M x V

= 0.22 x 1.3

= 0.286

benzoate moles = 0.21 x 1.3 = 0.273

HCl moles added = 0.056 ,

HCl gives H+ whcih adds to sodium benzoate to give benzoci acid

hence benzoate moles = 0.273 - 0.056 = 0.217

benzoic acid moles = 0.286 + 0.056 = 0.342

pH = 4.2 + log ( 0.217/0.342)

= 4.2 - 0.197 = 4.003

[H+] = 10^ -pH= 10^ -4.003

[H+] = 9.93 x 10^-4 M