Question

A buffer solution contains 0.259 M ammonium bromide and 0.394 M ammonia. If 0.0385 moles of hydrochloric acid are added to 250 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding hydrochloric acid)

Answer 1

mol of HCl added = 0.0385 mmol

NH3 will react with H+ to form NH4+

Before Reaction:

mol of NH3 = 0.394 M *250.0 mL

mol of NH3 = 98.5 mmol

mol of NH4+ = 0.259 M *250.0 mL

mol of NH4+ = 64.75 mmol

after reaction,

mol of NH3 = mol present initially - mol added

mmol of NH3 = (98.5 - 0.0385) mmol

mol of NH3 = 98.4615 mmol

mol of NH4+ = mol present initially + mol added

mol of NH4+ = (64.75 + 0.0385) mmol

mol of NH4+ = 64.7885 mmol

since volume is both in numerator and denominator, we can use mol instead of concentration

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.7447

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.7447+ log {64.7885/98.4615}

= 4.563

use:

PH = 14 - pOH

= 14 - 4.563

= 9.44

Answer: 9.44