Question

A buffer solution contains 0.225 M ammonium bromide and 0.318 M ammonia, If 0.0153 moles of hydrochloric acid are added to 125 mL of this buffer, what is the pH of the resulting solution? (Assume that the volume change does not change upon adding hydrochloric acid) pH =

Answer 1

mol of HCl added = 0.0153 mol

NH3 will react with H+ to form NH4+

Before Reaction:
mol of NH3 = 0.318 M *0.125 L
mol of NH3 = 0.0398 mol

mol of NH4+ = 0.225 M *0.125 L
mol of NH4+ = 0.0281 mol

after reaction,
mol of NH3 = mol present initially - mol added
mol of NH3 = (0.0398 - 0.0153) mol
mol of NH3 = 0.0244 mol

mol of NH4+ = mol present initially + mol added
mol of NH4+ = (0.0281 + 0.0153) mol
mol of NH4+ = 0.0434 mol
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)
= 4.745

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.745+ log {4.342*10^-2/2.445*10^-2}
= 4.994

use:
PH = 14 - pOH
= 14 - 4.9942
= 9.0058
Answer: 9.01