Question

A buffer solution contains 0.341 M ammonium chloride and 0.291 M ammonia.

If 0.0213 moles of potassium hydroxide are added to 150 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding potassium hydroxide)

pH=?

Answer 1

mol of KOH added = 21.3 mmol

NH4+ will react with OH- to form NH3

Before Reaction:

mol of NH3 = 0.291 M *150.0 mL

mol of NH3 = 43.65 mmol

mol of NH4+ = 0.341 M *150.0 mL

mol of NH4+ = 51.15 mmol

after reaction,

mol of NH3 = mol present initially + mol added

mol of NH3 = (43.65 + 21.3) mmol

mol of NH3 = 64.95 mmol

mol of NH4+ = mol present initially - mol added

mol of NH4+ = (51.15 - 21.3) mmol

mol of NH4+ = 29.85 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)

= 4.745

use:

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

= 4.745+ log {29.85/64.95}

= 4.407

use:

PH = 14 - pOH

= 14 - 4.4071

= 9.5929

Answer: 9.59