Question

A buffer solution contains 0.444 M NH4Br and 0.365 M NH3 (ammonia). Determine the pH change when 0.084 mol NaOH is added to 1.00 L of the buffer. pH after addition − pH before addition = pH change =

Answer 1

Step 1: calculate initial pH
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 {0.444/0.365}
= 4.83

use:
PH = 14 - pOH
= 14 - 4.8298
= 9.170

Step 2: calculate pH after adding NaOH

mol of NaOH added = 0.084 mol

NH4+ will react with OH- to form NH3

Before Reaction:
mol of NH3 = 0.365 M *1.0 L
mol of NH3 = 0.365 mol

mol of NH4+ = 0.444 M *1.0 L
mol of NH4+ = 0.444 mol

after reaction,
mol of NH3 = mol present initially + mol added
mol of NH3 = (0.365 + 0.084) mol
mol of NH3 = 0.449 mol

mol of NH4+ = mol present initially - mol added
mol of NH4+ = (0.444 - 0.084) mol
mol of NH4+ = 0.36 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 {0.36/0.449}
= 4.649

use:
PH = 14 - pOH
= 14 - 4.649
= 9.351

Step 3:
Change in pH = final pH - initial pH
= 9.351 - 9.170
= 0.181
Answer: 0.181