Question

Consider a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For ammonia, pKb=4.75.

Calculate the pH of 1.0 L upon addition of 0.050 mol of solid NaOH to the original buffer solution.

Thanks!

Answer 1

mol of NaOH added = 0.05 mol

NH4+ will react with OH- to form NH3

Before Reaction:

mol of NH3 = 0.5 M *1.0 L

mol of NH3 = 0.5 mol

mol of NH4+ = 0.2 M *1.0 L

mol of NH4+ = 0.2 mol

after reaction,

mol of NH3 = mol present initially + mol added

mol of NH3 = (0.5 + 0.05) mol

mol of NH3 = 0.55 mol

mol of NH4+ = mol present initially - mol added

mol of NH4+ = (0.2 - 0.05) mol

mol of NH4+ = 0.15 mol

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

use:

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

= 4.75+ log {0.15/0.55}

= 4.186

use:

PH = 14 - pOH

= 14 - 4.1857

= 9.8143

Answer: 9.81