Question

Consider a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For ammonia, pKb=4.75. You may want to reference (Pages 648 - 658) Section 16.2 while completing this problem. Part A Calculate the pH of 1.0 L of the original buffer, upon addition of 0.030 mol of solid NaOH. Express the pH to two decimal places.

Answer 1

mol of NaOH added = 0.03 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 + 0.03) mol

mol of NH3 = 0.53 mol

mol of NH4+ = mol present initially - mol added

mol of NH4+ = (0.2 - 0.03) mol

mol of NH4+ = 0.17 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 {0.17/0.53}

= 4.251

use:

PH = 14 - pOH

= 14 - 4.2509

= 9.7491

Answer: 9.75