Question

Calculate the pH at 25°C of 264.0 mL of a buffer solution that is 0.410 M NH4Cl and 0.410 M NH3 before and after the addition of 2.60 mL of 6.0 M HNO3. (The pKa for NH4+ = 9.75)

Part 1: pH before=

Part 2: pH after=

Answer 1

1)

use:

pH = pKa + log {[conjugate base]/[acid]}

= 9.75+ log {0.41/0.41}

= 9.75

Answer: 9.75

2)

mol of HNO3 added = 6.0M *2.6 mL = 15.6 mmol

NH3 will react with H+ to form NH4+

Before Reaction:

mol of NH3 = 0.41 M *264.0 mL

mol of NH3 = 108.24 mmol

mol of NH4+ = 0.41 M *264.0 mL

mol of NH4+ = 108.24 mmol

after reaction,

mol of NH3 = mol present initially - mol added

mmol of NH3 = (108.24 - 15.6) mmol

mol of NH3 = 92.64 mmol

mol of NH4+ = mol present initially + mol added

mol of NH4+ = (108.24 + 15.6) mmol

mol of NH4+ = 123.84 mmol

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

use:

pH = pKa + log {[conjugate base]/[acid]}

= 9.75+ log {92.64/1.238*10^2}

= 9.624

Answer: 9.62