Question

Calculate the pH at 25°C of 217.0 mL of a buffer solution that is 0.210 M NH4Cl and 0.210 M NH3 before and after the addition of 1.80 mL of 6.0 M HNO3. (The pKa for NH4+ = 9.75)

Answer 1

1)

Lets calculate the initial pH

use:

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

= 9.75+ log {0.21/0.21}

= 9.75

Answer: 9.75

2)

Lets calculate the final pH

mol of HNO3 added = 6.0M *1.8 mL = 10.8 mmol

NH3 will react with H+ to form NH4+

Before Reaction:

mol of NH3 = 0.21 M *217.0 mL

mol of NH3 = 45.57 mmol

mol of NH4+ = 0.21 M *217.0 mL

mol of NH4+ = 45.57 mmol

after reaction,

mol of NH3 = mol present initially - mol added

mmol of NH3 = (45.57 - 10.8) mmol

mol of NH3 = 34.77 mmol

mol of NH4+ = mol present initially + mol added

mol of NH4+ = (45.57 + 10.8) mmol

mol of NH4+ = 56.37 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 {34.77/56.37}

= 9.54

Answer: 9.54