Question

calculate the change in pH when 3.00 ml of 0.1 M HCL is added to 100 ml of a buffer solution that is 0.1 M in NH3 and 0.1 M in NH4Cl.
Calculate the change in pH when 3.00 ml of 0.1 M NaOH is added to the original buffer solution.

Answer 1

1)

mol of HCl added = 0.1M *3.0 mL = 0.300 mmol

NH3 will react with H+ to form NH4+

Before Reaction:

mol of NH3 = 0.1 M *100.0 mL

mol of NH3 = 10 mmol

mol of NH4+ = 0.1 M *100.0 mL

mol of NH4+ = 10 mmol

after reaction,

mol of NH3 = mol present initially - mol added

mmol of NH3 = (10 - 0.300) mmol

mol of NH3 = 9.7 mmol

mol of NH4+ = mol present initially + mol added

mol of NH4+ = (10 + 0.300) mmol

mol of NH4+ = 10.3 mmol

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 {10.3/9.7}

= 4.771

use:

PH = 14 - pOH

= 14 - 4.7708

= 9.2292

Answer: 9.23

2)

mol of NaOH added = 0.1M *3.0 mL = 0.300 mmol

NH4+ will react with OH- to form NH3

Before Reaction:

mol of NH3 = 0.1 M *100.0 mL

mol of NH3 = 10 mmol

mol of NH4+ = 0.1 M *100.0 mL

mol of NH4+ = 10 mmol

after reaction,

mol of NH3 = mol present initially + mol added

mol of NH3 = (10 + 0.300) mmol

mol of NH3 = 10.3 mmol

mol of NH4+ = mol present initially - mol added

mol of NH4+ = (10 - 0.300) mmol

mol of NH4+ = 9.7 mmol

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 {9.7/10.3}

= 4.719

use:

PH = 14 - pOH

= 14 - 4.7187

= 9.2813

Answer: 9.28