Question

Calculate the change in pH when 7.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). A list of ionization constants can be found here.

pH= ?

Calculate the change in pH when 7.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

pH=?

Answer 1

1)

mol of HCl added = 0.1M *7.0 mL = 0.70 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 - 0.70) mmol

mol of NH3 = 9.3 mmol

mol of NH4+ = mol present initially + mol added

mol of NH4+ = (10 + 0.70) mmol

mol of NH4+ = 10.7 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 {10.7/9.3}

= 4.806

use:

PH = 14 - pOH

= 14 - 4.8056

= 9.1944

Answer: 9.19

2)

mol of NaOH added = 0.1M *7.0 mL = 0.70 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 + 0.70) mmol

mol of NH3 = 10.7 mmol

mol of NH4+ = mol present initially - mol added

mol of NH4+ = (10 - 0.70) mmol

mol of NH4+ = 9.3 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 {9.3/10.7}

= 4.684

use:

PH = 14 - pOH

= 14 - 4.6838

= 9.3162

Answer: 9.32