Question

Calculate the change in pH when 3.00 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.

https://sites.google.com/site/chempendix/ionization

Calculate the change in pH when 3.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Answer 1

pKb of NH3 = 4.74

pKa = 9.26

pH = pKa + log [base / acid ]

    = 9.26 + log [0.100 / 0.100]

pH = 9.26

moles of HCl added = 3 x 0.100 = 0.300

moles of NH3 = NH4Cl = 100 x 0.100 = 10

pH = pKa + log [base - C / acid + C]

    = 9.26 + log [10 - 0.3 / 10 + 0.3]

   = 9.23

pH change = 9.23 - 9.26 = - 0.03

pH change = - 0.03

moles of NaOH added = 0.300

pH = pKa + log [base + C / acid - C]

    = 9.26 + log [10 + 0.3 / 10 - 0.3]

   = 9.29

pH change = 9.29 - 9.26

pH change = 0.03