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.

pH= ?

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

pH=?

Answer 1

Iniitla pH:

pH = pKa + log(NH3/NH4+)

pKa fo rNH4+, 9.25

pH = 9.25 + log(0.1/0.1) = 9.25 initially

after,

a)

addition of 3*0.1 = 0.3 mmol

mmol of NH4+ = MV = 0.1*100 = 10

mmol of NH3 = MV = 0.1*100 = 10

after reaction

mmol of NH4+ = 10+0.3 = 10.3

mmol of NH3 = 10-0.3 = 9.7

pH = 9.25 + log(9.7/10.3)

pH = 9.2239

pH change = 9.2239-9.25 = -0.0261 or decrease in 0.0261 units

b)

addition of 3*0.1 = 0.3 mmol

mmol of NH4+ = MV = 0.1*100 = 10

mmol of NH3 = MV = 0.1*100 = 10

after reaction

mmol of NH4+ = 10-0.3 = 9.7

mmol of NH3 = 10+0.3 = 10.3

pH = 9.25 + log(10.3/9.7)

pH = 9.276

pH change = 9.276-9.25 = 0.026 increase