Question

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

Answer 1

1)

pH when 3 ml of HCl are adde dto V = 100 ml buffer

pOH = pKb + log(NH4+/NH3)

pOH = 4.76 +log(0.1/0.1= 4.75

ph = 14-4.75 = 9.25

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

mmo of NH3 = MV = 100*0.1 = 10 mmol

after HCl added, MVM = 3*0.1 = 0.3 mmol of Hcl

then

mmol of NH4+ = 10 + 0.3 = 10.3

mmo of NH3 = 10 - 0.3 = 9.7

then

pOH = 4.75+ log(10.3/9.7) = 4.77606

2)

after mmol of base = MV = 3*0.1 = 0.3 is added

mmol of NH4+ = 10 - 0.3 = 9.7

mmo of NH3 = 10 + 0.3 = 10.3

then

pOH = 4.75+ log(9.7/10.3) = 4.723