Question

Calculate the change in pH when 6.00 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.

change in pH =

Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Answer 1

1) first calculate the initial pH

mixture of NH3 and NH4Cl act as basic buffer

pOH = pKb + log [NH4Cl] / [NH3]

pOH = 4.74 + log [0.1] /[0.1]

pOH = 4.74

pH = 14 - 4.74

pH = 9.26

if we add 0.1 M HCl

[NH3] = (100 x 0.1) - (6.0 x 0.1) = 9.4 / 106 = 0.089 M

[NH4Cl] = (100 x 0.1) + (6.0 x 0.1) = 10.6 / 106 = 0.1 M

pOH = 4.74 + log [0.1] / [0.089]

pOH = 4.79

pH = 14 - 4.79

pH = 9.21

pH = 9.26 - 9.21 = 0.05

2) after NaOH added

[NH3] = (100 x 0.1) + (6 x 0.1) = 10.6 / 106 = 0. 1M

[NH4Cl] = (100 x 0.1) - (6 x 0.1) = 9.4 / 106 = 0.089 M

pOH = 4.74 + log [0.089] / [0.1]

pOH = 4.69

pH = 14 - 4.69

pH = 9.31

pH = 9.31 - 9.26 = 0.05