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 has a 0.100 M concentration of NH3(aq) and a 0.100 M concentration of NH4Cl(aq).

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

On a second question,

500.0 mL of 0.100 M NaOH is added to 615 mL of 0.250 M weak monoprotic acid (Ka = 4.91 × 10-5). What is the pH of the resulting buffer?

Answer 1

Part A:

Initial pH

pH = pKa + log([NH3]/[NH4Cl]

      = 9.25 + log(0.1/0.1)

      = 9.25

When HCl = 0.1 x 6 = 0.6 mmol added

[NH3] = (0.1 x 100 - 0.6)/106 = 0.089 M

[NH4Cl] = (0.1 x 100 + 0.6)/106 = 0.1 M

New pH = 9.25 + log(0.089/0.1) = 9.20

change in pH = old pH - new pH = 0.05

Part B:

Initial pH

pH = pKa + log([NH3]/[NH4Cl]

      = 9.25 + log(0.1/0.1)

      = 9.25

When NaOH = 0.1 x 6 = 0.6 mmol added

[NH3] = (0.1 x 100 + 0.6)/106 = 0.1 M

[NH4Cl] = (0.1 x 100 - 0.6)/106 = 0.089 M

New pH = 9.25 + log(0.1/0.089) = 9.30

change in pH = old pH - new pH = -0.05

Part C :

moles of NaOH = 0.1 M x 500 ml = 50 mmol

moles of weak monoprotic acid = 0.250 M x 615 ml = 153.75 mmol

[HA] remaining = 103.75/1115 ml = 0.093 M

[A-] formed = 50 mmol/1115 ml = 0.045 M

pH = pKa + log([A-]/[HA])

     = 4.31 + log(0.045/0.093)

     = 3.99