Question

2. The composition of an aqueous solution is 0.1M acetic acid + 0.2M sodium acetate. (a) Calculate the pH of the solution (pKa of acetic acid is 4.74). 3 pts. (b) What would the pH be if 2 ml of 10M NaOH was added to 1 liter of this solution? 4 pts. (c). By comparison, what would the pH be if the same amount of NaOH were added to 1 L of pure water? 3 pts.

Answer 1

[CH3COOH] = 0.1 M

[CH3COO-] = 0.2 M

a)

For a buffer solution,

pH = pKa + log([CH3COO-]/[CH3COOH])

= 4.74 + log(0.2/0.1)

= 5.04

b)

Total Volume of solution = 1 L + 2/1000 L = 1.002 L

Initial moles of CH3COOH = 0.1 M * 1 L = 0.1

Initial moles of CH3COO- = 0.2 M * 1 L = 0.2

Moles of NaOH added = 2 mL / 1000 * 10 M = 0.02

CH3COOH --> CH3COO- + H+

0.1– 0.02 --> 0.2 + 0.02 + y

[CH3COOH] = (0.1 – 0.02)/1.002 = 0.08 M

[CH3COO-] = (0.2 + 0.02)/1.002 = 0.22 M

pH = pKa + log([CH3COO-]/[CH3COOH])

= 4.74 + log(0.22/0.08)

= 5.18

c)

Total Volume of solution = 1 L + 2/1000 L = 1.002 L

Moles of NaOH added to 1 L pure water

= 2/1000 * 10 M = 0.02

[NaOH] = [OH-] = 0.02/1.002

= 1.7