Question

A buffer solution contains 0.11 mol of acetic acid and 0.14 mol of sodium acetate in 1.00 L.

a) What is the pH of the buffer after the addition of 0.03 mol of KOH? Express your answer to two decimal places.

b) What is the pH of the buffer after the addition of 0.03 mol of HNO3? Express your answer to two decimal places.

Answer 1

The ionization of acetic acid can be viewed as

CH₃COOH (aq) <=======> H⁺ (aq) + CH₃COO^- (aq)

a) K_a of acetic acid = 1.7*10^-5

Moles acetic acid = 0.11 mole.

Moles acetate = 0.14 mole.

Moles OH^- (as NaOH) added = 0.03 mole

Set up the ICE chart as below.

CH₃COOH (aq) + OH^- (aq) <======> CH₃COO^- (aq) + H₂O (l)

initial                     0.11                  0.03                                0.14

change                  -0.03                 -0.03                               +0.03

equilibrium            0.08                  0.0                                   0.17

The concentrations at equilibrium are

[CH₃COOH] = (0.08 mole)/(1.00 L) = 0.08 M

[CH₃COO^-] = (0.17 mole)/(1.00 L) = 0.17 M.

Write down the expression for K_a as below.

K_a = [H⁺][CH₃COO^-]/[CH₃COOH]

=====> [H⁺] = K_a*[CH₃COOH]/[CH₃COO^-]

=====> [H⁺] = (1.7*10^-5)*(0.08 M)/(0.17 M)

=====> [H⁺] = 8.00*10^-6 M

pH = -log [H⁺]

= -log (8.00*10^-6 M)

= 5.0969 ≈ 5.10

The pH of the buffer solution is 5.10 (ans, correct to 2 decimal places).

b) Moles H⁺ (as HNO₃) added = 0.03 mole.

Set up the ICE chart as below.

CH₃COOH (aq) + H⁺ (aq) <======> CH₃COO^- (aq) + H₂O (l)

initial                     0.11                  0.03                                0.14

change                  +0.03                 -0.03                               -0.03

equilibrium            0.14                  0.0                                  0.11

The concentrations at equilibrium are

[CH₃COOH] = (0.14 mole)/(1.00 L) = 0.14 M

[CH₃COO^-] = (0.11 mole)/(1.00 L) = 0.11 M.

Write down the expression for K_a as below.

K_a = [H⁺][CH₃COO^-]/[CH₃COOH]

=====> [H⁺] = K_a*[CH₃COOH]/[CH₃COO^-]

=====> [H⁺] = (1.7*10^-5)*(0.14 M)/(0.11 M)

=====> [H⁺] = 2.16*10^-5 M

pH = -log [H⁺]

= -log (2.16*10^-5 M)

= 4.6655 ≈ 4.66

The pH of the buffer solution is 4.66 (ans, correct to 2 decimal places).