Question

Consider the titratkon of 50.0 mL of 0.200 M HC₂H₃O₂ (K_a = 1.8 x 10^-5) with 0.100 M NaOH.

(A.) Calculate the pH of the HC₂H₃O₂ before the addition of NaOH

(B.) Calculate the pH of the solution after the addition of 20.0 mL of 0.100 M NaOH.

(C.) Calculate the pH of the solution after the addition of 50.0 mL of 0.100 M NaOH.

(D.) Calculate the pH of the solution after the addition of 100.0 mL of 0.100 M NaOH.

(E.) Calculate the pH of the solution after the addition of 150.0 mL of 0.100 M NaOH.

Answer 1

(A)

HC2H3O2 is acetic acid, usually written CH3COOH; it dissociates as follows:

CH3COOH    CH3COO^- + H⁺, Ka = 1.8 x 10^-5

If x moles of CH3COOH dissociate, x moles of H⁺ and x moles of CH3COO^- are produced, and the concentration of CH3COOH is reduced by x:

Ka = [CH3COO^-] [H⁺] / [CH3COOH]

Ka = x² / (0.20 - x)

x will be small compared to 0.20, so

Ka = x² / 0.20

1.8 x 10^-5 = x² / 0.20

x² = 3.6 x 10^-6

x = 1.9 x 10^-3

[H+] = x = 1.9 x 10^-3

pH = - log [H+] = - log (1.9 x 10^-3) = 2.72

(B)

Molarity = Moles / Liter

50.0 mL of 0.200 M CH3COOH

Moles of CH3COOH = 0.05 L x 0.2 M = 0.01 mole

20.0 mL of 0.100 M NaOH

Moles of NaOH = 0.02 L x 0.1 M = 0.002 mole

0.002 mole of NaOH will neutralize 0.002 mole of CH3COOH.

So, remaining mole of CH3COOH = 0.01 – 0.002 = 0.008 mole

Total volume = (50 + 20) mL = 70 mL = 0.07 L

[CH3COOH] = 0.008 mole / 0.07 L = 0.11 M

CH3COOH    CH3COO^- + H⁺ ; Ka = 1.8 x 10^-5

If x moles of CH3COOH dissociate, x moles of H⁺ and x moles of CH3COO^- are produced, and the concentration of CH3COOH is reduced by x:

Ka = [CH3COO^-] [H⁺] / [CH3COOH]

Ka = x² / (0.11 - x)

x will be small compared to 0.11, so

Ka = x² / 0.11

1.8 x 10^-5 = x² / 0.11

x² = 2 x 10^-6

x = 1.4 x 10^-3

[H+] = x = 1.4 x 10^-3

pH = - log [H+] = - log (1.4 x 10^-3) = 2.85

(C)

50.0 mL of 0.200 M CH3COOH

Moles of CH3COOH = 0.05 L x 0.2 M = 0.01 mole

50.0 mL of 0.100 M NaOH

Moles of NaOH = 0.05 L x 0.1 M = 0.005 mole

0.005 mole of NaOH will neutralize 0.005 mole of CH3COOH.

So, remaining mole of CH3COOH = 0.01 – 0.005 = 0.005 mole

Total volume = (50 + 50) mL = 100 mL = 0.1 L

[CH3COOH] = 0.005 mole / 0.1 L = 0.05 M

CH3COOH    CH3COO^- + H⁺, Ka = 1.8 x 10^-5

If x moles of CH3COOH dissociate, x moles of H⁺ and x moles of CH3COO^- are produced, and the concentration of CH3COOH is reduced by x:

Ka = [CH3COO^-] [H⁺] / [CH3COOH]

Ka = x² / (0.05 - x)

x will be small compared to 0.05, so

Ka = x² / 0.05

1.8 x 10^-5 = x² / 0.05

x² = 9 x 10^-7

x = 9.5 x 10^-4

[H⁺] = x = 9.5 x 10^-4

pH = - log [H⁺] = - log (9.5 x 10^-4) = 3.02

(D)

50.0 mL of 0.200 M CH3COOH

Moles of CH3COOH = 0.05 L x 0.2 M = 0.01 mole

100.0 mL of 0.100 M NaOH

Moles of NaOH = 0.10 L x 0.1 M = 0.01 mole

0.01 mole of CH3COOH will react with 0.01 mole of NaOH, to produce 0.01 mole of CH3COONa. So, now only CH3COONa will be present.

Total volume = (50 + 100) mL = 150 mL = 0.15 L

[CH3COONa] = 0.01 mole / 0.15 L = 0.067 M

CH3COO^- + H2O   CH3COOH + OH^-,            Ka = 1.8 x 10^-5

If x moles of CH3COOH dissociate, x moles of H⁺ and x moles of CH3COO^- are produced, and the concentration of CH3COOH is reduced by x:

Kb=Kw/Ka

Kw=1 x 10^-14

Ka=1.8 x 10^-5

Kb=5.6 x 10^-10

Kb = [CH3COOH] [OH^-] / [CH3COONa]

Ka = x² / (0.067 - x)

x will be small compared to 0.067, so

Ka = x² / 0.067

5.6 x 10^-10 = x² / 0.067

x² = 3.8 x 10^-11

x = 6.16 x 10^-6

[OH^-] = x = 6.16 x 10^-6

pOH = - log [H+] = - log (6.16 x 10^-6) = 5.21

pH = 14 – pOH = 14 – 5.21 = 8.79

(E)

50.0 mL of 0.200 M CH3COOH

Moles of CH3COOH = 0.05 L x 0.2 M = 0.01 mole

150.0 mL of 0.100 M NaOH

Moles of NaOH = 0.15 L x 0.1 M = 0.015 mole

0.01 mole of CH3COOH will neutralize 0.01 mole of NaOH.

So, remaining mole of NaOH = 0.015 – 0.010 = 0.005 mole

Total volume = (50 + 150) mL = 200 mL = 0.2 L

[NaOH] = 0.005 mole / 0.2 L = 0.0025 M

NaOH is a strong base and it will dissociate completely.

So, [OH^-] = 0.0025 M

pOH = - log [OH^-] = - log (0.0025) = 2.6

pH = 14 – pOH = 14 – 2.6 = 11.4