Question

Calculate the pH change when 5.0 mL of 5.0-M NaOH is added to 0.500 L of a solution of:

(The pKa for acetic acid is 4.74.)

a) 0.50-M acetic acid and 0.50-M sodium acetate.

pH change = _____

b) 0.050-M acetic acid and 0.050-M sodium acetate.

pH change = ______

c) 0.0050-M acetic acid and 0.0050-M sodium acetate.

pH change = _______

Answer 1

a) Initial pH

Using Hendersen-Hasselbalck equation for pH,

pH = pKa + log(base/acid)

     = 4.74

moles of NaOH added = 5 M x 0.005 ml = 0.025 mols

Final concentration of,

[acetic acid] in solution = (0.5 M x 0.5 L - 0.025) mols/0.505 L = 0.445 M

[acetate-] in solution = (0.5 M x 0.5 L + 0.025) mols/0.505 L = 0.544 M

Using Hendersen-Hasselbalck equation for pH,

pH = pKa + log(base/acid)

     = 4.74 + log(0.544/0.445)

     = 4.83

Change in pH = (Final - initial)pH = 4.83 - 4.74 = 0.09

[pl. note If we to take initial pH - final pH = change in pH = -0.09]

b) moles of NaOH added = 5 M x 0.005 ml = 0.025 mols

Final concentration of,

[acetic acid] in solution = (0.05 M x 0.5 L - 0.025) mols/0.505 L = 0 M

[acetate-] in solution = (0.05 M x 0.5 L + 0.025) mols/0.505 L = 0.099 M

CH3COO- + H2O <==> CH3COOH + OH-

Kb = Kw/Ka = 1 x 10^-14/1.8 x 10^-5 = x^2/0.099

x = [OH-] = 7.42 x 10^-6 M

[H+] = 1 x 10^-14/7.42 x 10^-6 = 1.35 x 10^-9 M

pH = -log[H+] = 8.87

Change in pH = (Final - initial)pH = 8.87 - 4.74 = 4.13

[pl. note If we to take initial pH - final pH = change in pH = -4.13]

c) moles of NaOH added = 5 M x 0.005 ml = 0.025 mols

Final concentration of,

[acetic acid] in solution = (0.005 M x 0.5 L - 0.025) mols/0.505 L = 0 M

[acetate-] in solution = (0.005 M x 0.5 L + 0.025) mols/0.505 L = 0.054 M

CH3COO- + H2O <==> CH3COOH + OH-

Kb = Kw/Ka = 1 x 10^-14/1.8 x 10^-5 = x^2/0.054

x = [OH-] = 5.48 x 10^-6 M

[H+] = 1 x 10^-14/5.48 x 10^-6 = 1.82 x 10^-9 M

pH = -log[H+] = 8.74

Change in pH = (Final - initial)pH = 8.74 - 4.74 = 4.00

[pl. note If we to take initial pH - final pH = change in pH = -4.00]