Question

What is the pH of a solution containing 0.020 mol sodium hydroxide and 0.100 mol acetic acid dissolved in 100.0 mL of water, given that the pKa of acetic acid is 4.75?

Answer 1

In the solution, sodium hydroxide and acetic acid neutralize each other according to the equation:

CH₃COOH + NaOH → CH₃COONa + H₂O

As a result, as 0.020 mol sodium hydroxide interact with 0.100 mol acetic acid, 0.100 mol - 0.020 mol = 0.080 mol of acetic acid are left in the solution.

As the volume of the solution is 100.0 mL, the molar concentration of the acetic acid in a solution equals:

c = n / V = 0.080 mol / 100.0 mL = 0.080 mol / 0.100 L = 0.8 mol/L = 0.8 M

As acetic acid is a weak acid:

K_a = [H⁺][A^-]/[AH] = 10^-pKa

As [H⁺] = [A^-]:

10^-pKa = [H⁺]² / [AH]

From here:

[H⁺] = (10^-pKa × [AH])^1/2

As [H⁺] << [AH], [AH] = 0.8 M. As a result:

[H⁺] = ( 10^-4.75 × 0.8 M)^1/2 = 0.00377 M

From here, pH of the solution equals:

pH = -log[H⁺] = -log(0.00377) = 2.42

Answer: 2.42