Question

What is the pH at 25°C of a solution that results from mixing equal volumes of a 0.0500 M solution of acetic acid and a  0.0374 M solution of sodium hydroxide? The pK_a of acetic acid is 4.74.

Express the pH to at least two decimal places.

Answer 1

Let the volume of CH3COOH and NaOH be 1 mL each

Given:

M(CH3COOH) = 0.05 M

V(CH3COOH) = 1 mL

M(NaOH) = 0.0374 M

V(NaOH) = 1 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.05 M * 1 mL = 0.05 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.0374 M * 1 mL = 0.0374 mmol

We have:

mol(CH3COOH) = 0.05 mmol

mol(NaOH) = 0.0374 mmol

0.0374 mmol of both will react

excess CH3COOH remaining = 0.0126 mmol

Volume of Solution = 1 + 1 = 2 mL

[CH3COOH] = 0.0126 mmol/2 mL = 0.0063M

[CH3COO-] = 0.0374/2 = 0.0187M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.74+ log {1.87*10^-2/6.3*10^-3}

= 5.213

Answer: 5.21