Question

pH calculations, pKa of acetic acid = 4.76. A mixture of 100mL of 0.10 M acetic acid with 100mL of 0.05M NaOH

Answer 1

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 100 mL

M(NaOH) = 0.05 M

V(NaOH) = 100 mL

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

mol(CH3COOH) = 0.1 M * 100 mL = 10 mmol

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

mol(NaOH) = 0.05 M * 100 mL = 5 mmol

We have:

mol(CH3COOH) = 10 mmol

mol(NaOH) = 5 mmol

5 mmol of both will react

excess CH3COOH remaining = 5 mmol

Volume of Solution = 100 + 100 = 200 mL

[CH3COOH] = 5 mmol/200 mL = 0.025M

[CH3COO-] = 5/200 = 0.025M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

pKa = 4.76

use:

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

= 4.76+ log {0.025/0.025}

= 4.76

pH = 4.76