Question

25.00 mL of 0.1 M acetic acid are titrated with 0.05 M NaOH. Calculate the pH of the solution after addition of 25 mL of NaOH. Ka(CH3COOH) = 1.8 x 10-5.

Answer 1

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 25 mL

M(NaOH) = 0.05 M

V(NaOH) = 25 mL

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

mol(CH3COOH) = 0.1 M * 25 mL = 2.5 mmol

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

mol(NaOH) = 0.05 M * 25 mL = 1.25 mmol

We have:

mol(CH3COOH) = 2.5 mmol

mol(NaOH) = 1.25 mmol

1.25 mmol of both will react

excess CH3COOH remaining = 1.25 mmol

Volume of Solution = 25 + 25 = 50 mL

[CH3COOH] = 1.25 mmol/50 mL = 0.025M

[CH3COO-] = 1.25/50 = 0.025M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {2.5*10^-2/2.5*10^-2}

= 4.745

Answer: 4.75