Question

Consider a titration of 50.0mL sample of 0.500 M HC2H3O2 (a weak acid, which contains 0.0250 moles H3O+) with 0.400 M NaOH (a strong base). Determine the pH of the solution after 15.0 mL of NaOH (0.00600 moles of OH-) is added.

Answer 1

Given:
M(CH3COOH) = 0.5 M
V(CH3COOH) = 50 mL
M(NaOH) = 0.4 M
V(NaOH) = 15 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.5 M * 50 mL = 25 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.4 M * 15 mL = 6 mmol


We have:
mol(CH3COOH) = 25 mmol
mol(NaOH) = 6 mmol

6 mmol of both will react

excess CH3COOH remaining = 19 mmol
Volume of Solution = 50 + 15 = 65 mL
[CH3COOH] = 19 mmol/65 mL = 0.2923M

[CH3COO-] = 6/65 = 0.0923M

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 {9.231*10^-2/0.2923}
= 4.244


Answer: 4.24