Question

What is the pH of a mixture of 15mL of 0.25 M HC2H3O2 and 5.3mL of 0.18 M NaOH? Ka= 1.8*10^-5

Answer 1

Given:
M(CH3COOH) = 0.25 M
V(CH3COOH) = 15 mL
M(NaOH) = 0.18 M
V(NaOH) = 5.3 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.25 M * 15 mL = 3.75 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.18 M * 5.3 mL = 0.954 mmol


We have:
mol(CH3COOH) = 3.75 mmol
mol(NaOH) = 0.954 mmol

0.954 mmol of both will react

excess CH3COOH remaining = 2.796 mmol
Volume of Solution = 15 + 5.3 = 20.3 mL
[CH3COOH] = 2.796 mmol/20.3 mL = 0.137734M

[CH3COO-] = 0.954/20.3 = 0.046995M

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 {0.046995/0.137734}
= 4.28


pH = 4.28