Question

What is the pH of a solution made by mixing 30.00 mL of 0.10 M acetic acid with 50.00 mL of 0.100 M KOH? Assume that the volumes of the solutions are additive. Ka = 1.8 × 10-5 for CH3CO2H.

Answer 1

Given:
M(CH3COOH) = 0.1 M
V(CH3COOH) = 30 mL
M(KOH) = 0.1 M
V(KOH) = 50 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.1 M * 30 mL = 3 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 50 mL = 5 mmol


We have:
mol(CH3COOH) = 3 mmol
mol(KOH) = 5 mmol

3 mmol of both will react

excess KOH remaining = 2 mmol
Volume of Solution = 30 + 50 = 80 mL
[OH-] = 2 mmol/80 mL = 0.025 M

use:
pOH = -log [OH-]
= -log (2.5*10^-2)
= 1.60


use:
PH = 14 - pOH
= 14 - 1.60
= 12.40
Answer: 12.40