Question

12) What is the pH of the resulting solution if 30.00 mL of 0.10 M acetic acid is added to 10.00 mL of 0.10 M NaOH? Assume that the volumes of the solutions are additive. Ka = 1.8 × 10-5 for CH3CO2H. A) 9.56 B) 4.44 C) 5.05 D) 8.95

use the Henderson Hasselbalch equation to solve please thank you

Answer 1

Ka of acetic acid = 1.8*10^-5
pka = -log Ka
= -log (1.8*10^-5)
= 4.745

mol of CH3COOH present initially = M*V = 0.10 M * 30 mL = 3 mmol
mol of NaOH added = 0.10 M * 10 mL = 1 mmol

1 mmol of both will react to form 1 mmol of CH3COONa
after reaction,
mol of CH3COONa = 1 mmol
mol of CH3COOH remaning = (3-1) mmol = 2 mmol

total volume = 30 + 10 = 40 mL
[CH3COONa] = 1/40 = 0.025 M
[CH3COOH] = 2/40 = 0.05 M
use:
pH = pKa + log {[CH3COONa]/[CH3COOH]}
= 4.745 + log (0.025/0.05)
=4.745 - 0.301
= 4.444
Answer: B