- #1
Eclair_de_XII
- 1,083
- 91
Homework Statement
"A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) os titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 mL, (f) 50.0 mL."
Homework Equations
Ka = [H+][A-]/[HA]
Ka = 1.8 ⋅ 10-5
x = ½(-b ± √(b2 - 4ac)
pH = -log[H+]
pH = 2.78 (from back of book)
The Attempt at a Solution
0.15 mol/L ⋅ 1 L/1000 mL ⋅ 35.0 mL = 5.25 ⋅ 10-3 mol HAc
1.8 ⋅ 10-5 = (x)(x)/(5.25 ⋅ 10-3 - x)
x2 + 1.8 ⋅ 10-5x - 9.45 ⋅ 10-8 = 0
x = ½(-1.8 ⋅ 10-5 ± √3.24 ⋅ 10-10 - 37.8 ⋅ 10-8)
x = 2.99 ⋅ 10-4 mol H+
[H+] = (2.99 ⋅ 10-4 mol H+)/(0.0350 L) = 8.54 ⋅ 10-3 M
-log(8.54 ⋅ 10-3) = 2.069 ≠ 2.78
This is rather an unorthodox, and thus, unreliable method of finding the initial pH of the acetic acid solution. But I couldn't find any other way to do it... Could someone tell me if I'm doing this problem right?