Can someone help me with this question? ?
4. In a beaker Johan mixed the following things together:
a. 10 g dry acetic acid
b. 5.8 g Potassium acetate
c. 0.9M 250 mL acetic acid
d. 0.5M 680 mL potassium acetate
e. Half liter distilled water.
What was the pH of his mixture?
- pisgahchemistLv 710 months ago
The term "dry acetic acid" is misleading, because it might be construed as a solid. What you mean is "glacial acetic acid", or "anhydrous acetic acid."
10g HC2H3O2 x (1 mol HC2H3O2 / 60.05g HC2H3O2) = 0.165 mol HC2H3O2
0.250L x (0.9 mol HC2H3O2 / 1L) = 0.225 mol HC2H3O2
Total = 0.390 mol HC2H3O2
5.8g KC2H3O2 x (1 mol KC2H3O2 / 98.1g KC2H3O2) = 0.0591 mol KC2H3O2 ... 0.0591 mol C2H3O2^-
0.680L x (0.5 mol C2H3O2^- / 1L) = 0.349 mol C2H3O2^-
Total = 0.408 mol C2H3O2^-
Total volume = 1.43L ....... give or take, assuming volumes are additive
Concentration of HC2H3O2 = 0.390 mol / 1.43L = 0.273M
Concentration of C2H3O2^- = 0.408 mol / 1.43L = 0.285M
So, if you are happy with the assumptions of the Henderson-Hasselbalch equation, then
pH = pKa + log([base] / [acid]) ..... assume [base] = [acid]
pH = pKa = -log(1.76x10^-5) = 4.75
If you are not happy assuming that [base] = [acid], then pH = 4.77