Zahin asked in Science & MathematicsChemistry · 10 months ago

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?

Relevance
• 10 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