Question

1. The pH of 300 mL solution made of 0.59 M acetic acid and 1.07 M potassium acetate is (Ka=1.8 x 10^-5) after the addition of 0.74 moles NaOH?

Answer 1

13.7

Step-by-step explanation:

First we must calculate the moles of HC2H3O2 and KC2H3O2

300 mL = .300 L

.300 L x (.59 moles /L) = 0.18 moles of Acetic Acid

.300 L x (1.07 moles / L) = .321 moles of Potassium Acetate

Since more moles of NaOH is added than there are moles of Acid we will find the excess NaOH

.74 - .18 = .56 moles

Convert this to molarity .56 moles OH- / .300 L = 1.9 M

pH = pOH + 14

pH = -log(1.9) + 14 = 13.7