Question

Part A If 10.0 mL of glacial acetic acid (pure HC2H3O2) is diluted to 1.70 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL. Express your answer to two decimal places.

Answer 1

density of acetic acid = 1.05 g/ml

mol/L = g x 1000/molar mass x ml

molarity of AcOH = 1.05 x 1000/60.05 x 1 = 17.48 M

Molarity of diluted solution = 0.010 x 17.48/1.7 = 0.103 M

HOAc <==> H+ + AcO-

let x amount has dissociated then,

Ka = 1.8 x 10^-5 = x^2/0.103

x = [H+] = 1.36 x 10^-3 M

pH = -log[H+] = -log(1.36 x 10^-3) = 2.866