Question

a) A solution was prepared by dissolving 0.02 moles of acetic acid (HOAc; pKa = 4.8) in water to give 1 liter of solution. What is the pH? b) To this solution was then added 0.008 moles of concentrated sodium hydroxide (NaOH). What is the new pH? (In this problem, you may ignore changes in volume due to the addition of NaOH). c) An additional 0.012 moles of NaOH is then added. What is the pH?

Answer 1

a) concentration of acetic acid = moles/L = 0.02 moles/1L = 0.02 M

acetic acid disscoiates as,

CH3COOH <==> CH3COO- + H+

let x amount of acid has dissociated,

Ka = [CH3COO-][H+]/[CH3COOH]

pKa = -logKa = 4.8

Ka = 1.585 x 10^-5

1.585 x 10^-5 = x^2/0.02

x = [H+] = 5.63 x 10^-4 M

Thus, pH = -log[H+] = -log(5.63 x 10^-4) = 3.25

b) 0.008 moles of NaOH added

new concentration of CH3COOH = (0.02 - 0.008)/1 = 0.012 M

Ka = 1.585 x 10^-5 = x^2/0.012

x = [H+] = 4.36 x 10^-4 M

Thus new pH = -log(4.36 x 10^-4) = 3.36

c) additional 0.012 moles of NaOH is added

moles of salt formed CH3COONa = 0.012 mols

concentration of salt = 0.012mols/1L = 0.012 M

salt hydrolyzes,

CH3COO- + H2O <==> CH3COOH + OH-

let x amount of salt hydrolyzed then,

Kb = Kw/Ka = 1 x 10^-14/1.585 x 10^-5 = x^2/0.012

x = [OH-] = 2.75 x 10^-6 M

pOH = -log[OH-] = -log(2.75 x 10^-6) = 5.56

pH = 14 - pOH = 8.44