Question

In an experiment, 20.00 mL of 0.30 M acetic acid is titrated with 0.30 M NaOH. What is the pH when the following volumes of NaOH is added? Ka for acetic acid is 1.8 x 10^-5. Show work with ICE table.

a)5mL of NaOH added

b)20mL of NaOH added

c)35mL of NaOH added

Answer 1

ICE = initial, change, equilibrium

a)

V = 5 mL of base

initially (I)

mmol of acid = MV = 20*0.3 = 6 mmol

mmol of acetate = 0

Change

mmol of base = MV = 5*0.3 = 1.5 mmol

mmol of acid = MV = -1.5

mmol of acetate = 1.5

Equilibrium (E)

mmol of acid = 6-1.5 = 4.5

mmol of acetate = 0 + 1.5

then

pKa = -log(Ka) = -log(1.8*10^-5) = 4.75

pH = pKa + log(A-/HA)

pH = 4.75 + log(1.5/4.5)

pH = 4.272878

b)

initially (I)

mmol of acid = MV = 20*0.3 = 6 mmol

mmol of acetate = 0

Change

mmol of base = MV = 20*0.3 = 6 mmol

mmol of acid = MV = -6

mmol of acetate = +6

Equilibrium (E)

mmol of acid = 6

mmol of acetate =6

therefore

Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.55*10^-10

Kb = [HA][OH-]/[A-]

Kb = (x*x)/(0.15-x)

5.55*10^-10= (x*x)/(0.15-x)

x = IOH = 9.12*10^-6

pOH = -log(9.12*10^-6) = 5.0400

pH = 14-pÓH = 14-5.0400 = 8.96

c)

finally

initially

mmol of NaOH = MV = 35*0.3 = 10.5 mmol

mmol of acid = MV = 20*0.3 = 6 mmol

excess base in equilibrium

10.5-6 = 4.5 mmol

[OH-] = 4.5 / (20+35) = 0.08181

pOH = -log(0.08181) = 1.0871

pH = 14-1.0871 = 12.9129