Question

If 10 mL of 0.025 M benzoic acid (K_a = 6.5 * 10^-5) is titrated with 0.01 M NaOH, what is pH after 0 ml, 20 ml, 50 ml, and 90 ml of base have been added?

Answer 1

Titration

a) 0 ml NaOH added

PhCOOH + H2O <==> PhCOO- + H3O+

Ka = [PhCOO-][H3O+]/[PhCOOH]

6.5 x 10^-5 = x^2/0.025

x = [H3O+] = 1.27 x 10^-3 M

pH = -log[H3O+] = 2.89

b) 20 ml 0.01 M NaOH added

moles of PhCOOH = 0.025 m x 10 ml = 0.25 mmol

moles of NaOH added = 0.01 M x 20 ml = 0.2 mmol

[PhCOO-] formed = 0.2 mmol/30 ml = 0.0067 M

[PhCOOH] remained = 0.05 mmol/30 ml = 0.00167 M

pH = pKa + log(base/acid)

     = 4.19 + log(0.0067/0.00167)

     = 4.79

c) 50 ml 0.01 M NaOH added

moles of PhCOOH = 0.025 m x 10 ml = 0.25 mmol

moles of NaOH added = 0.01 M x 50 ml = 0.5 mmol

excess [NaOH] = 0.25 mmol/60 ml = 0.004167 M

pOH = -log[0.004167] = 2.38

pH = 14 - pOH = 11.62

d) 90 ml 0.01 M NaOH added

moles of PhCOOH = 0.025 m x 10 ml = 0.25 mmol

moles of NaOH added = 0.01 M x 90 ml = 0.9 mmol

excess [NaOH] = 0.65 mmol/100 ml = 0.0065 M

pOH = -log[0.004167] = 2.19

pH = 14 - pOH = 11.81