40.0 mL of 0.200 M benzoic acid is titrated with 0.200 M NaOH (aq). What is the pH...

Question

a) of the acid solution initially?

my answer: pH = 0.699

b) of the acid solution halfway to the equivalence point?

my answer: pH = 1.18

c) 5.0 mL before the equivalence point?

my answer: pH = 1.00

d) at the equivalence point?

my answer: pH = 12.5

e) 2.0 mL past the equivalence point?

my answer: pH = 12.0

Answer 1

Benzoic acid (C₆H₅COOH) is a weak acid. We can represent it as HA

NaOH is a strong base. Reaction is an acid-base neutralization reaction.

HA + NaOH ==> NaA + H2O

To answer this question, we need to know the Ka for benzoic acid. I found it to be 6.46x10^-5

(a) pH of initial solution:

HA ==> H⁺ + A^-

Ka = 6.46x10^-5 = (x)(x) / 0.200 - x (assume x is small relative to 0.200 and ignore it in the denominator)

6.46x10^-5 = x² / 0.200

x² = 1.29x10^-5

x = 3.59x10^-3 M = [H⁺] (note: this is less than 2% of 0.200 so above assumption was valid)

pH = -log [H⁺] = -log 3.59x10^-3

pH 2.44

(b) At half way to equivalence point, pH = pKa

pKa = -log Ka = -log 6.46x10^-5 = 4.19

pH = 4.19

(c) 5 ml before equivalence point

equivalence point would be when 40.0 ml NaOH was added to 40.0 ml benzoic acid

5 ml before = 35.0 ml NaOH added

35 ml x 1 L / 1000 ml x 0.200 mol / L = 0.007 moles NaOH

HA + NaOH ==> NaA + H2O

0.008.....0.007........0.............Initial

-0.007...-0.007......+0.007.....Change

0.001........0............0.007.....Equilibrium

Final volume = 60.0 ml = 0.060 L

Final [HA] = 0.001 mol / 0.060 L = 0.01667 M

Final [A^-] = 0.007 mol / 0.060 L = 0.1167 M

pH = pKa + log [A-]/[HA] = 4.19 + log (0.1167 / 0.01667)

pH = 4.19 + 0.85

pH = 5.04

(d) At the equivalence point, all HA is converted to A- and there is no HA and no NaOH

Volume of NaOH needed = 40.0 mls

Final volume = 80.0 mls = 0.080 L

Final moles A^- formed = 0.008 moles (40.0 ml x 1 L / 1000 ml x 0.200 mol / L = 0.00800 mols)

Final [A^-] = 0.008 mol / 0.080 L = 0.1 M

Hydrolysis of A^- : A^- + H2O ==> HA + OH^-

Kb for A^- = 1x10^-14 / Ka = 1x10^-14 / 6.46x10^-5 = 1.55x10^-10

1.55x10^-10 = (x)(x) / 0.1 - x (assume x is small and ignore it)

x² =1.55x10^-11

x = 3.93x10^-6 M = [OH^-]

pOH = -log 3.93x10^-6 = 5.40

pH = 14 - pOH

pH = 8.60

(e) 2 ml past equivalence = 42 ml NaOH added

To simplify things, assume the OH^- contribution from hydrolysis of A^- (3.93x10^-6 M) is minor compared to the [OH^-] from the 2 mls of excess NaOH.

moles excess OH^- = 2 ml x 1 L / 1000 ml x 0.200 mol / L = 4x10^-4 moles OH^-

final volume = 40 mls + 42 mls = 82 mls = 0.082 L

final [OH^-] from excess NaOH = 4x10^-4 moles OH^- / 0.082 L = 4.88x10^-3 M (this is more than 1000 x the value of 3.93x10^-6 M, so above assumption was valid)

pOH = -log 4.88x10^-3 = 2.31

pH = 11.69