Question

You titrate a 20.0 mL acid sample containing 1.104 g of ascorbic acid (molar mass=176.12 g/mol and pKa=4.10) with 0.200 M NaOH. Calculate the following

1. The pH at the begining of the titration, before any base was added

2. The volume of base (NaOH) required to reach the equivalence point

3. The pH at the equivalence point and the half-equivalence point

4. The pH after 35.5 mL of base have beedn added

Answer 1

molar concentration of ascorbic acid = moles/L = 1.104/176.12 x 0.02 = 0.313 M

pKa = 4.10

Ka = 7.94 x 10^-5

1. pH at the begining

Ka = [ascorbate-][H+]/[ascorbic acid]

let x amount has dissociated

7.94 x 10^-5 = x^2/0.313

x = [H+] = 4.98 x 10^-3 M

pH = 2.30

2. Volume of base to reach equivalence point = 6.63 x 10^-3mol/0.2M = 0.0313 L = 31.3 ml

3. pH at haf equivalence point = pKa = 4.10

ph at equivalence point

ascorbate- + H2O <==> ascorbic acid + OH-

[ascorbate-] = 6.63 x 10^-3/0.0513 = 0.13 M

let x amount reacted,

Kb = Kw/Ka = x^2/[ascorbate-]

1 x 10^-14/7.94 x 10^-5 = x^2/0.13

x = [OH-] = 4.05 x 10^-6 M

pOH = 5.40

pH = 14 - pOH = 8.61

4. pH after 35.5 ml of base is added

excess [NaOH] = (0.0355 - 0.0313) x 0.2/0.0557 = 0.0164 M

pOH = -log[OH-] = 1.78

pH = 14 - pOH = 12.21