Phosphorous acid is a diprotic oxyacid that is an important

Phosphorous acid is a diprotic oxyacid that is an important compount in industry and agriculture. Calculate the pH for each of the followin points in the titration of 50mL of 2.4M H3PO3 with 2.4M KOH

a. before addition of KOH
b. after addition 24 ml KOH

c. after addition 50 ml KOH

d. after addition 75 ml KOH

e. after addition 100 ml KOH

Solution

H3PO3 + 2KOH

a. before addition of KOH

[H3PO3] = 2.4 M

H3PO3 + H2O <==> H2PO3- + H3O+

let x amount dissociated

Ka1 = [H2PO3-][H3O+]/[H3PO3]

5 x 10^-2 = x^2/(2.4 - x)

x^2 + 5 x 10^-2x - 0.12 = 0

x = [H3O+] = 0.3223 M

pH = -log[H3O+] = 0.492

b. after 25 ml KOH added

initial H3PO3 = 2.4 M x 50 ml = 120 mmol

KOH added = 2.4 M x 25 ml = 60 mmol

half of H3PO3 is neutralized to its first equivalence point

H2PO3- formed = H3PO3 remained

pH = pKa1 + log(H2PO3-/H3PO3)

      = 1.30

c. after 50 ml KOH added

initial H3PO3 = 2.4 M x 50 ml = 120 mmol

KOH added = 2.4 M x 50 ml = 120 mmol

H3PO3 neutralized to its first equivalence point

pH = 1/2(pKa1 + pKa2)

     = 1/2(1.30 + 6.70)

     = 4.0

d. after 75 ml KOH added

initial H3PO3 = 2.4 M x 50 ml = 120 mmol

KOH added = 2.4 M x 75 ml = 180 mmol

1/2 of H2PO3- neutralized to HPO3^2-

second half-equivalence point

pH = pKa2 = 6.70

e. after 100 ml KOH added

all of H3PO3 is neutralized

[HPO3^2-] formed = 2.4 M x 50 ml/150 ml = 0.8 M

HPO3^2- + H2O <==> H2PO3- + OH-

let x amount reacted

Kb1 = [H2PO3-][OH-]/[HPO3^2-]

2 x 10^-7 = x^2/0.8

x = [OH-] = 4 x 10^-4 M

pOH = -log[OH-] = 3.40

pH = 14 - pOH = 10.60