Question

# An 80.9 mL sample of a 0.100 M solution of a diprotic acid, H2A, is titrated...

An 80.9 mL sample of a 0.100 M solution of a diprotic acid, H2A, is titrated with 0.100 M KOH. For H2A, Ka1 = 8.09 x 10^-5 and Ka2 = 8.09 x 10^-10. A) Calculate the concentration of A2- prior to the addition of any KOH. B) What is the pH of the initial solution? C) What is the pH exactly halfway to the first equivalence point? D) What is the pH exactly halfway between the first and second equivalence points? E) What is the pH at the second equivalence point?

Titration

Ka2 = [A^2-] = 8.09 x 10^-10

(b) pH of initial solution

H2A <==> H+ + HA-

Ka1 = 8.09 x 10^-5 = x^2/0.1

x = [H+] = 2.84 x 10^-3 M

pH = -log[H+] = 2.55

(c) pH at first half equivalence point

pH = pKa1 = -logKa1 = 4.09

(d) pH at second equivalence point

pH = pKa2 = -log[Ka2] = 9.09

(e) pH at second equivalence point

Volume of KOH added = 2 x 0.1 x 80.9/0.1 = 161.8 ml

[A^2-] formed = 0.1 x 80.9/(161.8 + 80.9) = 0.033 M

A^2- + H2O <==> HA- + OH-

let x amount has hydrolyzed

Kb2 = 1 x 10^-14/8.09 x 10^-10 = x^2/0.033

x = [OH-] = 6.39 x 10^-4 M

pOH = -log[OH-] = 3.19

pH = 14 - pOH = 10.81

#### Earn Coins

Coins can be redeemed for fabulous gifts.