Question

A 425 mg sample of a weak diprotic acid is dissolved in enough water to make 275.0 ml of solution. The pH of this solution is 3.08. A saturated solution of calcium hydroxide (K_sp= 1.3 x 10^-6) is prepared by adding excess calcium hydroxide to water and then removing the undissolved solid by filtration. Enough of the calcium hydroxide solution is added to the solution of the acid to reach the second equivalence point. The pH at the second equivalence is 11.7. The first dissociation constant for the acid is 8.0 x 10^-5. Assume that the volumes of solutions are additive and that the first K for the acid is at least 1000 times greater than the second K.

            a- Calculate the molar mass of the acid

            b- Calculate the second dissociation constant for the acid.

Answer 1

pH = 3.08 = -log([H+])

[H+] = 8.32 x 10^-4 M

Diprotic acid be represented as H₂N with initial concentration C_o

H₂N ---> H⁺ + HN^-

Co - x x x

x² / (Co -x) = 8 x 10^-5 (1)

Since Ka1 >> Ka2

x = [H+] = 8.32 x 10^-4 M

(1) gives: Co - x = (8.32 x 10^-4)² / 8 x 10^-5

Co = 9.48 x 10^-3 M

Moles of acid = Co * 275 = 2.61mmoles

a) Molecular weight of acid = 425 / 2.61 = 163 g/mol

Volume of Ca(OH)₂ required

Ca(OH)2 ---> Ca²⁺ + 2OH^-

x 2x

Ksp = x.(2x)² = 4x³ = 1.3 x 10^-6

x = 0.007 M

[OH^-] = 2x = 0.014 M

Volume of Ca(OH)₂ * [OH^-] = 2.61 mmol

Volume of Ca(OH)₂ required = 2.61 / 0.014 = 189.6 mL

Total volume = 189.6 + 275 = 464.6 mL

At second equivalence point pOH = 14 -11.7 = 2.3

[OH^-] = 10^-2.3 = 0.005 M

[N^2-] at second equivalence point = 2.61 / 464.6 = 5.61 x 10^-3 M

Kb of N^2- salt = Kw / Ka2

N^2- + H₂O ---> HN^- + OH^-

0.00561 - x x x ; x = 0.005

Kb = [OH^-]²/([N^2- - [OH^-])

= 0.005² / (0.00561 - 0.005) = 4.2 x 10^-2

b) Second dissociation constant, Ka2 = Kw / Kb

Ka2 = 10^-14 / 4.2 x 10^-2 = 2.4 x 10^-13