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 (Ksp= 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

weight of the weak acid = 425 mg = 0.425 g

v = 275 ml

Kg = 8E^-5

pH = 3.08

ksp = 1.3E^-6

Ca(OH)2 --------> Ca⁺² + 2OH^-

(s) (2s)

ksp = s * (2s)² = 4s³

s = = 6.875E^-3 M

s = c for a soluble salt.

Ca(OH)2 + H2A -------------> CaA + 2H20

6.875E^-3 |DIPROTIC ACID| 6.875E^-3| Water

pH = pka + log {[CaA]/[HA]}

log [CaA]/[HA] = pH - pka = 3.08 - [-log(8E^-5)] = -0.016

[CaA]/[HA] = 10^-1.016 = 0.0963

Then;

[H2A] = [CaA] / 0.0963 = 6.875E^-3/0.0963 = 0.0713 M

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a) Molarity (mol/L) = solute moles / lts sln

H2A moles = 275ml * 0.0713 mol H2A/1000ml = 0.0196 moles H2A

H2A gr = 0.0196 moles H2A * 0.425 gr / 1mol H2A = 0.0083 gr H2A

The molar mass of the acid is 0.0083 gr H2A

b) pH = pka2 + log [s/A]

11.7 = pka2 + log [6.875E^-3/0.0713]

pka2 = 11.7 - log (9.635E^-2) = 12.716

Ka2 = 10^-12.716 = 1.9225E^-13

The dissiociation constant (Ka2) will be 1.9225E^-13