A 425 mg sample of a weak diprotic acid is dissolved in enough water to make 275
ID: 939761 • Letter: A
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.
Explanation / Answer
pH = 3.08 = -log([H+])
[H+] = 8.32 x 10-4 M
Diprotic acid be represented as H2N with initial concentration Co
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
[N2-] at second equivalence point = 2.61 / 464.6 = 5.61 x 10-3 M
Kb of N2- salt = Kw / Ka2
N2- + H2O ---> HN- + OH-
0.00561 - x x x ; x = 0.005
Kb = [OH-]2/([N2- - [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
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