This problem is designed to incorporate several concepts and techniques into one

Question

This problem is designed to incorporate several concepts and techniques into one situation. A 225-mg sample of a diprotic acid is dissolved in enough water to make 250. mL of solution. The pH of this solution is 2.06. A 6.9 times 10^-3 M solution of calcium hydroxide is prepared. 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 point (as determined by a pH meter) is 7.96. The first dissociation constant for the acid (K_a1) is 5.90 times 10^-2. Assume that the volumes of the solutions are additive, that all solutions are at 25 degree C, and that K_a1 is at least 1000 times greater than K_a2. a. Calculate the molar mass of the acid. b. Calculate the second dissociation constant for the acid (K_a2).

Explanation / Answer

a.) pH = - log [H+]

[H+] = - antilog [2.06]

[H+] = 8.709 X 10-3

H2A ------> H+ + HA-

Write down the ice table : H+ and HA- are x moles / L and H2A is Z-x moles / L at equilibrium.

Ka1 = [H3O+] [HA-] / [H2A]

[H2A] = [8.709 X 10-3]2 / 5.90 X 10-2

[H2A] = 1.285 X 10-3 M

(Z - x) = 1.285 X 10-3 M

Z = 9.996 X 10-3 M

0.250 L of 9.996 X 10-3 M is 0.0025 moles.

Therefore, 1 mole consists of 0.225 g / 0.0025 moles = 90 g/mol

The molar mass of the unknown acid is 90 g/mol.

b.) At the second equivalence point, moles of acid = moles of base

Henderson-Hasselbach equation is given by : pH2 = pKa2 + log ( base / acid )

Therefore, for equivalence point, pH2 = pKa2

So pKa2 = 7.96

pKa2 = - log (Ka2)

Ka2 = 1.096 X 10-8

The second dissociation constant is 1.096 X 10-8.

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