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

Solution.

As the acid is quite weak, pH=(pK+pc)/2;

Neglecting the second-step dissociation we can say that

275 mL of solution contain 425 mg of the acid;

1000 mL of solution contain x mg of the acid;

x = 1000*425/275 = 1545 mg or 1.545 g.

As 1L of the solution contains

the molar mass can be found as

At the second dissociation point, the pH calculation can be performed as for a weak base:

;

To find a concentration we need to get a total volume.

The molar concentration or calcium hydroxide is

To reach the second equivalence point, the amount of hydroxide-ions required is

To deliver this amount one needs V = n/c = 0.004758/0.022 = 0.216 L.

The total volume of the solution therefore is 0.216+0.275 = 0.491 L; the concentration of a salt is (0.00865*0.275)/0.491 = 0.00485 M;

pc = -log(c) = 2.315;