A 425 mg sample of a weak diprotic acid is dissolved in enough water to make 275

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

Since pH of the initial solution is 3.08 the [H+] will be

10-pH = 10-3.08 = 8.31 x 10-4

For a polyprotic system the pH contribution is mainly from the first dissociation

Also for an acid

Ka = [H+]2/Ca-[H+]

8.0 x 10-5 = [8.31 x 10-4]2/Ca-8.31 x 10-4

8.0 x 10-5 Ca - 8.0 x 10-5 x 8.31 x 10-4 = 6.91 x 10-7

8.0 x 10-5 Ca = 6.91 x 10-7 +8.0 x 10-5 x 8.31 x 10-4

8.0 x 10-5 Ca = 7.58 x 10-7

Ca = 7.58 x 10-7/8.0 x 10-5

Ca = 9.47 x 10-3 M

475 mg in 275 mL = 9.47 x 10-3 M

In 1000 mL it is 1727 mg

Now 1727 mg is 9.47 x 10-3 moles

1.727g/9.47 x 10-3 moles = 182.39 g/mol is the molar mass of the acid

Second dissociation constant is reported as first K for the acid is at least 1000 times greater than the second K

so second dissociation constant is 8.0 x 10-8

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.