[KungKemi]

I have noticed that this question was previously asked on this forum 10 years ago, however, I noted that no solution was reached. As such, I was just wondering if my process for completing this question seems viable as I often get bogged now by all of the activities going on in such questions. Here is the question:

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 saturated solution of calcium hydroxide (K_sp = 1.3 ⨯ 10^-6) is prepared by adding excess calcium hydroxide to pure 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 point (as determined by a pH meter) is 7.96. The first dissociation constant for the acid (K_a1) is 5.90 ⨯ 10^-2. Assume that the volumes of the solutions are additive, all solutions are at 25°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).

Process for (a):

m_H2A = 0.225 g
V_H2A = 0.250 L
pH₁ = 2.06
K_a1 = 5.90 × 10^-2
K_a1 >> K_a2 the main contributor of ions for the original solution is from monoprotic dissociation of H₂A...

H₂A_(aq) + H₂O_(l) HA^-_(aq) + H₃O⁺_(aq)    K_a = 5.90 × 10^-2

Initial:    x  +  Φ   0  +  0
Change: -y +  Φ +y + +y
Final:    x-y + Φ     y + y

y = 10^-2.06 = 8.71 × 10^-3 M

K_a = [HA^-][H₃O⁺]/[H₂A]   0.059 = y²/(x - y)
x = 16.95y² + y = 16.95(10^-2.06)² + 10^-2.06 = 1.00 × 10^-2 M H₂A
nH₂A = M_H2A · V_H2A = 0.01 · 0.250 = 2.50 × 10^-3 mol H₂A

nH₂A · molar mass H₂A = m_H2A
0.00250 · molar mass H₂A = 0.225
0.00250 · molar mass H₂A = 90.04 gmol^-1

Process for (b):

n_H3O⁺ = 2.18 × 10^-3 mol
n_HA^- = 2.18 × 10^-3 mol
   V_i = 0.250 L
  pH_f = 7.96

K_sp = [Ca²⁺][OH^-]²
1.3 × 10^-6 = x · (2x)²
1.3 × 10^-6 = 4x³
              x = 6.88 × 10^-3 M
       [OH^-] = 2x = 2 · 6.88 × 10^-3 = 1.38 × 10^-2

Neutralise H₃O⁺...

H₃O⁺_(aq) + OH^-_(aq) 2H₂O_(l)

n_OH^- = n_H3O⁺ = 2.18 × 10^-3 mol

V_OH^- = n_OH^-/M_OH^- = 0.00218/0.0138 = 0.158 L

V_i2 = 0.408 L

At the equivalence point n_OH^- = n_HA^- = 2.18 × 10^-3

V_f = 0.408 + 0.158 = 0.567 L

M_HA^- = 0.00218/0.567 = 3.84 × 10^-3 M

A^-_(aq) + H₂O_(l) OH^-_(aq) + HA^-_(aq)   K_b = z

Initial:     0  +  Φ      0.00384  +  0.00384
Change: +x +  Φ                -x + -x
Final:       x + Φ 0.00384 - x  +  0.00384 - x

K_b = [HA^-][OH^-]/[A^-] = (0.00384 - x)²/x

[OH^-]_f = 0.00384 - x

pH_f = 14 - pOH
7.96 = 14 - pOH
pOH = 6.04

[OH^-]_f = 10^-6.04 = 0.00384 - x

x = 0.00384 - 10^-6.04...

K_b = (0.00384 - x)²/x = (10^-6.04)²/(0.00384 - 10^-6.04) = 2.17 × 10^-10

K_a2 = K_w/K_b = (1 × 10^-14)/(2.17 × 10^-10) = 4.62 × 10^-5

Any direction as to whether the following process accurately accounted for all non-negligible ion equilibria would be appreciated. I am fairly confident with (a), but a little uncertain with (b).

Thank you,
KungKemi

[Borek]

What does it mean

Enough of the calcium hydroxide solution is added to the solution of the acid to reach the second equivalence point.

?

[KungKemi]

What does it mean

Enough of the calcium hydroxide solution is added to the solution of the acid to reach the second equivalence point.

?

Isn't the equivalence point when you add an equal molar amount of hydroxide ions to the solution as that of the HA^-? But since it's a weak acid you'll still have OH^- in solution once equilibrium has been established resulting in a pH greater than 7. However, I just thought that since the best acid reacts with the best base that enough hydroxide ions would need to be added first in order to neutralise the presiding hydronium ions in solution, and then, after that, the necessary amount of hydroxide ions would be added in order to reach the equivalence point.

Did I forget to factor in something?

Thank you,
KungKemi

[Borek]

Isn't the equivalence point when you add an equal molar amount of hydroxide ions to the solution as that of the HA^-?

http://www.titrations.info/titration-basic-terms

[KungKemi]

Okay, so at the equivalence point no hydroxide ions or HA^- exist in solution - only A^-. However, this A^- hydrolyses to produce OH^- and HA^- once more.

OH^-_(aq) + HA^-_(aq) A^-_(aq) + H₂O_(l)   K_b = z

Initial:     0  +  0       0.00384  +  Φ
Change: +x + +x             -x    + Φ
Final:       x + x    0.00384 - x + Φ

K_b = [OH^-][HA^-]/[A^-] = x²/(0.00384 - x)

[OH^-]_f = x

pH_f = 14 - pOH
7.96 = 14 - pOH
pOH = 6.04

[OH^-]_f = 10^-6.04 = x

K_b = (10^-6.04)²/(0.00384 - 10^-6.04) = 2.17 × 10^-10

K_a2 = K_w/K_b = (1 × 10^-14)/(2.17 × 10^-10) = 4.62 × 10^-5

Funny how the answer is still the same, however, it appears that my initial working out reverted that incorrect philosophy to the correct one. Interesting.

I'm assuming that this is what you were hinting at Borek?

Thank you for the help,
KungKemi

[Borek]

No wonder you get the same answer if you repeat the calculations. To be honest I can't find a difference between what you did before and what you did now.

What I was pointing at was that your final volume - calculated based on the assumption you have to add enough base to neutralize H⁺ in the solution, and not all of the H₂A present - was wrong.

It is not always easy to follow your calculations, as you use plenty of variables, not always describing what they are. Sometimes it is not difficult to guess, sometimes they are ambiguous.

[KungKemi]

No wonder you get the same answer if you repeat the calculations. To be honest I can't find a difference between what you did before and what you did now.

What I was pointing at was that your final volume - calculated based on the assumption you have to add enough base to neutralize H⁺ in the solution, and not all of the H₂A present - was wrong.

It is not always easy to follow your calculations, as you use plenty of variables, not always describing what they are. Sometimes it is not difficult to guess, sometimes they are ambiguous.

Oh, of course, thank you Borek.

KungKemi