Question

22.0 mL of 0.122 M diprotic acid (H2A) was titrated with 0.1020 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5and Ka2=3.4×10−10.

Find: At what added volume of base does the first equivalence point occur?

Find:At what added volume of base does the second equivalence point occur?

Find:

What is the pH at the first midway point?

Find: What is the pH at the second midway point?

Find: What is the pH at the first equivalence point?

Answer 1

Moles of H2A = volume x concentration

= 22.0/1000 x 0.122 = 0.002684 mol

(A) At the first equivalence point:

H2A + KOH => K+ + HA- + H2O

Moles of KOH = moles of H2A = 0.002394 mol

Volume of KOH = moles/concentration of KOH

= 0.002684 / 0.1020

= 0.02631 L

= 26.31 mL ≈ 26.3 mL

(B) At the second equivalence point:

H2A + 2 KOH => 2 K+ + A2- + 2 H2O

Moles of KOH = 2 x moles of H2A

= 2 x 0.002684 = 0.005368 mol

Volume of KOH = moles/concentration of KOH

= 0.005368 / 0.1020

= 0.05263 L

= 52.63 mL ≈ 52.6 mL

In the first midway point, you have half the volume of equivalence point, which means is 26.3 / 2 = 13.15 mL. At this point, the pH equals the pKa so:

pKa₁ = -log(5.2x10^-5) = 4.28 = pH

At the second midway point, the pH is equals to the pKa2 so:

pKa₂ = -log(3.4x10^-10) = 9.47 = pH

At the first equivalence point:

pH = pKa₁ + pKa₂ / 2

pH = 9.47 + 4.28 / 2

pH = 6.88

Hope this helps