Question

A weak acid H3A has pKa values of 1.95 (pKa1), 3.08 (pKa2), and 9.54 (pKa3). The disodium salt of that weak acid was dissolved in deionized water to make 250.0 mL of a 0.025 M solution.  a) What was the pH of the resulting solution? b)What was the equilibrium concentration of H2A-?

Answer 1

250 ml of 0.025 M Na2HA

a) pH calculation

pKa3 = 9.54 = -logKa

Ka3 = 2.9 x 10^-10

HA^2- + H2O <==> A^3- + H3O+

let x amount has dissociated

Ka3 = [H3O+][A^3-]/[HA^2-]

2.9 x 10^-10 = x^2/0.025

x = [H3O+] = 2.69 x 10^-6 M

pH = -log[H3O+] = 5.57

b) Equilibrium concentration of H2A-

HA^2- + H2O <==> H2A- + OH-

let x amount has hydrolyzed

Kb2 = Kw/Ka2 = [H2A-][OH-]/[HA^2-]

1 x 10^-14/8.32 x 10^-4 = x^2/0.025

x = [H2A-] = 5.48 x 10^-7 M at equilibrium