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.

What was the equilibrium concentration of H2A-?

Answer 1

Use the relation pK_a = -log (K_a) to find the K_a values.

K_a = antilog (-pK_a)

pK_a1 = 1.95; ====> K_a1 = antilog (-1.95) = 0.0112

pK_a2 = 3.08; ====> K_a2 = antilog (-3.08) = 8.3176*10^-4

pK_a3 = 9.54; ====> K_a3 = antilog (-9.54) = 2.8840*10^-10

We have Na₂HA; the weak base is HA^2- and the molar concentration is 0.025 M

a) Consider the three dissociations:

H₃A <=====> H⁺ + H₂A^-

H₂A^- <=====> H⁺ + HA^2-

HA^2- <=====> H⁺ + A^3-

We must consider the last dissociation and write

K_a3 = [H⁺][A^3-]/[HA^2-] = (x)(x)/(0.025 – x)

===> 2.8840*10^-10 = x²/(0.025 – x)

Use the small x approximation, i.e, as K_a3 is very small, we can assume x << 0.025 and write

2.8840*10^-10 = x²/0.025

===> x² = 7.21*10^-12

===> x = 2.685*10^-6

Therefore, [H⁺] = 2.685*10^-6 M and pH = -log [H⁺] = -log (2.685*10^-6) = 5.57 (ans).

b) Define D = [H⁺]³ + K_a1[H⁺]² + K_a1K_a2[H⁺] + K_a1K_a2K_a3

= (2.685*10^-6)³ + (0.0112)*(2.685*10^-6)² + (0.0112)*(8.3176*10^-4)*(2.685*10^-6) + (0.0112)*(8.3176*10^-4)*(2.8840*10^-10)

= 1.9356*10^-17 + 8.0743*10^-14 + 2.5013*10^-11 + 2.6866*10^-11

≈ 5.1879*10^-11

The fraction of H₂A^- in the system is given by

α(H₂A^-) = K_a1*[H⁺]²/D = (0.0112)*(2.685*10^-6)²/(5.1879*10^-11) = 1.5564*10^-3

Equilibrium concentration of H₂A^- = α*(total concentration) = 1.5564*10^-3*0.025 M = 3.891*10^-5 M (ans).