Question

A diprotic acid, H, A, has acid dissociation constants of Kat = 3.78 x 10 and K₂ = 4.21 x 10⁻¹¹. Calculate the pH and molar

concentrations of H, A, HA, and A₂ at equilibrium for each of the solutions.
A 0.177 M solution of H, A.
pH =
[HA] =

Answer 1

The pH of a diprotic acid solution is calculated using the dissociation constants and an ICE table to find the equilibrium concentration of hydronium ions, from which pH is derived. For molar concentrations at equilibrium, the Henderson-Hasselbalch equation can be employed to determine the ratio of the conjugate base to the acid.

Step-by-step explanation:

To calculate the pH and molarconcentrations of H2A, HA-, and A2- at equilibrium for a 0.177 M solution of H2A, a diprotic acid with dissociation constants Ka1 = 3.78 × 10-3 and Ka2 = 4.21 × 10-11, one must perform a series of calculations using an ICE table (Initial, Change, Equilibrium) and the given Ka values. Due to the large difference between Ka1 and Ka2, we can assume that the first dissociation step will contribute significantly to the [H+], while the second dissociation step's contribution to additional [H+] is negligible.

Using the formula pH = -log[H+] where [H+] is the hydronium ion concentration, we can approximate the pH after considering the dissociation for the first hydrogen from H2A. As we know for weak acids, the concentration of hydronium ions is much less than the molar concentration of the acid itself. Hence, an ICE table can be used to determine the equilibrium concentrations:

• H2A → H+ + HA-
• HA- → H+ + A2-

To find [HA] at equilibrium, you can use the Henderson-Hasselbalch equation, which relates the pH to the ratio of concentrations of the conjugate base (A-) to the conjugate acid (HA).