o A diprotic acid (HA) has Kal = 4.60 × 10-7 and Ka2 = 5.24 × 10-¹¹. What is the pH of a 0.489 M H₂A solution? pH =

Answer 1

The diprotic acid is represented as H₂A (to ionizable hydrogens). The ionization looks like this:

H₂A ==> H⁺ + HA^- ... Ka1 = 4.60x10^-7

HA- ==> H⁺ + A^2- ... Ka2 = 5.24x10^-11

Since Ka2 is 10,000x less than Ka1, we generally do not consider it in contributing to the pH. That being the case, we will only consider Ka1. (note: if Ka2 is within 1000x Ka1, we generally do consider it).

H₂A ==> H⁺ + HA^-

Ka = [H⁺][HA^-] / [H₂A]

4.60x10^-7 = (x)(x) / 0.489 - x (ignore x in denominator since it will be small relative to 0.489 M)

4.60x10^-7 = x2 / 0.489

x² = 2.25x10^-7

x = [H⁺] = 4.74x10^-4 M (this is only ~0.1% of 0.489 so above assumption was valid)

pH = -log [H⁺]

pH = -log 4.74x10^-4

pH = 3.32