Question

Consider the triprotic acid, H3A; pKa1: 2.91; pKa2: 8.2; pKa3: 13.3. Calculate the pH and the concentrations of H3A, H2A-, HA2-, A3-, OH- and H3O+ at equilibrium in a 0.10 M solution of the triprotic acid.

Answer 1

Given,

pKa1 = 2.91, pKa2 = 8.2, pKa3 = 13.3

=> Ka1 = 1.23 x 10^-3

=> Ka2 = 6.31 x 10^-9

=> Ka3 = 5.01 x 10^-14

H3A ------> H+    +   H2A-

0.1 - X......X+Y+Z ......X - Y

Ka1 = 1.23 x 10^-3 = (X+Y+Z) (X-Y) / (0.1 - X)

Since Ka1 >> Ka2 >> Ka3 => X >> Y >> Z

=> 1.23 x 10^-3 = X^2 / 0.1 - X

=> X = 0.011 M

H2A- -----> H+   +   HA 2-

X - Y.......X+Y+Z........Y - Z

Ka2 = (X+Y+Z) (Y-Z) / (X-Y)

=> 6.31 x 10^-9 = Y

HA2- -----> H+   +   A 3-

Y - Z .....X+Y+Z......Z

Ka3 = 5.01 x 10^-14 = (X+Y+Z) (Z) / (Y-Z) = XZ / Y

=> 5.01 x 10^-14 = 0.011 x Z / 6.31 x 10^-9

=> Z = 2.87 x 10^-20 M

[H+] = X+Y+Z = X (approx.) = 0.011 M = [H3O+]

=> pH = - log [H3O+] = - log 0.011 = 1.96

[H3A] = 0.1 - X = 0.1 - 0.011 = 0.089 M

[H2A-] = X - Y = 0.011 M

[HA 2-] = Y - Z = 6.31 x 10^-9 M

[A 3-] = Z = 2.87 x 10^-20 M

[OH-] = 10^-14 / [H+] = 9.1 x 10^-13 M