Question

Consider a solution that is 0.10 M in a weak triprotic acid which is represented by the general formula H3A with ionization constants K1 = 1.0 × 10−3 , K2 = 1.0 × 10−8 , and K3 = 1.0 × 10−12. What is the pH of the solution?

Answer 1

H3A <-------------------------> H2A-   + H+

0.10                                    0             0

0.10-x                                  x             x

Ka1 = [H2A-][H+]/[H3A]

1.0 x 10^-3 = x^2 / 0.1 -x

x^2 + 1.0 x 10^-3 x - 1.0 x 10^-4 = 0

x = 9.51 x 10^-3

[H+] = x = 9.51 x 10^-3 M

pH = -log [H+]

pH = -log (9.51 x 10^-3)

pH = 2.02