Question

1) Boric acid, H3 BO3, is a triprotic acid that dissociates in three reactions. The first dissociation step is: H3BO3 ⇌ H^+ + H2BO3^-, K(a1) = 7.3 x 10^-10; the second dissociation step is: H2BO3^- ⇌ H^+ + HBO3^-2, K(a2) = 1.8 x 10^-13; and the third dissociation step is:
HBO3 ⇌ H^+ + BO3^-3, K(a3) = 1.6 x 10^-14. Calculate the pH of a 0.050 M solution of boric acid.

2) Explain any approximations or assumptions that you make in your calculation.

Answer 1

First you must recognize that k1 is the largest K of the group and the pH of the solution will be due largely to the first dissociation. The second and third don't produce enough to concern us too much.
.......H3PO3 ==> H^+ + H2BO3^-
I......0.05M.....0.......0
C......-x........x.......x
E......0.05-x....x.......x

Substitute the E line into the Ka1 expression and solve for x = (H^+), then convert to pH.

Answer 2

Boric acid, H3 BO3, is a triprotic acid that dissociates in three reactions. The first dissociation step is: H3BO3 ⇌ H+ + H2BO3, Ka1 = 7.3 x 1010; the second dissociation step is: H2BO3 ⇌ H+ + HBO32, Ka2 = 1.8 x 1013; and the third dissociation step is:
HBO3 ⇌ H+ + BO33, Ka3 = 1.6 x 1014. Calculate the pH of a 0.050 M solution of boric acid and Explain any approximations or assumptions that you make in your calculation.