Question

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 1010; the second dissociation step is: H2BO3 ⇌ H+ + HBO32, Ka2 = 1.8 x 1013; and the third dissociation step is:
HBO3 ⇌ H+ + BO33, Ka3 = 1.6 x 1014. Calculate the pH of a 0.050 M solution of boric acid.
Answer:
Since the quadratic formula must be used.

x=(-7.4x10^-4 (+-) 1.2 x10^-2)/2
x = 5.63 x 103 = [H+] pH = 2.25

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

Answer:
Assumption: [H+] in third dissociation step will be small enough not to affect the pH.

Answer 1

1. Check your data. I don't think the Ka values are correct.I might believe 7.3e-10. And is that 7.3 or 7.4?

2. I don't believe the quadratic formula must be used.

3. Finally,I don't believe the calculation for the quadratic formula.