Question

"Arsenic acid (H3AsO4) is a triprotic acid with Ka1 = 5 10-3, Ka2 = 8 10-8, and Ka3 = 6 10-10. Calculate [H+], [OH ‾ ], [H3AsO4], [H2AsO4‾ ], [HAsO42 ‾ ], and [AsO43 ‾ ] in a 0.14 M arsenic acid solution."

Okay, so far, using successive approximation, I have determined that [H+] = 2e-2M

[OH^‾] = 5e-13M

[H3AsO4] .12M

[H2AsO4^‾] = 2e-2M

[HAsO4^2‾] =8e-8M

But I cannot get the right answer for [AsO4^3-]. I thought since x was so negligible that with the amount of sigfigs it would be the same as the Ka3 (6e-10) but this is wrong. Can anyone help?

Answer 1

I think you are not using the number of significant figures allowed. For (H^+) I solved the quadratic and obtained 0.0243 which rounds to 0.024M. That changes OH^- and some of the others, particularly H2AsO4^- as well as H3AsO4. Likewise, I think HAsO4^2- should be reported as 8.1E-8.
For k3 = 6.1E-10 = (H+)(AsO4^3-)/(H2AsO4^2-) I would substitute 0.024 for (H^+) and 8.1E-8 for H2AsO4^2- and solve for AsO4^3- to at least two s.f. I don't think (H^+) = (HAsO4^2-) but you are setting them equal if you let k3 = AsO4^3-. Let me know if my reasoning is not correct.

Answer 2

@DrBob222 Thank you so much, you were right!