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 0.20 M arsenic acid solution.
i cant do this. i don't even know how to start.
First, you need to recognize that Ka1 is so much larger than Ka2 and Ka3 (10^5 larger than Ka2 and 10^7 larger than Ka3) that most of the (H^+) comes from Ka1. So, do (H^+) just as you would a monoprotic acid; i.e.,
H3AsO4 ==> H+ + H2AsO4^-
Then do the ICE chart (just like acetic acid ionizing) so
Ka1 = (H^+)(H2AsO4^-)/(H3AsO4^-)
Then plug in the values.
Ka1 you have. H is y, H2AsO4- = y, and H3AsO4 is 0.2-y and solve for y. That gives you (H^+) (and you ignore the H form Ka2 and Ka3). That allows you to calculate OH^- and H3AsO4.
Then you look at Ka2.
Ka2 = (H^+)(HAsO4^-2)/(H2AsO4^-)
Since you have already determined that, for all practical purposes, (H^+) = (H2AsO4^-), that makes (HAsO4^-2) = Ka2.
Now you know all but AsO4^-3 and you can plug in the values you know into Ka3 and solve for that.
Check my work. It's easy to miss a negative sign or an exponent plus typos.
i cant get the first part.
H+ i got .0316
ur saying to do
y^2/(.2-y) = 5e-3??
You didn't use the quadratic equation and you must. Instead of getting 0.0319 (I don't get your 0.0316), you will get 0.02949 which rounds to 0.0295 to 3 s.f.
i did quadratic and got .02922
and that's not the right answer for the H+ question...
i did quadratic and got .02922
and that's not the right answer for the H+ question...