chem

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.

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  1. 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.

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  2. i cant get the first part.
    H+ i got .0316

    ur saying to do

    y^2/(.2-y) = 5e-3??

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  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.

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  4. i did quadratic and got .02922
    and that's not the right answer for the H+ question...

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  5. i did quadratic and got .02922
    and that's not the right answer for the H+ question...

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  6. I think 0.0295 is the correct answer for H^+. The problem says Ka1= 5.10 x 10^-3. The smaller number of 0.02922 may be because you are showing 5 x 10^-3 in your post. And the data base to which your are typing in your answer may very well be able to tell the difference between 0.0295 and 0.0292 and if you are typing in 0.02922 then it definitely will tell you it is wrong because that's too many significant figures.

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  7. I tried .0295 and it didn't work...-___-

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  8. OK. I noticed that you have the molarity of 0.20 which is to two significant figures only so lets do the quadratic and round to to s.f. If I did it right, I obtained 0.029489 M which, to two places, would round to 0.030. That may be picking at straws but if you meant 0.20 and not 0.200, then we can have only two s.f.

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  9. Im slightly confused, don't all the answers that have been stated above^^^ violate the 5% rule?

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  10. I got how to solve H+, but little bit confused about OH-?

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  11. [H+][OH-]=1e-14

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  12. when i did the quadratic equation, i ended up with a negitive number under the radical. and you cant slove from then?

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  13. Make sure your equation is [b +-(b^2 - 4ac)^(1/2)]/2a

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  14. Doesn't using the quadratic to solve this violate the 5% rule? if I do this the way our teacher said to, I also get 0.0316

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