Friday

December 9, 2016
Posted by **Cab** on Sunday, February 15, 2009 at 10:27pm.

i cant do this. i don't even know how to start.

- chem -
**DrBob222**, Sunday, February 15, 2009 at 10:48pmFirst, 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. - chem -
**Cab**, Sunday, February 15, 2009 at 11:17pmi cant get the first part.

H+ i got .0316

ur saying to do

y^2/(.2-y) = 5e-3?? - chem -
**DrBob222**, Sunday, February 15, 2009 at 11:31pmYou 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.

- chem -
**Spencer**, Sunday, February 15, 2009 at 11:43pmi did quadratic and got .02922

and that's not the right answer for the H+ question... - chem -
**Cab**, Sunday, February 15, 2009 at 11:44pmi did quadratic and got .02922

and that's not the right answer for the H+ question... - chem -
**DrBob222**, Monday, February 16, 2009 at 12:18amI 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.

- chem -
**Cab**, Monday, February 16, 2009 at 10:34amI tried .0295 and it didn't work...-___-

- chem -
**DrBob222**, Monday, February 16, 2009 at 2:41pmOK. 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.

- chem -
**Charles425**, Wednesday, March 17, 2010 at 12:44amIm slightly confused, don't all the answers that have been stated above^^^ violate the 5% rule?

- chem -
**Askar**, Monday, November 29, 2010 at 12:41amI got how to solve H+, but little bit confused about OH-?

- chem -
**Anonymous**, Wednesday, February 16, 2011 at 10:23pm[H+][OH-]=1e-14

- chem -
**Caitlin**, Saturday, February 9, 2013 at 11:09pmwhen i did the quadratic equation, i ended up with a negitive number under the radical. and you cant slove from then?

- chem -
**James**, Friday, April 5, 2013 at 12:14amMake sure your equation is [b +-(b^2 - 4ac)^(1/2)]/2a

- chem -
**Ashley**, Thursday, February 4, 2016 at 8:52pmDoesn'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