# Thread: Very hard chemistry problem. >:(

1. Ok. First of all this is a homework problem so please provide your steps on how you solved this.

Calculate the for the following acids using the given information.

(a) 0.220 M solution of H3AsO4, pH = 1.50

so i know the concentration of H+ ions would be 10^-1.5 = 0.0316 M

I think the dilution of the H3AsO4 is:
H3AsO4 --> 3H + AsO4

now what?  2.

3. Well firstly we can say that this acid is a weak acid.

Therefore we can make the assumption that the dissacotiation of
the acid is low and therefore
[H+] = [H2AsO4-]

Ka = [H+][H2AsO4-]/[H3AsO4]

therefore using the above assumption

Ka = [H+]^2 / [H3AsO4]

you used the correct method to get [H+]

doing this i get an answer of 4.55 *10^-3 mol dm-3  4. I want to hijack this thread. Sorry. How can you calculate the pH of chloroform, for example?  5. Thats exactly what I put for my answer but I keep getting it wrong!

I'm entering my answer in a website called 'Webassign' where the teacher assigns us homework but it keeps saying my answer is wrong. Its so frustrating.  6. Originally Posted by georginho_juventusygr
I want to hijack this thread. Sorry. How can you calculate the pH of chloroform, for example?
I think the pH of any substance depends on its concentration.
For example a very concentrated amount of hydrochloric acid would have a lower pH than a not so concentrated one.  7. Thats exactly what I put for my answer but I keep getting it wrong!

I'm entering my answer in a website called 'Webassign' where the teacher assigns us homework but it keeps saying my answer is wrong. Its so frustrating.  8. if its a weak acid, i doubt that it would undergo such a great hydrogenation thus i think only 1 ion for the equation would be correct.  9. Originally Posted by oceanwave
if its a weak acid, i doubt that it would undergo such a great hydrogenation thus i think only 1 ion for the equation would be correct.

nope still didn't work

oh well..   10. Originally Posted by DivideByZero Originally Posted by georginho_juventusygr
I want to hijack this thread. Sorry. How can you calculate the pH of chloroform, for example?
I think the pH of any substance depends on its concentration.
For example a very concentrated amount of hydrochloric acid would have a lower pH than a not so concentrated one.
Aye ... but it seems I can't figure it out. It's chloroform, not a usual acid.  11. I get 0.0053

Ka = [H+] * [H2AsO4-]/[H3AsO4]

[H+] and [H2AsO4-] both equal 0.0316 M

[H3AsO4] = .22-0.0316

so Ka = 0.0316^2/(.22-.0316) = 0.0053

Edit: I suspect that you are forgetting to subtract the concentration of H+ from the concentration of H3AsO4 when you set up your equilibrium equation.  12. H3AsO4 --> 3H + AsO4
so I believe that [H+]=[AsO4,3-] isn't is right

Yeah, it's a weak acid and there're 3 H in such a acid, then comes 3 stages of ionization, someone just calculate the first two stages which is like:
Ka' = [H+] * [H2AsO4-]/[H3AsO4]
Ka'' = [H+][HAsO4,2-]/[H2AsO4-]
But i think the overall Ka should be

Ka = [H+][AsO4,3-]/[H3AsO4].  13. Originally Posted by ArezList
H3AsO4 --> 3H + AsO4
so I believe that [H+]=[AsO4,3-] isn't is right

Yeah, it's a weak acid and there're 3 H in such a acid, then comes 3 stages of ionization, someone just calculate the first two stages which is like:
Ka' = [H+] * [H2AsO4-]/[H3AsO4]
Ka'' = [H+][HAsO4,2-]/[H2AsO4-]
But i think the overall Ka should be

Ka = [H+][AsO4,3-]/[H3AsO4].
shldnt a weak acid have a 'double arrow' instead? since Ka means the acidity constant, what we are calculating for would be the constant for a weak acid. if it really did ionise 3 ways which i think is very improbable for weak acids, then the Ka would really be very hard a ascertain.  14. Only the first proton comes off.  Bookmarks
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