Thread: Very hard chemistry problem. >:(

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?

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

I want to hijack this thread. Sorry.

How can you calculate the pH of chloroform, for example?

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.

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.

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.

try changing the number of significant figures in your answer

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.

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..

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.

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.

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].

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.

Only the first proton comes off.