Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: 21385 on March 25, 2008, 11:22:00 PM
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Can I another question of this sort of nature?
Srry, I know I may be asking too much but I would really appreciate it if you look into it for me.....my brain is going crazy!
Question: Calculate [H+], [OH-], [HSO4-], [SO4(2-)] in a 1.0E-7 M solution of sulfuric acid. Kw=1.0E-14 and K2=1.2E-2 and assume that all H2SO4 goes to HSO4-
From what I learned, I would approach this question with 2 balance equations.
[H+] = [HSO4-] + 2[SO4(2-)] + [OH-]
Total concentration of acid (1.0E-7) = [HSO4-] + [SO4(2-)]
Then, I would solve for [HSO4-] in terms of [H+] for both equations and equate the two expressions together and solve for [H+]. However, I am not getting the correct answer. And the correct method is this::
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi28.tinypic.com%2F2dad6x0.jpg&hash=28a99f9833a2baae8dcb6968f3024e5ebca28644)
I don't really know why my method won't get the correct answer.
Thanks a lot
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Then, I would solve for [HSO4-] in terms of [H+] for both equations and equate the two expressions together and solve for [H+].
So far so good, but what you will get will be the equation in H+ and SO42-, it requires further work.
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For [H+] = [HSO4-] + 2[SO4(2-)] + [OH-]
Total concentration of acid (1.0E-7) = [HSO4-] + [SO4(2-)]
I will solve for HSO4- (not SO4(2-), although i think it'll yield the same result) in both equations, so...
k2=[H+][SO42-]/[HSO4-]
[H+] = [HSO4-] + 2[SO4(2-)] + [OH-]
[H+] - [OH-] = [HSO4-] + 2k2[HSO4-]/[H+]
[H+] - Kw/[H+] = [HSO4-](1 + 2k2/[H+])
([H+] - Kw/[H+])/(1 + 2k2/[H+]) = [HSO4-]
(1.0E-7) = [HSO4-] + k2[HSO4-]/[H+]
(1.0E-7) = [HSO4-](1 + k2/[H+])
(1.0E-7)/(1 + k2/[H+]) = [HSO4-]
Equate these two expressions together (1.0E-7)/(1 + k2/[H+])=([H+] - Kw/[H+])/(1 + 2k2/[H+])
Crossed multiplied =>
[H+]^2 + k2[H+] - Kw[H+] - k2kw - 1E-7[H+]^2 - (1E-7)(2Ka) =0
Did the quadratic formula, [H+]=2.0E-7, which is different from the official solutions
But I think the real question is why did the official solution make so many assumptions when with the charge balance and mass balance solutions, with 2 unknowns, you can clearly (in my view) get the answer.
Also, its final answer is different....so, i don't really know what's wrong
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(1.0E-7)/(1 + k2/[H+])=([H+] - Kw/[H+])/(1 + 2k2/[H+])
Product of extremes equals the product of the means (which you have called cross multiplication - no idea which version is correct, could be that both can be used)
(1 + k2/[H+])([H+] - Kw/[H+]) = (1.0E-7)(1 + 2k2/[H+])
LHS = [H+] + k2 - kw/[H+] - k2kw/[H+]2
Now, multiply by [H+]2 to get rid of [H+] in denominators - and you get cubic equation, not quadratic. So unless I am missing something you did something wrong when rearranging.
As for correct answer - see attachement.
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Hmm....So, because the equations is cubic, and if you only had a scientific calculator, you would have to make some assumptions like the ones in the method above?
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You can set an iterative procedure
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No need for iterations, there are procedures that let you analytically solve cubic equation (for example Cardano's method) - but they are long and rarely used, so almost nobody knows them.
Please note that having or not scientific calculator is irrelevant to the question. I don't have one, which doesn't stop me from solving quadratic equations. There are plenty of web pages that will tell you how.
Hmm....So, because the equations is cubic, (...) you would have to make some assumptions like the ones in the method above?
That's the way I see it.