Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: crazihouse on December 17, 2006, 09:51:40 PM
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I've been looking over this problem for an hour now..
This is the question: Calculate the pH of a 0.100 M ammonium sulfate solution (Kb = 1.8x10-5).
I know that (NH4)2SO4 dissociates into 2NH4+ + SO42-.
..Which is where my first question comes up: 1. Is it a full ionic dissociation or do I somehow have to use Kb to find the resulting [NH4+]?
So essentially, the reaction would be: (NH4)2SO4 <--> 2NH4+ + SO42-.
From there, I use the ICE table along with my Kb value in order to find [H+].
My second question is: 2. Is NH4 a polyprotic weak acid or do I just leave it at this?
Please tell me if I'm doing this right, I'd greatly appreciate it.
Thanks.
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1. Salt is fully dissociated into both NH4+ and SO42-
2. NH4+ is a monoprotic acid. You may safely assume it is the only source of H+.
Once you will be ready please show your answer, as I would like to comment on the question.
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From the dissociated NH4 ion, I made an ICE table in order to find [H3O+] with the help of Ka (Ka = Kw/Kb).
My [H3O+] was 7.45 x 10-6 M, and from that I found a pH of 5.13. Is that good? :P
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Use a correct concentration of NH4+ (0.2 M). Show not only your result, but also number put to the equation.
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So the equation would have to be
2NH4(aq) + 2H2O(l) <--> 2NH3-(aq) + 2H3O+(aq) ?
Why is the correct concentration of NH4+ 0.2 M? I'm not sure I understand that..
Edit: Doing it this way, I got pH = 3.11
My ICE table is:
Initial:
NH4(aq) = 0.100 M
Change:
NH4(aq) = 0
NH3-(aq) = +2x
H3O+(aq) = +2x
Equilibrium:
NH4(aq) = 0.100 M
NH3-(aq) = 2x
H3O+(aq) = 2x
To Find Ka:
Ka = Kw / Kb
Ka = (1 x 10-14) / (1.8x10-5)
Ka = 5.55 x 10-10
Equation to find x ( [H3O+] )
[NH3-]2[H3O+]2 = 5.55 x 10-10
[NH4]2
(2x)2(2x)2 / (0.100)2 = 5.55 x 10-10
16x4 / 0.010 = 5.55 x 10-10
[H3O+] = x = 7.67 x 10-4
pH = -log [H3O+] = 3.11
Good? :P
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Good?
Sigh. No :'(
So the equation would have to be
2NH4(aq) + 2H2O(l) <--> 2NH3-(aq) + 2H3O+(aq) ?
No. You start with NH4+, not some NH4(aq). (NH4)2SO4 is a salt - salt contains cation and anion.
Why is the correct concentration of NH4+ 0.2 M? I'm not sure I understand that..
Look at the ammonium sulfate formula - how many NH4+ cations per molecule? If you have 0.1 mole of ammonium sulfate, how much NH4+ does it contain?
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I can't believe I looked over such an important part of the problem! Thanks a lot.
Initial:
NH4+(aq) = 0.200 M
H2O(l) = Unaccounted for
Change:
NH4+(aq) = -x
NH3(aq) = +x
H3O+(aq) = +x
Equilibrium:
NH4+(aq) = 0.200 M
NH3(aq) = x
H3O+(aq) = x
My final formula is x2 / 0.2 = 5.55 x 10-10
x = 1.053 x 10-5 = [H3O+]
-log [H3O+] = pH = 4.98
:-[
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Initial:
NH4+(aq) = 0.200 M
Change:
NH4+(aq) = -x
Equilibrium:
NH4+(aq) = 0.200 M
0.200 - x = 0.200 ?
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Yeah, our teacher said we can ignore the x because it's so small. It's obviously better to have the x and use the general equation formula to find it.
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Yeah, our teacher said we can ignore the x because it's so small.
That's not always the case. To be correct you should check - after calculations - if x is less then 5% of the initial concentration. If it is, assumption that it could be ignored is OK, if not - you have to go through quadratic equation.
See pH of weak acid (http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base).
Now, you could ignore the x so your result is correct. But the pH of this solution is close to 5.49 (at the HS level, it is even different in reality ;) ). Do you have any idea why your result is off by 0.5 unit?
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A general que to go by as well, to see if you can ignore the H+ dissociated for NH4+ conc. at equilibrium, is the following rule:
To assume that Kb or Ka is relatively small relative to the orignal acid/base conc.,
if initial conc. > 400 , then you can safely assume that x is relatively small.
Ka
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Well, if we are talking about detailed rules you may as well take a look at the ultimate pH cheat sheet (http://www.chembuddy.com/?left=BATE&right=pH-cheat-sheet) ;)
But the question still remains - why the pH is 0.5 unit off?
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Well, if we are talking about detailed rules you may as well take a look at the ultimate pH cheat sheet (http://www.chembuddy.com/?left=BATE&right=pH-cheat-sheet) ;)
But the question still remains - why the pH is 0.5 unit off?
I shall bookmark that one. ;D
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Well, if we are talking about detailed rules you may as well take a look at the ultimate pH cheat sheet (http://www.chembuddy.com/?left=BATE&right=pH-cheat-sheet) ;)
But the question still remains - why the pH is 0.5 unit off?
Hi I know this is an old topic and I apologise for bringing it up. I need to calculate ammonium sulphate solution pH for work and I am not a chemist. Every time I try to google it this old thread turns up.
My problem is that I don't understand why the calculated and actual pH in this problem is 0.5 units off. Can anyone explain in a simple way please?
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SO42- is a base strong enough to shift the pH by half a unit up.
Note that actual pH is even different, as the solution has quite large ionic strength.
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Thanks for that.
Would it be possible for you to outline the calculation with SO4 please so that I am able to calculate ammonium sulphate solution pH in a practical way? It is for an agricultural application.
Do you have any opinion of the combined effect of bicarbonates and ammonium sulphate in water? Usually we predict resulting pH by ignoring everything but bicarb, and calculating effect of various added acids on bicarb equivalents per litre. I'm guessing this is too simplistic?
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It is not a trivial thing to do manually, all I can suggest is using a specialized software.
See for example http://www.chembuddy.com/?left=Buffer-Maker&right=pH-calculator
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Is it possible use mass and charge balance to calculate pH accurately in this question too (instead of using ice charts)
[SO42-] (initial)= [SO42-] (final) +[HSO42-] + [H2SO4] where you can assume [h2so4] to be negligible
[NH4+] (initial)= [NH4+] (final) + [NH3]
[NH4+] + [H+] = 2[SO42-] + [HSO41-] + [OH-]
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Is it possible use mass and charge balance to calculate pH accurately in this question too (instead of using ice charts)
Sure, go ahead. I would love to see you solving manually a 6th degree resulting polynomial. Even make it 5th, neglecting H2SO4 as you (correctly) suggested.
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Thanks for ideas.
I have tried a complex solution pH calculator before and I found I couldn't work it- was too complex and more suited to preparation of buffer solutions in a lab than natural waters. Is this one user friendly enough in a kind of bucket chemistry approach?
Application is fertiliser solutions in agriculture with different borehole water starting alkalinity. Typically fertilisers are added at pretty low concentrations and resulting concentrations in fertigation water are mM or less. We usually use nitric, sulphuric, or phosphoric acid to reduce bicarb to 20-40ppm which generally gives acceptable pH. Usually the approach is to ignore anything but milliequivalents of bicarb, treat acids like H2SO4 as complete single H dissociation only, and work out a rough approximation. pH is usually in the right ballpark but never correct. I was wondering about the effect of ammonium and whether we should be trying to take account?
Or any different ideas about what might be throwing it off? Waters I am using typically have a fair bit of HCO3 which we do take account of but usually also some silicates which we do not. EC predictions we make are usually about right which makes me think we have most of the important ions involved.
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For more complicated solutions you have programs like MINTEQ or PHREEQC (and some others). Last time I checked none of these was what I would call "user friendly". Hardly surprising, these calculations are not easy and require tons of parameters.
Ammonium ion is definitely one of important acids present in the solution.