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Topic: Ammonium Sulfate pH  (Read 30767 times)

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Offline crazihouse

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Ammonium Sulfate pH
« on: December 17, 2006, 09:51:40 PM »
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.
« Last Edit: December 17, 2006, 09:59:09 PM by crazihouse »

Offline Borek

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Re: Ammonium Sulfate pH
« Reply #1 on: December 18, 2006, 03:41:48 AM »
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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Offline crazihouse

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Re: Ammonium Sulfate pH
« Reply #2 on: December 18, 2006, 05:14:56 AM »
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

Offline AWK

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Re: Ammonium Sulfate pH
« Reply #3 on: December 18, 2006, 05:49:29 AM »
Use a correct concentration of NH4+ (0.2 M). Show not only your result, but also number put to the equation.
AWK

Offline crazihouse

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Re: Ammonium Sulfate pH
« Reply #4 on: December 18, 2006, 01:00:30 PM »
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
« Last Edit: December 18, 2006, 01:25:19 PM by crazihouse »

Offline Borek

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Re: Ammonium Sulfate pH
« Reply #5 on: December 18, 2006, 01:46:19 PM »
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.

Quote
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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Offline crazihouse

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Re: Ammonium Sulfate pH
« Reply #6 on: December 18, 2006, 02:07:34 PM »
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

 :-[

Offline Borek

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Re: Ammonium Sulfate pH
« Reply #7 on: December 18, 2006, 02:18:24 PM »
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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Offline crazihouse

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Re: Ammonium Sulfate pH
« Reply #8 on: December 18, 2006, 07:36:20 PM »
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.

Offline Borek

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Re: Ammonium Sulfate pH
« Reply #9 on: December 18, 2006, 07:51:59 PM »
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.

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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Offline english

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Re: Ammonium Sulfate pH
« Reply #10 on: December 19, 2006, 07:57:43 AM »
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

Offline Borek

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Re: Ammonium Sulfate pH
« Reply #11 on: December 19, 2006, 08:21:49 AM »
Well, if we are talking about detailed rules you may as well take a look at the ultimate pH cheat sheet  ;)

But the question still remains - why the pH is 0.5 unit off?
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Offline english

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Re: Ammonium Sulfate pH
« Reply #12 on: December 19, 2006, 08:24:55 AM »
Well, if we are talking about detailed rules you may as well take a look at the ultimate pH cheat sheet  ;)

But the question still remains - why the pH is 0.5 unit off?

I shall bookmark that one.    ;D

Offline ronmar5693

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Re: Ammonium Sulfate pH
« Reply #13 on: March 03, 2015, 06:46:27 AM »
Well, if we are talking about detailed rules you may as well take a look at the ultimate 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?

Offline Borek

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Re: Ammonium Sulfate pH
« Reply #14 on: March 03, 2015, 09:17:44 AM »
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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