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### Topic: Ammonium Sulfate pH  (Read 30767 times)

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#### crazihouse

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• Mole Snacks: +0/-0 ##### 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 »

#### Borek ##### 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+.

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#### crazihouse

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« 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? #### AWK

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« 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

#### crazihouse

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« 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? « Last Edit: December 18, 2006, 01:25:19 PM by crazihouse »

#### Borek ##### 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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#### crazihouse

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« 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 #### Borek ##### 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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#### crazihouse

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• Mole Snacks: +0/-0 ##### 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.

#### Borek ##### 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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#### english

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« 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

#### Borek ##### 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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#### english

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« 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. #### ronmar5693

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« 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?

#### Borek ##### 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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