[crazihouse]

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 (K_b = 1.8x10^-5).

I know that (NH₄)₂SO₄ dissociates into 2NH₄⁺ + SO₄^2-.

..Which is where my first question comes up: 1. Is it a full ionic dissociation or do I somehow have to use K_b to find the resulting [NH₄⁺]?

So essentially, the reaction would be: (NH₄)₂SO₄ <--> 2NH₄⁺ + SO₄^2-.

From there, I use the ICE table along with my K_b 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.

[Borek]

1. Salt is fully dissociated into both NH₄⁺ and SO₄^2-
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.

[crazihouse]

From the dissociated NH₄ ion, I made an ICE table in order to find [H₃O⁺] with the help of K_a (K_a = K_w/K_b).

My [H₃O⁺] was 7.45 x 10^-6 M, and from that I found a pH of 5.13. Is that good?

[AWK]

Use a correct concentration of NH4+ (0.2 M). Show not only your result, but also number put to the equation.

[crazihouse]

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 K_a:

K_a = K_w / K_b
K_a = (1 x 10^-14) / (1.8x10^-5)
K_a = 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?

[Borek]

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?

[crazihouse]

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]

Initial:
NH₄⁺(aq) = 0.200 M

Change:
NH₄⁺(aq) = -x

Equilibrium:
NH₄⁺(aq) = 0.200 M

0.200 - x = 0.200 ?

[crazihouse]

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]

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?

[english]

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 K_b or K_a is relatively small relative to the orignal acid/base conc.,

if initial conc.  > 400  , then you can safely assume that x is relatively small.
         K_a

[Borek]

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?

[english]

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]

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]

SO₄^2- 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.