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Author Topic: pH of (NH4)2HPO4  (Read 2121 times)

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Raderford

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pH of (NH4)2HPO4
« on: December 19, 2012, 06:58:00 AM »

Calculate the pH in a (NH4)2HPO4 solution (c=0.3M).

(NH4)2HPO4 is fully dissociated (there is two times more NH4+ than HPO42-) and I thought to solve this as one reaction:
NH4++HPO42-+2H2O ::equil:: NH3+H2PO4-+H3O++OH- whose K I can calculate using the Ka table, but this means that H+=OH- and pH=7, which is not correct.
I could really use some help here. How to solve?
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Borek

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Re: pH of (NH4)2HPO4
« Reply #1 on: December 19, 2012, 08:49:12 AM »

No idea how to solve without getting 4th degree polynomial. None of the usual assumptions will work here.
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Raderford

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Re: pH of (NH4)2HPO4
« Reply #2 on: December 20, 2012, 02:19:52 AM »

I actually got the right answer when I used the equation 13.16 from here http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified. I assumed that the hydrogen phosphate acts as a base, not as an ampholite .I did this assumption in the first post, too.

One more question: Is it more accurate when I want to calculate the pH of a solution that contains HPO42-, to assume that HPO42- is only a base or to use the formula for the ampholite pH? I think I should assume that it is a base as Ka3 is so little. Am I right?
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Borek

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Re: pH of (NH4)2HPO4
« Reply #3 on: December 20, 2012, 03:12:00 AM »

I actually got the right answer when I used the equation 13.16 from here http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified

This equation is derived assuming concentrations of acid and base are identical, this is not the case. So either the equation works by accident, or it can be derived using different assumptions - but using it blindly is not a good idea.

Quote
One more question: Is it more accurate when I want to calculate the pH of a solution that contains HPO42-, to assume that HPO42- is only a base or to use the formula for the ampholite pH? I think I should assume that it is a base as Ka3 is so little. Am I right?

Depends on the final pH. If the final pH is around 8 (as it is in this case), ratio of concentrations of PO43-/HPO42- is around 108-12.3 - which means acidic dissociation of HPO4- can be safely ignored.
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Raderford

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Re: pH of (NH4)2HPO4
« Reply #4 on: December 20, 2012, 03:42:49 AM »

It should be the same as equations 13.9 and 13.11 will look like this:
[OH-]+2[HPO42-]=[H+]+[NH4+] i.e. 2[HPO42-]=[NH4+] (1)
2([HPO42-]+[H2PO4-])=[NH4+]+[NH3]
Combining these I get:
[NH3]=2[H2PO4-] (2)
When I put (1) and (2) in equation 13.7, two and two cancel each other so I get again the equation 13.13, meaning that the final equation (13.16) works here.

I shouldn't use this blindly, so I will check next time when I want to use an equation made from many assumptions.
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Borek

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Re: pH of (NH4)2HPO4
« Reply #5 on: December 20, 2012, 04:39:48 AM »

Good that you checked - as I wrote, it could be possible to arrive to the same result using a different path.

Can you write what is the exact result you got? Using pKa3=12.35 and pKa=9.25 (for ammonia) I am not getting anything close to the correct answer (which is 8.06).
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Raderford

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Re: pH of (NH4)2HPO4
« Reply #6 on: December 20, 2012, 06:08:06 AM »

The book answer is 8.23.
Here, NH4+ is the acid and HPO42- is the base. The conjugate acid of HPO42- is H2PO4- so I used pKa2 (7.2) which gives the correct result.
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Borek

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Re: pH of (NH4)2HPO4
« Reply #7 on: December 20, 2012, 06:35:27 AM »

Not "correct result" but "result given by the book", as it is still off by 0.15 pH unit.

To my surprise looks like this approach yields a reasonably correct result.
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Re: pH of (NH4)2HPO4
« Reply #8 on: December 20, 2012, 06:53:28 AM »

Yes, I meant for the book answer.
Seems like that approach is valid not even for ampholites type AB, but also for A2B, AB2, AB3...
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Re: pH of (NH4)2HPO4
« Reply #9 on: December 26, 2012, 02:07:39 PM »

It should be the same as equations 13.9 and 13.11 will look like this:
[OH-]+2[HPO42-]=[H+]+[NH4+] i.e. 2[HPO42-]=[NH4+] (1)


I am confused on this line; wouldn't [H2PO4-] be in this charge balance equation on the left-hand side?

Edit: I guess since Kb2 for HPO4- is small, the concentration of that anion can be neglected in the same approximations. My bad :P
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