[Rutherford]

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

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

[Borek]

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

[Rutherford]

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 HPO₄^2-, to assume that HPO₄^2- is only a base or to use the formula for the ampholite pH? I think I should assume that it is a base as K_a3 is so little. Am I right?

[Borek]

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.

One more question: Is it more accurate when I want to calculate the pH of a solution that contains HPO₄^2-, to assume that HPO₄^2- is only a base or to use the formula for the ampholite pH? I think I should assume that it is a base as K_a3 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 PO₄^3-/HPO₄^2- is around 10^8-12.3 - which means acidic dissociation of HPO₄^- can be safely ignored.

[Rutherford]

It should be the same as equations 13.9 and 13.11 will look like this:
[OH^-]+2[HPO₄^2-]=[H⁺]+[NH₄⁺] i.e. 2[HPO₄^2-]=[NH₄⁺] (1)
2([HPO₄^2-]+[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.

[Borek]

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).

[Rutherford]

The book answer is 8.23.
Here, NH₄⁺ is the acid and HPO₄^2- is the base. The conjugate acid of HPO₄^2- is H₂PO₄^- so I used pK_a2 (7.2) which gives the correct result.

[Borek]

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.

[Rutherford]

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...

[XGen]

It should be the same as equations 13.9 and 13.11 will look like this:
[OH^-]+2[HPO₄^2-]=[H⁺]+[NH₄⁺] i.e. 2[HPO₄^2-]=[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 K_b2 for HPO₄^- is small, the concentration of that anion can be neglected in the same approximations. My bad