[XeLa.]

I was trying to calculate the pH of a mixture containing 0.35 M NH₃ and 0.20 M acetic acid. I first considered that acetic acid reacted with ammonia giving acetate ion and ammonium ion. Since acetic acid is the limiting reagent, at this stage of calculation we would have 0.15 M NH₃, 0.2 M NH₄⁺, and 0.2 M CH₃COO^-.

I then preceded to consider the pH of the ammonia-ammonium ion buffer (since this is now the strongest acid-base pair) using Henderson-Hasselbalch equation and found the pH of the solution (using pOH from H-H equation) to be 9.13.

Now, I went to check my solution using an online calculator and found the pH of the solution to be roughly 9.10 instead. Why would this be the case? Would I have to consider the hydrolysis by acetate ions as well?

Any help would be appreciated!

[Borek]

In my experience 9.10 and 9.13 is exactly the same answer (and as the ionic strength of the solution is 0.2 both are wrong, in reality pH will be a bit higher, something like 9.3 if I were to trust my calculators).

First thing to check is the dissociations constants used.

[XeLa.]

In my experience 9.10 and 9.13 is exactly the same answer (and as the ionic strength of the solution is 0.2 both are wrong, in reality pH will be a bit higher, something like 9.3 if I were to trust my calculators).

First thing to check is the dissociations constants used.

Okay, thanks, Borek!

I considered a pKb of 4.75 for ammonia and a pKb of 10.25 for acetic acid.

XeLa

[Borek]

I considered a pKb of 4.75 for ammonia and a pKb of 10.25 for acetic acid.

9.13 it is (ionic force ignored, exact equilibrium).

[XeLa.]

Okay, awesome! Maybe I should use a different pH calculator in future then? Which one did you use, Borek?

[Borek]

http://www.chembuddy.com/?left=BATE&right=pH-calculator

but this program is more general and much more powerful:

http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator