# Acid/Base pH after reaction

Gold Member

## Homework Statement

http://i.minus.com/jfUXIXvQJw8hw.png [Broken]

## Homework Equations

Ka of hydrogen sulfate ion will be taken as 1.2x10-2.

Ka of ammonium ion will be taken as 5.6x10-10

Kb of ammonia will be taken as 1.8x10-5

Kb of sulfate ion will be taken as something very tiny (Kw/Ka of hydrogen sulfate ion).

These values are consistent throughout the literature.

## The Attempt at a Solution

The reaction of hydrogen sulfate ion and bubbled in ammonia is a large extent reaction. We'll take it as going to completion. Also the pH controlling species here is obviously hydrogen sulfate ion since the magnitude of its Ka is far beyond that of ammonium or ammonia.

The problem is that with 0.25 moles of hydrogen sulfate ion left in the system there is NO way there can be a pH of 2.40. A pH of 1.3, yes. It seems that my teacher may have used the natural log instead of the base ten log because taking the natural log of the hydronium ion concentration yields a 2.90 (which may look like 2.40). The pH can easily be ascertained through a simple x^2/Mi = Ka calculation.

So am I wrong or is my teacher wrong?

Last edited by a moderator:

Borek
Mentor
pH of the 1M NH4HSO4 solution is around 1.0, pH of the 1M (NH4)2SO4 solution is around 5.47, 2.40 is definitely between these two values.

epenguin
Homework Helper
Gold Member
It's a salt of a strong acid and weak base essentially. You have already hit on an essential simplifying fact - the ammonia is essentially all protonated in the problem. So it is no different than if you replaced NH4+ everywhere by Na+, a fairly ordinary problem.

Gold Member
No, that's not quite right either; I know what I did wrong. I forgot to realize this was a buffer solution. We have an ammonia/ammonium buffer and a hydrogen sulfate ion/sulfate ion buffer. We use the Henderson-Hasselbach equation then to estimate pH. I should also have realized that with the sheer amount of sulfate ion in the system relative to hydrogen sulfate ion (ratio of 3:1 or 0.75 moles to 0.25 moles) should, by Le Chatlier's principle, reduce the production of hydronium ion (relative to if there were only hydrogen sulfate ion in the system, as I had erroneously supposed).

Borek
Mentor
To some extent it is a buffer solution, but using HH equation is quite dangerous. HSO4- is too strong for that. HH equation requires use of the equilibrium concentrations, you can't be sure typical assumption about the neutralization ("went to completion") is justified.

I agree with epenguin, ammonia dissociation can be safely ignored.

IOW equivalent question is "How much NaOH has to be added to 1M NaHSO4 to produce pH 2.4 solution". I would try to approach it with some variant of the ICE table.

Last edited: