[zmasterflex]

What is the pH of the solution formed by mixing 25mL of a .15M solution of NH3 with 25mL of .12M HCl? (Kb of NH3 = 1.8e-5)

Because this is a strong acid and a weak base, we need to calculate the moles of each one separately..
.12M of HCl x .025L = 3e-3 moles of H+ in solution.
.15M of NH3 ---> [NH4][OH]/[NH3] = Kb ---> .15 x 1.8e-5 = [NH4][OH]
[OH] = 1.643167673e-3, however this is molar conc. so multiply it by .025L = 4.107919e-5

There are a lot more moles of acid than base, so why does the given answer have a basic pH?
Am I doing the setup or calculation incorrectly? thanks

[Borek]

NH₃ + H⁺ NH₄⁺

[zmasterflex]

So if I combine all of the non-dissociated NH3 and combine it with the H+ in the solution, I would extract all of it and be left with remnant OH? Therefore the calculations were unnecessary.... Just confirming, when given a weak acid and strong base (and vice versa) all I need to is just match up equimolar amounts of each and then do the pH calculation from the remnant?
thanks

[Borek]

Not exactly, it is a buffer system.

http://www.chembuddy.com/?left=buffers&right=composition-calculation

[Adenosine TriBossphate]

Since HCl is a strong acid, it will completely dissociate in solution into H⁺ and Cl^-. This being said, all the H⁺ ions will donate themselves to the base to make a conjugate acid. So make up an ICE table to see what your ending concentrations are, and then plug those into the Henderson-Hasselbalch.

[Borek]

So make up an ICE table to see what your ending concentrations are, and then plug those into the Henderson-Hasselbalch.

While not incorrect, this advice is slightly misguiding. All you need to solve the problem is stoichiometry. When you suggest using ICE table, most people will try to use it to calculate equilibrium concentrations - once they know equilibrium, plugging numbers into HH equation is no longer necessary, as you already know [H⁺].

Compare

http://www.chembuddy.com/?left=buffers&right=composition-calculation

and

http://www.chembuddy.com/?left=buffers&right=with-ICE-table