Ammonia (NH3) is a weak base with a Kb= 1.8 X 10^-5 and hydrochloric acid (HCl) is a strong acid. What is the pH of a mixture made by adding 6.61 grams of NH4Cl to a 412 ml of a 0.180M solution of NH3?

Answer 1

Use the Henderson-Hasselbalch equation.
Post your work if you get stuck.

Answer 2

DrBob222, the pka value would be 9.26,. but how would you figure out the concentrations of the acid and conjugate base (M). You would have 0.124 mol NH4Cl and 0.07416 mol NH3, but how do you get them to concentrations. I know I could divide each by the total volume, but the problem does not give that info. Help Please

Answer 3

Re-read the problem. Looks like 412 mL to me. And both are in the same 412 mL. Actually, you don't need to know the volume for it cancels.
Note M = moles/L.
So M NH3 = 0.0742/0.412 L = ??
M NH4Cl = 0.124/0.412 L = ??
BUT since you are dividing base/acid that is
pH = pKa + log (b/a)
pH = pKa + log (moles b/0.412/moles a/0.412) and the 0.412 cancels leaving just
pH = pKa + log (mols b/mols a).

Answer 4

Thanks for poiting that out. I got a pH of 9.03. My only question is why the pH isn't acidic if it is a strong acid mixed with a weak base?

Answer 5

It isn't a strong acid mixed with a weak base. That is the salt. The salt, NH4Cl, IS the salt of a strong acid and a weak base, as you point out; however, you have that salt mixed with AMMONIA which is more of the weak base. And a weak base of ammonia is in excess. Furthermore, a weak base and it's salt makes a BASIC buffer.
The water solution of NH4Cl WOULD be acidic but adding ammonia to it makes it basic.