We've previously calculated the pH of the buffer that is described in this problem here. And we determined the pH to be 9.427. This was the concentration of ammonia that's actually given in the problem. But with that previous work, we determined that the ammonium, the NH4+ concentration, was 0.748. Now we're asked to calculate the pH when we take a small amount of a strong acid and add it to the buffer. The buffer will neutralize that strong acid. And take it out of the system, and that's why it resists change to pH. So to work this problem we have to first write the reaction between that strong acid, which I always let H3O+ represent my strong acid. And the portion of the buffer that will neutralize it, which would be the weak base. Ammonia is a weak base. And I will do a proton swap. The H+ from the acid is donated to the base and that will turn the base into ammonium and the H3O+ will be converted to water. So that is the reaction. Because the acid is strong, it is a one-way reaction and we will an ICF table, and we will plug moles into that table. Now we're ready to go determine the moles of each of these substances. Let's begin with the moles of H3O+. The moles of H3O+ would be the same as the moles of HCL. HCL. So find out the molarity of the HCL, which I do, it's given to me in the problem. And I know the volume of the HCL which I do, I will have the most being 0.0002 moles. And I can plug that in to the table. To get the moles of ammonia, we're going to take the molarity times the volume. The volume that we'll use is a volume of the buffer of this portion of the titration. So we're using a 75 milliliters, not the 500 milliliters, because we're not titrating with all of that. This is going to give to me 0.05625 moles, now that's an extra significant figure that I'm just going to carry along for now. And that's the moles of ammonia. Now before this reaction takes place, there's still some ammonium present. Because it's a buffer and it contains ammonium in a buffer. So I'm going to go ahead and calculate how much ammonia we're starting with. I'll take the molarity of the this molarity we determined in the previous example problem that we did in the last learning objective. And we'll multiply by the 75 milliliters, or the 0.075 liters here, and this is going to give me 0.0561 moles. And the water we can leave out. Now we're going to consume this smaller quantity, and that would be the strong acid. And I will subtract it from both sides. I mean from this side, and add it to this side. This is going to leave me with none of the H3O+. It's going to bring my ammonium down to 0.05605. It's going to increase my ammonium up to 0.0563. Now, because I have present a weak base, ammonia, and it's conjugate acid, I have a buffer. And so the Henderson–Hasselbalch equation will apply. pH equals pKa. Well pKa is the negative log of the Ka. The Ka is Kw divided by Kb, I did not write Kb in this problem. But it was in the previous problem. It's 1.76 x 10 to the -5. That's the Kb of ammonia. So I want to divide by 1.76 x 10 to the -5, and I'm going to do the moles of the base over the moles of the acid. The moles of the base are sitting right here, 0.05605. The moles of the acid is sitting right here, so at 0.0563. And that will give me a pH = 9 0.243. And as we look at this pH, 9.243, we see that it is slightly lower than what the pH was of just the buffer. It's resisting change. It's not completely not changing at all, but it is resisting it even so slightly. It is going down, and that is to be expected because you are after all adding an acid to the solution. So you expect the pH to be a little smaller or a little bit lower.
