Titration of a weak base with a strong acid (continued)

- [Voiceover] In the last video, we looked at the titration curve for the titration of a weak base with a strong acid. And we found the pH at two points on our titration curve. We found the pH before we'd added any of our acid. So that was right here. We did that in part a of our question. And part b of our question, we found the pH after we'd added 20 milliliters of our acid. So that point is right here on our titration curve. And we found the pH in part b. In part c our goal is to find the pH after we've aded 40 milliliters of our 0.1 molar solution of HCl. So how many moles of acid did we add at this time? So the concentration of HCl is equal to 0.1 molar, so that's equal to 0.1 molar, molarity is moles over liters. So how many moles of acid did we add? We added 40 milliliters which, if we move our decimal place one, two, three, that's 0.04 liters. So we added 0.04 liters and we can just solve for moles here. So the moles of HCl will be equal to 0.1 times 0.04. So 0.1 times 0.04 is equal to 0.004. So that's how many moles of HCl we have added. So you can think about this as H plus and Cl minus and, if you have 0.004 moles of HCl, that's how many moles of H plus you have too. Or you can think about H plus and H two O giving you H three O plus, or hydronium. So you can also say, 0.004 moles of H three O plus. Alright, how many moles of ammonia did we start with? Right, well our concentration was 0.1, so the concentration of ammonia that we started with was equal to 0.1 molar, and once again, molarity is moles over liters, we started with 40 milliliters of ammonia, which is 0.04 liters. So we started with 40 milliliters, or 0.04 liters, so to find moles of ammonia, it's the same calculation we just did. 0.1 times 0.04 gives us 0.004 moles of ammonia, moles of NH three. So the acid that we added, in our titration, is going to react with the base that we started with. So, let's go ahead and write out the neutralization reaction. So one way you can write that would be, NH three plus H three O plus. So, ammonia, NH three, plus hydronium, H three O plus, so H three O plus is our acid, right? It donates a proton to ammonia, so if NH three picks up a proton, it turns into NH four plus and if H three O plus donates a proton, we're left with H two O. So that's one way to represent our reaction. The other way we could write it would be, NH three plus HCl. So let's write it that way too. So ammonia plus HCl, HCl donates a proton to NH three giving us NH four plus and once HCl donates a proton, you have Cl minus, the chloride anion, the conjugate base. So this would be NH four plus Cl minus. Alright, let's write down how many moles of acid that we have. So we added 0.004 moles of our acid. So let's write down 0.004 here, for moles of acid that we added. And for moles of base that we started with, that's also equal to 0.004. Let's get some more room down here. So we started with 0.004 moles of our base. If you look at your molar ratios there, one to one mole ratio. So we have now added enough acid to completely react with the base that we originally had present. So this represents the equivalence point of our titration. This is the equivalence point for our titration, because we've now added enough acid to completely react with our base. So all of this acid is going to react, right? So we use up all of that and it uses up all of the base that we had present. So all this base is going to react with the acid we're gonna lose, all 0.004 moles of NH three. So we're left with zero of that. Ammonia turns into ammonium, NH four plus. And so, if we start with zero for NH four plus, whatever we lose for ammonia, was what we gain for NH four plus. So for losing 0.004 moles of ammonia, that's how many moles of ammonium we are gaining. So we gain 0.004 moles of NH four plus. So, after we add the acid, we're left with 0.004 moles of NH four plus. So this is at our equivalence point. So we have some ammonium present. What's the concentration of ammonium that we have present? Well, the concentration, once again, is moles over liters. So we have 0.004 moles of NH four plus, so let's write 0.004 moles of NH four plus. What is the volume? What's the total volume now? Well let's go back up to the problem. We started with 40 milliliters of our... We started with 40 milliliters and we added another 40. So 40 plus 40 is 80 milliliters. Or one, two, three, 0.08 liters. So our total volume now is 0.08 liters. So let's go ahead and write that in here. So 0.08 liters. So 0.004 divided by 0.08 is just equal to 0.05. So 0.05 molar is the concentration of NH four plus that is present. Alright, ammonium is a weak acid. Ammonium can react with the water that is present. So let's now write out that reaction. So NH four plus is going to react with the water. And ammonium is going to function as an acid. Ammonium donates a proton to water. So if NH four plus donates a proton to H two O, H two O turns into H three O plus, or hydronium. And if you take away a proton for NH four plus, you're left with NH three or ammonia. So ammonium reacts with water and that's going to affect your pH. If I go back up here, so NH four plus does react with water, but the chloride anion does not react appreciably with water. So you don't have to worry about this being present, but the NH four plus will affect the pH, which our goal, of course, is to figure out the pH. Alright, so let's get some more room and let's write in our initial concentration of NH four plus, which is 0.05. So the initial concentration of NH four plus is 0.05. And, once again, for these problems, these weak acid equilibrium problems, we assume that we're starting with zero for the concentration of our products and our change, we're going to lose a certain concentration of ammonium. A certain concentration of ammonium is going to react with water. So we're gonna lose x. And then whatever we lose for ammonium turns into ammonia. So if we lose x for the concentration of ammonium, that's the same concentration that we gain, for NH three, for ammonia. And therefore, that's also the same concentration of NH three O plus. So at equilibrium, we're gonna have 0.05 minus x for the concentration of NH four plus. For the concentration on H three O plus, we have x. And for the concentration of NH thee, we also have x. Next, this is functioning as an acid, alright? NH four plus is donating a proton to water, so for the equilibrium expression, we have to write Ka here. And Ka is equal to concentration of products over reactants, so we have concentration of H three O plus, which is x times concentration of NH three, which is x all over the concentration of NH four plus, which is 0.05 minus x. So 0.05 minus x. Ka for NH four plus, Ka for ammonium, we calculated in an earlier video and we found it to be 5.6 times ten to the negative ten. So this is equal to... This is equal to, x times x is x squared and 0.05 minus x, if x is a very small concentration, 0.05 minus x is approximately the same as 0.05. So we write 0.05 in here, to make our lives easier for the calculation and we can get out the calculator and solve for x. 5.6 times ten to the negative ten. Alright, we're gonna multiply that by 0.05 and then we're gonna take the square root of our answer and solve for x. X is equal to 5.3 times ten to the negative six. So let's go ahead and write that down. X is equal to 5.3 times ten to the negative six and we go back up here to see what x represents. X represents the concentration of hydronium ions at equilibrium. So this would be the concentration of H three O plus. We've done a lot of work and the last thing we have to do is find the pH. And now we've found the concentration of hydronium ions, the pH is just the negative log of what we just found. So negative log of 5.3 times ten to the negative six Let's get out the calculator and do it. So we have the negative log of 5.3 times ten to the negative six and we get a pH of 5.28 if we round that. So our pH is equal to 5.28. So at the equivalence point, our solution is acidic, right? A pH of 5.28. Let's find that on our titration curve here. So we've added 40 milliliters of our acid and our pH is 5.28. So. let's say that is 5.28 here. So this is part c) of our question. The pH is equal to 5.28 at the equivalence point. Finally in part d), our goal is to find the pH after we've added 60 milliliters of our 0.1 molar solution of HCl. So this time, how many moles of acid have we added? We can find that from the molarity since our concentration is 0.1 molar, of HCl. And molarity is moles over liters. So we're adding 60 milliliters, which is 0.06 liters. So that's 0.06 liters, we solve for moles. 0.1 times 0.06 is equal to 0.006. So that's how many moles of HCl we've added now. HCl is H plus and CL minus, so you can think about this and being 0.006 moles of H plus or add your proton onto water to form hydronium, H three O plus. So 0.006 moles of acid and remember from part c), we needed 0.004 moles to completely react with the ammonia. So, at this point, we're gonna lose 0.004 moles, because that much is necessary to react with our ammonia. So how much is left over? So after our acid has reacted with the ammonia, we're left with 0.002 moles of acid. So that's how much we have present in solution. So, next, since our goal is to find the pH, we need to find the concentration of hydronium ions. So the concentration of hydronium ions will be equal to the moles, which is 0.002 and we need to divide that by our volume. Well, our volume will be 40 milliliters plus 60 milliliters, which will be 100 millililters. So that's our total volume here. 100 milliliters will be 0.1 liters, so we have 0.1, we have 0.1 liters here. and we can solve for the concentration of hydronium ions. So 0.002 divided by 0.1 gives us a concentration of hydronium equal to, this is 0.0200 molar. So now we're ready to find the pH since the pH is equal to the negative log of the concentration of hydronium. So the negative log of 0.0200, let's get out the calculator, is equal to, so the negative log of 0.0200 is equal to 1.699, so our pH here, our pH is equal to 1.699, or however you want to think about rounding these things. You can just say 1.70 just to make things a little bit easier, a little bit easier to think about that number. So approximately 1.7. After that it's 60 milliliters of our acid, so on our titration curve, we've added 60 milliliters of acid, so here we are right here and our pH is pretty close to 1.7, so we're past the equivalent point on our titration curve now.