WEBVTT 00:00:00.000 --> 00:00:00.016 The following content is provided under a Creative 00:00:00.016 --> 00:00:00.022 Commons license. 00:00:00.022 --> 00:00:00.038 Your support will help MIT OpenCourseWare continue to 00:00:00.038 --> 00:00:00.054 offer high quality educational resources for free. 00:00:00.054 --> 00:00:00.072 To make a donation or view additional materials from 00:00:00.072 --> 00:00:00.088 hundreds of MIT courses, visit MIT OpenCourseWare at 00:00:00.088 --> 00:00:00.110 ocw.mit.edu. 00:00:00.110 --> 00:00:26.180 PROFESSOR: OK, so we're going to leave off where we were 00:00:26.180 --> 00:00:28.780 last time talking about buffers. 00:00:28.780 --> 00:00:33.570 And I'm a big fan of buffers, actually, and buffers are 00:00:33.570 --> 00:00:37.890 extremely useful in my research because they keep the 00:00:37.890 --> 00:00:39.910 p h of things constant. 00:00:39.910 --> 00:00:42.670 So if you're going to do any kind of biology research or 00:00:42.670 --> 00:00:46.510 biochemistry research, you need to know about buffers. 00:00:46.510 --> 00:00:51.270 And as I'll mentioned later, also your body needs to be 00:00:51.270 --> 00:00:55.450 buffered appropriately, you don't want a large changes in 00:00:55.450 --> 00:00:59.880 p h in your body or things won't function properly. 00:00:59.880 --> 00:01:02.580 So buffering is very important. 00:01:02.580 --> 00:01:06.290 So let's go over buffers. 00:01:06.290 --> 00:01:09.160 So if we talk about an acid buffer, that's something 00:01:09.160 --> 00:01:13.460 that's buffering on the acidic side of the p h range. 00:01:13.460 --> 00:01:18.070 And in that, you want to have play on both sides. 00:01:18.070 --> 00:01:21.720 So if you have a strong acid, it can be neutralized, the p h 00:01:21.720 --> 00:01:22.370 stays the same. 00:01:22.370 --> 00:01:25.110 If you had a strong base, that would be neutralized so the p 00:01:25.110 --> 00:01:26.570 h stays the same. 00:01:26.570 --> 00:01:29.750 So what happens, you have to have an acid and its conjugate 00:01:29.750 --> 00:01:32.840 base in the mixture to have a good buffer. 00:01:32.840 --> 00:01:36.260 So what the weak acid is going to do is it will transfer 00:01:36.260 --> 00:01:41.130 protons or hydrogen ions to hydroxide ions that are 00:01:41.130 --> 00:01:45.540 supplied by the strong base to neutralize that strong base 00:01:45.540 --> 00:01:46.740 that was added. 00:01:46.740 --> 00:01:49.450 So you need to have an acid in there that's capable of giving 00:01:49.450 --> 00:01:54.190 up a proton to neutralize the hydroxide ion that was added. 00:01:54.190 --> 00:01:57.230 You also need to have a conjugate base. 00:01:57.230 --> 00:02:00.600 The conjugate base is going to accept protons if an acid is 00:02:00.600 --> 00:02:06.050 added, thereby neutralizing the effect of that added acid. 00:02:06.050 --> 00:02:09.260 So again, you need a weak acid and its conjugate base, you 00:02:09.260 --> 00:02:12.000 need both to have a good buffer. 00:02:12.000 --> 00:02:14.750 So I've been talking about weak acids and weak conjugate 00:02:14.750 --> 00:02:21.740 bases, so why would a strong acid and the salt of its 00:02:21.740 --> 00:02:27.050 conjugate base not make a good buffer? 00:02:27.050 --> 00:02:29.330 What do strong acids do? 00:02:29.330 --> 00:02:34.620 They dissociate almost completely, and that's not 00:02:34.620 --> 00:02:35.960 going to do you any good. 00:02:35.960 --> 00:02:39.490 You want to be able to shift that equation both directions. 00:02:39.490 --> 00:02:42.600 So a strong acid goes to completion is going to drive 00:02:42.600 --> 00:02:45.700 it to the right, and for a good buffer you want to be 00:02:45.700 --> 00:02:49.380 able to switch things around that if you add a strong acid, 00:02:49.380 --> 00:02:51.630 then you neutralize that, you add a strong base, you 00:02:51.630 --> 00:02:52.720 neutralize it. 00:02:52.720 --> 00:02:56.620 So you need the conjugate to be effective as a base, and a 00:02:56.620 --> 00:02:59.960 strong acid has an ineffective conjugate base. 00:02:59.960 --> 00:03:04.190 It's not good as a base at all, so that won't work. 00:03:04.190 --> 00:03:07.770 So buffers are made up of weak acids and their conjugate 00:03:07.770 --> 00:03:11.210 bases, or weak bases and their conjugate acids. 00:03:11.210 --> 00:03:13.380 They need to be in the weak range or there's 00:03:13.380 --> 00:03:17.800 no buffering capacity. 00:03:17.800 --> 00:03:21.420 So let's look at a base buffer example. 00:03:21.420 --> 00:03:24.590 The only difference here is that a basic buffer is going 00:03:24.590 --> 00:03:28.340 to buffer on the basic end of the p h range. 00:03:28.340 --> 00:03:29.970 So here's an equation. 00:03:29.970 --> 00:03:33.490 We have a base in water forming a conjugate of that 00:03:33.490 --> 00:03:35.150 weak base and hydroxide ions. 00:03:35.150 --> 00:03:39.150 So let's think about what happens if a 00:03:39.150 --> 00:03:41.150 strong acid is added. 00:03:41.150 --> 00:03:46.000 Well if a strong acid is added, then this base, n h 3, 00:03:46.000 --> 00:03:50.080 can accept protons from the incoming acid and make more n 00:03:50.080 --> 00:03:55.390 h 4 plus, thereby removing the strong acid from the solution 00:03:55.390 --> 00:03:58.060 and neutralizing the p h. 00:03:58.060 --> 00:04:02.540 If on the other hand a strong base is added, n h 4 plus or 00:04:02.540 --> 00:04:06.960 ammonium ions can donate the proton forming its conjugate 00:04:06.960 --> 00:04:11.760 base in water, and again, the p h remains about the same. 00:04:11.760 --> 00:04:16.590 So the base and its conjugate acid can both react, 00:04:16.590 --> 00:04:21.050 neutralizing the p h. 00:04:21.050 --> 00:04:25.160 So in the base buffer acid, a weak base, b, will accept 00:04:25.160 --> 00:04:29.480 protons supplied by the strong acid, removing a strong acid 00:04:29.480 --> 00:04:30.760 from the solution. 00:04:30.760 --> 00:04:34.650 The conjugate acid of that weak base, which again, would 00:04:34.650 --> 00:04:39.640 be a weak acid, can transfer protons to the hydroxide ions 00:04:39.640 --> 00:04:41.840 added by the strong base, again, 00:04:41.840 --> 00:04:45.380 neutralizing the effect there. 00:04:45.380 --> 00:04:50.630 So a buffer is a mixture of a weak acid base conjugate pair 00:04:50.630 --> 00:04:55.370 that will stabilize the p h serving as a source or a sink 00:04:55.370 --> 00:04:59.350 for the added protons. 00:04:59.350 --> 00:05:03.880 So that's how a buffer works. 00:05:03.880 --> 00:05:06.910 So, I mentioned buffering is very important. 00:05:06.910 --> 00:05:09.410 It's very important in your blood. 00:05:09.410 --> 00:05:13.510 Your blood is buffered in the range of p h 7 . 00:05:13.510 --> 00:05:15.080 35 to 7 . 00:05:15.080 --> 00:05:22.710 45, and you have a buffering pair and a conjugate acid base 00:05:22.710 --> 00:05:25.820 pair that's in your blood that does to work. 00:05:25.820 --> 00:05:30.020 So nature has its own buffering system design to 00:05:30.020 --> 00:05:33.220 keep the p h of your blood constant. 00:05:33.220 --> 00:05:37.550 Well what happens if the p h of your blood changes? 00:05:37.550 --> 00:05:43.390 So let me just tell you about one disease that can change 00:05:43.390 --> 00:05:45.330 the p h of your blood. 00:05:45.330 --> 00:05:47.960 This is from a vitamin B12 dependent enzyme called 00:05:47.960 --> 00:05:54.070 methylmalonic coa mutase, and this enzyme is your friend. 00:05:54.070 --> 00:05:57.070 I think that most of you are excited about the idea of 00:05:57.070 --> 00:06:00.980 having all your fat go to energy, so you should be all 00:06:00.980 --> 00:06:02.790 fan of this enzyme. 00:06:02.790 --> 00:06:07.180 So it breaks down from fatty acids -- you get methylmalonyl 00:06:07.180 --> 00:06:11.270 coa, which is converted to succinyl coa and goes into the 00:06:11.270 --> 00:06:12.820 citric acid cycle. 00:06:12.820 --> 00:06:15.530 But if this step is blocked, if there's some kind of 00:06:15.530 --> 00:06:19.650 genetic disease associated with that particular enzyme, 00:06:19.650 --> 00:06:22.890 you get a condition called methylmalonic aciduria. 00:06:22.890 --> 00:06:25.910 So this is excreted in the body, it can't be converted to 00:06:25.910 --> 00:06:29.260 the succinyl coa, so it has to be excreted from the body, and 00:06:29.260 --> 00:06:31.010 it has an acidic p h. 00:06:31.010 --> 00:06:35.320 And so when it's excreted, it will actually, there's a large 00:06:35.320 --> 00:06:38.460 amount of it, and it can change the p h of blood. 00:06:38.460 --> 00:06:41.080 So even though you have this buffering system, it can 00:06:41.080 --> 00:06:45.360 overwhelm the buffering system and cause a change in the p h, 00:06:45.360 --> 00:06:51.340 which can lead to neurological disorders and sometimes death. 00:06:51.340 --> 00:06:55.880 So they can look for this in newborns, they can do a test, 00:06:55.880 --> 00:07:00.550 and I was -- one of the tests is -- it's not in every state, 00:07:00.550 --> 00:07:04.370 but it is in Massachusetts, so my daughter, Samantha, had 00:07:04.370 --> 00:07:08.940 this test. But they can simply look at the urine of a newborn 00:07:08.940 --> 00:07:11.290 and look at the p h and whether any of 00:07:11.290 --> 00:07:13.680 this is being excreted. 00:07:13.680 --> 00:07:16.390 And some infants that have this condition can excrete 00:07:16.390 --> 00:07:18.460 like a gram of this acid a day. 00:07:18.460 --> 00:07:21.930 So it's really large quantities of acid that are 00:07:21.930 --> 00:07:26.280 excreted and it goes beyond the buffering capacity. 00:07:26.280 --> 00:07:29.560 So, I'm going to tell you a little story and it sort of 00:07:29.560 --> 00:07:30.950 emphasizes some of the 00:07:30.950 --> 00:07:32.830 importance of genetic research. 00:07:32.830 --> 00:07:35.850 So before the human genome project, scientists were 00:07:35.850 --> 00:07:38.500 trying to find the gene to figure out what was wrong with 00:07:38.500 --> 00:07:40.710 these patients that had this condition. 00:07:40.710 --> 00:07:41.960 And here are all the amino acids 00:07:41.960 --> 00:07:45.670 translated from the gene. 00:07:45.670 --> 00:07:50.090 This is not a small protein, that's a lot of amino acids. 00:07:50.090 --> 00:07:53.040 And they found that the problem was right here, or one 00:07:53.040 --> 00:07:54.230 of the main problems is right here. 00:07:54.230 --> 00:07:57.250 There's a glycine residue, which glycine is the smallest 00:07:57.250 --> 00:08:00.450 amino acid and that had been changed to something else and 00:08:00.450 --> 00:08:02.240 that led to the disease. 00:08:02.240 --> 00:08:06.070 Now I want you to imagine if you were parents of a small 00:08:06.070 --> 00:08:10.080 child and you found out your child had this condition and 00:08:10.080 --> 00:08:12.600 then you had heard that there were advances, they now know 00:08:12.600 --> 00:08:15.920 what caused it, and so, they went to the doctor and the 00:08:15.920 --> 00:08:21.240 doctor said, yes, your child, the DNA is wrong, and so let's 00:08:21.240 --> 00:08:27.570 substitute a different amino acid for glycine here. 00:08:27.570 --> 00:08:29.040 OK. 00:08:29.040 --> 00:08:30.980 What does that do for you? 00:08:30.980 --> 00:08:32.760 Like why is that a problem? 00:08:32.760 --> 00:08:35.310 Well, they didn't know why it was a problem, they just knew 00:08:35.310 --> 00:08:38.930 that it had the wrong amino acid in there. 00:08:38.930 --> 00:08:43.340 And so, the next step was to try to understand what was 00:08:43.340 --> 00:08:45.210 actually going on. 00:08:45.210 --> 00:08:49.820 So what happened in this case was that the vitamin B12 00:08:49.820 --> 00:08:53.690 cofactor had changed its conformation when it bound to 00:08:53.690 --> 00:08:54.680 the protein. 00:08:54.680 --> 00:08:58.640 And instead of having this sort of loop structure here, 00:08:58.640 --> 00:09:00.400 it had more of a tail. 00:09:00.400 --> 00:09:05.350 So this base here had moved down, and so there was a 00:09:05.350 --> 00:09:10.520 different shape to the vitamin than had been expected. 00:09:10.520 --> 00:09:13.270 So when the first structure was solved of one of these 00:09:13.270 --> 00:09:17.890 enzymes, it found that here's kind of what the vitamin B12 00:09:17.890 --> 00:09:20.460 looked like, it had this extended region which no one 00:09:20.460 --> 00:09:24.100 was expecting, and so that had to fit into the protein. 00:09:24.100 --> 00:09:27.190 So this is the domain of the protein that binds the 00:09:27.190 --> 00:09:30.540 vitamin, and it had a large hole, and you put these 00:09:30.540 --> 00:09:33.970 together and that's great, it fits and 00:09:33.970 --> 00:09:35.710 you have happy enzyme. 00:09:35.710 --> 00:09:40.780 Well, how do you create a hole -- nature supposedly abhors a 00:09:40.780 --> 00:09:43.110 vacuum, so how do you have this hole? 00:09:43.110 --> 00:09:47.250 Well, what nature did was it put the smallest amino acids 00:09:47.250 --> 00:09:51.350 there are, glycines, right along here to make room for 00:09:51.350 --> 00:09:53.360 this to fit together. 00:09:53.360 --> 00:09:57.470 So, what was happening in this genetic case is that the 00:09:57.470 --> 00:10:02.060 switch of a glycine to an argenine did this to the hole. 00:10:02.060 --> 00:10:06.750 So, what people were trying to do is treat these patients 00:10:06.750 --> 00:10:10.570 with extra B12, you give them injections of it, you have a 00:10:10.570 --> 00:10:14.250 lot more B12, and you want to the B12 to bind and get the 00:10:14.250 --> 00:10:16.140 activity back going again. 00:10:16.140 --> 00:10:20.410 But really, no matter how much vitamin B12 you added to it, 00:10:20.410 --> 00:10:22.890 it wasn't really going to do much good, it just wasn't 00:10:22.890 --> 00:10:24.070 going to fit. 00:10:24.070 --> 00:10:27.410 So instead, there was a suggestion that instead of 00:10:27.410 --> 00:10:30.850 giving the whole vitamin, perhaps you could just give a 00:10:30.850 --> 00:10:32.920 truncated version of the vitamin, which was 00:10:32.920 --> 00:10:37.530 commercially available, and that that might bind and 00:10:37.530 --> 00:10:41.020 restore activity of the protein. 00:10:41.020 --> 00:10:45.500 So this is what was suggested once you knew what the problem 00:10:45.500 --> 00:10:47.140 actually was. 00:10:47.140 --> 00:10:51.380 So this is just an example of a genetic disease, and it's 00:10:51.380 --> 00:10:54.770 not that common, but as I said, it is tested for here in 00:10:54.770 --> 00:10:55.900 Massachusetts. 00:10:55.900 --> 00:11:00.150 And how knowing something about what the problem is can 00:11:00.150 --> 00:11:02.610 lead to a potential solution. 00:11:02.610 --> 00:11:05.920 But for most of you, your blood is doing just fine, you 00:11:05.920 --> 00:11:09.350 don't have a genetic condition which is pumping too much acid 00:11:09.350 --> 00:11:13.820 into your bloodstream, so that nature's buffering capacity is 00:11:13.820 --> 00:11:18.040 working quite well. 00:11:18.040 --> 00:11:23.530 So buffers are important. 00:11:23.530 --> 00:11:26.980 It's great to have my important statements 00:11:26.980 --> 00:11:29.560 emphasized up there. 00:11:29.560 --> 00:11:32.570 We've been rehearsing at home. 00:11:32.570 --> 00:11:33.220 OK. 00:11:33.220 --> 00:11:36.780 So let's do a sample buffer problem. 00:11:36.780 --> 00:11:40.670 All right, suppose we have an acid, 0.1 moles of an acid, 00:11:40.670 --> 00:11:44.460 and 0.5 moles of its conjugate base added in 00:11:44.460 --> 00:11:46.090 the form of a salt. 00:11:46.090 --> 00:11:50.390 And so they're put into water and diluted to 1.0 liter. 00:11:50.390 --> 00:11:54.480 And you're given information about the k a of the acid, 1 . 00:11:54.480 --> 00:11:58.380 77 times 10 to the minus 4, and you're asked to 00:11:58.380 --> 00:12:00.840 calculate the p h. 00:12:00.840 --> 00:12:02.450 So what do you do? 00:12:02.450 --> 00:12:05.060 All right, first, it's always good to write the equation 00:12:05.060 --> 00:12:06.380 that you're talking about. 00:12:06.380 --> 00:12:12.810 And so you have acid in water forming hydronium ions and the 00:12:12.810 --> 00:12:14.390 conjugate base. 00:12:14.390 --> 00:12:19.050 So here, the acid is giving up a proton to the water, forming 00:12:19.050 --> 00:12:24.480 h 3 o plus and it's conjugate base. 00:12:24.480 --> 00:12:28.570 Now we want to think about what happens at equilibrium. 00:12:28.570 --> 00:12:31.020 So we know how many moles we have and we know what the 00:12:31.020 --> 00:12:32.300 total volume is. 00:12:32.300 --> 00:12:35.650 And in this table we should be using molarity, but the math 00:12:35.650 --> 00:12:40.910 here is pretty easy because we have 1 mole and 1 liter or 1, 00:12:40.910 --> 00:12:46.030 and 0.5 moles in 1 liter or 0.5 for molarity. 00:12:46.030 --> 00:12:50.900 And now at equilibrium what's going to happen, well, when 00:12:50.900 --> 00:12:54.480 this equilibrates some of this will go away and more of these 00:12:54.480 --> 00:12:56.430 are going to be formed. 00:12:56.430 --> 00:13:01.520 So we have minus x over here, plus x and plus x over there, 00:13:01.520 --> 00:13:05.480 and we add it all together you get 1 minus x plus x 00:13:05.480 --> 00:13:08.290 and 0.5 plus x. 00:13:08.290 --> 00:13:10.710 So the important thing to remember is in buffering 00:13:10.710 --> 00:13:14.440 problems, you have the acid and the conjugate. 00:13:14.440 --> 00:13:17.170 So we're used to seeing these tables where we only have 00:13:17.170 --> 00:13:21.130 something over here and it's all zero on the other side. 00:13:21.130 --> 00:13:24.390 But in a buffering problem, you have things on both sides 00:13:24.390 --> 00:13:25.710 from the very beginning. 00:13:25.710 --> 00:13:28.320 And that's a really important point, and that alone, if you 00:13:28.320 --> 00:13:30.730 can remember that, will get you a long way 00:13:30.730 --> 00:13:31.510 through this unit. 00:13:31.510 --> 00:13:35.090 So you remember that in a buffer you got to have both an 00:13:35.090 --> 00:13:39.600 acid in its conjugate base or a base in its conjugate acid. 00:13:39.600 --> 00:13:42.630 So here we're talking about an acid in water, it's an acidic 00:13:42.630 --> 00:13:46.800 buffer, and so we can use our k a value -- we want to know 00:13:46.800 --> 00:13:49.720 what the concentrations are at equilibrium to 00:13:49.720 --> 00:13:51.530 calculate our p h. 00:13:51.530 --> 00:13:57.500 So we can use our k a and we can set it up, so we have on 00:13:57.500 --> 00:14:01.510 the top products over reactants, and we we're not 00:14:01.510 --> 00:14:04.360 including water in this because it's dilute in 00:14:04.360 --> 00:14:06.840 solution and so it's not changing very much. 00:14:06.840 --> 00:14:10.710 And now we can plug in our values, so we have 0.5 plus x, 00:14:10.710 --> 00:14:15.780 x over 1 minus x. 00:14:15.780 --> 00:14:19.550 All right, now this is just written again from the last 00:14:19.550 --> 00:14:24.180 slide, we can try an approximation, which is it 00:14:24.180 --> 00:14:28.190 that x is small compared to 1 molar or 0.5 molar. 00:14:28.190 --> 00:14:32.400 And so we can get rid of the plus x and the minus x down 00:14:32.400 --> 00:14:37.100 here and just have one x term, which makes it simpler to 00:14:37.100 --> 00:14:39.400 solve, but we'll have to go back and check that 00:14:39.400 --> 00:14:42.010 approximation in a few minutes. 00:14:42.010 --> 00:14:46.160 So making that approximation, we can calculate x as 3 . 00:14:46.160 --> 00:14:51.040 54 times 10 to the minus 4 molar, and now we need to 00:14:51.040 --> 00:14:52.720 check that assumption. 00:14:52.720 --> 00:14:56.740 Well, you can probably guess it's going to be OK, because 00:14:56.740 --> 00:15:00.740 something times 10 to the minus 4 compared to 0.5 and 1, 00:15:00.740 --> 00:15:03.860 that's probably going to be OK that it's small. 00:15:03.860 --> 00:15:08.060 But our assumption here is that it needs to be less than 00:15:08.060 --> 00:15:11.320 5%, and here it's 0.1%. 00:15:11.320 --> 00:15:14.930 So you could just take this value of x, divide it by the 00:15:14.930 --> 00:15:20.840 smaller of the two, 0.5 , times 100, and calculate the 00:15:20.840 --> 00:15:21.500 percent ionization. 00:15:21.500 --> 00:15:26.310 If it's less than 5% you're OK, if it's more you need to 00:15:26.310 --> 00:15:30.740 use the quadratic equation to solve the problem. 00:15:30.740 --> 00:15:35.410 So x here is our concentration of hydronium ion, which is 00:15:35.410 --> 00:15:39.720 great, because then we can easily calculate the p h. 00:15:39.720 --> 00:15:42.920 So p h again is minus log of the hydronium ion 00:15:42.920 --> 00:15:47.230 concentration, and so the p h here is 3 . 00:15:47.230 --> 00:15:48.700 45. 00:15:48.700 --> 00:15:51.380 And again, significant figures, we'll be talking 00:15:51.380 --> 00:15:53.580 about this as we go through. 00:15:53.580 --> 00:15:55.830 The volume had two, it was 1 . 00:15:55.830 --> 00:15:59.620 0 liters, and so we're going to have two significant 00:15:59.620 --> 00:16:02.760 figures after the decimal point, because the volume was 00:16:02.760 --> 00:16:08.110 limiting insignificant figures. 00:16:08.110 --> 00:16:14.250 All right, now let's consider what happens if some strong 00:16:14.250 --> 00:16:17.730 acid had been added in the solution. 00:16:17.730 --> 00:16:20.520 So the volume is still going to be 1 liter, because the 00:16:20.520 --> 00:16:28.690 acid was added in before we went up to the 1 liter mark. 00:16:28.690 --> 00:16:32.810 So strong acids, they go to completion. 00:16:32.810 --> 00:16:37.150 So, if we added 0.1 moles of that strong acid, it will 00:16:37.150 --> 00:16:41.480 react with equal number of moles of the conjugate base to 00:16:41.480 --> 00:16:44.870 form the conjugate acid. 00:16:44.870 --> 00:16:47.070 So we can just do subtractions here. 00:16:47.070 --> 00:16:49.100 We don't have to worry about setting up any kind of 00:16:49.100 --> 00:16:52.920 equilibrium, we assume it goes completely. 00:16:52.920 --> 00:16:58.420 So for the conjugate base we had 0.5 moles before, and it's 00:16:58.420 --> 00:17:03.120 going to be reacting then with a strong acid, so 0.1 of those 00:17:03.120 --> 00:17:08.110 will react, leaving us with 0.4 moles of the conjugate, 00:17:08.110 --> 00:17:10.710 and that's in 1 liter. 00:17:10.710 --> 00:17:13.530 So now we have 0.4 molar. 00:17:13.530 --> 00:17:17.550 For the acid, we had 1 mole to begin with, but now we formed 00:17:17.550 --> 00:17:22.010 more as they reaction has taken place, and 00:17:22.010 --> 00:17:24.130 so we have 0.1 more. 00:17:24.130 --> 00:17:25.220 Now we have 1 . 00:17:25.220 --> 00:17:29.270 1 moles again in 1 liter. 00:17:29.270 --> 00:17:33.190 So now we can do the same thing again. 00:17:33.190 --> 00:17:36.390 So after this has happened, after we have added the strong 00:17:36.390 --> 00:17:40.920 acid, then a new equilibrium is going to be reached. 00:17:40.920 --> 00:17:46.340 And so we can plug in this table as well. 00:17:46.340 --> 00:17:47.970 So we have 1 . 00:17:47.970 --> 00:17:55.250 1 now over here, and 0.4 on the other side. 00:17:55.250 --> 00:17:59.880 As equilibrium is approached, you lose some of the acid as 00:17:59.880 --> 00:18:04.510 it reacts with water and ionizes and you form more of 00:18:04.510 --> 00:18:08.040 the h 3 o plus and more of the conjugate. 00:18:08.040 --> 00:18:09.290 So we have 1 . 00:18:09.290 --> 00:18:14.810 1 minus x, x, and 0.4 plus x. 00:18:14.810 --> 00:18:18.030 So again, the trick here is just to remember that there is 00:18:18.030 --> 00:18:21.810 a reaction with a strong acid that has been added, and you 00:18:21.810 --> 00:18:25.410 need to figure out the new molarities, and then go back 00:18:25.410 --> 00:18:33.720 to your equilibrium table with those new molarities. 00:18:33.720 --> 00:18:37.790 So here we can set up with a k a again, and do the same 00:18:37.790 --> 00:18:42.030 problem that we did before. 00:18:42.030 --> 00:18:45.390 And so, we can make the assumption again that x is 00:18:45.390 --> 00:18:49.410 small, try it out and see if it works. 00:18:49.410 --> 00:18:52.060 X turns out to be 4 . 00:18:52.060 --> 00:18:55.820 87 times 10 to the minus 4 molars -- so again, it's a 00:18:55.820 --> 00:18:57.440 small number. 00:18:57.440 --> 00:19:03.300 And so, it turns out to be 1 . -- less than 1% of 0.4 . 00:19:03.300 --> 00:19:06.840 If it's less than 5% of the smaller number, you don't have 00:19:06.840 --> 00:19:10.100 to worry about the bigger number, and so your assumption 00:19:10.100 --> 00:19:13.630 is OK, you don't need to use the quadratic equation. 00:19:13.630 --> 00:19:17.580 And so, then we can calculate at the end that the 00:19:17.580 --> 00:19:19.260 p h is now 3 . 00:19:19.260 --> 00:19:22.260 31, again, two significant figures after the decimal 00:19:22.260 --> 00:19:24.300 place because of the volume. 00:19:24.300 --> 00:19:27.990 So addition of 0.1 moles of a strong acid changed 00:19:27.990 --> 00:19:29.420 our p h from 3 . 00:19:29.420 --> 00:19:32.200 45 to 3 . 00:19:32.200 --> 00:19:33.290 31. 00:19:33.290 --> 00:19:34.750 So we buffered pretty well. 00:19:34.750 --> 00:19:38.540 There was a change in p h, we added acid and the p h did go 00:19:38.540 --> 00:19:41.630 down, but not by an enormous amount. 00:19:41.630 --> 00:19:45.040 And so, that's because this turned out to be a pretty 00:19:45.040 --> 00:19:46.890 decent buffer here, so we didn't have 00:19:46.890 --> 00:19:54.740 a really huge effect. 00:19:54.740 --> 00:20:07.530 So, those are -- that's an example of a buffer problem. 00:20:07.530 --> 00:20:10.720 So, let's think for a minute about designing a buffer. 00:20:10.720 --> 00:20:17.690 Suppose you wanted to create a buffer at a particular p h, 00:20:17.690 --> 00:20:19.680 what do you need to think about? 00:20:19.680 --> 00:20:22.870 Well, you need to think about the ratio of your acid to the 00:20:22.870 --> 00:20:26.570 conjugate, and you need to think about the p k a of the 00:20:26.570 --> 00:20:32.660 acid, and the p h that's desired. 00:20:32.660 --> 00:20:38.940 So here is a generic equation for an acid in water, so we 00:20:38.940 --> 00:20:42.060 have our acid, h a, in water forming hydronium ions and our 00:20:42.060 --> 00:20:44.340 conjugate base. 00:20:44.340 --> 00:20:47.300 And now we're going to do a little derivation to come up 00:20:47.300 --> 00:20:50.010 with an equation that would be useful to you in thinking 00:20:50.010 --> 00:20:53.420 about buffers and designing buffers. 00:20:53.420 --> 00:20:57.910 So, we can write a generic term for k a. 00:20:57.910 --> 00:21:01.420 K a equals products over reactants, and so we have our 00:21:01.420 --> 00:21:05.220 hydronium ions are conjugate base over our acid. 00:21:05.220 --> 00:21:09.400 And now we can take this term and rearrange it. 00:21:09.400 --> 00:21:11.330 So we can pull a hydronium ion 00:21:11.330 --> 00:21:14.380 concentration over to one side. 00:21:14.380 --> 00:21:18.860 And now we're going to take the logs of both sides, so we 00:21:18.860 --> 00:21:24.090 get the log of hydronium ion concentration, the log of k a, 00:21:24.090 --> 00:21:27.560 and the log of h a concentration over a minus 00:21:27.560 --> 00:21:31.680 concentration, and now we're going to multiply everything 00:21:31.680 --> 00:21:36.950 by negative sign, so we have minus logs of things. 00:21:36.950 --> 00:21:40.070 And I'm just going to move this now up to the top of the 00:21:40.070 --> 00:21:43.950 screen, and we have this equation. 00:21:43.950 --> 00:21:49.160 So what is minus log of hydronium ion concentration? 00:21:49.160 --> 00:21:49.850 P h. 00:21:49.850 --> 00:21:53.310 What's minus log of k a? 00:21:53.310 --> 00:21:55.430 P k a. 00:21:55.430 --> 00:22:00.110 So, we have p h equals p k a minus the log of a 00:22:00.110 --> 00:22:03.970 concentration of your acid over the concentration of your 00:22:03.970 --> 00:22:05.120 conjugate base. 00:22:05.120 --> 00:22:08.030 And these are equilibrium concentrations for that, 00:22:08.030 --> 00:22:10.570 because remember, we derived it from our equilibrium 00:22:10.570 --> 00:22:13.230 expression for k a. 00:22:13.230 --> 00:22:16.400 But a lot of times when you're working buffer problems, you 00:22:16.400 --> 00:22:19.710 know the concentrations that you added of these things, not 00:22:19.710 --> 00:22:23.780 necessarily the concentrations of equilibrium. 00:22:23.780 --> 00:22:27.550 And if we rewrite this expression in terms of the 00:22:27.550 --> 00:22:31.990 initial concentrations or original or o concentrations, 00:22:31.990 --> 00:22:36.550 then we have p h is approximately equal to p k a 00:22:36.550 --> 00:22:40.310 minus log of the initial concentrations of your acid 00:22:40.310 --> 00:22:41.810 over your conjugate base. 00:22:41.810 --> 00:22:42.900 And this is known as a 00:22:42.900 --> 00:22:45.140 Henderson Hasselbalch equation. 00:22:45.140 --> 00:22:48.490 And people love the Henderson Hasselbalch equation, and it 00:22:48.490 --> 00:22:51.870 can be incredibly useful to you in solving these problems. 00:22:51.870 --> 00:22:55.090 But I'm going to emphasize when it's OK to use it and 00:22:55.090 --> 00:22:58.560 when it's not OK to use it, because people try to apply it 00:22:58.560 --> 00:23:02.300 to everything, and it's for buffer problems, and people 00:23:02.300 --> 00:23:04.340 apply it to all sorts of things that are not buffer 00:23:04.340 --> 00:23:07.160 problems. So I want to make sure that it's clear when you 00:23:07.160 --> 00:23:10.060 can use it and when you can't use it. 00:23:10.060 --> 00:23:14.110 And it's fine to use it and you should use it when it's 00:23:14.110 --> 00:23:17.150 acceptable, but not at other times. 00:23:17.150 --> 00:23:19.800 It's one of the few equations in acid base, so people get 00:23:19.800 --> 00:23:22.400 very excited by it. 00:23:22.400 --> 00:23:26.680 All right, so remember that it's really equal when they're 00:23:26.680 --> 00:23:29.330 at equilibrium concentrations, and this is just an 00:23:29.330 --> 00:23:32.520 approximation that we're saying that those are initial 00:23:32.520 --> 00:23:34.180 concentrations. 00:23:34.180 --> 00:23:39.080 So when is it OK to say well, the equilibrium concentration 00:23:39.080 --> 00:23:43.240 is more or less the same as the initial concentration. 00:23:43.240 --> 00:23:47.400 And that's true when x, which is in this particular example 00:23:47.400 --> 00:23:50.500 your hydronium ion concentration, is small 00:23:50.500 --> 00:23:54.110 compared to your initial concentration of the acid and 00:23:54.110 --> 00:23:57.000 the conjugate base that you added to the buffer. 00:23:57.000 --> 00:24:01.030 So, and when x is small, then pretty much, the equilibrium 00:24:01.030 --> 00:24:04.110 concentration equals the initial concentration. 00:24:04.110 --> 00:24:08.110 So there's not really very much change if x is a really 00:24:08.110 --> 00:24:10.050 small number. 00:24:10.050 --> 00:24:13.510 So, a lot of the time this is true. 00:24:13.510 --> 00:24:15.740 Remember, we're making buffers from weak acids and weak 00:24:15.740 --> 00:24:19.120 conjugate bases, and the definition of a weak acid is 00:24:19.120 --> 00:24:22.390 that it only loses a tiny fraction of its protons. 00:24:22.390 --> 00:24:26.070 It only ionizes slightly in water. 00:24:26.070 --> 00:24:29.960 And a weak base typically only accepts a fraction of the 00:24:29.960 --> 00:24:31.740 protons it can accept. 00:24:31.740 --> 00:24:36.330 It's a weak base, so it's not doing a whole lot 00:24:36.330 --> 00:24:37.650 of chemistry there. 00:24:37.650 --> 00:24:41.440 So this approximation is good a lot of the time. 00:24:41.440 --> 00:24:44.590 So just you know when you can and can not use it, we're 00:24:44.590 --> 00:24:47.870 going to give a rule and say it's the same rule as before. 00:24:47.870 --> 00:24:52.040 When we say when x is small or your hydronium ion 00:24:52.040 --> 00:24:55.200 concentration is small compared to the acid or the 00:24:55.200 --> 00:24:58.360 conjugate base, when it's less than 5%, that's what we're 00:24:58.360 --> 00:25:02.380 going to call small, so the same rule we've 00:25:02.380 --> 00:25:03.800 been using all along. 00:25:03.800 --> 00:25:06.880 When x is small, this approximation 00:25:06.880 --> 00:25:08.170 works pretty well. 00:25:08.170 --> 00:25:11.800 And you can plug in your concentrations right there -- 00:25:11.800 --> 00:25:14.380 your initial concentrations not your equilibrium 00:25:14.380 --> 00:25:19.060 concentrations. 00:25:19.060 --> 00:25:22.060 All right, so let's use Henderson Hasselbalch and 00:25:22.060 --> 00:25:26.490 design a buffer system if we want a p h of 4 . 00:25:26.490 --> 00:25:27.480 6. 00:25:27.480 --> 00:25:32.060 And a good rule to remember is that a buffering solution is 00:25:32.060 --> 00:25:36.420 most effective when it's plus or minus 1 away 00:25:36.420 --> 00:25:38.300 from the p k a. 00:25:38.300 --> 00:25:40.540 So that's a good thing too I keep in mind 00:25:40.540 --> 00:25:43.050 for research as well. 00:25:43.050 --> 00:25:46.490 If you're really far away from the p k a, it's not going to 00:25:46.490 --> 00:25:49.200 be a good buffering system, and this is a mistake the 00:25:49.200 --> 00:25:52.300 people make in the lab sometimes, so you may run into 00:25:52.300 --> 00:25:55.580 that in some of your laboratory courses. 00:25:55.580 --> 00:25:59.270 So we want a buffer about p h 4 . 00:25:59.270 --> 00:26:00.130 6. 00:26:00.130 --> 00:26:03.870 So we can look at our ionization constants of acid, 00:26:03.870 --> 00:26:05.830 and of course, we're always doing everything at room 00:26:05.830 --> 00:26:09.840 temperature in this unit, and so we can look at what p k a's 00:26:09.840 --> 00:26:14.680 are of some weak acids, and acetate, acetate is an easy 00:26:14.680 --> 00:26:19.650 buffer to get your hands on, acetate is quite available, 00:26:19.650 --> 00:26:21.860 and it has a good p k a 4 . 00:26:21.860 --> 00:26:22.980 75. 00:26:22.980 --> 00:26:27.340 So that's great, we can use that in designing our buffer. 00:26:27.340 --> 00:26:32.780 So, acetic acid has the right p k a, and we're all set, we 00:26:32.780 --> 00:26:34.490 can prepare a buffer. 00:26:34.490 --> 00:26:38.010 But we'll need to add the acid and the conjugate, and we need 00:26:38.010 --> 00:26:42.890 to know how much acid and how much of the conjugate to add. 00:26:42.890 --> 00:26:46.410 So we can use Henderson Hasselbalch equation for this. 00:26:46.410 --> 00:26:50.140 We're creating a buffer, and so this is an equation we can 00:26:50.140 --> 00:26:51.510 use for buffers. 00:26:51.510 --> 00:26:55.840 We know what p h we want, we know the p k a of the thing 00:26:55.840 --> 00:26:58.730 we're going to use, and we want to figure out how much of 00:26:58.730 --> 00:27:01.700 the acid to use and how much of the conjugate to use. 00:27:01.700 --> 00:27:05.660 We need to know the ratio of one to the other to set up our 00:27:05.660 --> 00:27:07.820 buffer ideally. 00:27:07.820 --> 00:27:12.200 So we can plug our numbers in and calculate that. 00:27:12.200 --> 00:27:15.660 So, we can rearrange this if you want so that the unknown 00:27:15.660 --> 00:27:18.560 is on one side and we have to p k a and the p h 00:27:18.560 --> 00:27:21.300 on the other side. 00:27:21.300 --> 00:27:26.560 And so, we have here our p k a minus our p h and we get 0 . 00:27:26.560 --> 00:27:27.710 15. 00:27:27.710 --> 00:27:32.500 Now we need to inverse log and come up with the ratio, and 00:27:32.500 --> 00:27:34.640 the ratio that we want is 1 . 00:27:34.640 --> 00:27:35.510 4. 00:27:35.510 --> 00:27:38.530 So we want to have a ratio from our acid to the 00:27:38.530 --> 00:27:40.260 conjugate of 1 . 00:27:40.260 --> 00:27:40.910 4. 00:27:40.910 --> 00:27:43.480 Well how much exactly are we going to add them, we know the 00:27:43.480 --> 00:27:46.460 ratio now, how much are we going to add? 00:27:46.460 --> 00:27:48.690 Well, you could use 1 . 00:27:48.690 --> 00:27:53.130 4 molar and 1 molar, for example, that would have the 00:27:53.130 --> 00:27:54.860 correct ratio. 00:27:54.860 --> 00:27:59.010 But how do you know if that's going to be good? 00:27:59.010 --> 00:28:02.310 Well, the ratio is more important than the exact 00:28:02.310 --> 00:28:07.150 amounts, however, the amounts get to this issue of sort of 00:28:07.150 --> 00:28:11.190 the capacity of the buffer -- how resistant it is to changes 00:28:11.190 --> 00:28:12.170 in the p h. 00:28:12.170 --> 00:28:15.440 So if you use too low concentrations of both, it 00:28:15.440 --> 00:28:17.490 won't be all that resistant. 00:28:17.490 --> 00:28:21.490 So if you need to have a better buffering capacity, the 00:28:21.490 --> 00:28:24.880 higher concentrations that you should use. 00:28:24.880 --> 00:28:27.630 And you can use Henderson Hasselbalch then to calculate 00:28:27.630 --> 00:28:31.350 what sort of the minimum amount you use, the minimum 00:28:31.350 --> 00:28:35.050 amount to have that 5% rule work. 00:28:35.050 --> 00:28:39.520 So in the example I told you about this condition, the 00:28:39.520 --> 00:28:42.730 buffer in the blood was pretty good, but the acid with so 00:28:42.730 --> 00:28:46.120 much that it overwhelmed buffering capacity. 00:28:46.120 --> 00:28:48.790 So if there had been larger concentrations of the buffer, 00:28:48.790 --> 00:28:52.760 that would have certainly helped in that case. 00:28:52.760 --> 00:28:55.240 So the higher the concentrations you use, the 00:28:55.240 --> 00:28:57.760 more resistant to change. 00:28:57.760 --> 00:29:02.010 And that's the idea of buffering capacity. 00:29:02.010 --> 00:29:04.870 So, and if you use too low concentrations, then your 00:29:04.870 --> 00:29:08.770 Henderson Hasselbalch equation won't even be valid. 00:29:08.770 --> 00:29:11.520 So we can go back and calculate what is sort of the 00:29:11.520 --> 00:29:15.500 minimum concentrations we need to use for Henderson 00:29:15.500 --> 00:29:18.830 Hasselbalch to be valid to meet that 5% rule. 00:29:18.830 --> 00:29:20.630 So for a p h of 4 . 00:29:20.630 --> 00:29:24.830 6, a hydronium ion concentration is a 2 . 00:29:24.830 --> 00:29:26.880 5 times 10 to the minus 5. 00:29:26.880 --> 00:29:31.040 And so then if we work our equation backwards, we want to 00:29:31.040 --> 00:29:35.010 be less than 5%, so we have this over the concentration of 00:29:35.010 --> 00:29:39.310 either one of those times 100% gives you 5%. 00:29:39.310 --> 00:29:42.890 So the concentrations need to be greater than 5 times 10 to 00:29:42.890 --> 00:29:46.690 the minus 4 moles or you won't meet this 5% rule. 00:29:46.690 --> 00:29:49.710 So at least to be greater than that and they need to be in 00:29:49.710 --> 00:29:50.880 the right ratio. 00:29:50.880 --> 00:29:55.210 Other than that you're somewhat free to decide what 00:29:55.210 --> 00:29:56.910 you're going to do with that, and there may be other 00:29:56.910 --> 00:30:01.930 considerations involved if you're working with a certain 00:30:01.930 --> 00:30:04.930 concentration of protein, you might not want your buffer 00:30:04.930 --> 00:30:07.810 concentration to be too high, it might start interfering 00:30:07.810 --> 00:30:10.700 with enzyme assay, for example. 00:30:10.700 --> 00:30:12.540 And often buffers are somewhere around 100 00:30:12.540 --> 00:30:15.440 millimolar, that's a pretty common 00:30:15.440 --> 00:30:19.170 concentration for buffers. 00:30:19.170 --> 00:30:24.810 All right, so this slide we're doing really well. 00:30:24.810 --> 00:30:28.790 So we talked about weak acids in water, weak bases in water, 00:30:28.790 --> 00:30:31.410 and I'm trying to convince you that these are the same as the 00:30:31.410 --> 00:30:34.320 salt and water problems. So we're going to talk much more 00:30:34.320 --> 00:30:36.510 about salt and water in a few minutes. 00:30:36.510 --> 00:30:39.560 And I told you about buffers, so we only have 2 00:30:39.560 --> 00:30:40.950 more things to go. 00:30:40.950 --> 00:30:43.890 We need to talk about strong acids in water, and strong 00:30:43.890 --> 00:30:47.340 bases in water, and then you'll have all of the 5 types 00:30:47.340 --> 00:30:50.900 of problems to do the problem-set. 00:30:50.900 --> 00:30:55.690 And as I mentioned last time, it looks deceivingly short, 00:30:55.690 --> 00:30:58.840 the number of questions on them, but the titration 00:30:58.840 --> 00:31:00.700 problems are long. 00:31:00.700 --> 00:31:04.400 So don't leave those to the last minute or there will not 00:31:04.400 --> 00:31:06.860 be a whole lot of sleep involved. 00:31:06.860 --> 00:31:11.460 So just a word of warning from the past. Everyone looks and 00:31:11.460 --> 00:31:13.300 go, "Oh, this is an easy problem-set, there aren't many 00:31:13.300 --> 00:31:19.500 problems." So, titration problems do 00:31:19.500 --> 00:31:20.950 take a lot of time. 00:31:20.950 --> 00:31:24.790 So let's talk about titration, so that as soon as class is 00:31:24.790 --> 00:31:27.650 over today, you can go work on those questions on the 00:31:27.650 --> 00:31:28.140 problem-set. 00:31:28.140 --> 00:31:29.530 All right. 00:31:29.530 --> 00:31:31.410 So acid base titrations. 00:31:31.410 --> 00:31:37.300 How many of you in high school titrated an acid with a base? 00:31:37.300 --> 00:31:42.630 Large fraction of you, OK. 00:31:42.630 --> 00:31:47.740 So usually what titration problems are meant to do, that 00:31:47.740 --> 00:31:52.810 you have something known about an acid or a base, base say of 00:31:52.810 --> 00:31:58.390 known concentration, and and acid of unknown concentration 00:31:58.390 --> 00:32:01.660 or maybe unknown molecular weight, and you're titrating 00:32:01.660 --> 00:32:04.790 them out, titrating them together to find missing 00:32:04.790 --> 00:32:06.030 information. 00:32:06.030 --> 00:32:08.750 So you can determine concentration, sometimes you 00:32:08.750 --> 00:32:11.580 can determine a molecular weight. 00:32:11.580 --> 00:32:14.690 All right, so here's what a plot looks like and some of 00:32:14.690 --> 00:32:19.090 the key terms involved in acid base titrations. 00:32:19.090 --> 00:32:24.350 So you have a p h on one side, and then volume of either base 00:32:24.350 --> 00:32:26.690 or acid added on the other side. 00:32:26.690 --> 00:32:28.940 So p h versus a volume. 00:32:28.940 --> 00:32:33.220 And the actual experiment here, you have a base that 00:32:33.220 --> 00:32:38.440 you're dripping, one drip at a time supposedly into your 00:32:38.440 --> 00:32:40.340 strong acid. 00:32:40.340 --> 00:32:44.600 And then you're looking to figure out when you're going 00:32:44.600 --> 00:32:47.050 to reach the either equivalence point or the end 00:32:47.050 --> 00:32:50.620 point, these are basically the same terms. It's also called 00:32:50.620 --> 00:32:54.650 the stoichiometric point, and often equivalence point is 00:32:54.650 --> 00:33:00.010 used as a more theoretical amount of volume is added, 00:33:00.010 --> 00:33:03.860 where as end point is the experimentally measured. 00:33:03.860 --> 00:33:06.500 So we're going to be talking a lot about equivalence points, 00:33:06.500 --> 00:33:09.110 because we have no lab associated with this course, 00:33:09.110 --> 00:33:12.360 so everything will be theorectical in terms of 00:33:12.360 --> 00:33:15.970 these, but they should be the same. 00:33:15.970 --> 00:33:19.190 So, many times you will measure p h with the p h meter 00:33:19.190 --> 00:33:21.400 -- this just shows a p h meter. 00:33:21.400 --> 00:33:25.540 And for those of you who have done these experiments, this 00:33:25.540 --> 00:33:26.890 should look familiar. 00:33:26.890 --> 00:33:30.090 And for those of you who haven't, often what you're 00:33:30.090 --> 00:33:32.330 doing when you're dripping something in, you're waiting 00:33:32.330 --> 00:33:35.970 for an indicator dye to change color as an indication that 00:33:35.970 --> 00:33:38.710 you have reached the end point, and often you're 00:33:38.710 --> 00:33:41.870 adding, and it's clear, clear, clear for what seems to be 00:33:41.870 --> 00:33:45.920 forever, and just as you get incredibly impatient, you 00:33:45.920 --> 00:33:47.960 start adding it faster and you go all the way 00:33:47.960 --> 00:33:49.250 to this dark color. 00:33:49.250 --> 00:33:54.800 And what you want is this very, very, very light changed 00:33:54.800 --> 00:33:57.000 color that indicates the end point. 00:33:57.000 --> 00:34:02.030 So you usually go too slow and then go too fast and then have 00:34:02.030 --> 00:34:03.950 to do the experiment over again. 00:34:03.950 --> 00:34:06.880 Those of you who know how to calculate theoretical values 00:34:06.880 --> 00:34:09.770 could sit down and do a calculation first and then go 00:34:09.770 --> 00:34:12.970 really fast right until you get around this point and add 00:34:12.970 --> 00:34:16.500 it really slowly, and be done with lab sooner. 00:34:16.500 --> 00:34:20.110 So, we'll talk about how you do the theoretical value, 00:34:20.110 --> 00:34:23.810 which could save you a lot of time if you ever run across 00:34:23.810 --> 00:34:30.280 these labs in any future class. 00:34:30.280 --> 00:34:34.140 So, here are the two curves that you would see for either 00:34:34.140 --> 00:34:36.920 a strong acid with a strong base, or strong base with a 00:34:36.920 --> 00:34:38.210 strong acid. 00:34:38.210 --> 00:34:41.410 So if you're titrating a strong acid with a strong 00:34:41.410 --> 00:34:44.630 base, you're going to start down being very acidic, 00:34:44.630 --> 00:34:47.450 because all you have is your strong acid. 00:34:47.450 --> 00:34:51.640 And then as you add in base, the p h will go up. 00:34:51.640 --> 00:34:55.260 You'll reach this point -- s is for the stoichiometric 00:34:55.260 --> 00:34:58.060 point, we also call this, again, the equivalence point, 00:34:58.060 --> 00:35:01.840 and then it'll continue to go up and then start to level off 00:35:01.840 --> 00:35:03.880 a bit more up here. 00:35:03.880 --> 00:35:05.960 If you're going the other direction, you're starting 00:35:05.960 --> 00:35:09.790 with strong base, your p h is going to be high, and the p h 00:35:09.790 --> 00:35:12.120 will decrease as you add the acid. 00:35:12.120 --> 00:35:14.750 And then you get to the stoichiometric or equivalence 00:35:14.750 --> 00:35:20.090 point in the middle of this curve here. 00:35:20.090 --> 00:35:23.350 And then as you go down farther, more acidic, it 00:35:23.350 --> 00:35:24.680 starts to level off. 00:35:24.680 --> 00:35:29.880 So those are what the titration curves look like. 00:35:29.880 --> 00:35:32.190 So let's do an example. 00:35:32.190 --> 00:35:36.040 We're going to have a strong base being titrated with a 00:35:36.040 --> 00:35:42.095 strong acid first. And our strong base is n a o h, and 00:35:42.095 --> 00:35:47.040 our strong acid is h c l. 00:35:47.040 --> 00:35:56.560 So let's calculate the p h at the equivalence point. 00:35:56.560 --> 00:36:00.730 Well, let's just calculate the p h at 5 mils -- 00:36:00.730 --> 00:36:03.450 I don't know whether it's equivalence point, actually. 00:36:03.450 --> 00:36:08.200 5 mils of this have been added to the amount of base. 00:36:08.200 --> 00:36:12.920 So we have 25 mils of our strong base at 0.15 molar 00:36:12.920 --> 00:36:17.710 concentration, and we're adding 5 mils of our 0.34 00:36:17.710 --> 00:36:21.520 molar acid. 00:36:21.520 --> 00:36:24.580 So what we want to do is figure out how many moles of 00:36:24.580 --> 00:36:28.280 base we had to start with. 00:36:28.280 --> 00:36:33.050 So how many moles of o h minus -- again, n a o h is a strong 00:36:33.050 --> 00:36:37.300 base, so however much n a o h you add, is the amount of o h 00:36:37.300 --> 00:36:42.390 minus you get, and so we just put in the number of liters 00:36:42.390 --> 00:36:44.780 times the concentration to get moles. 00:36:44.780 --> 00:36:48.760 We don't need any equilibrium table here. 00:36:48.760 --> 00:36:52.290 So then we want to figure out the amount of acid added when 00:36:52.290 --> 00:36:55.660 you're adding 5 mils of it. 00:36:55.660 --> 00:36:59.950 So again, h c l is a strong acid, so the amount of h c l 00:36:59.950 --> 00:37:05.040 added equals the amount of h 3 o plus, that's formed, goes to 00:37:05.040 --> 00:37:09.220 completion, and so we added 5 mils, the concentration was 00:37:09.220 --> 00:37:12.480 0.34 molar, so we have 1 . 00:37:12.480 --> 00:37:17.480 7 times 10 to the minus 3 moles. 00:37:17.480 --> 00:37:21.100 So now, the strong acid is going to react with the strong 00:37:21.100 --> 00:37:24.850 base, and we want to figure out how much of the base is 00:37:24.850 --> 00:37:28.980 left, how much hydroxide ion is left after it reacts with 00:37:28.980 --> 00:37:31.210 the hydronium ions. 00:37:31.210 --> 00:37:35.650 So it'll react 1:1, so we had 6 . 00:37:35.650 --> 00:37:41.900 25 times 10 to the minus 3 moles, and we added 1 . 00:37:41.900 --> 00:37:45.260 7 times 10 to the minus 3 moles of the acid, so we're 00:37:45.260 --> 00:37:47.350 going to have 4 . 00:37:47.350 --> 00:37:53.430 55 times 10 the minus 3 moles left. 00:37:53.430 --> 00:37:59.270 And now we can calculate the molarity here, so we have the 00:37:59.270 --> 00:38:03.900 number of moles and the new volume, so we had 25 mils to 00:38:03.900 --> 00:38:04.990 the base to begin with. 00:38:04.990 --> 00:38:08.600 We've added 5 mils to the acid, so our new volume is 30, 00:38:08.600 --> 00:38:12.980 and so we have our new concentration. 00:38:12.980 --> 00:38:15.280 And now we can calculate p h. 00:38:15.280 --> 00:38:20.380 First we'll calculate p o h, and then use 14 minus to get 00:38:20.380 --> 00:38:23.180 the new p h. 00:38:23.180 --> 00:38:29.630 So if you were making your own little titration curve, we 00:38:29.630 --> 00:38:34.480 could have volume of acid on one side and p h on the other 00:38:34.480 --> 00:38:39.080 side, and you have a point of 13 . 00:38:39.080 --> 00:38:40.430 18. 00:38:40.430 --> 00:38:49.360 That's how much we have after you have 5 mils added. 00:38:49.360 --> 00:38:52.560 So we didn't, in this curve, we have not measured the zero 00:38:52.560 --> 00:39:00.420 point, but we know what one point after 5 mils of the acid 00:39:00.420 --> 00:39:02.760 has been added. 00:39:02.760 --> 00:39:06.250 So now let's go right to the equivalence point -- so the 00:39:06.250 --> 00:39:08.180 way I draw it should indicate that was not 00:39:08.180 --> 00:39:09.510 an equivalence point. 00:39:09.510 --> 00:39:12.850 So we want to figure out how much we need to add of the 00:39:12.850 --> 00:39:15.150 acid to reach the equivalence point. 00:39:15.150 --> 00:39:18.290 So the equivalence point or the stoichiometric point means 00:39:18.290 --> 00:39:22.900 that you add the same amount of moles of your titrate as 00:39:22.900 --> 00:39:26.740 you had, and so they're going to be equal to each other. 00:39:26.740 --> 00:39:30.250 So if you're adding acid to a strong base, you've added 00:39:30.250 --> 00:39:33.370 enough acid that you now have equal number of moles to the 00:39:33.370 --> 00:39:36.380 number of moles of base that you had to begin with. 00:39:36.380 --> 00:39:37.640 So we know we have 6 . 00:39:37.640 --> 00:39:41.030 25 times 10 to the minus 3 moles of 00:39:41.030 --> 00:39:42.590 the base were present. 00:39:42.590 --> 00:39:46.400 So at the equivalence point, you need to have that exact 00:39:46.400 --> 00:39:49.560 same number of moles of acid. 00:39:49.560 --> 00:39:53.700 So we know the concentration of the acid we have and the 00:39:53.700 --> 00:39:56.590 number of moles we need, and so we can calculate the 00:39:56.590 --> 00:39:59.160 volume, which is 18 . 00:39:59.160 --> 00:40:04.480 4 millimeters, or 0.0184 liters. 00:40:04.480 --> 00:40:12.860 So what would the p h then be of the equivalence point? 00:40:12.860 --> 00:40:15.980 We're titrating a strong acid with a strong base, what 00:40:15.980 --> 00:40:19.830 should the p h be? 00:40:19.830 --> 00:40:22.270 The p h should be 7. 00:40:22.270 --> 00:40:27.120 So somewhere around here we're going to be at 18, so you can 00:40:27.120 --> 00:40:36.650 just try to draw a curve, and here we have a p h of 7 when 00:40:36.650 --> 00:40:39.770 we've gone to about 18 . 00:40:39.770 --> 00:40:44.300 4 milliliters added. 00:40:44.300 --> 00:40:47.620 So when you titrate a strong acid with a strong base, you 00:40:47.620 --> 00:40:49.610 form a salt that's neutral. 00:40:49.610 --> 00:40:52.320 Because the conjugates of a strong acid and a strong base 00:40:52.320 --> 00:40:53.380 are not basic or acidic. 00:40:53.380 --> 00:40:56.740 They're ineffective as acids or bases, so you're going to 00:40:56.740 --> 00:40:58.640 get a neutral salt. 00:40:58.640 --> 00:41:01.470 So if you're doing a problem with a strong acid and strong 00:41:01.470 --> 00:41:04.130 base, it's not a lot of calculations that need to be 00:41:04.130 --> 00:41:06.910 done at this point, you just need to recognize that your 00:41:06.910 --> 00:41:09.420 salt should be neutral. 00:41:09.420 --> 00:41:12.920 So now, we can talk about what happens if you add an extra 00:41:12.920 --> 00:41:17.170 milliliter of your acid, after you've reached that 00:41:17.170 --> 00:41:19.820 equivalence point, which is something that people probably 00:41:19.820 --> 00:41:22.820 have done in doing these titrations not meaning to, 00:41:22.820 --> 00:41:27.550 gone well beyond the equivalence point. 00:41:27.550 --> 00:41:31.140 So first we can find out the number of moles of the strong 00:41:31.140 --> 00:41:36.250 acid that we've added extra, so we added one mil of the 00:41:36.250 --> 00:41:39.660 strong acid, and so again, it goes to completion. 00:41:39.660 --> 00:41:44.480 The concentration, the acid, is 0.34 molar times our 1 mil, 00:41:44.480 --> 00:41:45.940 so we've added 3 . 00:41:45.940 --> 00:41:52.100 4 times 10 to the minus 4 extra moles of our acid, and 00:41:52.100 --> 00:41:55.650 so let's calculate then what the molarity is that we have 00:41:55.650 --> 00:42:03.660 added that was extra. 00:42:03.660 --> 00:42:05.330 And even if you don't have a calculator, 00:42:05.330 --> 00:42:41.980 we've helped you out. 00:42:41.980 --> 00:43:05.530 Let's just take 10 more seconds. 00:43:05.530 --> 00:43:10.650 I think this is a first, isn't it? 00:43:10.650 --> 00:43:12.300 Did I mention it's a good idea to start 00:43:12.300 --> 00:43:15.660 the problem-set early. 00:43:15.660 --> 00:43:20.060 So the trick here was about the volume. 00:43:20.060 --> 00:43:25.960 So we had 25 mils to begin with, then we added 18 . 00:43:25.960 --> 00:43:30.110 4 to get to the equivalence point, and then we're 1 mil 00:43:30.110 --> 00:43:32.310 beyond the equivalence point. 00:43:32.310 --> 00:43:35.360 And the tricks to these parts of the problems are to 00:43:35.360 --> 00:43:38.160 remember all the things that have been added 00:43:38.160 --> 00:43:39.630 to get to the volume. 00:43:39.630 --> 00:43:43.150 So the problem itself is not very tricky of having just an 00:43:43.150 --> 00:43:47.640 acid in water, but the one trick is in calculating the 00:43:47.640 --> 00:43:50.530 molarity, remembering all of the things that were added to 00:43:50.530 --> 00:43:52.490 get to this point. 00:43:52.490 --> 00:43:56.060 And so by the end of the problem-set this should be 00:43:56.060 --> 00:43:59.230 pretty familiar to you, but it's also something that you 00:43:59.230 --> 00:44:02.670 want to make sure that you check on an exam that you're 00:44:02.670 --> 00:44:06.530 not making any volume mistakes. 00:44:06.530 --> 00:44:10.620 So again, here are the things that you want to remember to 00:44:10.620 --> 00:44:13.640 add in in calculating this. 00:44:13.640 --> 00:44:17.120 And then, if you calculate the p h of that, you 00:44:17.120 --> 00:44:19.620 get a p h of 2 . 00:44:19.620 --> 00:44:24.450 1106, so that's down somewhere in here. 00:44:24.450 --> 00:44:33.080 We've added 1 more mill, and we're at a p h of 2 . 00:44:33.080 --> 00:44:38.140 1106, and this can be a check as well. 00:44:38.140 --> 00:44:41.100 If you forgot to add some of your volume, you might get a p 00:44:41.100 --> 00:44:43.230 h that doesn't make a lot of sense. 00:44:43.230 --> 00:44:45.150 So that can be a double check. 00:44:45.150 --> 00:44:49.290 Always ask yourself when you have a strong base and you're 00:44:49.290 --> 00:44:52.160 adding acid, before you've added very much, your p h 00:44:52.160 --> 00:44:53.830 should be basic. 00:44:53.830 --> 00:44:56.800 At the equivalence point of a strong acid, strong base 00:44:56.800 --> 00:44:58.710 titration, it'll be 7. 00:44:58.710 --> 00:45:01.660 And if you continue to add acid, then you should have a 00:45:01.660 --> 00:45:04.590 pretty low p h, you'll notice the curve drops off pretty 00:45:04.590 --> 00:45:07.580 fast. So it should be a dramatic change in p h by 00:45:07.580 --> 00:45:09.750 adding extra of your acid. 00:45:09.750 --> 00:45:11.680 So always check your work. 00:45:11.680 --> 00:45:14.980 OK, so let's stop there, and we're going to weak acid, weak 00:45:14.980 --> 00:45:17.760 base titrations next time.