WEBVTT 00:00:00.030 --> 00:00:02.400 The following content is provided under a Creative 00:00:02.400 --> 00:00:03.780 Commons license. 00:00:03.780 --> 00:00:06.020 Your support will help MIT OpenCourseWare 00:00:06.020 --> 00:00:10.090 continue to offer high quality educational resources for free. 00:00:10.090 --> 00:00:12.660 To make a donation or to view additional materials 00:00:12.660 --> 00:00:16.580 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.580 --> 00:00:17.250 at ocw.mit.edu. 00:00:25.872 --> 00:00:28.580 PROFESSOR: So again, the equivalence point 00:00:28.580 --> 00:00:33.660 is where you've added all of the moles of your strong base 00:00:33.660 --> 00:00:36.240 that you need to convert all the moles you 00:00:36.240 --> 00:00:39.803 had of the weak acid to its conjugate base. 00:00:43.180 --> 00:00:46.610 So if we have this type of problem, 00:00:46.610 --> 00:00:49.780 a strong base titrating a weak acid, 00:00:49.780 --> 00:00:52.600 the pH is going to be greater than 7 00:00:52.600 --> 00:00:55.660 at this equivalence or stoichiometric point. 00:00:55.660 --> 00:00:57.090 And we can see that in the plot. 00:00:57.090 --> 00:00:59.880 Here is pH 7, little arrow going up 00:00:59.880 --> 00:01:02.590 indicating it's going to be greater than 7. 00:01:02.590 --> 00:01:06.080 And that's because we have just the conjugate base left 00:01:06.080 --> 00:01:07.630 at this point. 00:01:07.630 --> 00:01:11.170 So again, pH depends on the property 00:01:11.170 --> 00:01:15.070 of the salt that's formed at the equivalence point. 00:01:15.070 --> 00:01:20.620 And when it is a weak acid being titrated with a strong base-- 00:01:20.620 --> 00:01:23.940 and that's what we have here, weak acid, strong base-- 00:01:23.940 --> 00:01:25.970 you're going to get a salt and water. 00:01:25.970 --> 00:01:30.600 But this salt, now, is going to have basic properties. 00:01:30.600 --> 00:01:35.440 And we saw this before that sodium has no effect on pH. 00:01:35.440 --> 00:01:38.780 Things in group 1 are not going to have any effect on pH. 00:01:38.780 --> 00:01:42.810 But HCO2- is a conjugate base of a weak acid. 00:01:42.810 --> 00:01:44.700 It is a weak base itself. 00:01:44.700 --> 00:01:46.580 So this is going to be basic. 00:01:46.580 --> 00:01:50.420 And so that's why the pH is going to be greater than 7. 00:01:50.420 --> 00:01:52.290 Remember, salt and water problems 00:01:52.290 --> 00:01:55.690 really break down to weak acid and water or weak base 00:01:55.690 --> 00:01:57.820 and water problems, depending on what 00:01:57.820 --> 00:02:00.760 went into a form that particular salt. 00:02:00.760 --> 00:02:03.180 So when you're doing these problems, 00:02:03.180 --> 00:02:06.300 and if you're on the exam and you get to the end 00:02:06.300 --> 00:02:09.539 and you know that the pH should be greater than 7, 00:02:09.539 --> 00:02:11.695 but the pH you calculated, for some reason 00:02:11.695 --> 00:02:14.000 and you don't know where you made the mistake, 00:02:14.000 --> 00:02:15.440 is less than 7. 00:02:15.440 --> 00:02:18.120 If you say to me, this doesn't make sense. 00:02:18.120 --> 00:02:21.690 It should be greater than 7, because it should 00:02:21.690 --> 00:02:23.460 be basic at this point, should just 00:02:23.460 --> 00:02:26.300 have conjugate base around, you will get points back 00:02:26.300 --> 00:02:32.360 for recognizing that the answer you gave me can't be right. 00:02:32.360 --> 00:02:35.200 So always pay attention to what answer you're getting. 00:02:35.200 --> 00:02:37.470 Does that answer make sense? 00:02:37.470 --> 00:02:40.850 I really care more about that people understand what's 00:02:40.850 --> 00:02:42.990 going on, than they can do the math perfectly 00:02:42.990 --> 00:02:46.210 in a very short amount of time. 00:02:46.210 --> 00:02:48.560 But you'll often be asked to calculate the pH. 00:02:48.560 --> 00:02:50.940 So let's think about how we would calculate the pH. 00:02:50.940 --> 00:02:53.260 We know it's greater than 7, but what is it exactly 00:02:53.260 --> 00:02:55.020 for this problem. 00:02:55.020 --> 00:02:56.840 And to do that, we need to first know 00:02:56.840 --> 00:02:58.660 what the volume is going to be. 00:02:58.660 --> 00:03:01.100 What volume of that strong base do we 00:03:01.100 --> 00:03:05.130 need to add to reach the stoichiometric or equivalence 00:03:05.130 --> 00:03:06.130 point? 00:03:06.130 --> 00:03:09.450 So we had in the beginning 2.5 times 10 00:03:09.450 --> 00:03:13.190 to the minus third moles of our weak acid. 00:03:13.190 --> 00:03:15.990 So that means that at the stoichiometric point, 00:03:15.990 --> 00:03:19.200 we're going to form that number of moles of conjugate base. 00:03:19.200 --> 00:03:20.900 And that also means that to do that we 00:03:20.900 --> 00:03:24.560 need to add that number of moles of the strong base. 00:03:24.560 --> 00:03:28.250 So we need to add 2.5 times 10 to the minus third moles 00:03:28.250 --> 00:03:29.960 of our strong base. 00:03:29.960 --> 00:03:34.150 We know the concentration in the strong base, 0.15 molar. 00:03:34.150 --> 00:03:36.420 So we can calculate that the volume we need 00:03:36.420 --> 00:03:39.390 is 16.7 milliliters. 00:03:39.390 --> 00:03:43.370 Now, the total volume to get to the stoichiometric point 00:03:43.370 --> 00:03:46.310 was 25-- that's what we had originally-- 00:03:46.310 --> 00:03:56.730 plus this 16.74, so 41.7 milliliters or 0.0417 liters. 00:03:56.730 --> 00:04:01.190 We can calculate the molarity of the conjugate base that's 00:04:01.190 --> 00:04:02.020 formed. 00:04:02.020 --> 00:04:06.090 How many moles in this total volume gives us a molarity 00:04:06.090 --> 00:04:09.740 of 0.0600 molar. 00:04:09.740 --> 00:04:12.670 And now, we can go ahead and solve the problem. 00:04:12.670 --> 00:04:15.274 But why don't you tell me how we're 00:04:15.274 --> 00:04:16.399 going to solve the problem. 00:04:16.399 --> 00:04:18.493 What are we going to use to solve this problem? 00:04:33.826 --> 00:04:35.075 Let's just do 10 more seconds. 00:04:50.880 --> 00:04:58.690 Yep, so Kb, and so if we look at this problem for a minute, 00:04:58.690 --> 00:05:04.690 so this is, again, the type of problem, 00:05:04.690 --> 00:05:07.130 it's a weak base problem. 00:05:07.130 --> 00:05:09.090 We've converted all the weak acid 00:05:09.090 --> 00:05:11.490 we had to its conjugate base, because we 00:05:11.490 --> 00:05:13.440 added enough moles of the strong base 00:05:13.440 --> 00:05:16.150 to convert all the weak acid to the conjugate base. 00:05:16.150 --> 00:05:19.150 So the equivalence point, all we have is the weak base. 00:05:19.150 --> 00:05:21.610 So this is a base and water problem. 00:05:21.610 --> 00:05:25.210 So we write our base plus water going to the conjugate acid 00:05:25.210 --> 00:05:26.810 plus hydroxide. 00:05:26.810 --> 00:05:28.610 And when it's a base and water problem, 00:05:28.610 --> 00:05:31.100 you should have hydroxide on the other side. 00:05:31.100 --> 00:05:32.920 So we can set up this expression. 00:05:32.920 --> 00:05:37.400 We get 0.06 molar minus x, x plus x. 00:05:37.400 --> 00:05:40.650 And we can use Kb to solve the problem. 00:05:40.650 --> 00:05:44.750 And Kb, in this case, is 5.6 times 10 to the minus 11th. 00:05:44.750 --> 00:05:52.910 That's going to be equal to x squared over 0.0600 minus x. 00:05:52.910 --> 00:05:56.540 So if I was given Ka and I now need Kb, 00:05:56.540 --> 00:06:00.616 what do I use to convert Ka and Kb? 00:06:00.616 --> 00:06:03.460 AUDIENCE: [INAUDIBLE]. 00:06:03.460 --> 00:06:04.350 PROFESSOR: Right. 00:06:04.350 --> 00:06:07.470 And you can use Kw to solve it, yep. 00:06:07.470 --> 00:06:10.460 So we can easily convert between these two. 00:06:10.460 --> 00:06:12.770 Not a problem. 00:06:12.770 --> 00:06:14.501 So this is how you would do the problem. 00:06:14.501 --> 00:06:16.750 And always remember, ask yourself what type of problem 00:06:16.750 --> 00:06:17.290 it is. 00:06:17.290 --> 00:06:20.530 If it's weak base and water, you want a Kb. 00:06:20.530 --> 00:06:23.200 So can I use Henderson-Hasselbalch for this? 00:06:23.200 --> 00:06:23.930 AUDIENCE: No. 00:06:23.930 --> 00:06:26.260 PROFESSOR: No, I can't. 00:06:26.260 --> 00:06:29.060 And you should not as well. 00:06:29.060 --> 00:06:30.680 So what do you do after this? 00:06:30.680 --> 00:06:32.300 We should be able to go from here. 00:06:32.300 --> 00:06:34.600 We're not going to go through all the steps. 00:06:34.600 --> 00:06:38.710 We can simplify, pretend x is small, make sure x is small. 00:06:38.710 --> 00:06:42.780 And in this case, x is quite small of 1.83 times 10 00:06:42.780 --> 00:06:45.000 to the minus sixth molar. 00:06:45.000 --> 00:06:47.380 From that, we have to remember that x, now, 00:06:47.380 --> 00:06:50.410 is the hydroxide ion concentrations. 00:06:50.410 --> 00:06:55.500 So we're going to calculate pOH first and then calculate pH 00:06:55.500 --> 00:07:01.730 by subtract 14.00 minus pOH to get us the pH. 00:07:01.730 --> 00:07:03.680 And now, it's 8.26. 00:07:03.680 --> 00:07:04.960 That's above 7. 00:07:04.960 --> 00:07:06.670 That number makes sense. 00:07:06.670 --> 00:07:08.350 If I had stopped and forgotten what 00:07:08.350 --> 00:07:12.290 x was and realized the pH was 5.74, 00:07:12.290 --> 00:07:15.950 I should have realized there was a problem there. 00:07:15.950 --> 00:07:18.470 So weak base in water problem. 00:07:21.090 --> 00:07:22.070 So we go up here. 00:07:22.070 --> 00:07:25.190 That's Point S, the stoichiometric point. 00:07:25.190 --> 00:07:28.030 We have a pH of 8.26. 00:07:28.030 --> 00:07:29.947 Now, we've added this little E at the end. 00:07:29.947 --> 00:07:32.280 We're just going to think about the last type of problem 00:07:32.280 --> 00:07:36.020 you might see, which is past the equivalence point. 00:07:36.020 --> 00:07:39.870 So here, you are at a volume beyond the volume 00:07:39.870 --> 00:07:42.580 needed to get you to the equivalence point. 00:07:42.580 --> 00:07:46.770 And at this point, you have your conjugate base in solution. 00:07:46.770 --> 00:07:52.040 But you're adding a lot of concentrated base of sodium 00:07:52.040 --> 00:07:54.730 hydroxide to that solution. 00:07:54.730 --> 00:07:58.720 So the amount of the conjugate base that you have there 00:07:58.720 --> 00:08:01.450 is really not going to contribute to pH anymore, 00:08:01.450 --> 00:08:04.320 not compared to the pH change that's 00:08:04.320 --> 00:08:07.860 going to be caused by just putting extra strong base 00:08:07.860 --> 00:08:09.140 into solution. 00:08:09.140 --> 00:08:13.420 So the pH is really going to be determined by the excess amount 00:08:13.420 --> 00:08:15.160 of any OH you have. 00:08:15.160 --> 00:08:17.790 And you can pretty much forget all the work you just 00:08:17.790 --> 00:08:21.422 did in calculating what the pH was due to the conjugate base, 00:08:21.422 --> 00:08:23.380 because that's going to be overwhelmed by this. 00:08:23.380 --> 00:08:25.380 And I'll show you that that's true. 00:08:25.380 --> 00:08:30.330 So what this is then is a strong base in water problem. 00:08:30.330 --> 00:08:32.450 So beyond the equivalence point now, 00:08:32.450 --> 00:08:34.820 we're a strong base and water problem. 00:08:34.820 --> 00:08:35.860 So how do we do this? 00:08:35.860 --> 00:08:37.570 And we saw this already, but we'll just 00:08:37.570 --> 00:08:39.470 review it one more time. 00:08:39.470 --> 00:08:43.840 So we're 5 mills past the equivalence point. 00:08:43.840 --> 00:08:47.280 So we're going to figure out how many extra moles of OH 00:08:47.280 --> 00:08:48.580 we added. 00:08:48.580 --> 00:08:51.320 5 mills times our concentration, so we 00:08:51.320 --> 00:08:56.380 have 7.5 times 10 to the minus fourth moles that are extra. 00:08:56.380 --> 00:09:00.710 And now, we need to calculate the concentration. 00:09:00.710 --> 00:09:03.230 And we have to remember our volume. 00:09:03.230 --> 00:09:05.970 So we added 5 extra mills. 00:09:05.970 --> 00:09:08.090 We had 25 to begin with. 00:09:08.090 --> 00:09:12.280 And we use 16.7 to get to the equivalence point. 00:09:12.280 --> 00:09:14.540 And so when you have the whole volume in there, 00:09:14.540 --> 00:09:19.580 you can calculate that your concentration is 0.016 molar 00:09:19.580 --> 00:09:20.600 OH. 00:09:20.600 --> 00:09:24.230 Then from that, we can calculate pOH. 00:09:24.230 --> 00:09:27.140 And from that, we calculate pH. 00:09:27.140 --> 00:09:30.340 And it's 12.21. 00:09:30.340 --> 00:09:33.130 And just to convince you that it was OK 00:09:33.130 --> 00:09:36.180 that I forgot all about that conjugate base-- 00:09:36.180 --> 00:09:37.870 remember, the concentration that we 00:09:37.870 --> 00:09:43.840 had calculated of OH that is due to that weak base in solution? 00:09:43.840 --> 00:09:47.420 This number, really small compared to that number. 00:09:47.420 --> 00:09:49.880 And if you want to be very particular, 00:09:49.880 --> 00:09:53.236 you can add this to this before calculating this. 00:09:53.236 --> 00:09:54.860 But it's really not going to give you-- 00:09:54.860 --> 00:09:57.060 to the number of significant figures you have, 00:09:57.060 --> 00:09:59.420 it's not going to make any difference whatsoever. 00:09:59.420 --> 00:10:02.500 So this is a strong base in water problem. 00:10:02.500 --> 00:10:05.970 We're only going to think about how many moles extra of OH 00:10:05.970 --> 00:10:08.350 do we have, and what is the total volume, 00:10:08.350 --> 00:10:11.650 and then you're done. 00:10:11.650 --> 00:10:14.590 So now, we've done this whole curve. 00:10:14.590 --> 00:10:15.940 We started at the beginning. 00:10:15.940 --> 00:10:18.860 It was a weak acid problem down here. 00:10:18.860 --> 00:10:20.850 We went into the buffering region. 00:10:20.850 --> 00:10:23.260 We can use Henderson-Hasselbalch here. 00:10:23.260 --> 00:10:25.810 We can do a very simple calculation 00:10:25.810 --> 00:10:29.460 at the half equivalence point, pH equals pKa. 00:10:29.460 --> 00:10:31.370 Then at the stoichiometric point, 00:10:31.370 --> 00:10:33.300 we're a weak base problem. 00:10:33.300 --> 00:10:36.980 And then we're a strong base problem. 00:10:36.980 --> 00:10:39.920 So you can now do the same thing the other way. 00:10:39.920 --> 00:10:40.684 Yeah, question. 00:10:40.684 --> 00:10:44.925 AUDIENCE: What if you were given a problem like, 00:10:44.925 --> 00:10:46.865 figure out what the buffering region was? 00:10:46.865 --> 00:10:48.805 We never calculated E. 00:10:48.805 --> 00:10:50.620 PROFESSOR: Yes, right. 00:10:50.620 --> 00:10:54.330 So we skipped E, because there was a lot of different points. 00:10:54.330 --> 00:11:03.560 Yeah, so if you are in a region where 00:11:03.560 --> 00:11:06.220 you have both conjugate acid, conjugate bass, 00:11:06.220 --> 00:11:08.180 you can assume that's in the buffering region. 00:11:08.180 --> 00:11:10.550 So you need to have both to be in the buffering region, 00:11:10.550 --> 00:11:14.270 if the problem only has your weak acid in water 00:11:14.270 --> 00:11:17.020 and initially you have sort of zero of the other. 00:11:17.020 --> 00:11:19.940 But if you've added some of the strong acid, strong base, 00:11:19.940 --> 00:11:21.600 that means you've converted some, 00:11:21.600 --> 00:11:24.620 but you know you're not at the equivalence point yet. 00:11:24.620 --> 00:11:26.120 You can assume a buffer problem. 00:11:26.120 --> 00:11:30.190 And then pretty much, it's kind of in the buffering region, 00:11:30.190 --> 00:11:32.040 unless it's right here or right there. 00:11:32.040 --> 00:11:34.250 So pretty much anywhere in here, you 00:11:34.250 --> 00:11:36.310 can assume it's going to be a buffer problem. 00:11:36.310 --> 00:11:37.820 And when you do the subtraction, you 00:11:37.820 --> 00:11:40.994 should see that you had your weak acid. 00:11:40.994 --> 00:11:42.785 And you've converted some to the conjugate. 00:11:42.785 --> 00:11:45.280 And you can see that you have amounts of both 00:11:45.280 --> 00:11:47.090 when you do that subtraction. 00:11:47.090 --> 00:11:49.080 And when you have concentrations of both, 00:11:49.080 --> 00:11:50.580 then you're in the buffering region. 00:11:50.580 --> 00:11:52.284 AUDIENCE: So you would go from B to F? 00:11:52.284 --> 00:11:55.060 PROFESSOR: So you would do D the same way 00:11:55.060 --> 00:11:58.080 that you did B. Yeah, that would be the same. 00:11:58.080 --> 00:12:01.175 And then if you want to make sure that you're not there-- 00:12:01.175 --> 00:12:02.800 but you should know that when you start 00:12:02.800 --> 00:12:07.010 the problem, because when you start doing these problems, 00:12:07.010 --> 00:12:09.840 you're going to calculate, say, how much moles of the weak acid 00:12:09.840 --> 00:12:10.540 you had. 00:12:10.540 --> 00:12:13.815 And then you calculate how many moles of the base you added. 00:12:13.815 --> 00:12:15.440 And if they're equal, then you're like, 00:12:15.440 --> 00:12:17.170 oh, I'm not in the buffering region. 00:12:17.170 --> 00:12:18.496 I'm at the equivalence point. 00:12:18.496 --> 00:12:20.870 If they're not equal, if the number of moles you've added 00:12:20.870 --> 00:12:23.510 are lesser the strong base but non-zero, 00:12:23.510 --> 00:12:25.010 then you're in the buffering region. 00:12:28.690 --> 00:12:30.314 Any other good questions? 00:12:30.314 --> 00:12:32.230 And we're not going to go the other direction. 00:12:32.230 --> 00:12:34.479 We're not going to do all the points in the other one. 00:12:34.479 --> 00:12:37.380 But there's a problem set for that. 00:12:37.380 --> 00:12:40.850 But I'm not going to leave acid-base quite yet. 00:12:40.850 --> 00:12:42.700 We're almost there, but not quite yet, 00:12:42.700 --> 00:12:46.550 because I've got to say a little more about pKa's. 00:12:46.550 --> 00:12:52.570 So pKa's are not just important in titration problems. 00:12:52.570 --> 00:12:56.160 And I want to share with you one of my favorite videos 00:12:56.160 --> 00:12:58.705 about why pKa's are important. 00:13:05.689 --> 00:13:06.355 [VIDEO PLAYBACK] 00:13:06.355 --> 00:13:08.320 - My name is Samuel Thompson. 00:13:08.320 --> 00:13:09.774 I'm a rising senior at MIT. 00:13:09.774 --> 00:13:11.190 And for the past three years, I've 00:13:11.190 --> 00:13:13.090 been working with Alice Ting. 00:13:13.090 --> 00:13:15.660 My project is in the field of chemical biology, which 00:13:15.660 --> 00:13:17.490 means that I'm a tool maker. 00:13:17.490 --> 00:13:19.290 And I've been making tools to allow 00:13:19.290 --> 00:13:20.790 people to look at proteins. 00:13:20.790 --> 00:13:22.900 We want to see where they are, what they're doing, 00:13:22.900 --> 00:13:25.410 what they're interacting with. 00:13:25.410 --> 00:13:27.360 But all proteins and all cells are 00:13:27.360 --> 00:13:29.110 completely transparent whenever you look 00:13:29.110 --> 00:13:31.790 at them under a microscope. 00:13:31.790 --> 00:13:35.090 So we use an enzyme to attach an organic molecule-- a very, very 00:13:35.090 --> 00:13:37.760 small molecule-- to the protein, so that we get 00:13:37.760 --> 00:13:38.940 the same fluorescent output. 00:13:38.940 --> 00:13:41.410 We see a bright light whenever we shine laser light on it 00:13:41.410 --> 00:13:44.330 and look at it under a microscope. 00:13:44.330 --> 00:13:46.100 After some very complicated research, 00:13:46.100 --> 00:13:49.210 we came up with a very basic problem in our design. 00:13:49.210 --> 00:13:53.170 We had a mismatch between our probe and the pH of the cells. 00:13:53.170 --> 00:13:57.989 Cells typically exist at about 7.2 to 7.5 pH. 00:13:57.989 --> 00:13:59.280 That's what's healthy for them. 00:13:59.280 --> 00:13:59.905 They need that. 00:13:59.905 --> 00:14:01.640 And if you go outside those boundaries, 00:14:01.640 --> 00:14:03.890 they're very unhealthy, and they behave abnormally. 00:14:03.890 --> 00:14:07.770 And the pKa of our probe is also 7.5. 00:14:07.770 --> 00:14:10.670 pKa is the point where any sort of species 00:14:10.670 --> 00:14:13.580 is half protonated, half deprotonated. 00:14:13.580 --> 00:14:16.070 And our probe needs to be deprotonated. 00:14:16.070 --> 00:14:18.300 It needs to be in some sort of basic solution 00:14:18.300 --> 00:14:20.740 compared to its pKa in order for it 00:14:20.740 --> 00:14:24.450 to be visible, in order for it to glow. 00:14:24.450 --> 00:14:28.020 Since our pKa of our fluorescent molecule 00:14:28.020 --> 00:14:31.320 and the pH of the cells exactly match up, 00:14:31.320 --> 00:14:34.530 that means that our probe is half protonated, half 00:14:34.530 --> 00:14:36.160 deprotonated. 00:14:36.160 --> 00:14:37.920 This is a huge problem for our labeling, 00:14:37.920 --> 00:14:41.050 because it immediately means that we're at 50% efficiency. 00:14:41.050 --> 00:14:43.990 We couldn't get more than half of these molecules to glow. 00:14:43.990 --> 00:14:46.390 And we couldn't see more than half of our proteins. 00:14:46.390 --> 00:14:49.087 Our solution was to change the pKa of the probe. 00:14:49.087 --> 00:14:50.920 So we changed to a different molecule that's 00:14:50.920 --> 00:14:54.440 very, very similar and we hoped would work with our system 00:14:54.440 --> 00:14:57.610 but has a much lower pKa, 3.5, which 00:14:57.610 --> 00:15:00.650 allows us to work through all these neutral pHs 00:15:00.650 --> 00:15:04.070 and these physiological pHs and still be very bright 00:15:04.070 --> 00:15:07.300 and still be completely deprotonated. 00:15:07.300 --> 00:15:10.100 A lot of diseases are caused by mutations, 00:15:10.100 --> 00:15:12.460 which change where proteins go. 00:15:12.460 --> 00:15:14.910 It either locks them into a specific compartment 00:15:14.910 --> 00:15:18.290 or puts them in places that they don't need to be. 00:15:18.290 --> 00:15:21.630 I hope that my work can be used by other people 00:15:21.630 --> 00:15:23.590 to study their own systems. 00:15:23.590 --> 00:15:26.080 They can use my process to label that protein 00:15:26.080 --> 00:15:28.920 and then look at disease cells and find out 00:15:28.920 --> 00:15:31.450 where that protein is and what it's doing. 00:15:31.450 --> 00:15:33.360 Hopefully, this can be used to unlock 00:15:33.360 --> 00:15:36.410 the keys to new therapeutic methods and medicine. 00:15:36.410 --> 00:15:38.690 [END PLAYBACK] 00:15:38.690 --> 00:15:41.860 PROFESSOR: So that was Samuel-- he graduated; 00:15:41.860 --> 00:15:45.950 he's now at UCSF in graduate school-- 00:15:45.950 --> 00:15:49.470 and talking about his research in Alice Ting's lab. 00:15:49.470 --> 00:15:52.600 Samuel was my academic advisee. 00:15:52.600 --> 00:15:54.020 He's a chemistry major. 00:15:54.020 --> 00:15:56.200 And then in the later part of graduate school, 00:15:56.200 --> 00:15:58.570 he actually switched and did research in my lab. 00:15:58.570 --> 00:16:01.000 And so that was one of my favorite videos. 00:16:01.000 --> 00:16:02.650 And I miss Samuel. 00:16:02.650 --> 00:16:06.360 Anyway so it shows an undergraduate, just like you, 00:16:06.360 --> 00:16:11.110 caring about pKa's So that's why it's one of my favorites. 00:16:11.110 --> 00:16:14.870 So let's do a couple of clicker questions related 00:16:14.870 --> 00:16:17.550 to Samuel's video. 00:16:17.550 --> 00:16:21.370 And so now, consider, what if the pKa of the probe 00:16:21.370 --> 00:16:23.130 had been 10? 00:16:23.130 --> 00:16:27.490 How much of it would glow at physiological pH? 00:16:27.490 --> 00:16:30.045 What can you say about that probe? 00:16:41.720 --> 00:16:42.460 10 more seconds. 00:16:58.920 --> 00:17:01.170 So let's take a look at the answer. 00:17:01.170 --> 00:17:05.990 So most of it-- not much of it is going to be glowing 00:17:05.990 --> 00:17:07.660 is the answer to that part. 00:17:07.660 --> 00:17:11.250 And let's just look at why that is true for a minute. 00:17:11.250 --> 00:17:17.020 So this is what Samuel talked about in his video 00:17:17.020 --> 00:17:22.089 that, at pH 7-- the physiological pH, 00:17:22.089 --> 00:17:25.520 they had one probe where the pKa was equal to the pH they 00:17:25.520 --> 00:17:26.640 were using. 00:17:26.640 --> 00:17:29.800 And so as Samuel told you, if we think 00:17:29.800 --> 00:17:32.050 about Henderson-Hasselbalch, that's 00:17:32.050 --> 00:17:35.810 going to mean that the ratio of HA to A 00:17:35.810 --> 00:17:37.440 is going to be equal to 1. 00:17:37.440 --> 00:17:38.710 It's one to one. 00:17:38.710 --> 00:17:42.680 And so that's going to mean only 50% maximum efficiency. 00:17:42.680 --> 00:17:46.070 And so here you see that when the pH is equal to the pKa, 00:17:46.070 --> 00:17:50.150 you have equal moles of HA as you have A minus. 00:17:50.150 --> 00:17:53.280 So you're not going to have more-- you can't possibly 00:17:53.280 --> 00:17:58.450 have more than 50% efficiency, 50% more glowing. 00:17:58.450 --> 00:18:00.820 But then in their design strategy, 00:18:00.820 --> 00:18:04.050 what they did was they used another molecule 00:18:04.050 --> 00:18:06.960 with a different pKa, one of 3.5. 00:18:06.960 --> 00:18:10.465 And now, the pH is much greater than the pKa. 00:18:10.465 --> 00:18:14.500 And so as you go above, pH above the pKa, 00:18:14.500 --> 00:18:18.150 you get more and more deprotonated species. 00:18:18.150 --> 00:18:20.820 So if you use Henderson-Hasselbalch here, 00:18:20.820 --> 00:18:24.410 you could calculate the ratio is 1 to 8,000. 00:18:24.410 --> 00:18:27.890 So you're going to have a lot of glowing probe 00:18:27.890 --> 00:18:30.300 under those circumstances. 00:18:30.300 --> 00:18:32.870 And then the clicker question that I just asked you is, 00:18:32.870 --> 00:18:36.070 what happens then if your pKa was 10? 00:18:36.070 --> 00:18:41.470 So now, you have a situation where the pH is below the pKa. 00:18:41.470 --> 00:18:44.750 And if we did the math, we would see 00:18:44.750 --> 00:18:49.830 that the ratio now of protonated to deprotonated is 400 to 1, 00:18:49.830 --> 00:18:51.800 so very little glowing. 00:18:51.800 --> 00:18:54.050 So this is an important thing to think about 00:18:54.050 --> 00:18:56.460 in doing these other kinds of problems. 00:18:56.460 --> 00:19:00.790 When the pH equals the pKa, you have equal amounts 00:19:00.790 --> 00:19:04.940 of HA and A minus; pHs above, more deprotonated; 00:19:04.940 --> 00:19:08.510 pHs below, more protonated. 00:19:08.510 --> 00:19:14.010 So let's try one more clicker question. 00:19:14.010 --> 00:19:20.870 And see now thinking about the pKa's of three different groups 00:19:20.870 --> 00:19:25.040 here for this amino acid, which structure would you get? 00:19:25.040 --> 00:19:30.475 Should this amino group be protonated or deprotonated, NH3 00:19:30.475 --> 00:19:35.800 or NH2, OH or O minus, and OH or O minus here? 00:19:46.460 --> 00:19:47.794 Let's just take 10 more seconds. 00:19:47.794 --> 00:19:49.251 I know a lot of people aren't done, 00:19:49.251 --> 00:19:50.990 but we have a demo that we need to get to 00:19:50.990 --> 00:19:52.630 and a t-shirt vote to get to. 00:20:03.450 --> 00:20:05.139 Not bad. 00:20:05.139 --> 00:20:06.620 [LAUGHTER] 00:20:06.620 --> 00:20:07.960 Oh, we have no answer to it. 00:20:07.960 --> 00:20:14.940 But it is D. So this is the structure. 00:20:14.940 --> 00:20:18.220 So here, for these two, we have pKa's that are high. 00:20:18.220 --> 00:20:23.210 They're above the pH, so we expect them to be protonated. 00:20:23.210 --> 00:20:25.870 And here, we have a pKa that's below the pH, 00:20:25.870 --> 00:20:28.510 so we expect that to be deprotonated. 00:20:28.510 --> 00:20:30.270 So again, you want to think about, how 00:20:30.270 --> 00:20:35.130 does the pKa relate to the pH. 00:20:35.130 --> 00:20:40.590 So that is now the end of the acid-bases, 00:20:40.590 --> 00:20:43.420 end of Exam 3 material.