1 00:00:00,030 --> 00:00:02,400 The following content is provided under a Creative 2 00:00:02,400 --> 00:00:03,780 Commons license. 3 00:00:03,780 --> 00:00:06,020 Your support will help MIT OpenCourseWare 4 00:00:06,020 --> 00:00:10,090 continue to offer high quality educational resources for free. 5 00:00:10,090 --> 00:00:12,660 To make a donation or to view additional materials 6 00:00:12,660 --> 00:00:16,580 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,580 --> 00:00:17,250 at ocw.mit.edu. 8 00:00:25,872 --> 00:00:28,580 PROFESSOR: So again, the equivalence point 9 00:00:28,580 --> 00:00:33,660 is where you've added all of the moles of your strong base 10 00:00:33,660 --> 00:00:36,240 that you need to convert all the moles you 11 00:00:36,240 --> 00:00:39,803 had of the weak acid to its conjugate base. 12 00:00:43,180 --> 00:00:46,610 So if we have this type of problem, 13 00:00:46,610 --> 00:00:49,780 a strong base titrating a weak acid, 14 00:00:49,780 --> 00:00:52,600 the pH is going to be greater than 7 15 00:00:52,600 --> 00:00:55,660 at this equivalence or stoichiometric point. 16 00:00:55,660 --> 00:00:57,090 And we can see that in the plot. 17 00:00:57,090 --> 00:00:59,880 Here is pH 7, little arrow going up 18 00:00:59,880 --> 00:01:02,590 indicating it's going to be greater than 7. 19 00:01:02,590 --> 00:01:06,080 And that's because we have just the conjugate base left 20 00:01:06,080 --> 00:01:07,630 at this point. 21 00:01:07,630 --> 00:01:11,170 So again, pH depends on the property 22 00:01:11,170 --> 00:01:15,070 of the salt that's formed at the equivalence point. 23 00:01:15,070 --> 00:01:20,620 And when it is a weak acid being titrated with a strong base-- 24 00:01:20,620 --> 00:01:23,940 and that's what we have here, weak acid, strong base-- 25 00:01:23,940 --> 00:01:25,970 you're going to get a salt and water. 26 00:01:25,970 --> 00:01:30,600 But this salt, now, is going to have basic properties. 27 00:01:30,600 --> 00:01:35,440 And we saw this before that sodium has no effect on pH. 28 00:01:35,440 --> 00:01:38,780 Things in group 1 are not going to have any effect on pH. 29 00:01:38,780 --> 00:01:42,810 But HCO2- is a conjugate base of a weak acid. 30 00:01:42,810 --> 00:01:44,700 It is a weak base itself. 31 00:01:44,700 --> 00:01:46,580 So this is going to be basic. 32 00:01:46,580 --> 00:01:50,420 And so that's why the pH is going to be greater than 7. 33 00:01:50,420 --> 00:01:52,290 Remember, salt and water problems 34 00:01:52,290 --> 00:01:55,690 really break down to weak acid and water or weak base 35 00:01:55,690 --> 00:01:57,820 and water problems, depending on what 36 00:01:57,820 --> 00:02:00,760 went into a form that particular salt. 37 00:02:00,760 --> 00:02:03,180 So when you're doing these problems, 38 00:02:03,180 --> 00:02:06,300 and if you're on the exam and you get to the end 39 00:02:06,300 --> 00:02:09,539 and you know that the pH should be greater than 7, 40 00:02:09,539 --> 00:02:11,695 but the pH you calculated, for some reason 41 00:02:11,695 --> 00:02:14,000 and you don't know where you made the mistake, 42 00:02:14,000 --> 00:02:15,440 is less than 7. 43 00:02:15,440 --> 00:02:18,120 If you say to me, this doesn't make sense. 44 00:02:18,120 --> 00:02:21,690 It should be greater than 7, because it should 45 00:02:21,690 --> 00:02:23,460 be basic at this point, should just 46 00:02:23,460 --> 00:02:26,300 have conjugate base around, you will get points back 47 00:02:26,300 --> 00:02:32,360 for recognizing that the answer you gave me can't be right. 48 00:02:32,360 --> 00:02:35,200 So always pay attention to what answer you're getting. 49 00:02:35,200 --> 00:02:37,470 Does that answer make sense? 50 00:02:37,470 --> 00:02:40,850 I really care more about that people understand what's 51 00:02:40,850 --> 00:02:42,990 going on, than they can do the math perfectly 52 00:02:42,990 --> 00:02:46,210 in a very short amount of time. 53 00:02:46,210 --> 00:02:48,560 But you'll often be asked to calculate the pH. 54 00:02:48,560 --> 00:02:50,940 So let's think about how we would calculate the pH. 55 00:02:50,940 --> 00:02:53,260 We know it's greater than 7, but what is it exactly 56 00:02:53,260 --> 00:02:55,020 for this problem. 57 00:02:55,020 --> 00:02:56,840 And to do that, we need to first know 58 00:02:56,840 --> 00:02:58,660 what the volume is going to be. 59 00:02:58,660 --> 00:03:01,100 What volume of that strong base do we 60 00:03:01,100 --> 00:03:05,130 need to add to reach the stoichiometric or equivalence 61 00:03:05,130 --> 00:03:06,130 point? 62 00:03:06,130 --> 00:03:09,450 So we had in the beginning 2.5 times 10 63 00:03:09,450 --> 00:03:13,190 to the minus third moles of our weak acid. 64 00:03:13,190 --> 00:03:15,990 So that means that at the stoichiometric point, 65 00:03:15,990 --> 00:03:19,200 we're going to form that number of moles of conjugate base. 66 00:03:19,200 --> 00:03:20,900 And that also means that to do that we 67 00:03:20,900 --> 00:03:24,560 need to add that number of moles of the strong base. 68 00:03:24,560 --> 00:03:28,250 So we need to add 2.5 times 10 to the minus third moles 69 00:03:28,250 --> 00:03:29,960 of our strong base. 70 00:03:29,960 --> 00:03:34,150 We know the concentration in the strong base, 0.15 molar. 71 00:03:34,150 --> 00:03:36,420 So we can calculate that the volume we need 72 00:03:36,420 --> 00:03:39,390 is 16.7 milliliters. 73 00:03:39,390 --> 00:03:43,370 Now, the total volume to get to the stoichiometric point 74 00:03:43,370 --> 00:03:46,310 was 25-- that's what we had originally-- 75 00:03:46,310 --> 00:03:56,730 plus this 16.74, so 41.7 milliliters or 0.0417 liters. 76 00:03:56,730 --> 00:04:01,190 We can calculate the molarity of the conjugate base that's 77 00:04:01,190 --> 00:04:02,020 formed. 78 00:04:02,020 --> 00:04:06,090 How many moles in this total volume gives us a molarity 79 00:04:06,090 --> 00:04:09,740 of 0.0600 molar. 80 00:04:09,740 --> 00:04:12,670 And now, we can go ahead and solve the problem. 81 00:04:12,670 --> 00:04:15,274 But why don't you tell me how we're 82 00:04:15,274 --> 00:04:16,399 going to solve the problem. 83 00:04:16,399 --> 00:04:18,493 What are we going to use to solve this problem? 84 00:04:33,826 --> 00:04:35,075 Let's just do 10 more seconds. 85 00:04:50,880 --> 00:04:58,690 Yep, so Kb, and so if we look at this problem for a minute, 86 00:04:58,690 --> 00:05:04,690 so this is, again, the type of problem, 87 00:05:04,690 --> 00:05:07,130 it's a weak base problem. 88 00:05:07,130 --> 00:05:09,090 We've converted all the weak acid 89 00:05:09,090 --> 00:05:11,490 we had to its conjugate base, because we 90 00:05:11,490 --> 00:05:13,440 added enough moles of the strong base 91 00:05:13,440 --> 00:05:16,150 to convert all the weak acid to the conjugate base. 92 00:05:16,150 --> 00:05:19,150 So the equivalence point, all we have is the weak base. 93 00:05:19,150 --> 00:05:21,610 So this is a base and water problem. 94 00:05:21,610 --> 00:05:25,210 So we write our base plus water going to the conjugate acid 95 00:05:25,210 --> 00:05:26,810 plus hydroxide. 96 00:05:26,810 --> 00:05:28,610 And when it's a base and water problem, 97 00:05:28,610 --> 00:05:31,100 you should have hydroxide on the other side. 98 00:05:31,100 --> 00:05:32,920 So we can set up this expression. 99 00:05:32,920 --> 00:05:37,400 We get 0.06 molar minus x, x plus x. 100 00:05:37,400 --> 00:05:40,650 And we can use Kb to solve the problem. 101 00:05:40,650 --> 00:05:44,750 And Kb, in this case, is 5.6 times 10 to the minus 11th. 102 00:05:44,750 --> 00:05:52,910 That's going to be equal to x squared over 0.0600 minus x. 103 00:05:52,910 --> 00:05:56,540 So if I was given Ka and I now need Kb, 104 00:05:56,540 --> 00:06:00,616 what do I use to convert Ka and Kb? 105 00:06:00,616 --> 00:06:03,460 AUDIENCE: [INAUDIBLE]. 106 00:06:03,460 --> 00:06:04,350 PROFESSOR: Right. 107 00:06:04,350 --> 00:06:07,470 And you can use Kw to solve it, yep. 108 00:06:07,470 --> 00:06:10,460 So we can easily convert between these two. 109 00:06:10,460 --> 00:06:12,770 Not a problem. 110 00:06:12,770 --> 00:06:14,501 So this is how you would do the problem. 111 00:06:14,501 --> 00:06:16,750 And always remember, ask yourself what type of problem 112 00:06:16,750 --> 00:06:17,290 it is. 113 00:06:17,290 --> 00:06:20,530 If it's weak base and water, you want a Kb. 114 00:06:20,530 --> 00:06:23,200 So can I use Henderson-Hasselbalch for this? 115 00:06:23,200 --> 00:06:23,930 AUDIENCE: No. 116 00:06:23,930 --> 00:06:26,260 PROFESSOR: No, I can't. 117 00:06:26,260 --> 00:06:29,060 And you should not as well. 118 00:06:29,060 --> 00:06:30,680 So what do you do after this? 119 00:06:30,680 --> 00:06:32,300 We should be able to go from here. 120 00:06:32,300 --> 00:06:34,600 We're not going to go through all the steps. 121 00:06:34,600 --> 00:06:38,710 We can simplify, pretend x is small, make sure x is small. 122 00:06:38,710 --> 00:06:42,780 And in this case, x is quite small of 1.83 times 10 123 00:06:42,780 --> 00:06:45,000 to the minus sixth molar. 124 00:06:45,000 --> 00:06:47,380 From that, we have to remember that x, now, 125 00:06:47,380 --> 00:06:50,410 is the hydroxide ion concentrations. 126 00:06:50,410 --> 00:06:55,500 So we're going to calculate pOH first and then calculate pH 127 00:06:55,500 --> 00:07:01,730 by subtract 14.00 minus pOH to get us the pH. 128 00:07:01,730 --> 00:07:03,680 And now, it's 8.26. 129 00:07:03,680 --> 00:07:04,960 That's above 7. 130 00:07:04,960 --> 00:07:06,670 That number makes sense. 131 00:07:06,670 --> 00:07:08,350 If I had stopped and forgotten what 132 00:07:08,350 --> 00:07:12,290 x was and realized the pH was 5.74, 133 00:07:12,290 --> 00:07:15,950 I should have realized there was a problem there. 134 00:07:15,950 --> 00:07:18,470 So weak base in water problem. 135 00:07:21,090 --> 00:07:22,070 So we go up here. 136 00:07:22,070 --> 00:07:25,190 That's Point S, the stoichiometric point. 137 00:07:25,190 --> 00:07:28,030 We have a pH of 8.26. 138 00:07:28,030 --> 00:07:29,947 Now, we've added this little E at the end. 139 00:07:29,947 --> 00:07:32,280 We're just going to think about the last type of problem 140 00:07:32,280 --> 00:07:36,020 you might see, which is past the equivalence point. 141 00:07:36,020 --> 00:07:39,870 So here, you are at a volume beyond the volume 142 00:07:39,870 --> 00:07:42,580 needed to get you to the equivalence point. 143 00:07:42,580 --> 00:07:46,770 And at this point, you have your conjugate base in solution. 144 00:07:46,770 --> 00:07:52,040 But you're adding a lot of concentrated base of sodium 145 00:07:52,040 --> 00:07:54,730 hydroxide to that solution. 146 00:07:54,730 --> 00:07:58,720 So the amount of the conjugate base that you have there 147 00:07:58,720 --> 00:08:01,450 is really not going to contribute to pH anymore, 148 00:08:01,450 --> 00:08:04,320 not compared to the pH change that's 149 00:08:04,320 --> 00:08:07,860 going to be caused by just putting extra strong base 150 00:08:07,860 --> 00:08:09,140 into solution. 151 00:08:09,140 --> 00:08:13,420 So the pH is really going to be determined by the excess amount 152 00:08:13,420 --> 00:08:15,160 of any OH you have. 153 00:08:15,160 --> 00:08:17,790 And you can pretty much forget all the work you just 154 00:08:17,790 --> 00:08:21,422 did in calculating what the pH was due to the conjugate base, 155 00:08:21,422 --> 00:08:23,380 because that's going to be overwhelmed by this. 156 00:08:23,380 --> 00:08:25,380 And I'll show you that that's true. 157 00:08:25,380 --> 00:08:30,330 So what this is then is a strong base in water problem. 158 00:08:30,330 --> 00:08:32,450 So beyond the equivalence point now, 159 00:08:32,450 --> 00:08:34,820 we're a strong base and water problem. 160 00:08:34,820 --> 00:08:35,860 So how do we do this? 161 00:08:35,860 --> 00:08:37,570 And we saw this already, but we'll just 162 00:08:37,570 --> 00:08:39,470 review it one more time. 163 00:08:39,470 --> 00:08:43,840 So we're 5 mills past the equivalence point. 164 00:08:43,840 --> 00:08:47,280 So we're going to figure out how many extra moles of OH 165 00:08:47,280 --> 00:08:48,580 we added. 166 00:08:48,580 --> 00:08:51,320 5 mills times our concentration, so we 167 00:08:51,320 --> 00:08:56,380 have 7.5 times 10 to the minus fourth moles that are extra. 168 00:08:56,380 --> 00:09:00,710 And now, we need to calculate the concentration. 169 00:09:00,710 --> 00:09:03,230 And we have to remember our volume. 170 00:09:03,230 --> 00:09:05,970 So we added 5 extra mills. 171 00:09:05,970 --> 00:09:08,090 We had 25 to begin with. 172 00:09:08,090 --> 00:09:12,280 And we use 16.7 to get to the equivalence point. 173 00:09:12,280 --> 00:09:14,540 And so when you have the whole volume in there, 174 00:09:14,540 --> 00:09:19,580 you can calculate that your concentration is 0.016 molar 175 00:09:19,580 --> 00:09:20,600 OH. 176 00:09:20,600 --> 00:09:24,230 Then from that, we can calculate pOH. 177 00:09:24,230 --> 00:09:27,140 And from that, we calculate pH. 178 00:09:27,140 --> 00:09:30,340 And it's 12.21. 179 00:09:30,340 --> 00:09:33,130 And just to convince you that it was OK 180 00:09:33,130 --> 00:09:36,180 that I forgot all about that conjugate base-- 181 00:09:36,180 --> 00:09:37,870 remember, the concentration that we 182 00:09:37,870 --> 00:09:43,840 had calculated of OH that is due to that weak base in solution? 183 00:09:43,840 --> 00:09:47,420 This number, really small compared to that number. 184 00:09:47,420 --> 00:09:49,880 And if you want to be very particular, 185 00:09:49,880 --> 00:09:53,236 you can add this to this before calculating this. 186 00:09:53,236 --> 00:09:54,860 But it's really not going to give you-- 187 00:09:54,860 --> 00:09:57,060 to the number of significant figures you have, 188 00:09:57,060 --> 00:09:59,420 it's not going to make any difference whatsoever. 189 00:09:59,420 --> 00:10:02,500 So this is a strong base in water problem. 190 00:10:02,500 --> 00:10:05,970 We're only going to think about how many moles extra of OH 191 00:10:05,970 --> 00:10:08,350 do we have, and what is the total volume, 192 00:10:08,350 --> 00:10:11,650 and then you're done. 193 00:10:11,650 --> 00:10:14,590 So now, we've done this whole curve. 194 00:10:14,590 --> 00:10:15,940 We started at the beginning. 195 00:10:15,940 --> 00:10:18,860 It was a weak acid problem down here. 196 00:10:18,860 --> 00:10:20,850 We went into the buffering region. 197 00:10:20,850 --> 00:10:23,260 We can use Henderson-Hasselbalch here. 198 00:10:23,260 --> 00:10:25,810 We can do a very simple calculation 199 00:10:25,810 --> 00:10:29,460 at the half equivalence point, pH equals pKa. 200 00:10:29,460 --> 00:10:31,370 Then at the stoichiometric point, 201 00:10:31,370 --> 00:10:33,300 we're a weak base problem. 202 00:10:33,300 --> 00:10:36,980 And then we're a strong base problem. 203 00:10:36,980 --> 00:10:39,920 So you can now do the same thing the other way. 204 00:10:39,920 --> 00:10:40,684 Yeah, question. 205 00:10:40,684 --> 00:10:44,925 AUDIENCE: What if you were given a problem like, 206 00:10:44,925 --> 00:10:46,865 figure out what the buffering region was? 207 00:10:46,865 --> 00:10:48,805 We never calculated E. 208 00:10:48,805 --> 00:10:50,620 PROFESSOR: Yes, right. 209 00:10:50,620 --> 00:10:54,330 So we skipped E, because there was a lot of different points. 210 00:10:54,330 --> 00:11:03,560 Yeah, so if you are in a region where 211 00:11:03,560 --> 00:11:06,220 you have both conjugate acid, conjugate bass, 212 00:11:06,220 --> 00:11:08,180 you can assume that's in the buffering region. 213 00:11:08,180 --> 00:11:10,550 So you need to have both to be in the buffering region, 214 00:11:10,550 --> 00:11:14,270 if the problem only has your weak acid in water 215 00:11:14,270 --> 00:11:17,020 and initially you have sort of zero of the other. 216 00:11:17,020 --> 00:11:19,940 But if you've added some of the strong acid, strong base, 217 00:11:19,940 --> 00:11:21,600 that means you've converted some, 218 00:11:21,600 --> 00:11:24,620 but you know you're not at the equivalence point yet. 219 00:11:24,620 --> 00:11:26,120 You can assume a buffer problem. 220 00:11:26,120 --> 00:11:30,190 And then pretty much, it's kind of in the buffering region, 221 00:11:30,190 --> 00:11:32,040 unless it's right here or right there. 222 00:11:32,040 --> 00:11:34,250 So pretty much anywhere in here, you 223 00:11:34,250 --> 00:11:36,310 can assume it's going to be a buffer problem. 224 00:11:36,310 --> 00:11:37,820 And when you do the subtraction, you 225 00:11:37,820 --> 00:11:40,994 should see that you had your weak acid. 226 00:11:40,994 --> 00:11:42,785 And you've converted some to the conjugate. 227 00:11:42,785 --> 00:11:45,280 And you can see that you have amounts of both 228 00:11:45,280 --> 00:11:47,090 when you do that subtraction. 229 00:11:47,090 --> 00:11:49,080 And when you have concentrations of both, 230 00:11:49,080 --> 00:11:50,580 then you're in the buffering region. 231 00:11:50,580 --> 00:11:52,284 AUDIENCE: So you would go from B to F? 232 00:11:52,284 --> 00:11:55,060 PROFESSOR: So you would do D the same way 233 00:11:55,060 --> 00:11:58,080 that you did B. Yeah, that would be the same. 234 00:11:58,080 --> 00:12:01,175 And then if you want to make sure that you're not there-- 235 00:12:01,175 --> 00:12:02,800 but you should know that when you start 236 00:12:02,800 --> 00:12:07,010 the problem, because when you start doing these problems, 237 00:12:07,010 --> 00:12:09,840 you're going to calculate, say, how much moles of the weak acid 238 00:12:09,840 --> 00:12:10,540 you had. 239 00:12:10,540 --> 00:12:13,815 And then you calculate how many moles of the base you added. 240 00:12:13,815 --> 00:12:15,440 And if they're equal, then you're like, 241 00:12:15,440 --> 00:12:17,170 oh, I'm not in the buffering region. 242 00:12:17,170 --> 00:12:18,496 I'm at the equivalence point. 243 00:12:18,496 --> 00:12:20,870 If they're not equal, if the number of moles you've added 244 00:12:20,870 --> 00:12:23,510 are lesser the strong base but non-zero, 245 00:12:23,510 --> 00:12:25,010 then you're in the buffering region. 246 00:12:28,690 --> 00:12:30,314 Any other good questions? 247 00:12:30,314 --> 00:12:32,230 And we're not going to go the other direction. 248 00:12:32,230 --> 00:12:34,479 We're not going to do all the points in the other one. 249 00:12:34,479 --> 00:12:37,380 But there's a problem set for that. 250 00:12:37,380 --> 00:12:40,850 But I'm not going to leave acid-base quite yet. 251 00:12:40,850 --> 00:12:42,700 We're almost there, but not quite yet, 252 00:12:42,700 --> 00:12:46,550 because I've got to say a little more about pKa's. 253 00:12:46,550 --> 00:12:52,570 So pKa's are not just important in titration problems. 254 00:12:52,570 --> 00:12:56,160 And I want to share with you one of my favorite videos 255 00:12:56,160 --> 00:12:58,705 about why pKa's are important. 256 00:13:05,689 --> 00:13:06,355 [VIDEO PLAYBACK] 257 00:13:06,355 --> 00:13:08,320 - My name is Samuel Thompson. 258 00:13:08,320 --> 00:13:09,774 I'm a rising senior at MIT. 259 00:13:09,774 --> 00:13:11,190 And for the past three years, I've 260 00:13:11,190 --> 00:13:13,090 been working with Alice Ting. 261 00:13:13,090 --> 00:13:15,660 My project is in the field of chemical biology, which 262 00:13:15,660 --> 00:13:17,490 means that I'm a tool maker. 263 00:13:17,490 --> 00:13:19,290 And I've been making tools to allow 264 00:13:19,290 --> 00:13:20,790 people to look at proteins. 265 00:13:20,790 --> 00:13:22,900 We want to see where they are, what they're doing, 266 00:13:22,900 --> 00:13:25,410 what they're interacting with. 267 00:13:25,410 --> 00:13:27,360 But all proteins and all cells are 268 00:13:27,360 --> 00:13:29,110 completely transparent whenever you look 269 00:13:29,110 --> 00:13:31,790 at them under a microscope. 270 00:13:31,790 --> 00:13:35,090 So we use an enzyme to attach an organic molecule-- a very, very 271 00:13:35,090 --> 00:13:37,760 small molecule-- to the protein, so that we get 272 00:13:37,760 --> 00:13:38,940 the same fluorescent output. 273 00:13:38,940 --> 00:13:41,410 We see a bright light whenever we shine laser light on it 274 00:13:41,410 --> 00:13:44,330 and look at it under a microscope. 275 00:13:44,330 --> 00:13:46,100 After some very complicated research, 276 00:13:46,100 --> 00:13:49,210 we came up with a very basic problem in our design. 277 00:13:49,210 --> 00:13:53,170 We had a mismatch between our probe and the pH of the cells. 278 00:13:53,170 --> 00:13:57,989 Cells typically exist at about 7.2 to 7.5 pH. 279 00:13:57,989 --> 00:13:59,280 That's what's healthy for them. 280 00:13:59,280 --> 00:13:59,905 They need that. 281 00:13:59,905 --> 00:14:01,640 And if you go outside those boundaries, 282 00:14:01,640 --> 00:14:03,890 they're very unhealthy, and they behave abnormally. 283 00:14:03,890 --> 00:14:07,770 And the pKa of our probe is also 7.5. 284 00:14:07,770 --> 00:14:10,670 pKa is the point where any sort of species 285 00:14:10,670 --> 00:14:13,580 is half protonated, half deprotonated. 286 00:14:13,580 --> 00:14:16,070 And our probe needs to be deprotonated. 287 00:14:16,070 --> 00:14:18,300 It needs to be in some sort of basic solution 288 00:14:18,300 --> 00:14:20,740 compared to its pKa in order for it 289 00:14:20,740 --> 00:14:24,450 to be visible, in order for it to glow. 290 00:14:24,450 --> 00:14:28,020 Since our pKa of our fluorescent molecule 291 00:14:28,020 --> 00:14:31,320 and the pH of the cells exactly match up, 292 00:14:31,320 --> 00:14:34,530 that means that our probe is half protonated, half 293 00:14:34,530 --> 00:14:36,160 deprotonated. 294 00:14:36,160 --> 00:14:37,920 This is a huge problem for our labeling, 295 00:14:37,920 --> 00:14:41,050 because it immediately means that we're at 50% efficiency. 296 00:14:41,050 --> 00:14:43,990 We couldn't get more than half of these molecules to glow. 297 00:14:43,990 --> 00:14:46,390 And we couldn't see more than half of our proteins. 298 00:14:46,390 --> 00:14:49,087 Our solution was to change the pKa of the probe. 299 00:14:49,087 --> 00:14:50,920 So we changed to a different molecule that's 300 00:14:50,920 --> 00:14:54,440 very, very similar and we hoped would work with our system 301 00:14:54,440 --> 00:14:57,610 but has a much lower pKa, 3.5, which 302 00:14:57,610 --> 00:15:00,650 allows us to work through all these neutral pHs 303 00:15:00,650 --> 00:15:04,070 and these physiological pHs and still be very bright 304 00:15:04,070 --> 00:15:07,300 and still be completely deprotonated. 305 00:15:07,300 --> 00:15:10,100 A lot of diseases are caused by mutations, 306 00:15:10,100 --> 00:15:12,460 which change where proteins go. 307 00:15:12,460 --> 00:15:14,910 It either locks them into a specific compartment 308 00:15:14,910 --> 00:15:18,290 or puts them in places that they don't need to be. 309 00:15:18,290 --> 00:15:21,630 I hope that my work can be used by other people 310 00:15:21,630 --> 00:15:23,590 to study their own systems. 311 00:15:23,590 --> 00:15:26,080 They can use my process to label that protein 312 00:15:26,080 --> 00:15:28,920 and then look at disease cells and find out 313 00:15:28,920 --> 00:15:31,450 where that protein is and what it's doing. 314 00:15:31,450 --> 00:15:33,360 Hopefully, this can be used to unlock 315 00:15:33,360 --> 00:15:36,410 the keys to new therapeutic methods and medicine. 316 00:15:36,410 --> 00:15:38,690 [END PLAYBACK] 317 00:15:38,690 --> 00:15:41,860 PROFESSOR: So that was Samuel-- he graduated; 318 00:15:41,860 --> 00:15:45,950 he's now at UCSF in graduate school-- 319 00:15:45,950 --> 00:15:49,470 and talking about his research in Alice Ting's lab. 320 00:15:49,470 --> 00:15:52,600 Samuel was my academic advisee. 321 00:15:52,600 --> 00:15:54,020 He's a chemistry major. 322 00:15:54,020 --> 00:15:56,200 And then in the later part of graduate school, 323 00:15:56,200 --> 00:15:58,570 he actually switched and did research in my lab. 324 00:15:58,570 --> 00:16:01,000 And so that was one of my favorite videos. 325 00:16:01,000 --> 00:16:02,650 And I miss Samuel. 326 00:16:02,650 --> 00:16:06,360 Anyway so it shows an undergraduate, just like you, 327 00:16:06,360 --> 00:16:11,110 caring about pKa's So that's why it's one of my favorites. 328 00:16:11,110 --> 00:16:14,870 So let's do a couple of clicker questions related 329 00:16:14,870 --> 00:16:17,550 to Samuel's video. 330 00:16:17,550 --> 00:16:21,370 And so now, consider, what if the pKa of the probe 331 00:16:21,370 --> 00:16:23,130 had been 10? 332 00:16:23,130 --> 00:16:27,490 How much of it would glow at physiological pH? 333 00:16:27,490 --> 00:16:30,045 What can you say about that probe? 334 00:16:41,720 --> 00:16:42,460 10 more seconds. 335 00:16:58,920 --> 00:17:01,170 So let's take a look at the answer. 336 00:17:01,170 --> 00:17:05,990 So most of it-- not much of it is going to be glowing 337 00:17:05,990 --> 00:17:07,660 is the answer to that part. 338 00:17:07,660 --> 00:17:11,250 And let's just look at why that is true for a minute. 339 00:17:11,250 --> 00:17:17,020 So this is what Samuel talked about in his video 340 00:17:17,020 --> 00:17:22,089 that, at pH 7-- the physiological pH, 341 00:17:22,089 --> 00:17:25,520 they had one probe where the pKa was equal to the pH they 342 00:17:25,520 --> 00:17:26,640 were using. 343 00:17:26,640 --> 00:17:29,800 And so as Samuel told you, if we think 344 00:17:29,800 --> 00:17:32,050 about Henderson-Hasselbalch, that's 345 00:17:32,050 --> 00:17:35,810 going to mean that the ratio of HA to A 346 00:17:35,810 --> 00:17:37,440 is going to be equal to 1. 347 00:17:37,440 --> 00:17:38,710 It's one to one. 348 00:17:38,710 --> 00:17:42,680 And so that's going to mean only 50% maximum efficiency. 349 00:17:42,680 --> 00:17:46,070 And so here you see that when the pH is equal to the pKa, 350 00:17:46,070 --> 00:17:50,150 you have equal moles of HA as you have A minus. 351 00:17:50,150 --> 00:17:53,280 So you're not going to have more-- you can't possibly 352 00:17:53,280 --> 00:17:58,450 have more than 50% efficiency, 50% more glowing. 353 00:17:58,450 --> 00:18:00,820 But then in their design strategy, 354 00:18:00,820 --> 00:18:04,050 what they did was they used another molecule 355 00:18:04,050 --> 00:18:06,960 with a different pKa, one of 3.5. 356 00:18:06,960 --> 00:18:10,465 And now, the pH is much greater than the pKa. 357 00:18:10,465 --> 00:18:14,500 And so as you go above, pH above the pKa, 358 00:18:14,500 --> 00:18:18,150 you get more and more deprotonated species. 359 00:18:18,150 --> 00:18:20,820 So if you use Henderson-Hasselbalch here, 360 00:18:20,820 --> 00:18:24,410 you could calculate the ratio is 1 to 8,000. 361 00:18:24,410 --> 00:18:27,890 So you're going to have a lot of glowing probe 362 00:18:27,890 --> 00:18:30,300 under those circumstances. 363 00:18:30,300 --> 00:18:32,870 And then the clicker question that I just asked you is, 364 00:18:32,870 --> 00:18:36,070 what happens then if your pKa was 10? 365 00:18:36,070 --> 00:18:41,470 So now, you have a situation where the pH is below the pKa. 366 00:18:41,470 --> 00:18:44,750 And if we did the math, we would see 367 00:18:44,750 --> 00:18:49,830 that the ratio now of protonated to deprotonated is 400 to 1, 368 00:18:49,830 --> 00:18:51,800 so very little glowing. 369 00:18:51,800 --> 00:18:54,050 So this is an important thing to think about 370 00:18:54,050 --> 00:18:56,460 in doing these other kinds of problems. 371 00:18:56,460 --> 00:19:00,790 When the pH equals the pKa, you have equal amounts 372 00:19:00,790 --> 00:19:04,940 of HA and A minus; pHs above, more deprotonated; 373 00:19:04,940 --> 00:19:08,510 pHs below, more protonated. 374 00:19:08,510 --> 00:19:14,010 So let's try one more clicker question. 375 00:19:14,010 --> 00:19:20,870 And see now thinking about the pKa's of three different groups 376 00:19:20,870 --> 00:19:25,040 here for this amino acid, which structure would you get? 377 00:19:25,040 --> 00:19:30,475 Should this amino group be protonated or deprotonated, NH3 378 00:19:30,475 --> 00:19:35,800 or NH2, OH or O minus, and OH or O minus here? 379 00:19:46,460 --> 00:19:47,794 Let's just take 10 more seconds. 380 00:19:47,794 --> 00:19:49,251 I know a lot of people aren't done, 381 00:19:49,251 --> 00:19:50,990 but we have a demo that we need to get to 382 00:19:50,990 --> 00:19:52,630 and a t-shirt vote to get to. 383 00:20:03,450 --> 00:20:05,139 Not bad. 384 00:20:05,139 --> 00:20:06,620 [LAUGHTER] 385 00:20:06,620 --> 00:20:07,960 Oh, we have no answer to it. 386 00:20:07,960 --> 00:20:14,940 But it is D. So this is the structure. 387 00:20:14,940 --> 00:20:18,220 So here, for these two, we have pKa's that are high. 388 00:20:18,220 --> 00:20:23,210 They're above the pH, so we expect them to be protonated. 389 00:20:23,210 --> 00:20:25,870 And here, we have a pKa that's below the pH, 390 00:20:25,870 --> 00:20:28,510 so we expect that to be deprotonated. 391 00:20:28,510 --> 00:20:30,270 So again, you want to think about, how 392 00:20:30,270 --> 00:20:35,130 does the pKa relate to the pH. 393 00:20:35,130 --> 00:20:40,590 So that is now the end of the acid-bases, 394 00:20:40,590 --> 00:20:43,420 end of Exam 3 material.