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 --> 00:00:16,405 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,405 --> 00:00:17,030 at ocw.mit.edu. 8 00:00:25,744 --> 00:00:27,160 CATHERINE DRENNAN: So where are we 9 00:00:27,160 --> 00:00:30,270 in our grand list of our five types of problems? 10 00:00:30,270 --> 00:00:33,090 We've done weak acid in water, weak base in water, 11 00:00:33,090 --> 00:00:34,610 salt, and buffer. 12 00:00:34,610 --> 00:00:37,172 We only have two left. 13 00:00:37,172 --> 00:00:38,630 And that's what we're going to talk 14 00:00:38,630 --> 00:00:42,950 about right now-- strong acids and strong bases. 15 00:00:42,950 --> 00:00:46,570 And we're going to talk about them in terms of acid base 16 00:00:46,570 --> 00:00:47,760 titrations. 17 00:00:47,760 --> 00:00:51,930 So switching to today's handout, so 18 00:00:51,930 --> 00:00:54,220 titrations-- how many of you have done 19 00:00:54,220 --> 00:00:57,040 a titration in a laboratory? 20 00:00:57,040 --> 00:00:59,660 OK, a lot of people. 21 00:00:59,660 --> 00:01:03,840 If you haven't, I have to say, you're not missing anything. 22 00:01:03,840 --> 00:01:07,030 It is really-- it's like watching 23 00:01:07,030 --> 00:01:11,270 paint dry for most of the time during the titration. 24 00:01:11,270 --> 00:01:15,260 So here in an acid base type titration, 25 00:01:15,260 --> 00:01:17,660 you have the addition of, say, a base 26 00:01:17,660 --> 00:01:23,110 of known concentration to an acid of unknown concentration. 27 00:01:23,110 --> 00:01:25,940 And you can use this by figuring out 28 00:01:25,940 --> 00:01:29,880 what volume you need to reach the equivalence point. 29 00:01:29,880 --> 00:01:32,140 You can figure out the concentration 30 00:01:32,140 --> 00:01:33,800 of that acid or base. 31 00:01:33,800 --> 00:01:36,080 If you know the concentration of two things you, 32 00:01:36,080 --> 00:01:38,270 can often use it to figure out a Ka 33 00:01:38,270 --> 00:01:40,680 or calculate a molecular weight. 34 00:01:40,680 --> 00:01:43,920 But that's the purpose of the titration. 35 00:01:43,920 --> 00:01:47,330 So let's look at the shape of these titration curves. 36 00:01:47,330 --> 00:01:50,430 So here we have a strong acid with a strong base. 37 00:01:50,430 --> 00:01:51,380 So we start out. 38 00:01:51,380 --> 00:01:54,310 All we have at point 0 is the strong acid. 39 00:01:54,310 --> 00:01:57,530 So we should have a very low pH here. 40 00:01:57,530 --> 00:01:59,610 And then it should change slowly. 41 00:01:59,610 --> 00:02:04,880 Then it changes very rapidly and then changes slowly again. 42 00:02:04,880 --> 00:02:07,260 So what is this point S? 43 00:02:07,260 --> 00:02:10,030 So that point S has many, many names, which 44 00:02:10,030 --> 00:02:11,240 are all equivalent. 45 00:02:11,240 --> 00:02:13,330 It's called the equivalence point, 46 00:02:13,330 --> 00:02:16,410 and we had the clicker question in the beginning 47 00:02:16,410 --> 00:02:17,760 was mentioning that. 48 00:02:17,760 --> 00:02:22,000 The S here really stands for the stoichiometric point. 49 00:02:22,000 --> 00:02:23,730 Both of these are the same thing. 50 00:02:23,730 --> 00:02:25,920 And it's the theoretical volume and which 51 00:02:25,920 --> 00:02:28,620 that base that you've added, the number of moles of that base 52 00:02:28,620 --> 00:02:30,980 is equal to the number of moles of acid you had. 53 00:02:30,980 --> 00:02:34,700 So it's the point of which the moles are equal to each other. 54 00:02:34,700 --> 00:02:38,680 And so that stoichiometric 1-to-1, or equivalence. 55 00:02:38,680 --> 00:02:40,260 The moles are equivalent. 56 00:02:40,260 --> 00:02:41,942 So all these names are the same. 57 00:02:41,942 --> 00:02:43,150 They all mean the same thing. 58 00:02:43,150 --> 00:02:46,050 Don't get tricked if problem say one thing or the other. 59 00:02:46,050 --> 00:02:48,200 You also might hear end point. 60 00:02:48,200 --> 00:02:52,860 That's often the result of the experiment that you do, 61 00:02:52,860 --> 00:02:55,240 where you figure out what volume you needed 62 00:02:55,240 --> 00:02:56,920 to add to have a color change. 63 00:02:56,920 --> 00:02:59,490 So the way these work usually is you have this, 64 00:02:59,490 --> 00:03:02,780 and you go drip, drip, drip, and you look for a color change. 65 00:03:02,780 --> 00:03:04,680 And at that point of the color change, 66 00:03:04,680 --> 00:03:07,900 the first little hint of the color change 67 00:03:07,900 --> 00:03:10,700 is where you want to stop and measure the volume. 68 00:03:10,700 --> 00:03:12,360 But often you get very frustrated, 69 00:03:12,360 --> 00:03:15,430 and you go from totally clear, you're very, very slow, 70 00:03:15,430 --> 00:03:16,660 and you're right in here. 71 00:03:16,660 --> 00:03:18,160 And then you can't stand it anymore. 72 00:03:18,160 --> 00:03:19,660 You jump right up here, and you have 73 00:03:19,660 --> 00:03:21,130 to do it over and over again. 74 00:03:21,130 --> 00:03:23,820 But it should be that the end point should 75 00:03:23,820 --> 00:03:25,440 equal the equivalence point. 76 00:03:25,440 --> 00:03:27,550 So if you're good at these kinds of problems, 77 00:03:27,550 --> 00:03:30,690 you might be able to do a quick and dirty, estimate what 78 00:03:30,690 --> 00:03:33,380 the value is, and then figure out exactly how much you can 79 00:03:33,380 --> 00:03:36,730 add before you have to go slow, i.e. you can cheat, 80 00:03:36,730 --> 00:03:39,870 which I highly recommend in doing these kinds of laboratory 81 00:03:39,870 --> 00:03:40,890 exercises. 82 00:03:40,890 --> 00:03:42,840 So knowing how to do the math is really 83 00:03:42,840 --> 00:03:45,810 valuable of these problems. 84 00:03:45,810 --> 00:03:48,530 So we add a strong acid and a strong base. 85 00:03:48,530 --> 00:03:51,060 Here's the same curve for a strong base 86 00:03:51,060 --> 00:03:54,450 with a strong acid, where we're adding the strong acid 87 00:03:54,450 --> 00:03:56,550 into the strong base. 88 00:03:56,550 --> 00:03:59,770 Again, now we're going to start in the beginning at point 0 89 00:03:59,770 --> 00:04:01,330 before we've added any acid. 90 00:04:01,330 --> 00:04:02,640 We have a basic pH. 91 00:04:02,640 --> 00:04:05,310 It goes slow, goes fast through point 92 00:04:05,310 --> 00:04:08,590 S, the equivalence point or stoichiometric point, 93 00:04:08,590 --> 00:04:10,470 and then goes slow again. 94 00:04:10,470 --> 00:04:11,980 So how do you know? 95 00:04:11,980 --> 00:04:14,630 And the TSs are going to come down and do 96 00:04:14,630 --> 00:04:18,579 a little demo for you now about pH indicators. 97 00:04:18,579 --> 00:04:19,380 How do you know? 98 00:04:19,380 --> 00:04:25,640 And for those of you who have done these experiments in lab, 99 00:04:25,640 --> 00:04:27,610 again, you have this color change. 100 00:04:27,610 --> 00:04:29,180 And usually you're clear. 101 00:04:29,180 --> 00:04:30,502 You go a little bit more. 102 00:04:30,502 --> 00:04:32,460 You have a little bit more, and you go to this. 103 00:04:32,460 --> 00:04:34,490 But as I was saying, what happens most of time 104 00:04:34,490 --> 00:04:37,110 is you go from this clear, all the way to the really dark, 105 00:04:37,110 --> 00:04:39,110 and you have to start all over again because you 106 00:04:39,110 --> 00:04:41,380 missed the end point. 107 00:04:41,380 --> 00:04:45,520 So these pH indicators are able to tell you 108 00:04:45,520 --> 00:04:48,530 that there's this change in pH. 109 00:04:48,530 --> 00:04:56,011 So the TAs are going to show you, using a pH-- come on in. 110 00:04:56,011 --> 00:04:56,510 Set up. 111 00:04:56,510 --> 00:04:59,810 I'm just talking while you set up. 112 00:04:59,810 --> 00:05:03,480 Over there-- you can use the document camera if you need it. 113 00:05:03,480 --> 00:05:12,860 So our demo TAs have created a natural dye from red cabbage. 114 00:05:12,860 --> 00:05:17,110 And this dye, when it's in acid, has a slightly different 115 00:05:17,110 --> 00:05:20,080 structure then in base, which causes it to have these two 116 00:05:20,080 --> 00:05:21,490 different colors. 117 00:05:21,490 --> 00:05:24,350 So we can tell which things are acidic 118 00:05:24,350 --> 00:05:27,910 and which things are basic by the color of the indicator. 119 00:05:27,910 --> 00:05:35,342 And I'm just going to hand over a mic if it's on. 120 00:05:35,342 --> 00:05:36,550 TEACHING ASSISTANT: Is it on? 121 00:05:36,550 --> 00:05:37,530 Yes? 122 00:05:37,530 --> 00:05:38,030 Yes? 123 00:05:38,030 --> 00:05:38,860 OK, great. 124 00:05:38,860 --> 00:05:42,900 So we've got our solutions here. 125 00:05:42,900 --> 00:05:46,070 This is lemon juice. 126 00:05:46,070 --> 00:05:47,070 Did I put water in here? 127 00:05:47,070 --> 00:05:48,420 Right? 128 00:05:48,420 --> 00:05:51,110 Yeah, water, vinegar, I think. 129 00:05:51,110 --> 00:05:53,866 Maybe it vinegar-- this is vinegar. 130 00:05:56,697 --> 00:05:57,280 Don't do that. 131 00:05:57,280 --> 00:05:58,130 Don't do that in lab. 132 00:05:58,130 --> 00:05:59,120 CATHERINE DRENNAN: In the lab you should always 133 00:05:59,120 --> 00:06:00,807 label your containers. 134 00:06:00,807 --> 00:06:02,340 TEACHING ASSISTANT: Yes, indeed. 135 00:06:02,340 --> 00:06:06,725 So we've got lemon juice, water, vinegar, and ammonia. 136 00:06:06,725 --> 00:06:08,100 Did you add some water into that? 137 00:06:08,100 --> 00:06:09,330 Can you? 138 00:06:09,330 --> 00:06:12,130 So we're going to add a little bit of the indicator. 139 00:06:12,130 --> 00:06:14,310 The indicator came from red cabbage. 140 00:06:14,310 --> 00:06:17,170 So the dye in the red cabbage, it 141 00:06:17,170 --> 00:06:19,240 is a molecule that can be protonated. 142 00:06:19,240 --> 00:06:22,930 And when the molecule gets protonated, 143 00:06:22,930 --> 00:06:24,930 basically it's double bonds start moving around. 144 00:06:24,930 --> 00:06:28,700 And when you do that, it absorbs light differently. 145 00:06:28,700 --> 00:06:32,180 So what we're going to see when we pour into there 146 00:06:32,180 --> 00:06:36,150 is that you'll see this, which is the acid forms 147 00:06:36,150 --> 00:06:38,960 a nice pinkish color. 148 00:06:38,960 --> 00:06:39,752 This was the water. 149 00:06:39,752 --> 00:06:42,251 And when you pour it into the water, nothing really happens. 150 00:06:42,251 --> 00:06:44,800 And you can kind of understand that because the indicator 151 00:06:44,800 --> 00:06:46,840 itself is a water-based solution. 152 00:06:46,840 --> 00:06:48,370 So it's going to be purple. 153 00:06:48,370 --> 00:06:49,450 This was the vinegar. 154 00:06:49,450 --> 00:06:54,520 And as we did last time, the vinegar is acidic. 155 00:06:54,520 --> 00:06:56,450 And this was the ammonia. 156 00:06:56,450 --> 00:06:58,210 And I'm going to add little bit of water. 157 00:06:58,210 --> 00:06:59,420 Can you add a little bit of water into this one? 158 00:06:59,420 --> 00:06:59,961 You added it? 159 00:06:59,961 --> 00:07:01,090 OK, great. 160 00:07:01,090 --> 00:07:04,910 And then we're going to kind of do a little excursion here. 161 00:07:04,910 --> 00:07:07,635 So what this, this is a piece of-- oh, 162 00:07:07,635 --> 00:07:12,750 it's cold-- a piece of dry ice, which is just solid CO2. 163 00:07:12,750 --> 00:07:15,730 And what I'm going to do is I'm going to drop it into there. 164 00:07:15,730 --> 00:07:18,744 So this is kind of a bit of a-- 165 00:07:18,744 --> 00:07:20,035 CATHERINE DRENNAN: [INAUDIBLE]? 166 00:07:20,035 --> 00:07:21,826 TEACHING ASSISTANT: Yeah, put it over here. 167 00:07:21,826 --> 00:07:24,520 So this is an interesting demo because-- oh, that's not useful 168 00:07:24,520 --> 00:07:25,020 at all. 169 00:07:25,020 --> 00:07:27,061 CATHERINE DRENNAN: Yeah, actually we can change-- 170 00:07:27,061 --> 00:07:28,429 AUDIENCE: Can we change that? 171 00:07:28,429 --> 00:07:29,470 CATHERINE DRENNAN: Maybe. 172 00:07:29,470 --> 00:07:30,130 TEACHING ASSISTANT: Oh, maybe not. 173 00:07:30,130 --> 00:07:30,620 Oh, well, it's fine. 174 00:07:30,620 --> 00:07:31,070 Oh, man. 175 00:07:31,070 --> 00:07:31,110 CATHERINE DRENNAN: CATHERINE DRENNAN: 176 00:07:31,110 --> 00:07:32,120 I think we can do that. 177 00:07:32,120 --> 00:07:33,870 But this is beyond my ability [INAUDIBLE]. 178 00:07:33,870 --> 00:07:35,953 TEACHING ASSISTANT: Well, what I was talking about 179 00:07:35,953 --> 00:07:38,880 was basically when you drop the CO2 into this mixture, what's 180 00:07:38,880 --> 00:07:41,685 happening is that the CO2 actually decomposes, 181 00:07:41,685 --> 00:07:44,630 and it forms something called carbonic acid. 182 00:07:44,630 --> 00:07:46,920 Good idea. 183 00:07:46,920 --> 00:07:50,740 Carbonic acid is an acid, so it will actually 184 00:07:50,740 --> 00:07:54,600 change the pH of the solution as we're talking. 185 00:07:54,600 --> 00:07:58,250 And what this is relevant for is really for global warming. 186 00:07:58,250 --> 00:08:01,800 So we know about companies when they make-- I mean, it's true. 187 00:08:01,800 --> 00:08:05,160 When we have companies and they make 188 00:08:05,160 --> 00:08:07,650 lots of kinds of chemical compounds, 189 00:08:07,650 --> 00:08:09,770 they'll release CO2 into the atmosphere. 190 00:08:09,770 --> 00:08:12,997 And when the CO2 gets released-- is it working or no? 191 00:08:16,899 --> 00:08:18,190 CATHERINE DRENNAN: Upside down! 192 00:08:18,190 --> 00:08:19,149 That's even cooler. 193 00:08:19,149 --> 00:08:20,690 TEACHING ASSISTANT: Oh, look at that. 194 00:08:20,690 --> 00:08:23,535 So when the CO2 gets released into the atmosphere, 195 00:08:23,535 --> 00:08:25,160 basically, by Le Chatelier's principle, 196 00:08:25,160 --> 00:08:26,802 you add all this CO2. 197 00:08:26,802 --> 00:08:29,690 It will force the reaction to start making carbonic acid. 198 00:08:29,690 --> 00:08:31,910 And that will react with the water vapor 199 00:08:31,910 --> 00:08:37,210 and start making the water in the air very acidic. 200 00:08:37,210 --> 00:08:38,929 Wow, that's not useful at all. 201 00:08:38,929 --> 00:08:40,640 And the color's not really changing. 202 00:08:40,640 --> 00:08:42,477 But it was supposed to. 203 00:08:42,477 --> 00:08:43,177 (LAUGHING) 204 00:08:43,177 --> 00:08:44,510 CATHERINE DRENNAN: We have time. 205 00:08:44,510 --> 00:08:45,180 We can keep an eye on it. 206 00:08:45,180 --> 00:08:45,640 TEACHING ASSISTANT: I think we may have added too much base. 207 00:08:45,640 --> 00:08:48,100 But we can keep an eye on it and watch the color change 208 00:08:48,100 --> 00:08:49,620 while it's upside down, which is really useful. 209 00:08:49,620 --> 00:08:51,085 CATHERINE DRENNAN: Yes, because that's really cool. 210 00:08:51,085 --> 00:08:51,750 TEACHING ASSISTANT: But anyway, yeah, that's 211 00:08:51,750 --> 00:08:52,930 how the indicator works. 212 00:08:52,930 --> 00:08:56,010 And, yeah, that's that, I guess. 213 00:08:56,010 --> 00:08:57,300 CATHERINE DRENNAN: OK, great. 214 00:08:57,300 --> 00:08:57,815 [APPLAUSE] 215 00:08:57,815 --> 00:08:59,690 CATHERINE DRENNAN: So we can leave this here. 216 00:08:59,690 --> 00:09:03,700 I'll switch the document the camera back. 217 00:09:03,700 --> 00:09:09,030 And you might add a little more. 218 00:09:09,030 --> 00:09:11,530 It's very concentrated still. 219 00:09:11,530 --> 00:09:16,730 See if we can get a little bit more without too much overflow. 220 00:09:16,730 --> 00:09:18,630 I think it's changing a little bit actually. 221 00:09:18,630 --> 00:09:19,129 Yeah. 222 00:09:22,670 --> 00:09:25,560 Well, we'll keep an eye on it. 223 00:09:25,560 --> 00:09:30,120 So let's go through and calculate some points 224 00:09:30,120 --> 00:09:32,390 on a strong acid and base curve. 225 00:09:32,390 --> 00:09:39,420 So we can check off those two final types of pH problems. 226 00:09:39,420 --> 00:09:44,000 So here, when we're calculating strong acid and strong bases, 227 00:09:44,000 --> 00:09:49,490 say we have a strong base that's 2.5 molar, NaOH 228 00:09:49,490 --> 00:09:57,570 being traded with a strong acid, HCl at 0.34 molar. 229 00:09:57,570 --> 00:10:00,370 So first, we want to calculate the pH 230 00:10:00,370 --> 00:10:03,030 before the equivalence point, when we've just added, 231 00:10:03,030 --> 00:10:08,250 say, 5 mLs of our strong acid to the solution which 232 00:10:08,250 --> 00:10:12,720 was 25 mLs of our strong base. 233 00:10:12,720 --> 00:10:14,280 So the first thing we're going to do 234 00:10:14,280 --> 00:10:17,790 is calculate how many moles of base that is. 235 00:10:17,790 --> 00:10:20,390 So how many did we have in here? 236 00:10:20,390 --> 00:10:22,440 And since this base is strong, the number 237 00:10:22,440 --> 00:10:24,660 of moles of sodium hydroxide added 238 00:10:24,660 --> 00:10:28,730 equals the amount of hydroxide formed. 239 00:10:28,730 --> 00:10:30,820 That's the definition of a strong base. 240 00:10:30,820 --> 00:10:34,990 And it is hydroxide, so pretty much that's how much you get. 241 00:10:34,990 --> 00:10:37,230 So all we have to do is know how to calculate 242 00:10:37,230 --> 00:10:38,630 the number of moles. 243 00:10:38,630 --> 00:10:41,480 And that is the volume that we had, 244 00:10:41,480 --> 00:10:45,860 and we're converting that to liters times our concentration. 245 00:10:45,860 --> 00:10:51,200 And so that gives us 6.25 times 10 to the minus 3 moles. 246 00:10:51,200 --> 00:10:54,060 So we know how many moles of OH we have. 247 00:10:54,060 --> 00:10:56,630 Now we need to figure out how many moles of acid 248 00:10:56,630 --> 00:11:01,310 we added when we added 5 milliliters of the acid. 249 00:11:01,310 --> 00:11:03,160 So it's a strong acid. 250 00:11:03,160 --> 00:11:06,290 So the moles of acid added equal the number 251 00:11:06,290 --> 00:11:10,900 of moles of our hydronium ions that are formed-- definition 252 00:11:10,900 --> 00:11:12,360 of strong acid. 253 00:11:12,360 --> 00:11:15,080 And so we had 5 milliliters. 254 00:11:15,080 --> 00:11:18,160 We convert to liters times our concentration. 255 00:11:18,160 --> 00:11:23,470 That gives us 1.7 times 10 to the minus 3 moles. 256 00:11:23,470 --> 00:11:26,750 So now we need to figure out what 257 00:11:26,750 --> 00:11:29,830 the new amount of hydroxide is. 258 00:11:29,830 --> 00:11:32,690 So we're going to convert. 259 00:11:32,690 --> 00:11:36,250 Some of our hydroxide will react with the amount of acid 260 00:11:36,250 --> 00:11:39,640 added because the stoichiometry is one-to-one, 261 00:11:39,640 --> 00:11:45,250 you're going to react equal numbers of OH with H3O-plus. 262 00:11:45,250 --> 00:11:49,460 So we had 6.25 times 10 to the minus 3 moles. 263 00:11:49,460 --> 00:11:51,900 We've added this number of moles of acid. 264 00:11:51,900 --> 00:11:57,870 So that many moles are going to be formed into salt. 265 00:11:57,870 --> 00:12:02,700 And we're left with this minus this, or 4.55 times 266 00:12:02,700 --> 00:12:06,580 10 to the minus 3 moles of OH-minus left. 267 00:12:06,580 --> 00:12:10,300 So at this point, this is a strong base and water problem. 268 00:12:10,300 --> 00:12:14,770 All we have as hydroxide ions left in our solution. 269 00:12:14,770 --> 00:12:17,870 We've converted all the acid that we 270 00:12:17,870 --> 00:12:21,500 added into salt because it reacted with the base. 271 00:12:21,500 --> 00:12:23,760 And so this is a strong base now. 272 00:12:23,760 --> 00:12:28,250 So for strong base, all we have to do is calculate the molarity 273 00:12:28,250 --> 00:12:30,940 and then calculate pOH and pH. 274 00:12:30,940 --> 00:12:34,190 So for molarity, We have the number 275 00:12:34,190 --> 00:12:40,380 of moles left, 4.55 times 10 to the minus 3, in our new volume. 276 00:12:40,380 --> 00:12:42,340 And this is the part where people forget. 277 00:12:42,340 --> 00:12:43,880 We had 25 mLs. 278 00:12:43,880 --> 00:12:45,090 We added five more. 279 00:12:45,090 --> 00:12:47,920 So our new volume is 30 mLs. 280 00:12:47,920 --> 00:12:50,470 And so we have a new concentration. 281 00:12:50,470 --> 00:12:53,010 And then from that, we calculate pH. 282 00:12:53,010 --> 00:12:59,440 First we do pOh, plugging in our hydroxide ion concentration. 283 00:12:59,440 --> 00:13:02,959 and then calculate pH, using the fact 284 00:13:02,959 --> 00:13:05,000 that at room temperature-- and all these problems 285 00:13:05,000 --> 00:13:09,440 are room temperature-- that would be 14 minus the pOH 286 00:13:09,440 --> 00:13:11,070 is the pH. 287 00:13:11,070 --> 00:13:14,980 And we have a point somewhere around here, point B, 288 00:13:14,980 --> 00:13:19,460 on our titration curve after we've add 5 mLs of strong acid. 289 00:13:19,460 --> 00:13:24,910 Our pH is still pretty basic-- 13.18 left. 290 00:13:24,910 --> 00:13:28,420 So strong base in water problem-- all you have to do 291 00:13:28,420 --> 00:13:33,310 is calculate the concentration of OH-minus, then the pOH, 292 00:13:33,310 --> 00:13:34,730 and then the pH. 293 00:13:34,730 --> 00:13:38,300 So you don't have to have any of these minus and pluses in terms 294 00:13:38,300 --> 00:13:41,720 of setting up our tables and using Ka's and Kb's 295 00:13:41,720 --> 00:13:43,780 For a weak base in water problem, 296 00:13:43,780 --> 00:13:50,380 it's a concentration of hydroxide ion-- pOH pH. 297 00:13:50,380 --> 00:13:53,770 So let's move on now and think about what 298 00:13:53,770 --> 00:13:57,970 volume we would need to get to the stoichiometric point. 299 00:13:57,970 --> 00:13:59,870 So you're often asked to calculate 300 00:13:59,870 --> 00:14:01,820 the volume of the strong acid needed 301 00:14:01,820 --> 00:14:05,186 to get to the stoichiometric or the equivalence point. 302 00:14:05,186 --> 00:14:06,560 And to do this, you need to think 303 00:14:06,560 --> 00:14:09,310 about how many moles of OH you have, 304 00:14:09,310 --> 00:14:11,000 because at the equivalence point, 305 00:14:11,000 --> 00:14:13,450 you have equal numbers of moles of acid. 306 00:14:13,450 --> 00:14:18,780 So we had 6.25 times 10 to the minus 3 moles of OH originally. 307 00:14:18,780 --> 00:14:22,780 And at the equivalence point, then, how many moles of acid 308 00:14:22,780 --> 00:14:23,705 do we need to add? 309 00:14:26,230 --> 00:14:30,360 We have to add that exact same number, by definition. 310 00:14:30,360 --> 00:14:33,210 So the equivalence point, it's the stoichiometric point. 311 00:14:33,210 --> 00:14:36,500 So if you know how many moles of the base you had, 312 00:14:36,500 --> 00:14:39,690 that's going to be equal to the moles of acid you need to add. 313 00:14:39,690 --> 00:14:42,670 So now we want to figure out the volume that that's needed. 314 00:14:42,670 --> 00:14:44,790 We know the number of moles that we need, 315 00:14:44,790 --> 00:14:47,320 and we know the concentration, so now we just 316 00:14:47,320 --> 00:14:49,460 have to calculate the volume. 317 00:14:49,460 --> 00:14:50,640 So we have the moles. 318 00:14:50,640 --> 00:14:52,960 We use the concentration of the acid 319 00:14:52,960 --> 00:14:57,440 to figure out the volume that has to be added. 320 00:14:57,440 --> 00:15:01,900 And what would be the pH at this point? 321 00:15:01,900 --> 00:15:02,615 It would be 7. 322 00:15:05,210 --> 00:15:09,570 So this is a really fast question to ask on an exam, 323 00:15:09,570 --> 00:15:12,420 because all you have to do is say, 324 00:15:12,420 --> 00:15:14,910 if it was a strong base in a strong acid, 325 00:15:14,910 --> 00:15:19,797 they're going to form a neutral salt in water, so the pH is 7. 326 00:15:19,797 --> 00:15:20,380 And that's it. 327 00:15:20,380 --> 00:15:21,990 No calculations needed. 328 00:15:21,990 --> 00:15:25,970 You can answer that question very quickly-- so strong acid, 329 00:15:25,970 --> 00:15:30,320 strong base, at the equivalence point, equal number of moles. 330 00:15:30,320 --> 00:15:34,050 You're going to get a salt that's neutral. 331 00:15:34,050 --> 00:15:41,000 In the salt here, we have HCl and NaOH is just NaCl, salt. 332 00:15:41,000 --> 00:15:45,110 So remember, at the equivalence point, strong acid, 333 00:15:45,110 --> 00:15:50,020 strong base, the pH is going to be 7. 334 00:15:50,020 --> 00:15:55,030 So now let's go down here and try point D so just 335 00:15:55,030 --> 00:15:58,450 beyond the equivalence point. 336 00:15:58,450 --> 00:16:03,860 So here we want to calculate the pH after 1 mL of strong acid 337 00:16:03,860 --> 00:16:07,240 has been added after you have reached the equivalence point. 338 00:16:07,240 --> 00:16:10,160 So you went 1 mL too far in your titration, 339 00:16:10,160 --> 00:16:13,090 and this is a strong acid in water problem, 340 00:16:13,090 --> 00:16:16,540 so our last type of problem here. 341 00:16:16,540 --> 00:16:18,280 So first what you want to do is figure 342 00:16:18,280 --> 00:16:25,590 out how many moles of this acid were added due to this 1 mL 343 00:16:25,590 --> 00:16:26,930 extra. 344 00:16:26,930 --> 00:16:29,720 And again, it's a strong acid, so the amount of acid 345 00:16:29,720 --> 00:16:32,830 added is going to be equal to the hydronium ion 346 00:16:32,830 --> 00:16:36,330 concentrations formed-- definition of strong acid. 347 00:16:36,330 --> 00:16:41,420 So we had a concentration of 0.34 of our strong acid. 348 00:16:41,420 --> 00:16:46,580 We added 1 mL extra, so we have 3.4 times 10 349 00:16:46,580 --> 00:16:52,540 to the minus 4 moles of H3O-plus. 350 00:16:52,540 --> 00:16:55,765 Now we need to calculate the molarity of that, 351 00:16:55,765 --> 00:16:57,015 and that's a clicker question. 352 00:17:17,359 --> 00:17:22,780 So the trick here is just to think about what 353 00:17:22,780 --> 00:17:26,140 is the volume at this point. 354 00:17:26,140 --> 00:17:29,804 And if we look over here-- do you 355 00:17:29,804 --> 00:17:31,820 want to just-- you can it out. 356 00:17:31,820 --> 00:17:34,360 We're trying a couple of different variations 357 00:17:34,360 --> 00:17:36,350 of the experiment over here. 358 00:17:36,350 --> 00:17:42,535 But we'll just see if we can get it over time. 359 00:17:42,535 --> 00:17:44,330 Maybe we'll come back to that at the end. 360 00:17:44,330 --> 00:17:46,290 We'll just leave it there and say, OK. 361 00:17:46,290 --> 00:17:49,395 Or did you want to say something? 362 00:17:49,395 --> 00:17:51,520 Why don't you talk about it briefly, and then we'll 363 00:17:51,520 --> 00:17:52,061 look at the-- 364 00:17:52,150 --> 00:17:54,300 TEACHING ASSISTANT: We did the experiment again 365 00:17:54,300 --> 00:17:56,610 with a less concentrated solution, added several pieces 366 00:17:56,610 --> 00:17:57,110 of dry ice. 367 00:17:57,110 --> 00:18:00,364 And as you can see, the color-- yeah, 368 00:18:00,364 --> 00:18:01,530 we can put it on the camera. 369 00:18:01,530 --> 00:18:02,250 I don't know if we can. 370 00:18:02,250 --> 00:18:02,880 It's fine. 371 00:18:02,880 --> 00:18:05,890 But the color's changing very slowly, 372 00:18:05,890 --> 00:18:08,200 which is a lot slower than we expected it to change. 373 00:18:08,200 --> 00:18:10,060 But you dilute the concentration, 374 00:18:10,060 --> 00:18:11,960 and this is turning blue. 375 00:18:11,960 --> 00:18:15,140 And it will eventually turn into a yellowish-pink, 376 00:18:15,140 --> 00:18:17,770 which is when it will officially become an acid. 377 00:18:17,770 --> 00:18:19,062 So we'll just leave that there. 378 00:18:19,062 --> 00:18:20,811 CATHERINE DRENNAN: We'll keep an eye on it 379 00:18:20,811 --> 00:18:21,990 for the next few minutes. 380 00:18:21,990 --> 00:18:25,070 So the trick here was to figure out the volume. 381 00:18:25,070 --> 00:18:27,370 So you had 25 mLs to begin with. 382 00:18:27,370 --> 00:18:30,530 You needed 18.4 to get to the stoichiometric 383 00:18:30,530 --> 00:18:33,050 or the equivalence point. 384 00:18:33,050 --> 00:18:36,450 Then you added one more past the equivalence point. 385 00:18:36,450 --> 00:18:39,160 And if you remembered to add those three volumes together, 386 00:18:39,160 --> 00:18:41,860 then you will get the right concentration. 387 00:18:41,860 --> 00:18:43,630 And when you get the right concentration, 388 00:18:43,630 --> 00:18:46,840 the rest of the problem is very easy. 389 00:18:46,840 --> 00:18:50,680 And I'll just mention also, significant figures here. 390 00:18:50,680 --> 00:18:55,080 We have three here because we had three there. 391 00:18:55,080 --> 00:18:59,230 And so then you just take the pH of this. 392 00:18:59,230 --> 00:19:02,210 This is the hydronium ion concentration. 393 00:19:02,210 --> 00:19:03,550 And that will give your answer. 394 00:19:03,550 --> 00:19:08,147 And it'll be pH 2.116 with all of those significant figures, 395 00:19:08,147 --> 00:19:09,980 which you can never measure that accurately. 396 00:19:09,980 --> 00:19:13,210 But nonetheless, that's what the rules would tell us. 397 00:19:13,210 --> 00:19:16,940 So a strong acid in water problem 398 00:19:16,940 --> 00:19:19,720 is you just calculate the hydronium ion 399 00:19:19,720 --> 00:19:21,490 concentration and then pH. 400 00:19:21,490 --> 00:19:24,170 So this is a pretty simple part. 401 00:19:24,170 --> 00:19:30,310 So if we go back now to our diagram, 402 00:19:30,310 --> 00:19:31,860 we've measured a couple of parts. 403 00:19:31,860 --> 00:19:38,220 We measured point B. And so we had a pH of 13.18. 404 00:19:38,220 --> 00:19:42,070 This was like a strong base in water problem. 405 00:19:42,070 --> 00:19:44,420 At a point S, we calculated that we 406 00:19:44,420 --> 00:19:50,040 needed 18.4 mLs of strong acid, and our pH is 7. 407 00:19:50,040 --> 00:19:53,080 And so the trick to doing this problem 408 00:19:53,080 --> 00:19:55,430 was just to remember that the moles of strong acid 409 00:19:55,430 --> 00:19:58,670 equal the moles of strong base at this point. 410 00:19:58,670 --> 00:20:03,500 And then from point D, 1 mL past the equivalence point, the pH 411 00:20:03,500 --> 00:20:09,820 was 2.116, and this is like a strong acid in water problem. 412 00:20:09,820 --> 00:20:11,800 And notice how dramatic it is. 413 00:20:11,800 --> 00:20:13,260 Up here you're at pH 7. 414 00:20:13,260 --> 00:20:17,490 You add 1 mL too much, and you're already at pH 2. 415 00:20:17,490 --> 00:20:19,610 So this is a very rapid change. 416 00:20:19,610 --> 00:20:23,450 This is why a lot of these experiments are so frustrating. 417 00:20:23,450 --> 00:20:24,940 That you add a little bit too much, 418 00:20:24,940 --> 00:20:27,230 and you're just way, way, way beyond 419 00:20:27,230 --> 00:20:29,780 the stoichiometric point. 420 00:20:29,780 --> 00:20:32,670 So in one titration curve here, we 421 00:20:32,670 --> 00:20:34,630 have three types of problems. 422 00:20:34,630 --> 00:20:37,770 But as we get into weak acid in weak base, which we'll do now, 423 00:20:37,770 --> 00:20:40,720 you'll see that a titration curve is actually 424 00:20:40,720 --> 00:20:43,800 all five types of problems. 425 00:20:43,800 --> 00:20:49,640 So this is why one wants to start problems set 7 early. 426 00:20:49,640 --> 00:20:51,780 So we've done all of these now. 427 00:20:51,780 --> 00:20:55,200 And now we can apply all of these types of problems 428 00:20:55,200 --> 00:20:58,500 while doing weak acid strong base part. 429 00:20:58,500 --> 00:21:00,900 And we'll just start this, and we'll continue it later. 430 00:21:00,900 --> 00:21:04,520 But let's go through the curves now. 431 00:21:04,520 --> 00:21:07,080 So this is in your handout. 432 00:21:07,080 --> 00:21:09,430 This is the curve for the weak acid in strong base. 433 00:21:09,430 --> 00:21:12,300 And I just want to remind you what we just talked about, 434 00:21:12,300 --> 00:21:14,830 which is strong acid in strong base. 435 00:21:14,830 --> 00:21:17,630 So for weak acid in strong base, the curves 436 00:21:17,630 --> 00:21:19,790 look quite a bit different. 437 00:21:19,790 --> 00:21:24,470 So here we start at a higher pH, because it's a weak acid. 438 00:21:24,470 --> 00:21:25,220 Here's lower. 439 00:21:25,220 --> 00:21:26,580 It's a strong acid. 440 00:21:26,580 --> 00:21:28,240 Then we go up. 441 00:21:28,240 --> 00:21:31,320 The pH levels off. 442 00:21:31,320 --> 00:21:34,080 And this is called the buffering region. 443 00:21:34,080 --> 00:21:37,870 There's no such buffering region with a strong acid strong base, 444 00:21:37,870 --> 00:21:42,210 because strong acids and strong bases don't form good buffers. 445 00:21:42,210 --> 00:21:46,120 But weak acids and strong bases can generate good buffering 446 00:21:46,120 --> 00:21:49,180 region because a strong base can convert some of the weak acid 447 00:21:49,180 --> 00:21:51,940 to its conjugate weak base, and that creates 448 00:21:51,940 --> 00:21:53,860 a buffer in this region. 449 00:21:53,860 --> 00:21:55,930 Then the pH starts to change again. 450 00:21:55,930 --> 00:21:59,100 It's constant in the buffering region, starts to change again. 451 00:21:59,100 --> 00:22:00,940 You get to the equivalence point. 452 00:22:00,940 --> 00:22:02,870 The definition of equivalence points 453 00:22:02,870 --> 00:22:06,060 are the same-- equal number of moles of acids 454 00:22:06,060 --> 00:22:08,360 and base at the equivalence point. 455 00:22:08,360 --> 00:22:12,200 But now the equivalence point is not at pH 7 anymore. 456 00:22:12,200 --> 00:22:16,790 It's greater than pH 7 because we've converted our weak acid 457 00:22:16,790 --> 00:22:18,370 to its conjugate base. 458 00:22:18,370 --> 00:22:20,300 So up here we have conjugate base, 459 00:22:20,300 --> 00:22:22,340 which makes it more basic. 460 00:22:22,340 --> 00:22:24,560 And so the pH is greater than 7. 461 00:22:24,560 --> 00:22:27,530 And then the curve goes up again. 462 00:22:27,530 --> 00:22:32,615 We also have something that's special for a weak with strong 463 00:22:32,615 --> 00:22:35,670 titration that doesn't exist with strong -strong, 464 00:22:35,670 --> 00:22:38,370 and that's called the half-equivalence point. 465 00:22:38,370 --> 00:22:40,260 The half-equivalence point is the point 466 00:22:40,260 --> 00:22:43,130 where half the volume and half the number of moles 467 00:22:43,130 --> 00:22:46,690 needed to reach the equivalence point have been added. 468 00:22:46,690 --> 00:22:49,470 So that's the half-equivalence point. 469 00:22:49,470 --> 00:22:52,300 It's half of the equivalence point. 470 00:22:52,300 --> 00:22:56,250 So let's just look at these-- a couple of different points 471 00:22:56,250 --> 00:22:57,530 here. 472 00:22:57,530 --> 00:23:02,550 And look at this figure here, and we'll end with this. 473 00:23:05,630 --> 00:23:07,940 And I'm just going to have my TAs come down 474 00:23:07,940 --> 00:23:12,110 and just help me with this last part. 475 00:23:12,110 --> 00:23:16,110 So let's see if we can get two more. 476 00:23:16,110 --> 00:23:17,640 We already have someone with a hat. 477 00:23:17,640 --> 00:23:19,070 You have a hat. 478 00:23:19,070 --> 00:23:20,597 We have two more TAs. 479 00:23:23,730 --> 00:23:29,030 So these are all weak acids over here. 480 00:23:29,030 --> 00:23:32,390 So we have just our weak acids, and we have four of them. 481 00:23:35,000 --> 00:23:37,200 Other TAs who want to come down, can help out. 482 00:23:37,200 --> 00:23:39,060 This is volume equals 0. 483 00:23:39,060 --> 00:23:41,650 So this is like a weak acid in water problem. 484 00:23:41,650 --> 00:23:46,290 So now let's bring our moles of strong base over here. 485 00:23:46,290 --> 00:23:50,360 So now, Amanda, you're a strong base, so you're 1. 486 00:23:50,360 --> 00:23:56,960 So you're going to react just with one of these weak acids 487 00:23:56,960 --> 00:23:59,300 deprotonate it and form a conjugate base. 488 00:23:59,300 --> 00:23:59,850 There we go. 489 00:23:59,850 --> 00:24:03,260 And then you can go off and form your salt and water. 490 00:24:03,260 --> 00:24:05,780 So now we have a buffer situation. 491 00:24:05,780 --> 00:24:08,150 And we have a buffer situation when 492 00:24:08,150 --> 00:24:12,260 we have some conjugate acid and some conjugate base. 493 00:24:12,260 --> 00:24:15,210 We have a special type of category of buffer 494 00:24:15,210 --> 00:24:17,300 at the half-equivalence point, where 495 00:24:17,300 --> 00:24:20,630 you have half the conjugate S and half the conjugate A. 496 00:24:20,630 --> 00:24:24,020 So another mole of strong base comes in and deprotonates 497 00:24:24,020 --> 00:24:24,880 one of our acid. 498 00:24:24,880 --> 00:24:28,800 So why don't you be-- yeah, and you go off and form a salt. 499 00:24:28,800 --> 00:24:32,310 And so then we have half the weak acid and half 500 00:24:32,310 --> 00:24:34,070 the conjugate base. 501 00:24:34,070 --> 00:24:37,680 Now, as we go on, we're going to reach the equivalence point. 502 00:24:37,680 --> 00:24:40,990 And now we've added an amount of strong base 503 00:24:40,990 --> 00:24:43,850 that will fully deprotonate our weak acid. 504 00:24:43,850 --> 00:24:47,100 So why don't we have two more strong bases come along 505 00:24:47,100 --> 00:24:47,840 and deprotonate. 506 00:24:47,840 --> 00:24:52,890 And so now we have only weak bases left. 507 00:24:52,890 --> 00:24:55,200 There are just weak bases left because we 508 00:24:55,200 --> 00:24:58,830 added equal numbers of moles of our strong base 509 00:24:58,830 --> 00:25:00,890 as we had weak acid. 510 00:25:00,890 --> 00:25:02,670 So now we have conjugate base. 511 00:25:02,670 --> 00:25:05,820 So this is like a weak base in water problem. 512 00:25:05,820 --> 00:25:08,284 And then at the end, if we add more strong base, 513 00:25:08,284 --> 00:25:10,450 and a strong base comes over, there's nothing to do. 514 00:25:10,450 --> 00:25:12,570 He's just strong base in solution. 515 00:25:12,570 --> 00:25:15,320 And so this is a strong base in water problem. 516 00:25:15,320 --> 00:25:18,070 So the point is one titration problem 517 00:25:18,070 --> 00:25:22,730 is really five different types of acid base problems. 518 00:25:22,730 --> 00:25:26,930 So start problem set 7 early. 519 00:25:26,930 --> 00:25:29,080 See you on Wednesday. 520 00:25:32,560 --> 00:25:34,150 Let's just take 10 more seconds. 521 00:25:48,210 --> 00:25:50,430 I have a feeling that more people understand 522 00:25:50,430 --> 00:25:52,610 how to do this, but it's just a matter 523 00:25:52,610 --> 00:25:55,690 of thinking about the sign of things. 524 00:25:55,690 --> 00:26:04,190 So the trick here-- you have a pH of 4, pKa of 4.19. 525 00:26:04,190 --> 00:26:06,540 And so, obviously, it's the difference. 526 00:26:06,540 --> 00:26:09,610 But you have to think about if you 527 00:26:09,610 --> 00:26:14,820 want the ratio to be A minus 2 A, which expression you're 528 00:26:14,820 --> 00:26:18,620 going to use here to give you the correct one. 529 00:26:18,620 --> 00:26:21,720 So just pay attention to your sign. 530 00:26:21,720 --> 00:26:25,520 Always show all your work on an exam. 531 00:26:25,520 --> 00:26:27,330 So today, we're going to finish the lecture 532 00:26:27,330 --> 00:26:29,780 notes from last time on titrations, 533 00:26:29,780 --> 00:26:31,627 continue with titrations. 534 00:26:31,627 --> 00:26:33,210 And toward the end of the class, we're 535 00:26:33,210 --> 00:26:36,990 going to start the next unit, which is oxidation reduction. 536 00:26:36,990 --> 00:26:40,140 So we want to take out lecture note 537 00:26:40,140 --> 00:26:45,770 23, top of page 4, where we had this curve. 538 00:26:45,770 --> 00:26:48,340 So as we were finishing last time, 539 00:26:48,340 --> 00:26:52,000 we saw the curve for a titration of a weak acid 540 00:26:52,000 --> 00:26:53,630 with a strong base. 541 00:26:53,630 --> 00:26:56,530 And this is the curve for a weak base 542 00:26:56,530 --> 00:26:59,310 that's being titrated traded with a strong acid. 543 00:26:59,310 --> 00:27:03,120 And drawing curves on an exams-- that's one of my favorite exam 544 00:27:03,120 --> 00:27:06,850 questions So make sure you know how to draw 545 00:27:06,850 --> 00:27:09,070 these types of titrations. 546 00:27:09,070 --> 00:27:12,600 So things that you should pay attention to in drawing them, 547 00:27:12,600 --> 00:27:15,120 if it is a weak base that's being titrated 548 00:27:15,120 --> 00:27:17,730 with a strong acid, you would expect 549 00:27:17,730 --> 00:27:22,350 that before you've added any acid, that you're 550 00:27:22,350 --> 00:27:25,390 going to have a basic pH. 551 00:27:25,390 --> 00:27:29,950 So at point 0 here, you would expect a basic pH. 552 00:27:32,540 --> 00:27:36,410 So then the pH starts decreasing as you're 553 00:27:36,410 --> 00:27:38,560 titrating in the strong acid. 554 00:27:38,560 --> 00:27:43,000 And you go into this region of constant pH, 555 00:27:43,000 --> 00:27:44,870 where that's the buffering region. 556 00:27:44,870 --> 00:27:46,970 So the pH is changing very little. 557 00:27:46,970 --> 00:27:47,930 It plateaus. 558 00:27:47,930 --> 00:27:49,070 It's flat. 559 00:27:49,070 --> 00:27:51,140 So when you're drawing a region like this, 560 00:27:51,140 --> 00:27:54,100 make sure that in the curve it looks kind of flat, 561 00:27:54,100 --> 00:27:55,710 because the idea of a buffering region 562 00:27:55,710 --> 00:27:57,460 is that the pH isn't changing much. 563 00:27:57,460 --> 00:28:01,340 Here you have the conjugate acid and the conjugate base, 564 00:28:01,340 --> 00:28:03,590 so it's a buffer-like problem. 565 00:28:03,590 --> 00:28:05,560 Then when you get out of that buffering region, 566 00:28:05,560 --> 00:28:07,880 the pH drops rapidly, which is often 567 00:28:07,880 --> 00:28:10,519 one of the funs in doing these titrations in lab. 568 00:28:10,519 --> 00:28:11,310 Nothing's changing. 569 00:28:11,310 --> 00:28:11,980 Nothing's changing. 570 00:28:11,980 --> 00:28:12,771 Nothing's changing. 571 00:28:12,771 --> 00:28:14,630 And all of a sudden, you're down here. 572 00:28:14,630 --> 00:28:17,300 So at the equivalence point here, 573 00:28:17,300 --> 00:28:20,470 the equivalence point is going to be below pH 7 574 00:28:20,470 --> 00:28:23,750 because at this point, you have added the same number of moles 575 00:28:23,750 --> 00:28:25,410 of acid as the weak base. 576 00:28:25,410 --> 00:28:27,410 So you've converted all of the weak base 577 00:28:27,410 --> 00:28:28,890 to its conjugate acid. 578 00:28:28,890 --> 00:28:32,640 So the pH is below 7, and then it drops off rapidly. 579 00:28:32,640 --> 00:28:35,220 So again, in these titration problems, 580 00:28:35,220 --> 00:28:38,870 there are really five different types of problems within them. 581 00:28:38,870 --> 00:28:41,140 When you have the volume of zero, 582 00:28:41,140 --> 00:28:44,130 it's a weak base in water problem. 583 00:28:44,130 --> 00:28:48,270 In this buffering region here, it's a buffer problem. 584 00:28:48,270 --> 00:28:50,550 And there's a special kind of buffer problem 585 00:28:50,550 --> 00:28:52,550 at the half-equivalence point, where 586 00:28:52,550 --> 00:28:55,250 the number of moles of the conjugate acid 587 00:28:55,250 --> 00:28:58,740 formed equal the number of moles of baes. 588 00:28:58,740 --> 00:29:01,020 And then we get to the equivalence point, 589 00:29:01,020 --> 00:29:03,150 and now we've converted all the weak base 590 00:29:03,150 --> 00:29:04,390 to its conjugate acid. 591 00:29:04,390 --> 00:29:07,160 So this is just a weak acid in water problem. 592 00:29:07,160 --> 00:29:10,870 And then down here, it's a strong acid in water problem. 593 00:29:10,870 --> 00:29:13,510 So again, the trick in doing these 594 00:29:13,510 --> 00:29:16,030 is to recognize where you are in the curve 595 00:29:16,030 --> 00:29:18,130 and figure out what type of problem it is. 596 00:29:18,130 --> 00:29:21,070 Once you figure out what type of problem it is, 597 00:29:21,070 --> 00:29:23,240 it's usually not so bad to solve it 598 00:29:23,240 --> 00:29:25,590 because you can all figure out how to solve these five 599 00:29:25,590 --> 00:29:26,770 types of problems. 600 00:29:26,770 --> 00:29:30,520 It's recognizing what type of problem it is at various times 601 00:29:30,520 --> 00:29:31,990 during the titration. 602 00:29:31,990 --> 00:29:33,820 and in some of the problem set problems, 603 00:29:33,820 --> 00:29:35,970 you'll have all of the various points to do. 604 00:29:35,970 --> 00:29:37,470 But in an exam, you're going to just 605 00:29:37,470 --> 00:29:39,300 be thrown into the middle of a titration 606 00:29:39,300 --> 00:29:43,620 and have to think about where it is in the midst of titration 607 00:29:43,620 --> 00:29:46,030 what type of problem is it. 608 00:29:46,030 --> 00:29:49,260 So now we're going to go through and do an example, work 609 00:29:49,260 --> 00:29:52,030 point by point for a titration. 610 00:29:52,030 --> 00:29:54,040 And the titration we're going to do 611 00:29:54,040 --> 00:29:57,750 is a weak acid in a strong base. 612 00:29:57,750 --> 00:29:59,060 So that's the curve here. 613 00:29:59,060 --> 00:30:02,310 So we start at low pH and then go up. 614 00:30:02,310 --> 00:30:04,280 So the particular problem we're going to work-- 615 00:30:04,280 --> 00:30:05,880 and we're going to work all these points. 616 00:30:05,880 --> 00:30:07,630 And then we're going to have a bonus point 617 00:30:07,630 --> 00:30:11,140 E at the end-- is formic acid. 618 00:30:11,140 --> 00:30:14,820 And we have 25 mLs of 0.1 molar formic acid. 619 00:30:14,820 --> 00:30:16,600 And we're titrating with a strong base, 620 00:30:16,600 --> 00:30:19,440 sodium hydroxide, 0.15 molar. 621 00:30:19,440 --> 00:30:24,320 And you're often going to be given the Ka of the acid. 622 00:30:24,320 --> 00:30:28,100 So the first point, point A, is volume zero. 623 00:30:28,100 --> 00:30:32,310 You haven't added any of your strong base yet. 624 00:30:32,310 --> 00:30:36,430 So you can write the expression for what's going on there. 625 00:30:36,430 --> 00:30:39,490 You have your acid plus water going 626 00:30:39,490 --> 00:30:42,750 to hydronium ion concentration, and you're 627 00:30:42,750 --> 00:30:46,510 forming the conjugate base of that weak acid. 628 00:30:46,510 --> 00:30:50,340 So this is, again, a weak acid problem. 629 00:30:50,340 --> 00:30:52,600 So you know how to do weak acid problems. 630 00:30:52,600 --> 00:30:55,330 So titrations are just a collection 631 00:30:55,330 --> 00:30:58,560 of problems we've already talked about how to do. 632 00:30:58,560 --> 00:31:01,820 So just very briefly, we'll go through this pretty fast 633 00:31:01,820 --> 00:31:03,520 at this point. 634 00:31:03,520 --> 00:31:06,140 So we have our weak acid in water. 635 00:31:06,140 --> 00:31:11,400 We had 0.1 molar, and we had none of the other 636 00:31:11,400 --> 00:31:12,800 to begin with. 637 00:31:12,800 --> 00:31:15,340 So we're at zero volume here. 638 00:31:15,340 --> 00:31:19,970 So as the equilibrium is reached, we have 0.10 minus x, 639 00:31:19,970 --> 00:31:22,050 x, and x. 640 00:31:22,050 --> 00:31:25,580 So because this is a weak acid in water problem, 641 00:31:25,580 --> 00:31:29,170 we can use Ka to solve the problem, which is our acid 642 00:31:29,170 --> 00:31:30,920 ionization constant. 643 00:31:30,920 --> 00:31:36,280 It's 1.77 times 10 to the minus 4 for this particular one. 644 00:31:36,280 --> 00:31:40,640 We can set this up it's products over reactants at equilibrium. 645 00:31:40,640 --> 00:31:45,090 So x squared over 0.10 minus x. 646 00:31:45,090 --> 00:31:46,540 Again, water doesn't fit in. 647 00:31:46,540 --> 00:31:49,290 That's our solvent, so it's not in our expression. 648 00:31:49,290 --> 00:31:51,470 We can solve this with the quadratic, 649 00:31:51,470 --> 00:31:53,889 or we can assume that x is small, 650 00:31:53,889 --> 00:31:55,930 which, again, is a pretty good assumption usually 651 00:31:55,930 --> 00:31:58,940 for a weak acid problem. 652 00:31:58,940 --> 00:32:03,440 And by assuming x is small, that means this minus x drops out. 653 00:32:03,440 --> 00:32:06,810 So we're saying x is small compared to the concentration 654 00:32:06,810 --> 00:32:08,330 that we started with. 655 00:32:08,330 --> 00:32:10,430 And then we can solve for x. 656 00:32:10,430 --> 00:32:19,490 If we do that here, we get 0.00421, and that we can check, 657 00:32:19,490 --> 00:32:24,740 and we find that it is 4.2% of 0.1, so that's under 5%. 658 00:32:24,740 --> 00:32:27,200 So that assumption is OK. 659 00:32:27,200 --> 00:32:30,300 Again, I'm using the 5% rule. 660 00:32:30,300 --> 00:32:32,060 And then we can solve this. 661 00:32:32,060 --> 00:32:34,720 x is our hydronium ion concentration. 662 00:32:34,720 --> 00:32:38,010 So we can plug that in and get an answer. 663 00:32:38,010 --> 00:32:41,190 But now I want you to tell me how many sig figs this has. 664 00:32:41,190 --> 00:32:43,060 And to do that, you have to figure out 665 00:32:43,060 --> 00:32:45,914 how many sig figs were actually in this number here. 666 00:32:45,914 --> 00:32:47,830 So you have to go back and look at the problem 667 00:32:47,830 --> 00:32:50,290 and figure out the total number of sig figs. 668 00:33:02,310 --> 00:33:04,560 See how easy that was to fix? 669 00:33:04,560 --> 00:33:07,630 So it is two, because this had two, 670 00:33:07,630 --> 00:33:09,950 and the concentration was limiting here 671 00:33:09,950 --> 00:33:11,800 to two significant figures. 672 00:33:11,800 --> 00:33:13,330 So there were two here, which means 673 00:33:13,330 --> 00:33:16,400 there are two after the decimal point here. 674 00:33:16,400 --> 00:33:18,920 And I have to say I was looking a problem set 7, 675 00:33:18,920 --> 00:33:20,970 and some of those problems are just 676 00:33:20,970 --> 00:33:23,970 like if you love significant figures, 677 00:33:23,970 --> 00:33:27,620 they're a joy because you've got adding and subtracting. 678 00:33:27,620 --> 00:33:31,660 And then you had division, and then you have logs. 679 00:33:31,660 --> 00:33:37,900 It is the triple whammy of sig fig love-- so two 680 00:33:37,900 --> 00:33:41,200 after the decimal point. 681 00:33:41,200 --> 00:33:44,240 So we have 0.1. 682 00:33:44,240 --> 00:33:46,260 We have a pH that's acetic. 683 00:33:46,260 --> 00:33:48,450 That's good because at the beginning, 684 00:33:48,450 --> 00:33:50,380 all we have is a weak acid. 685 00:33:50,380 --> 00:33:54,110 So we have a pH of 2.38, so that's point A. 686 00:33:54,110 --> 00:33:58,080 So now let's look at this region here in the buffering region. 687 00:33:58,080 --> 00:33:59,830 And again, the buffering region is 688 00:33:59,830 --> 00:34:02,120 when you have added volumes of your acid 689 00:34:02,120 --> 00:34:04,760 that are greater than zero but less than the volume needed 690 00:34:04,760 --> 00:34:06,780 to reach equilibrium. 691 00:34:06,780 --> 00:34:08,310 And so in this particular problem, 692 00:34:08,310 --> 00:34:10,070 we're going to calculate what happens 693 00:34:10,070 --> 00:34:13,460 when 5 mLs of our strong base are added. 694 00:34:13,460 --> 00:34:16,219 So this is basically point B here. 695 00:34:16,219 --> 00:34:18,300 And because it's a strong base, it's 696 00:34:18,300 --> 00:34:22,290 going to react with all of the same number of moles 697 00:34:22,290 --> 00:34:23,679 of our weak acid. 698 00:34:23,679 --> 00:34:26,177 It's much stronger than the conjugate base. 699 00:34:26,177 --> 00:34:28,510 So it's going to react pretty much completely with this. 700 00:34:28,510 --> 00:34:32,460 So we can write out that the weak acid plus our base 701 00:34:32,460 --> 00:34:36,760 are going to be converted over to the conjugate base 702 00:34:36,760 --> 00:34:37,920 of this weak acid. 703 00:34:37,920 --> 00:34:41,070 So we can just figure out the moles 704 00:34:41,070 --> 00:34:45,190 and then subtract them to understand how much is left 705 00:34:45,190 --> 00:34:47,320 and how much is being formed. 706 00:34:47,320 --> 00:34:48,900 So first, then, we need to figure out 707 00:34:48,900 --> 00:34:51,020 how many moles we had initially. 708 00:34:51,020 --> 00:34:56,300 So for our acid, we had 25 mLs of 0.10 molar. 709 00:34:56,300 --> 00:35:00,360 So we have 2.5 times 10 to the minus 3 moles. 710 00:35:00,360 --> 00:35:04,770 For our hydroxide, our strong base, we've added 5 mLs. 711 00:35:04,770 --> 00:35:07,890 And it was a 0.15 molar solution. 712 00:35:07,890 --> 00:35:11,460 So we 0.75 times 10 to the minus 3 moles. 713 00:35:11,460 --> 00:35:13,890 So now there's going to be this reaction. 714 00:35:13,890 --> 00:35:15,990 So we want to think about what happens 715 00:35:15,990 --> 00:35:19,420 after you have the reaction of the strong base 716 00:35:19,420 --> 00:35:21,060 with the weak acid. 717 00:35:21,060 --> 00:35:23,710 And so just to emphasize this point, 718 00:35:23,710 --> 00:35:27,720 say we have 8 moles of our acid here 719 00:35:27,720 --> 00:35:31,970 and we added just 1 mole of our strong base. 720 00:35:31,970 --> 00:35:36,160 That will react, pull off, deprotonate, and form water, 721 00:35:36,160 --> 00:35:38,190 and form the conjugate base. 722 00:35:38,190 --> 00:35:41,040 So net we had eight. 723 00:35:41,040 --> 00:35:43,130 We added one of the strong base. 724 00:35:43,130 --> 00:35:44,380 So we have seven left. 725 00:35:44,380 --> 00:35:47,960 And we've formed one of the conjugated base. 726 00:35:47,960 --> 00:35:53,400 And so I'm just showing this here because so often on exams, 727 00:35:53,400 --> 00:35:56,110 I see people just plug the initial moles 728 00:35:56,110 --> 00:35:59,200 into the later part, and they forget that there's 729 00:35:59,200 --> 00:36:00,770 a reaction happening. 730 00:36:00,770 --> 00:36:03,190 And so I was hoping if I would show this, 731 00:36:03,190 --> 00:36:06,840 I'm going to get people used to thinking about, OK, I'm 732 00:36:06,840 --> 00:36:07,900 adding strong base. 733 00:36:07,900 --> 00:36:08,670 What's happening? 734 00:36:08,670 --> 00:36:10,470 How much is being converted? 735 00:36:10,470 --> 00:36:12,680 And if you're not subtracting things, 736 00:36:12,680 --> 00:36:16,040 there's a problem with the way you're doing the problem. 737 00:36:16,040 --> 00:36:20,680 So here, these are less easy numbers to visualize. 738 00:36:20,680 --> 00:36:23,760 2.5 times 10 to the minus 3 moles, how much 739 00:36:23,760 --> 00:36:25,500 we had of our weak acid. 740 00:36:25,500 --> 00:36:30,760 We have added a strong base at 0.75 times 10 to the minus 3. 741 00:36:30,760 --> 00:36:35,660 So we have 1.75 times 10 to the minus 3 moles of our acid left. 742 00:36:35,660 --> 00:36:39,050 So we subtract the amount that's converted due to the addition 743 00:36:39,050 --> 00:36:40,310 of the base. 744 00:36:40,310 --> 00:36:42,860 And so why don't you tell me how much, then, 745 00:36:42,860 --> 00:36:46,510 we're going to have of our conjugate base being formed. 746 00:37:11,759 --> 00:37:13,300 So I kind of tried to fool you there. 747 00:37:13,300 --> 00:37:18,440 But 70% of you were not having it. 748 00:37:18,440 --> 00:37:24,350 So here, it's going to be equal to the number of moles of a OH 749 00:37:24,350 --> 00:37:28,170 that's added because that's how much is going to be converted. 750 00:37:28,170 --> 00:37:31,420 That's how much of the conjugate acid 751 00:37:31,420 --> 00:37:34,230 is converted to the conjugate base. 752 00:37:37,140 --> 00:37:40,220 Now, we have moles. 753 00:37:40,220 --> 00:37:44,030 We need molarity if we're going to use a Ka 754 00:37:44,030 --> 00:37:46,970 to solve the problem, because the expression for equilibrium 755 00:37:46,970 --> 00:37:49,480 constant has molarity, not moles. 756 00:37:49,480 --> 00:37:52,120 So then in converting to molarity, 757 00:37:52,120 --> 00:37:55,650 the thing you have to remember is the volume. 758 00:37:55,650 --> 00:38:00,220 So we had 25 mLs to start with, and now we've added 5. 759 00:38:00,220 --> 00:38:03,630 So we need to make sure that we have the correct number 760 00:38:03,630 --> 00:38:06,520 of liters in our expressions. 761 00:38:06,520 --> 00:38:10,060 So we can calculate 1.75 times 10 to the minus 3 moles 762 00:38:10,060 --> 00:38:14,970 for our weak acid, volume concentration of weak acid. 763 00:38:14,970 --> 00:38:19,090 We have 0.75 times 10 to the minus 3 of our conjugate base. 764 00:38:19,090 --> 00:38:21,180 And then we have a concentration there. 765 00:38:21,180 --> 00:38:25,360 Now, there are two ways to solve this problem. 766 00:38:25,360 --> 00:38:28,970 And option one uses Ka. 767 00:38:28,970 --> 00:38:31,280 It doesn't involve Henderson-Hasselbalch. 768 00:38:31,280 --> 00:38:33,830 Option two will involve Henderson-Hasselbalch, 769 00:38:33,830 --> 00:38:36,690 and it's OK to use this because this is a buffer problem. 770 00:38:36,690 --> 00:38:39,550 So option one now for calculating point B-- 771 00:38:39,550 --> 00:38:41,340 so we can set up this expression. 772 00:38:41,340 --> 00:38:44,780 And always remember you're talking about molarity here, 773 00:38:44,780 --> 00:38:46,820 so don't put in your moles in this expression. 774 00:38:46,820 --> 00:38:47,890 Only put in molarity. 775 00:38:47,890 --> 00:38:50,230 So that's why the volume's really important. 776 00:38:50,230 --> 00:38:55,040 So we can put in our concentration 0.0583, 777 00:38:55,040 --> 00:38:58,840 and it'll will be minus x for equilibrium plus x 778 00:38:58,840 --> 00:39:01,940 for our hydronioum ion concentration and also 779 00:39:01,940 --> 00:39:07,970 our concentration of the conjugate base 0.0250 plus x. 780 00:39:07,970 --> 00:39:12,900 So this is like a buffer problem, using option one. 781 00:39:12,900 --> 00:39:16,520 And so we can use Ka, plug in our values. 782 00:39:16,520 --> 00:39:19,390 Here we have our conjugate base concentration 783 00:39:19,390 --> 00:39:21,880 at equilibrium times our hydronium 784 00:39:21,880 --> 00:39:28,120 ion, x over the equilibrium molarity of our weak base. 785 00:39:28,120 --> 00:39:30,760 We can, again, assume x is small, 786 00:39:30,760 --> 00:39:34,550 which drops out the plus x and the minus x. 787 00:39:34,550 --> 00:39:35,750 We solve for x. 788 00:39:35,750 --> 00:39:39,740 We get 4.13 times 10 to the minus 4. 789 00:39:39,740 --> 00:39:41,600 We can check the assumption. 790 00:39:41,600 --> 00:39:46,030 And sure enough, it's much smaller, 1.7%, 791 00:39:46,030 --> 00:39:49,380 0.7% of these two. 792 00:39:49,380 --> 00:39:51,920 And you really only need to check one. 793 00:39:51,920 --> 00:39:54,820 The smaller one-- assumptions OK. 794 00:39:54,820 --> 00:39:57,170 And we can calculate then by putting 795 00:39:57,170 --> 00:39:59,710 in x, which is hydronium ion concentration, 796 00:39:59,710 --> 00:40:03,840 into the equation pH equals minus log of the hydronium ion 797 00:40:03,840 --> 00:40:08,500 concentration equals the pH 3.38. 798 00:40:08,500 --> 00:40:12,650 We can also use Henderson-Hasselbalch, 799 00:40:12,650 --> 00:40:14,080 because it's a buffer problem. 800 00:40:14,080 --> 00:40:15,960 It's allowed. 801 00:40:15,960 --> 00:40:17,320 You were given the Ka. 802 00:40:17,320 --> 00:40:24,130 You can calculate pKa minus the log of Ka and put that number 803 00:40:24,130 --> 00:40:28,310 in, put in the concentrations that you determined. 804 00:40:28,310 --> 00:40:30,300 So you still need to subtract the moles. 805 00:40:30,300 --> 00:40:32,740 You can skip that step. 806 00:40:32,740 --> 00:40:34,950 But you can skip the volume step. 807 00:40:34,950 --> 00:40:37,250 Because when you use Henderson-Hasselbalch, 808 00:40:37,250 --> 00:40:40,230 the volume would cancel out if you had moles 809 00:40:40,230 --> 00:40:42,560 per volume moles per volume. 810 00:40:42,560 --> 00:40:44,060 The volume's the same. 811 00:40:44,060 --> 00:40:47,300 So you can skip the calculating molarity step here-- 812 00:40:47,300 --> 00:40:50,040 that simplifies the problem-- and solve. 813 00:40:50,040 --> 00:40:52,130 And you should get the exact same number, 814 00:40:52,130 --> 00:40:53,330 or something was wrong. 815 00:40:53,330 --> 00:40:57,810 Both of these should work equally well-- 3.38. 816 00:40:57,810 --> 00:41:00,300 But you'll often be asked to check 817 00:41:00,300 --> 00:41:03,360 the Henderson-Hasselbalch assumption. 818 00:41:03,360 --> 00:41:05,770 And remember there was an assumption here, 819 00:41:05,770 --> 00:41:08,370 because we solved the Henderson-Hasselbalch 820 00:41:08,370 --> 00:41:12,270 from an equation that use Ka, the equilibrium constant. 821 00:41:12,270 --> 00:41:15,360 So really this expression is at equilibrium, 822 00:41:15,360 --> 00:41:19,580 but we're plugging in values that are not at equilibrium. 823 00:41:19,580 --> 00:41:21,300 They're the initial values that we're 824 00:41:21,300 --> 00:41:22,960 putting into this equation. 825 00:41:22,960 --> 00:41:26,590 So we still need to check, and we're using the 5% rule. 826 00:41:26,590 --> 00:41:29,280 So to check the Henderson-Hasselbalch, 827 00:41:29,280 --> 00:41:34,200 then, we have to back calculate pH of 3.38 828 00:41:34,200 --> 00:41:38,260 equals a hydronium ion concentration of 4.2 times 10 829 00:41:38,260 --> 00:41:40,010 to the minus 4. 830 00:41:40,010 --> 00:41:43,650 And then you want to check that that number is less than 5% 831 00:41:43,650 --> 00:41:44,810 of the smaller number. 832 00:41:44,810 --> 00:41:48,160 And we checked that assumption when we looked at option A. 833 00:41:48,160 --> 00:41:51,260 So it's still good for option B. And we can 834 00:41:51,260 --> 00:41:53,100 use the HendersonHasselbalch. 835 00:41:53,100 --> 00:41:57,110 Now, if it was greater than 5%, then you 836 00:41:57,110 --> 00:41:59,360 can only use option one. 837 00:41:59,360 --> 00:42:02,150 And you must use the quadratic equation. 838 00:42:02,150 --> 00:42:04,570 But this doesn't happen very often 839 00:42:04,570 --> 00:42:07,600 because a buffer is usually not a very good buffer 840 00:42:07,600 --> 00:42:09,700 if x is not small. 841 00:42:09,700 --> 00:42:12,840 A buffer's a good boss for when you have a weak conjugate acid 842 00:42:12,840 --> 00:42:13,540 pair. 843 00:42:13,540 --> 00:42:17,010 And when you have weak conjugate acids and bases, then 844 00:42:17,010 --> 00:42:18,700 usually x is small. 845 00:42:18,700 --> 00:42:21,110 It doesn't ionized much in water. 846 00:42:21,110 --> 00:42:23,660 So most of the time these assumptions hold. 847 00:42:23,660 --> 00:42:25,770 But if you're asked to check an assumption, 848 00:42:25,770 --> 00:42:28,290 that's what you need to do. 849 00:42:28,290 --> 00:42:31,870 So we have point B two ways, and we got the same answer. 850 00:42:31,870 --> 00:42:33,560 It's 3.38. 851 00:42:33,560 --> 00:42:37,890 And that's point B, and we're in the buffering region there. 852 00:42:37,890 --> 00:42:40,160 What about point C? 853 00:42:40,160 --> 00:42:44,430 Point C is the half-equivalence point. 854 00:42:44,430 --> 00:42:46,640 So let's think about what the pH would be 855 00:42:46,640 --> 00:42:48,400 at the half-equivalence point. 856 00:42:48,400 --> 00:42:49,860 So at the half-equivalence point, 857 00:42:49,860 --> 00:42:52,730 by definition, you've added half the number 858 00:42:52,730 --> 00:42:55,930 of moles you need to reach the equivalence point, or half 859 00:42:55,930 --> 00:42:58,510 the volume of the strong base that you need 860 00:42:58,510 --> 00:43:00,360 to reach the equivalence point. 861 00:43:00,360 --> 00:43:05,010 And at this point, the number of moles of your weak acid 862 00:43:05,010 --> 00:43:08,610 equal the number of moles of your conjugate base. 863 00:43:08,610 --> 00:43:12,860 So when that is true and we look at the Henderson-Hasselbalch 864 00:43:12,860 --> 00:43:18,530 expression, we see that this term becomes 1, 865 00:43:18,530 --> 00:43:24,290 and so therefore pH equals pKa. 866 00:43:24,290 --> 00:43:29,200 So if you weren't given the pKa and you were just given the Ka, 867 00:43:29,200 --> 00:43:32,340 that's no problem because you can calculate it-- 868 00:43:32,340 --> 00:43:35,320 minus log of Ka. 869 00:43:35,320 --> 00:43:39,530 And so that's minus log of 1.77 times 10 to the minus 4 870 00:43:39,530 --> 00:43:46,300 for this problem, which gives us a pH about 3.75. 871 00:43:46,300 --> 00:43:51,950 So if you were just given the pKa, which is for this 3.75, 872 00:43:51,950 --> 00:43:53,940 and it's really not much more significant 873 00:43:53,940 --> 00:43:54,912 than that, honestly. 874 00:43:54,912 --> 00:43:57,370 And there actually not a lot of significant figures when it 875 00:43:57,370 --> 00:44:00,960 comes to measuring pH or pKa's. 876 00:44:00,960 --> 00:44:05,390 If on a test it said, the pKa of this is 3.75. 877 00:44:05,390 --> 00:44:08,940 What is the pH at the half-equivalence point? 878 00:44:08,940 --> 00:44:12,750 You could write down 3.75 and nothing else. 879 00:44:12,750 --> 00:44:16,540 You do not need to prove to me that this expression here 880 00:44:16,540 --> 00:44:18,730 is true at the half-equivalence point. 881 00:44:18,730 --> 00:44:21,440 It's just a way to ask a question that tests 882 00:44:21,440 --> 00:44:23,820 your knowledge of titrations. 883 00:44:23,820 --> 00:44:26,520 So this is a good problem that involves very little work, 884 00:44:26,520 --> 00:44:29,240 so you should be extremely excited if I'm asking you 885 00:44:29,240 --> 00:44:31,330 about the pH at a half-equivalence point, 886 00:44:31,330 --> 00:44:33,910 because that's going to be a real easy problem. 887 00:44:33,910 --> 00:44:37,110 Don't spend a lot of time on it. 888 00:44:37,110 --> 00:44:41,250 So we got point C, 3.75-- very close. 889 00:44:41,250 --> 00:44:43,650 We're sort of in this region where the pH 890 00:44:43,650 --> 00:44:47,510 is changing not that much. 891 00:44:47,510 --> 00:44:50,990 So that's the end of that lecture. 892 00:44:50,990 --> 00:44:56,130 But the beginning of the next one starts where we left off. 893 00:44:56,130 --> 00:45:00,780 As I said, acid base titration problems take a long time. 894 00:45:00,780 --> 00:45:03,900 They're in two handouts, and we're just 895 00:45:03,900 --> 00:45:08,680 going to continue now with the next handout and point 896 00:45:08,680 --> 00:45:11,580 S, the equivalence point.