Transcript for:
AP Chemistry Review Summary

Hi there, this is Jeremy Kug. And if you're watching this video, then you are in the final days before the AP Chemistry exam, maybe even the final hours before walking into that exam room. It's time to cram. This is a long video, but I think you'll find it's worth it. Don't forget to print out the free guided notes PDF linked in the description down below. That will help you follow along with me as I review and work the problems in this video. And if you want a more comprehensive review, then head over to ultimaterreviewpacket.com where you'll get much more than just a cram session. You'll get every review resource you could ever need to score that five study guides, unit summary videos, two full practice exams with answer explanations and walkthroughs for every single question. So head over to ultimaterreviewpacket.com and click on apchemistry to get access today. Now, if you've printed off your free PDF of notes for this video, you're ready to sit back, get your calculator ready, relax. Well, don't get too relaxed because we're about to cram. Let's get started. Now, generally speaking, I'm going to be working through the entire course, units 1 through nine. And so, we're starting kind of at the beginning here at unit one. And so, mass percent is an important concept to be able to grasp. For example, if we have this chemical compound magnesium chloride and you want to find out what percentage of that compound is magnesium and what percentage of that compound is chlorine by mass, it's really a very simple process. You take the atomic mass of each of the elements. So in this case, magnesium is about 24.31. And then we have chlorine which is about 35.45. But notice we have two atoms of that in the compound. So we have to times that by two which gives us about 70.90. Now we add these values together. So the total uh molar mass of this compound is 95.21 atomic mass units. And then we take the individual total value for each element and divide it by the total molar mass. So in the case of magnesium, we'd take this 24.31 right here and we divide that by the total, which is 95.21. So when you divide that out, and of course times it by 100, you find that the mass percent of magnesium is about 25.53%. And we can do the same thing for chlorine. We just take the 70.90 and divide that by the total 95.21 and times it by 100. and you find that its percentage is about 74.47%. That's all you have to do to find the mass percent. And that works for any compound. Now, let's try a different one. Let's try calcium chloride this time. Now, once again, calcium has an atomic mass of 40.08 atomic mass units. Chlorine is still 35.45 and we have two of those atoms, of course, so that's still 70.90. And once you add those together, we get that the total molar mass here is 110.98 atomic mass units amu. And so to find the mass percent of calcium, we just take the 40.08 divided by the total and we get 36.11%. And then the same thing for chlorine. We take the 70.90 and divide that by the total which is 110.98. We find that that's 63.89. 89%. So that's how you'd find the mass percent of any element in a compound. Now, of course, if we do everything right, the percentages should add up to 100. And it looks like they do here. Now, one little detail that you might have noticed. Notice that we have two different chlorides here. And there's a little pattern that you might notice that everything else being equal, the lighter the other element in the chloride, the greater the percentage of chloride in that compound is. So magnesium was a fairly light metal at 24.31 and so it made up less of a percentage in the total. Calcium is a little bit heavier, 40.08. 08. So it makes up a somewhat larger percent in the chloride and then the chloride would be less there. Now what if we had burillium chloride? How would that compare? Well, you might realize that burillium is a lighter metal than magnesium or calcium. In that case, the burillium would have a mass percent of somewhere less than 25% and the chloride would be higher than you know 74 or 75%. What about strontium chloride though? Well, strontium is heavier than magnesium or calcium as far as it its atomic mass goes. So that means that the strontium would make up a greater percentage than this 36% and the chlorine would be less than 63% in strontium chloride. So think about those patterns because sometimes they like to ask multiple choice questions about mass percent comparing the amount of chloride or some other element in different compounds. Now let's move on and take a look at mass spectrometry. Sometimes this is called mass spectra of elements. For any pure element we can basically get a graph from a mass spectrometer that that looks essentially like this. Now whenever you have two bars here this tells us that that element has two isotopes. And so in this case we have the two isotopes and one of the isotopes has a mass of about 85 atomic mass units and the other one has an atomic mass of about 87 atomic mass units. Now each of the heights of those bars represents the percentage abundance. And so in the case of the one that has a mass of 85, its abundance is about 75% or at least that's what it looks like on this bar graph here. And then the one that has a mass of 87 is about 25%. So if we take those individual atomic masses and multiply them by the decimal equivalents of their percentage abundances, we get these numbers here. If we add those together, that should give us the average atomic mass of the element. So in this case, we have an element that has an average atomic mass of around 85.50 atomic mass units. All we have to do is compare that to the periodic table and we can see that this is most likely the element rubidium because it has an atomic mass very very close to that. So this is the idea that you want to have whenever you're presented with a graph or a mass uh spectrum that looks something like this. Now let's take a look at electron configurations. It's important to be able to write electron configurations. Now of course they're not going to give you this graphic on the exam, but if you're familiar with that, you're going to be able to write an electron configuration for any element that they could throw at you on this exam. For example, if we're trying to write the configuration for the chlorine atom, remember we always start at the beginning of the table and we work our way through every suble through every section until we get to our goal. So in the case of chlorine which is right here its electron configuration is 1 s2 then we have 2 s2 then 2p6 we have 3s2 and finally 3 p5 because it's 1 2 3 4 five boxes into that suble. Now what if we change the question? What if we ask instead for the electron configuration for the chloride ion CL negative? In that case, we want to realize that an annion has gained electrons. And in this case, since it's Cl negative, it's gained exactly one electron. And that electron is going to be added into that veence shell, that last shell. So, we're going to change the 3p5 to 3p6. And so that's how you'd write the electron configuration for the chloride ion. Now, how about the aluminum atom? Well, the same idea here. We start with 1 s2, then 2 s2, 2p6, 3 s2, and finally 3 p1. And so that's the electron configuration for aluminum. Now, what about the aluminum 3+ ion? Well, since it has a positive charge, it's going to lose three electrons and it's going to lose those three electrons out of the veence shell. So that's the outermost shell represented by the largest coefficient there. So that means that these three electrons right here 3s2 and 3p1 are basically going to go away. And so here we have the electron configuration for the aluminum cation. It ends with 2 s2 2p6. Now let's try another one. How about the florine atom? Well, florine is right here and it's electron configuration is 1 s2 2 s2 2 p5 in this case. So that's the electron configuration for florine. Now what if we do this? What if we have the f positive ion? Now, if you have worked your way through AP chemistry, you probably will look at that charge and say that that looks kind of funny. Florine normally has a negative one charge. But what if it did have a positive charge? Well, it would remove one electron and it would remove one electron from that last suble which is 2p. So, it would actually be 2p4 in this case. So, that's the electron configuration for f positive. Now, how about germanmanium? What about this somewhat longer configuration? Well, we start from the top. It's 1 s2 2 s2 2 p6 3s2 3p6 4s2 3d10 and it's 4 p2. So that's the electron configuration for germanmanium using this method. You can write the electron configuration for pretty much any element that they could throw at you on this exam. Now, speaking of electrons and electron configurations, let's take a look at a periodic table trend. And probably one of the most important ones is atomic radius. It's important to realize that the largest atoms as far as radius will be at the left and the bottom of the table. So that would be these atoms kind of down here closer to cium and francium and barerium and radium kind of in that corner of the periodic table. That's where you're going to have the largest atoms. Now the reason for this is that atoms at the bottom of the table have more electron shells and as a result that last electron shell uh tends to be farther away from the nucleus. And so that's why your atoms at the bottom of the periodic table have a greater radius. Now, how about on the left? Well, on the left side, we want to talk about it in terms of protons, an effective nuclear charge. Atoms on the left side of the periodic table have fewer protons than atoms on the right side of the table. And so because of that, that smaller or that less effective nuclear charge means that those atoms on the left side are going to have fewer protons to pull in those electron shells or those energy levels. And so that's why atoms on the left side tend to be larger. Atoms on the right side tend to be smaller. Now, one mistake that students often make on the AP exam is they will try to explain why something is larger or smaller in terms of where something is on the periodic table. Don't make that mistake. Don't say that, you know, for example, cesium is larger than lithium because cium is farther down on the periodic table. That's not an explanation. That's not a reason. The reason is, you know, cesium has more electron shells and so its last shell is farther away from the nucleus. So be very careful as to how you make those explanations. Never use location on the periodic table as an explanation for anything that doesn't explain that just is a trend. Now another important trend is ionic radius. And so here we have uh a a general trend that as you can see here positive ions these these cations ions which I have in blue tend to be smaller than annions or those negative ions which we have in pink. And if you look at this it kind of makes sense as you think about why that is. Let's just take six ions that all have the same number of electrons. And so the nitride ion, which is right here, and the oxide ion and the fluoride ion, as well as sodium, magnesium, and aluminum ions all have the same number of electrons. And so if you were to write the electron configuration for all six of those, you'd find that they all have the same electron configuration. They're what's called iso electronic. And so what that means is because of that the atom that has more protons is going to be able to pull in those electrons more effectively. And so the one that has the most protons here is aluminum. And so aluminum is going to be able to pull in those 10 electrons much more tightly. And that's why the aluminum cation has a radius of a very very small amount something like 57 I think is what that says 57 pometers. whereas nitride has a lot fewer protons. And so that's why nitrite is something like three times as large as the aluminum ion. So once again, generally speaking, the more negative an annion is, the larger it is. And the more positive a cation is, the smaller it's going to be. And that's ionic radius. Now another important trend is first ionization energy. Now, generally speaking, first ionization energy, which by the way is how much energy it takes to remove that very last electron from an atom, it tends to be the greatest toward the right and the top of the table. So that would be these atoms over here, things like helium and neon and florine. Those are going to have the highest first ionization energy. And basically the reason is the same as it was before. At the top of the periodic table, we have fewer electron shells and that means that that last shell is closer into the nucleus and it has stronger attractions to the outermost electrons and so it's going to require more energy to pull that last electron since that last electron is closer to the nucleus and has that stronger attraction. On the other hand, the ones at the bottom of the table, you have more electron shells and so that last electron is just farther away from the nucleus and so it's not held in as tightly. The kulomic attractions are weaker and so it requires less energy to pull that last electron. Now, if we're talking about left and right, we want to think about this in terms of the number of protons and the effective nuclear charge. The ones that are farther toward the right have more protons and greater effective nuclear charge. And so the protons are pulling in those electron shells more tightly. And so it's going to require more energy to remove that last electron from oh say neon over here than it would from lithium over here. So you know more protons means a stronger uh kulomic attraction a stronger effective nuclear charge. Now let's take a look at one more concept here in unit one and that is photoeleron spectroscopy. So they like to ask these questions. I I can pretty much guarantee that you'll see at least a graph or two like this on the exam. And the way that you read a photoeleron spectroscopy diagram is you just read from left to right. Each peak represents a suble. So this first one is 1s for example and then it's 2s and then 2 p and then 3s and then 3 p. So if you can write an electron configuration, you can read a photoeleron spectroscopy diagram. The height of the peak has everything to do with how many electrons are in that suble. So we have three peaks that are these s peaks here that all have the same height. So it's safe to assume that they all have two electrons in them. So I just have two there. Now this 2p peak is three times higher than the s peak. So that tells me that this is 2p6. And this 3p peak is more than twice as high but not quite three times as high as the s peak. So that tells me that it has five electrons. So it ends with 3 p5. So if I think about the electron configuration for this, this tells me that this is the electron configuration or the the pes diagram for chlorine. Now we can do something different here. Let's take a look at another pes diagram. Do the same thing. We have 1 S, 2 S, 2 P, 3 S, and 3 P. And then we have two, two, the next one is six, and then we have two, and then that last one is six as well. So this element seems to be argon. Now, one little detail that sometimes they will ask about on the AP exam is the left and right. I want you to notice not just the the height or where the peaks are, but notice where they are on the x axis, the relative binding energy per electron. Generally speaking, the more protons an atom has, the greater the binding energy of its core electrons. And so if you look at argon here, notice that the 1s and the 2s and those 2p electrons are just a little bit scooted over to the left on this diagram. That tells me that that has more protons and it's pulling in those electrons more tightly. It has more binding energy. And so anytime you see a peak farther to the left, it probably has more protons. And so even if I did not have the um numbers and the letters and all the labels on here, I would be able to tell that the second diagram here has more protons than the first diagram. So that's a little detail that they like to ask about on the AP exam as well. So just be aware of that. Let's jump right into unit two here and we're going to take a look at Lewis electron dot diagrams. That's kind of the heart of unit two here and we're going to start with sulfur doulloride. So once again you always put the central atom in well in the center. So sulfur is in the middle. Personally I like to start with the outside and work my way in. I find it reduces mistakes. And so every florine atom has seven veence electrons. So I'll put seven dots around that florine and seven dots around that florine. And then the sulfur is going to have six. And so we have six dots there. And everything is trying to have eight for the most part. So everything has eight. And so I'm going to replace these shared pairs with a little line here. And so here's how I would draw the finished Lewis electron dot diagram for SF2. Looks like this. Now, how about CO2? Once again, I draw the central atom in the middle. And normally they will tell you or make it fairly obvious what the central atom is. In this case, it is carbon. So, that goes in the middle. We have seven veence electrons for every chlorine. So, we have that. We have the six veence electrons for oxygen up there. Carbon brings in four veence electrons. And once again, the goal is for everything to have eight. And the way I have this drawn, everything does not have eight. And so I need to bring maybe two veence electrons from the top of this oxygen down to the middle here, just like this. And now everything has eight. So this time I have a total of four shared pairs. There's a double bond here between the carbon and the oxygen. And we have two single bonds, one between each of the chlorines and the central carbon atom. So that is our finished Lewis electron dot diagram for CO2. Now if we take a look at some Lewis electron dot diagrams, it is important for you to be able to understand the types of bonds that it has. Sigma and pi bonds for example. Every single bond is a sigma bond. Every double bond has one sigma and one pi. Every triple bond has one sigma and two pi. So in this molecule here since we have two single bonds we have two sigma bonds and then the hybridization is sp3. We have to think about that because there are four electron regions or four electron domains around the central atom. It has a steeric number of four. These are all different ways of saying the same thing. There are four of these electron regions or electron domains around the sulfur which corresponds to a hybridization of sp3. Now in this one here we only have three electron regions or electron domains around that central carbon atom right there. So because it has three electron domains it has a hybridization of sp2. Now how many sigma and pi bonds does it have? Well, we have two singles and one double bond. So, every double bond has one sigma and one pi. So, that's why it's three sigma and one pi bond. The hybridization is sp2. Like we said, sometimes you'll have molecules that have multiple central atoms. In this molecule here, we actually have three central atoms. is each carbon is a central atom and then the oxygen acts as a central atom as well. So when you add up the types of bonds here we have looks like one two three four five six seven sigma bonds and I see one pi bond because there is a double bond over there and the hybridization each central atom has its own hybridization. So carbon number one has four electron regions or electron domains. So that's sp3. Carbon number two has three electron domains. So that's sp2. And then the oxygen atom has four electron domains. One, two, three, four. So that would be sp3. There's no shortcut around that. You have to know the hybridizations and be able to give the hybridization of every central atom in a molecule when there are multiple central atoms. Now let's take a look at another detail here and this is the concept of molecular geometry and bond angles. So, we'll go back to that first molecule that we looked at earlier. And notice that since it has two bonding regions here, or it's bonded to two atoms, and it has two lone pairs, two and two is bent, and it has a bond angle of about 104.5 degrees. Now, if you look at the structure, the way we have it drawn here, you might think its bond angle is 180 degrees. Well, it's not. It's a bent structure and that's why most chemists will draw it like more like this actually. Now the second structure it looks like we have a total of three bonding regions one two three it's bonded to three atoms zero lone pairs so three and zero is tragonal planer and that bond angle is 120°. Now on the next one we have three central atoms. So there are actually three molecular geometries. You have to be able to describe all of those. The first one is four bonding regions and zero lone pairs. So that's tetrahedral. And then the next one is trional planer. Again we have, you know, bonded three atoms, no lone pairs. So that's trigonal planer. The oxygen is bent because it has a bond to two different atoms and two lone pairs. So that's two and two. So it's bent. So we actually have three molecular geometries in the same molecule here. You have to be able to do that. It will have three different bond angles as well. Tetrahedral is 109.5°. Trigonal planer is 120° and bent is 104.5 degrees. Now, if you have a little trouble getting all these bond angles straight, uh, on the AP exam, they're a bit more lenient than your teacher might be because anything that has a hybridization of sp3 and has four electron domains, they're okay with it if you just call it 109.5 degrees. And so, if you say bent is about 109.5 degrees, that's fine. They're not going to uh count off because of that. Now, let's take a look at one other concept. This is the concept of polarity. Sometimes they'll ask the question, does this molecule have a dipole moment? And what that means is, is it a polar molecule or a non-polar molecule? Well, anytime you have a bent structure like this, it is a polar molecule. It does have a dipole moment. And one thing that you need to remember here is that it's not about symmetry. Okay? I noticed that some students want to talk about this in terms of symmetry. And that has nothing to do with this. It's not symmetry, folks. It has everything to do with is there an unbalanced region of negative charge in the molecule. And so all three of these molecules, you could make a case they're all symmetrical, but that doesn't mean they're all non-polar. In fact, obviously this is polar because we have a an unbalanced region of negative charge. We have these two lone pairs of electrons hanging out on that sulfur atom right there. So that is a polar molecule. Every molecule has London dispersion forces. So you need to know that. Now if it's a polar molecule, it also has dipole dipole forces. So that first molecule has both of those. Now the next one, this is also a polar molecule and that's because there is this unbalanced region of negative charge on that oxygen. Now this may not be quite as polar as the first one was but it is a polar molecule because it has a lopsidedness in its electron distribution and because of that it also has London dispersion forces and dipole dipole forces. In the third one there is no unbalanced region of negative charge at all. In fact there are no lone pairs to speak of at all on this molecule. So this is a nonpolar molecule and because of that it has only London dispersion forces. Now we need to realize there is this thing called hydrogen bonding as well. If you have a molecule that has a hydrogen bonded to an FO or N atom, it's going to have hydrogen bonding. Hydrogen bonding uh can be classified as a kind of a type of dipole dipole force. If you have a molecule like water, you want to say it has London dispersion forces and hydrogen bonding. You don't have to say that it also has dipole dipole forces. That's kind of understood if you say that. Some people would say that hydrogen bonding is completely different from dipole dipole. I guess the jury is still out on that. But if you say something has hydrogen bonding and London dispersion forces, you don't have to say that it also has dipole dipole. That's just a little a little clarification on that one. Now, let's take a look at a different concept here from unit two, and that's the concept of melting points here. And this goes right back to Kulom's law, which is kind of a thread that we see quite a bit, especially in the first half of this course. So, let's say that we're trying to compare the melting points of sodium chloride and sodium sulfide. Well, you want to think about this first of all, in terms of charge. sodium and chloride that's a + one minus1 pair and sodium sulfide is a + one minus2 pair. So we would expect the ones that have the greater magnitude of charge to have the higher melting point and sure enough that's exactly what happens. Sodium sulfide has a higher melting point than sodium chloride. Now, what if we have two ionic compounds that have the same ion charge in them? So, like aluminum chloride, aluminum fluoride. If they're the same, the tiebreaker is to look at the sizes of the ions. Because we know that chloride is a larger ion than fluoride. That means that the chloride ion is going to be larger and and the ion centers will be farther from each other than they are in aluminum and fluoride. So in aluminum and fluoride, since the ions are closer to each other, the attraction is higher. And so it's going to require more energy to melt that. And so you'd expect that the smaller the ions, the higher the melting point. And if you look at the melting points, that's exactly what you have. Aluminum chloride melts at about 192 degrees. Aluminum fluoride melts at about 1,291 degrees. That's a lot higher because the ion sizes are much smaller. Now, let's compare the melting points of sodium chloride and potassium broomemide. Well, once again, the first thing we do is look at the charges. + one minus one + one minus one. Well, since it's a tie, then we look at the ion sizes. So, we see that sodium and chloride ions are smaller than potassium and broomemide ions. And so that tells us that sodium chloride should have a higher melting point and potassium broomemide should have a lower melting point. And sure enough that is exactly what you have. Sodium chloride is I mean several degrees higher in its melting point. And that's because sodium and chloride are smaller ions. So their colombic attractions are going to be stronger. Now, as we think about how ionic compounds uh are arranged, there's this threedimensional lattice. Now, I don't have it drawn in three dimensions here, but you kind of get the idea that every ion is surrounded by ions of the other type of charge. We have positives and negatives attracting each other. So, here we have sodium fluoride. Now when this dissolves in water, we need to realize that the water molecule is a very polar molecule. So we have the negative pole of water, which is the oxygen, and we have the positive pole of water, which is the side that the hydrogen is on. Well, the positive poles of water will surround these negative ions and literally drag them off into solution. And the other side of water, the negative pole, is going to surround those positive cations ions and literally drag those off into solution. So imagine all these water molecules doing that to these other ions. And as long as those ion dipole forces between the water molecules and the ions are stronger than the ionic forces holding the crystal together, it's going to be soluble in water. and sodium fluoride is soluble in water. Now, let's take a look at unit three here. We'll jump right into a gas law question. And we have a problem. We have a gas mixture that's collected in a 3 L rigid container at a temperature of 315 Kelvin. It says that the pressure inside the container is measured to be 1.14 atmospheres. How many moles of gas are in the mixture? So, on a problem like this, you want to use the equation PV= NRT. We're basically just going to plug and chug right into that equation. P, the pressure is 1.14 atmospheres. V, the volume is 3.00 L. We're trying to solve for N, the number of moles. R is the universal gas constant of 0.08206 liter atmospheres per mole Kelvin. And the temperature is 315 Kelvin. So all we have to do is use algebra to solve for N. and you find that the number of moles is.132 moles in this mixture. Now what if we go a step further? What if it is determined that 0.045 moles of the mixture is oxygen? The other gas would compose how many moles? Well, when you have a gas mixture, we know that the two individual mole values are going to have to add up to the total number of moles in the container. So in this case if the total number of moles is.132 and we subtract out the 0.045 moles of oxygen we find that the other gas is 0.087 moles. Now what if we go even a step further? What if we determine that that other gas that we just found the number of moles of has a mass of 3.47 g. What's the molar mass of that gas? Well, remember molar mass is in units of grams per mole. So, grams per mole implies grams divided by moles. So, we just have to take the number of grams, which is 3.47 g, and divide it by the number of moles, which we just said was 087 moles. And when you divide that out on your calculator, you find that the answer is very close to 40 g per mole. And so you could take a guess as to what that gas was. It's probably it sounds like it's that there's a good chance it's argon. So it might be that or possibly some other gas that has a molar mass of 40 gram per mole. Now let's take a look at spectrophotometry because they're going to ask you something about this. The beer Lambert law is used to talk about spectrophotometry very often. Uh whenever you have a spectrophotometry problem, you're very likely going to have a graph like this. And we always run a blank and then we run several known concentrations that we record those absorbances. So we just plot those there on the XY axis and then we draw a best fit line. So a line that kind of best connects those or best approximates those points there. And then we determine the absorbance of the unknown and match it up to the concentration. So for example, if we have an absorbance of 0.2, all you'd have to do is match that up to the line like this, and you'd find that its concentration is very close to 0.55 moles per liter. And so that's how you match something up to a calibration curve using spectrophotometry here. That is literally how that's done in the lab. Now, one thing you want to be careful about is outliers. Occasionally, they'll ask you a question about outliers. If we have a dot like this one right here that's a little bit too high, that most likely means that the sample got contaminated with a concentration that was too high. If you have something uh like this, a dot that's too low on the xy axis here, that tells us that the sample was probably contaminated with something that uh had a concentration too low or more likely with distilled water or something that had basically a contaminant in there. So, just be aware of that. I have a much more detailed description of that in my daily videos there for unit three topic 13 if you want to see something in more detail about that. Now an important thing we have to be able to do in unit four is understand net ionic equations. So for example if you have this sentence here a piece of magnesium metal is added to a solution of copper 2 chloride. We need to remember that there are three things that are basically in that solution. We have magnesium metal that's been added and we have two ions in solution. We have copper 2 ions and we have chloride ions. And we need to understand that metals generally react with metal ions and non-metals generally react with non-metal ions. And so here we have a metal which is magnesium reacting with a metal ion which is copper 2+. So what's the chloride doing there? Well, the answer is it's not doing anything. Chloride is the spectator ion. It's just sitting there looking on like a spectator basically. So it's not going to be included in the net ionic equation. One of the most common mistakes that students make when writing net ionic equations is they try to put those spectator ions in there somewhere or they completely forget to omit those. And so make sure that you leave out spectator ions in net ionic equation. So Mg is going to react with Cu2+. Now generally speaking metals get oxidized that means the charge goes up. So magnesium is plus2 and then metal ions normally get reduced down into their metallic form. So down to Cu. Now we have to balance these half reactions in terms of charge. We see that we have a zero for magnesium. Mg2+ of course is 2+. So to balance that in terms of charge we have to add two electrons over here. And then kind of the same thing for copper. we have to add two electrons on the left side over here to make that work. And so now we can add these two half reactions together. The two electrons cancel out and we have our overall net ionic equation here for a redux process. It's important to understand that since in this first half reaction we're losing electrons. That's called oxidation. And in the second half reaction since we're gaining electrons that is reduction. So be aware of the difference between oxidation and reduction. Some people like to say Leo the lion goes gr. Lose electrons is oxidation. Gain electrons is reduction. As we move on, we can take a look at some other uh basic ideas here. Sodium metal is burned in air. Well, we know that that's Na. And anytime you burn something, you're adding oxygen to it. So the product is going to be something made of Na N O which is Na2O when you look at the charges. If you're ever asked to write an equation, you want to balance the equation. So we can balance the equation by adding a two here to balance the oxygens and then a four over here to balance the sodiums. If we have a sample of propane gas C3 H8 that's burned in air, well we just write that out. C3 H8, add it to O2, and the products of every complete combustion are carbon dioxide and water. And then you want to balance that equation. We have three carbons here. Looks like we need to balance the hydrogens's by putting a four right here. And then for our oxygen, we have six plus four. So that's 10 oxygens versus two right here. So a five right there will now balance that equation. Let's take a look at one more here. This is a good example of a precipitation reaction. Solutions of lead 3 nitrate and sodium carbonate are mixed in a test tube. So once again we have solutions lead 2 nitrate. We write them in their ion form. Sodium carbonate we write sodium and carbonate in its ion form. And we have to realize that when these ions try to swap partners only one combination is going to make a precipitate. The sodium and the nitrate are spectator ions. Anytime you see something with a sodium or a potassium or a nitrate, that combination will be the spectator ions. The other compound is actually going to make your precipitate. So in this case, it's Pb2+ reacting with CO32 negative and that's going to make lead 2 carbonate, which is Pb3. So there is your product. And here is your net ionic equation. I like to put the state symbols in here. Ions are always aquous. The precipitate of course is always a solid. Uh normally you do not have to put the state symbols in there. They're optional unless they specifically ask you to write them in there. Now let's move on to a stoeometry problem. In chemical reactions, it's very common that we're asked to work a stoeometry problem. And one thing that I always tell my students is the first step is convert to moles. The second step is use the mole ratio. Usually those numbers come straight out of the coefficients for the balanced equation. And the third step is to convert to some other final unit, usually grams, if you're asked to do that. So let's do that. Here we have a a balanced equation and it says in a chemical analysis excess silver ions are added to a solution containing carbonate ions. After the reaction, the total amount of silver carbonate solid is weighed and found to have a mass of 347 grams. How many grams of carbonate ions were present in the original sample? So once again, I'm going to start by writing down what's given to me. It says that we have a mass of 347 grams of silver carbonate. And the question is asking how many grams of carbonate ions were present in that original sample. So way down here at the end, we're trying to convert to grams of carbonate. So like in any stochometry problem that we do, step one is convert to moles. So that means I have to put grams on the bottom and one mole on top. And we consult the periodic table and see that there are 275.75 g in one mole of Ag2 CO3. Now grams are out top and bottom. We're now in moles of silver carbonate. Step two is the mole ratio. So I take Ag2 CO3 and put that on the bottom since that's what we're trying to get rid of. And we're converting to carbonate. So carbonate goes on the top. And I can look in the balanced equation here and see that this is a one:one ratio. There's one carbonate consumed for every one silver carbonate produced. So one one. So I can cancel silver carbonate top and bottom. I'm now in moles of carbonate. I want to be in grams of carbonate. So step three is to convert to that final unit, which is grams. So moles go on the bottom, grams go on the top. And consulting the periodic table, it looks like there are about 60.01 grams in one mole of the carbonate ion. So I can cancel moles top and bottom. And when I do my math, the 347 divided by 275.75 times 60.01, I find that my answer is about 0.0755 gram of carbonate. So there's a very typical stoeometry problem that we have there. Now let's take this a step further. Let's say that we have a student that adds 150 milliliters of 0.120 molar silver nitrate to a solution of excess sodium carbonate. Assuming the reaction goes to completion, how many moles of silver carbonate should be produced? Well, let's think about this first uh in terms of the moles of silver ion that we added to the solution. It says 150 milliliters of.120 molar silver nitrate solution. Well, the question is how many moles of silver ions were used? Now we can use the equation marity equals moles divided by liters to figure that out. If you rearrange that you find that the number of moles is just liters times marity. So I take 0.150 L times.120 molar silver ions and I find that I had 0180 moles of silver ions added to the solution. So now I'm going to take that value and I'm going to determine how many moles of silver carbonate should be produced because that's what the question is asking. How many moles of silver carbonate should be produced? So, you know, in my three-step process, I've just converted to moles. That's what I just did right there. So, I can jump right into step two, which is the mole ratio. So, silver goes on the bottom, silver carbonate goes on the top since that's what I'm converting to. And this time it looks like it's a one:2 ratio because those are my coefficients for the balanced equation. So, I can cancel silver and this time I take 0.0180 0180 divided by 2 and I find that the answer is about 9.00 * 10 -3rd moles of silver carbonate. So a couple of very straightforward stochometry problems. That last one there is a good example of solution stochometry. Now, as we jump into unit five, which is kinetics and rate laws, this is a very typical kinetics problem where you're given an equation and some sort of a table or a chart that looks like this and you're asked to determine the orders of the reaction with respect to the different reactants. So, in the in this case here, let's determine the order of the reaction with respect to the A2 molecule. So the key here is to find two trials where A2 is the only reactant that's changing concentration. And so I would recommend trials one and two because it looks like A2 is doubling from trial one to trial 2 whereas uh D2 is staying the same. So it looks like A2 is doubling while the rate is basically doubling as well. So we have to ask ourselves two raised to what power is equal to two? Well the power is one. So that's why we would say that it is first order with respect to A2. Now let's do the same thing with the other reactant D2. We have to find two trials where D2 is the only substance changing concentration. I would recommend trials one and three because it looks like A2 stays constant from trial one to three, but D2 is doubling. So D2 is doubling from trial one to trial three, but the rate seems to be going up by a factor of four. So once again, we have to ask ourselves what is the power here? If the concentration is doubling and the rate is quadrupling, what power is that? Two to what power is four? Well, it's the second power. And so that means that the reaction is second order with respect to D2. Now the overall order of this process is just the sum of those two individual orders. So 1 plus 2 of course is three. And so this is overall a third order process. Now let's take this a step further. If we're asked to write the rate law for this reaction, it's written in the format rate equals K times the first reactant, which is A2, raised to its order, which is 1, times the concentration of the next reactant, which is D2, raised to its order, which is 2. So the rate law is rate equals K * A2 * D2 squared. Now, the part of this that trips up students sometimes is determining the rate constant because we're asked to determine the rate constant with proper units. And that's the tough part for some students. Let's just work through this. The way that you determine the rate constant is you just take the rate law and you plug and chug the values from any one of those three trials. Any one of those three trials should work. So, I'm going to use trial one just because, well, it's the first trial. So in the first trial the rate is 1.12 * 10 -3rd marity/ second. Always conserve those units because the rate constant has to have proper units. Now K the rate constant is what we're trying to solve for. The A2 concentration in trial one is 0.040 molar and the D2 concentration there is 0.030 030 molar and that's got to be squared because it's raised to the second power. Now once you plug this into your calculator, you take 1.12 * 10 -3rd and divide it by 0.04 and divided by 003 squared. You get about 31. Now that's the numerical value, but you want to have the units on here too. Now here's how you do this. You want to isolate the variable. We're trying to isolate that K right there. So I have a marity times a marity squared. So in order to isolate that I have to multiply this side by marity to the -3rd to get rid of that marity cubed. Now if I'm going to do something on one side of the equation I have to do it on the other side of the equation too. So on this left side I'm going to also multiply by marity to the -3rd. And this marity time marity to the -3rd is marity to the -2. So the units are marity to the -2 over seconds. We have that seconds we still have to uh put in there too. So that is our value for K with correct units. So be very careful as you're making that calculation. Now there's another way to determine the rate law for a reaction. We can do this graphically. So if we have this reaction right here, 2 HI yields H2 plus I2. You want to make three graphs. One graph is going to be concentration versus time. Another graph is going to be natural log of concentration versus time. And the third graph is one over concentration versus time. One and only one of those graphs is going to be a straight line. And I think we can all tell that it's this this graph down here at the bottom. Well, if it's the one over concentration versus time, that is going to be second order. Now, if this graph up here, if concentration versus time had been a straight line, it would have been zero order. And if natural log of concentration versus time had been the straight line, it would have been first order. But it was one over concentration versus time. That's second order. Now the rate law just like it's been done before. Rate equals K times the concentration of HI squared. That squared of course comes from the fact that this is a second order process. And how would you determine the rate constant? It's equal to the slope of that straight line. if it's zero order or first order. Notice that those two plots are kind of going uh downhill, aren't they? So that means it would be negative slope if you were trying to find the rate constant of those. Just as a reminder, rate constant is always a positive value. Now, one last thing we have to realize for uh rate laws and kinetics is being able to work with reaction mechanisms. So let's say we have this reaction mechanism right here. Let's first of all identify the reaction intermediate. Now once again a reaction intermediate is a substance that is produced in an early step and then used up in a later step. So to me it looks like it's BF2 because it's produced in step one and then it's consumed in step two. It does not appear in the overall balanced equation. So BF2 is your reaction intermediate. Now, what's the overall balanced equation? Well, we just have to add these two equations together. It's kind of like adding two equations in algebra. Basically, we have B F3 plus these two oxygen atoms yields B plus O2F2 plus O2F. And so there we have our overall balanced equation. Now if we add a little bit of information in here and we say which of the steps is the rate determining step, we have to remember that the slow step is always the rate determining step. And so that would be step one because it is the slow step. And what's the rate law for the reaction? Well, essentially the rate law of the slow step is going to be the rate law of the overall reaction. So it's just rate equals K * BF3 time O2 like this. Now if the slow step happens to be the second step or a later step then we'd have to use a process called pre-equilibrium approximation in order to find the rate law. I have a detailed video about that in my daily videos but uh pretty good chunk of the time it'll be the first step and you won't have to worry about that. Moving on to thermochemistry. We have a question here where we're going to use the equation Q= MC delta T. Now, this problem says, a student adds 050 grams of warm copper metal to a 52.77 g sample of water. After stirring, the temperature of the copper metal drops from 356° C down to 23° C. The specific heat capacity of copper metal is 385 JW per g° C. And the specific heat capacity for water is 4.18 Jew per gram degree Celsius. Assume that no heat is lost to the surrounding. So the first question here is how many jewels of heat were lost by the copper. So whenever we have a question like this involving specific heat capacity and jewels and mass and temperature, we're going to use the equation Q= MC deltat T. Now the Q is representing jewels and that's what we're trying to solve for. So Q is going to be our unknown in this question. Now M represents the mass of the copper. It says this is 050 gram of copper. So that goes in there for M. Now C represents the specific heat capacity and the problem says that that value for copper is 385 JW per gram degrees Celsius. So that goes in for C. Now, the change in temperature, it says it went from 356° C down to 23° C. So, when you subtract that, you find that that's a drop of 333° C. So, that's why I have 333 plugged in there for delta T. Now, once you multiply this across, you find that the value for Q is about -64.1 Jewels. Now, that negative sign right there basically just implies that heat is being lost by the copper. That's what that negative sign means. Now, part B says, how many jewels of heat are gained by the water? Well, the first law of thermodynamics tells us that we never destroy or create heat. It's just transferred in a chemical or physical process. And the problem tells us that we're assuming that no heat is lost to the surroundings. So if 64.1 jewels of heat are lost by the copper metal, that means that 64.1 jewels of heat must be gained by the water. Now part C says, what was the original temperature of the water? Well, once again, we're going to use Q= MC delta T. And this time the Q is 64.1 jewels because that's the amount of heat gained by the water. Now m is the mass of water and it tells us that it is 52.77 g of water. So that goes in there for m. Now the value of the specific heat for water, the c is 4.18 jewels per gram degrees CC. So that goes in for C and we're solving for delta t. So now we just use algebra to solve for delta t. And you find that the value of that is about 29° C. So that tells us that the temperature went up by about 29° C. And we know that the final temperature of the copper and the final temperature of the water had to be the same. Whenever you drop something that has a very high temperature like the hot copper in this case into something much colder like the cold water, the temperature of the copper drops and the temperature of the water rises until they both equilibrate. And so the final temperature of both the copper and the water are going to be the same. That's called thermal equilibrium. And it says that the final temperature of the copper is 23°. So the final temperature of the water also has to be 23°. So the initial temperature of the water had to be about 29° less than the 23°. So that tells us that the initial temperature of the water is somewhere around 22.7° C. So there's a very good and typical example of thermochemistry using Q= MC delta T working with calorimetry. Now speaking of heat and how things change temperature and essentially just change as we add heat to them. Let's imagine we have this heating curve. So here we start with a solid at a fairly low temperature and as we add heat to that notice that the temperature goes up. That makes sense. You add heat to a solid, the temperature is going to go up. And once it gets to the melting point though, once it undergoes that phase change, the temperature is going to stay constant during that phase change. So while the object is melting, perhaps this is ice melting into liquid water, that temperature stays constant during the melting process. Now, once everything has melted and it's all liquid and you keep adding heat, the temperature is going to start going up again as it rises as a liquid. And then once again, once you get to the boiling point, it's going to stop its temperature change. And as it's vaporizing, the temperature is going to stay constant once again. And then once it's all boiled away and has turned into a gas, if you keep adding heat, that temperature in is going to rise some more. So this is what we call a heating curve. And we need to know that this bottom region down here is a solid. The one here in the middle is a liquid. And of course the one up here is a gas. And we have melting and boiling. Very often on the AP exam, they'll have this uh diagram except of course they won't label the solid liquid gas. They won't label the melting or the boiling. You have to know that part of it. Now, sometimes they actually draw this curve in reverse where it starts off at a high temperature and then it it drops. That's called a cooling curve essentially because you're going in reverse. Now, let's take a look at another example of thermmochemistry here. And this is where we can calculate the value of delta H. And whenever we have a question like this, we have to use the equation delta H is equal to the sum of all the enalpies of formation of the products minus the sum of all the enalpies of formation of the reactants. It's right side minus left side. And in this question, it says that the delta H of the reaction is equal to 149.7 kJ per mole. determine the heat of formation of sodium hydroxide. Now in this case they don't tell us what the enthalpy of formation of sodium hydroxide is but they are going to tell us the values for the sodium oxide Na2 and for water H2O. So if we don't know sodium hydroxide we can call that X and since there are two moles of that we have to multiply that value by two. So X time two of course is just 2X. Now perhaps they tell us that the value of sodium oxide is -416 kJ per mole. There's only one mole of that. So it's still -416. And then they might tell us that the heat of formation or enthalpy of formation, same thing basically of water liquid is -285.8 kJ per mole. One mole of that. So it's still the same. And so when you add up the reactants, well, we just have 2x. Whenever you add up the products, we can punch this into a calculator and see that it's about -71.8 kJ. Now once again the equation is delta H equals the sum of the products minus the sum of the reactants. The delta H is given to us in the problem up here as 149.7. The sum of the products is -701.8 and the reactants would be 2x. And so now we just have an algebra equation that we have to solve. If I add 701.8 to both sides, I get -2x = 851.5. If I divide both sides by -2x, I find that the enaly of formation of the unknown, the sodium hydroxide is somewhere very close to 426 K per mole. So we can calculate the enalpy of formation of a missing value. Sometimes they might give you the enalpies of formation of all the reactants and products and all you have to do is solve for the overall delta H. And that's probably the way that most students learn how to do these problems. Now, let's move right on to equilibrium, which is unit 7. And the most fundamental idea here about equilibrium is being able to write an equilibrium constant expression. At some point, you're going to have to do this on the exam. And remember, it's products over reactants raised to the power of the coefficients. You only put in gases and aquous solutions. Liquids and solids are omitted from equilibrium constant expression. So if we have this equation right here as an example and we're asked to write the expression for KC. K is in terms of concentration. That's what that C stands for. So we're going to use the brackets in that case. So it's going to be the concentration of Al3+ all over the concentration of Ag+ cubed is equal to KC. So there's your equilibrium constant expression. Whenever you're writing these, don't forget the KC equals part of it. An expression is just that. It is an equation. If you don't write it as an equation, you're not going to get the point. Once again, leave out the solids. Notice that the silver solid and the aluminum solid are not a part of this. Now, what if we have this expression here, this equation, and we're asked to write the KP. Now, P is in terms of pressure. So, we don't use brackets in this case. We use the parentheses in the partial pressure notation. So in this case, KP is equal to the partial pressure of N O quantity squared because there's a two right there times the partial pressure of N2 all over the partial pressure of N2O quantity squared. And we're using gases here. So everything gets included in that equilibrium constant expression. Now let's say we have an equilibrium problem. In this case, it says at a certain temperature, the reaction below occurs and the chemist investigates the process by filling a container with N2O gas to a pressure of 1.00 atmospheres. After the mixture attains equilibrium, the pressure of N2O gas has dropped to 0.80 atmospheres. Calculate the equilibrium constant KP at this temperature. So whenever we have initial values and final values, I would strongly recommend using an ice box for this type of problem. Now it says that the initial pressure of N2O is 1.00 atmosphere. So that's going to go in right here for the initial pressure of N2O. There's nothing else in the container at the initial state. So we can put in zeros for those other two initial values. The problem tells us that at equilibrium N2O has dropped to 0.80 atmospheres. So I'm going to put 0.80 atmospheres in right there. Now in my change row I have.20. That's the change from 1 to.8. Now on the other side of the arrow we know that N and N2 are going to have to increase. If the reactants went down, the products have to go up and the N O is going to go up by 0.20 because it is a two to two mole ratio. So, however much N2O goes down by is the same that N O is going to go up by. Now, N2, it's a 2:1 ratio. So, that means that the N2 only goes up by half as much as the N O went up by. So, if N is a plus.2 20 N2 is going to have to be plus10. So that tells us that our equilibrium pressures are going to be 0.20 for N O and 10 for N2. Now that's all very good, but that's not what the question is asking for. It says calculate the equilibrium constant KP at this temperature. So, so what I'm going to have to do is write the equilibrium constant expression products over reactants raised to the power of the coefficients just like we did earlier and then plug in those equilibrium values into the expression and we can use our calculator and find that the answer is 6.3 * 10 -3rd and so that's the value of KP. Now notice that this is a fairly small number. When I say small, I mean less than about 0.1. And generally speaking, the smaller the value of an equilibrium constant, the less amount of product you're going to have. And that's basically what we have here. We have a moderately small equilibrium constant. So that means we have a whole lot more reactants than we have products. We have 0.80 80 atmospheres of the reactant and much less product than that. And so that's the significance of the small equilibrium constant. Likewise, if we had had a very large equilibrium constant, larger than say about, you know, 10 or something like that, then we would have more product and less reactant. That's usually how that works. Now, let's move on here and take a look at another equilibrium problem. This time it says at the same temperature as the previous example, a mixture of three gases was added to a closed container. The partial pressure of N2O gas was 72 atmospheres. The partial pressure of N O gas was 0.12 atmospheres and the partial pressure of N2 gas was.16 atmospheres. We have the same equation. Part A says what is the total pressure in the container? Well, once again, the total pressure is just the sum of all the individual partial pressures. We just add those three together. 72 plus.12 plus.16 atmospheres and the answer is about 1.00 atmosphere. That's our total pressure. Now, part B says as the reaction proceeds, which gases will increase in pressure and which gases will decrease in pressure? And so, on this problem, we have to use the reaction quotient. Now the reaction quotient is products over reactants raised to the power of the coefficients. But we call it Q because it's not necessarily at equilibrium. So we plug these pressures in here. The N O is 0.12 that's got to be squared. The N2 is.16 and the N2O is 72 and that's got to be squared. And when you calculate the value for Q on your calculator, you find that it's equal to about 4.4* 4 * 10 -3rd. Now we have to compare Q with K. And it looks like Q is less than K because on our last problem we just calculated that K was 6.3 * 10 -3rd. So Q is less than K. Now, if you have trouble remembering what that means, I would strongly recommend drawing in the Pac-Man just like this, just like we saw in the uh daily videos and in the unit summary video there on the ultimate review packet. So, it looks like the Pac-Man is eating toward the right and so it's eating toward the product side. And so since it's going toward the product side, that means that the products N O and N2 are going to increase in concentration or in pressure in this case. And the N2O, the reactant is going to decrease in pressure. And let's take a look at another equilibrium question. This time we have Lhatier's principle and it says after the reaction has returned to equilibrium, how would each of the following changes affect the partial pressure of N2 gas in the reaction vessel? So let's say we have some N O gas removed. Well, anytime we remove an object from equilibrium, the reaction is going to adjust itself to replenish whatever we just removed. So in this case if we remove the N O then N2O is going to react in order to increase the value of N O and that's going to increase the N2 as well. So N2 pressure will increase in this case. How about if we add some N2O? Well once again when you add something to a system at equilibrium the system is going to adjust itself to deplete whatever you just added. So if we add some N2O, the reaction is going to use up some of that N2O and we're going to make more N and more N2. So the partial pressure of N2 will also increase in that case. What if we add some argon gas? Well, once again, argon gas has nothing to do with this reaction. It is basically an inert gas in this reaction. So the answer is nothing is going to happen. And there's not going to be any change when we add some inert gas like argon or neon or helium something like that. What if we increase the pressure by decreasing the volume? Well, when you decrease the volume, increase the pressure, the reaction adjusts itself to move toward the side that takes up less space, which is the side that has fewer moles of gas. So in this case that is the reactant side has two moles of gas versus 2 + 1 3 moles of gas. So that means that the reaction is going to shift toward the N2O. So that means that we're going to increase N2O and N O and N2 are both going to decrease. And so that's why we say the partial pressure of nitrogen will decrease in this example or in this part anyway. How about part E? the temperature of the container is increased. This is the one that trips up some students. We have to think about this in terms of delta H. The question says that delta H is a positive value. That means it's endothermic. Remember that endothermic processes imply that heat is a reactant. Heat is being absorbed. And so that's why I write heat as a reactant here. If we increase the temperature, that's the same thing as adding heat. So if you add heat then it's going to shift toward the product side isn't it? So we're going to increase the amount of N. We're also going to increase the amount of N2. So that will increase while the N2O of course would decrease. And so that's how you answer these questions involving Leadier's principle. Now let's move right on to unit 8 which is acids and bases. Let's take a look briefly at the relationships between the hydrronium ion concentration, the hydroxide ion concentration, pH and p. Now, we need to know that H30O plus the hydrronium ion concentration times the hydroxide ion concentration equals KW. And as long as we're at 25 degrees C, that value for KW is about 1* 10 -14th. So those are some very important values uh to use. Now they'll give you these equations on the equation packet, but really it's in your best interest to know these ahead of time as well. This is telling us that at 25° C, the concentration of hydrronium ions times the concentration of hydroxide ions equals 1 * 10 -14. And we have another pair of important equations here. PH is equal to the negative log of the hydrronium ion concentration and PO is equal to the negative log of the hydroxide ion concentration. So with this set of equations here we can answer all kinds of things about pH p hydroxide concentration hydrronium concentration. Let's say that at 25 degrees C a solution has a pH of 4.12. determine its p, its hydrronium ion concentration, and its hydroxide ion concentration. Well, the p part is probably the easiest. We know that pH plus p equals 14. So to solve for p, we just have to take 14 minus that pH. So 14 minus 4.12 is about 9.88. So that's the first part of this. Now the hydrronium ion concentration we're going to use this equation right here. And so to find the hydrronium ion concentration from the pH we have to take the negative antilogue which means we have to take 10 to the negative pH power in order to find hydrronium. So if the pH is 4.12 we have to take 10 to the -4.12 power on our calculator. And so that's equal to about 7.6. 6 * 10 - 5th moles per liter. Now the hydroxide ion concentration is done pretty much the same way. It's 10^ the PO power. So we just calculated earlier that the p was 9.88. So the hydroxide ion concentration is just 10^ the 9.88 power. So that's about 1.3* 10^ the -10th. Now, if you take the hydrronium ion concentration and the hydroxide ion concentration and multiply them together, you'll find that the answer is pretty close to 1 * 10 -14th. Not exactly because of rounding, but it is pretty close. Now, speaking of acids and bases, it's important that you're able to work with strong acids and strong bases. You have to recognize the strong acids. The AP curriculum sets out six strong acids that you have to know. Now, some textbooks and some teachers may uh include some others. That's great. These are the six that you have to know. Hydrochloric acid, hydro bromic, hydroiotic, nitric, sulfuric, and perchloric. And the concentration of an acid if it's a strong acid is equal to the concentration of the hydrronium ions. And so for example, if we want to find the pH of a strong acid, all we have to do is take the negative log of that strong acid's concentration. So for example, if you have a question where it's 0.15 molar nitric acid, you have to remember that nitric acid is a strong acid. So to find the pH, you just take the negative log of 0.15 and you find that the answer is 0.82. Now, how about the strong bases? Well, the strong bases are the group one and two hydroxides. Now, some of these are not very soluble, so we don't work with them very often, but these are the most common strong bases that you would probably see on the AP exam. Remember that for any strong base, the concentration of the hydroxide ion is equal to the concentration of that compound. Now don't forget that if you have the group two hydroxides, you have to multiply it by two because there are two hydroxides uh for every one formula unit in the compound. So for example, if you have 0.012 molar calcium hydroxide and the question asks you to find the pH of that, well you have to remember that the hydroxide concentration is two times that value. So the hydroxide is 0.024. So when you find the p you take the negative log of 024 not 0012 and when you do that you find that the p is 1.62 and to find the pH you take 14 minus that value. So you find that the pH is about 12.38 in that case. Now let's take a look at what happens in the case of weak acids. So for weak acids and bases, remember these are compounds that do not dissociate completely in water. And so you have to work them as an equilibrium problem. Basically an ice box is my preferred method, my recommended method of solving these. So let's say we have a question where we're asked to determine the pH of 0.4 40 molar hypob bromis acid HBr and it tells us the Ka the equilibrium constant here 2.3 * 109th well it would be helpful to write the equation so we have the HBr plus water and we know that the products are going to be the conjugate acid of the base the base is water in this case so the conjugate acid is hydrronium and the other product is the conjugate base of the acid ID the acid is HBr. So the conjugate base would have to be Br negative. Now as you write the ice box here we start with the 0.40 molar as the initial concentration of HBr. Of course the products are zero and we don't know any of the other values. So the change is minus x and on the product side it's plus x. So our equilibrium values are 040 minus x x and x. And now we have to write the equilibrium constant expression. So that's K a equals hydrronium concentration times the BR negative concentration all over the concentration of HBRO. Now we can plug and chug. The Ka is given to us in the problem as 2.3 * 109th. We have X and X all over 40 minus X. There's this little math trick that I recommend that you do so that you can avoid the quadratic equation and some extra steps and you just ignore that minus x and we can cross multiply. We find that x^2 = 9.2 * 10 -10th and now we take the square root and you find that x is equal to about 3.0 * 10 - 5th moles per liter. X in the ice box is shown to be the concentration of hydrronium ions. So to find the pH, we take the negative log of the hydrronium ion. So negative log of 3.0 * 105th. So the pH is equal to about 4.52 once you plug that into your calculator. Now sometimes they're going to ask you to calculate the percent dissociation. The percent dissociation for any weak acid is just the value of x that you got here divided by the initial concentration of the acid times 100 of course. So in this case it would be that 3.0 * 105th that was our x value divided by 0.40 and then times 100 to get the percent as about 0.0075%. That's a very minuscule percentage of these HBRO molecules that dissociated. But that makes sense because notice the Ka value is very very small. That's an exceedingly small Ka value. The smaller the value for Ka, generally speaking, the smaller or the lower your percent dissociation is going to be. And likewise, the larger the Ka value, the higher your percent dissociation is normally going to be. And of course, we can work weak bases in a similar way, except of course, we have KB and we solve for the hydroxide value and we find the P and then get the pH from that. Now, sometimes we have questions involving salts, ionic compounds. And this is something that I've I've kind of taken for granted in my daily videos. And sometimes we have uh questions about this. I'll have NAF and I assume that students know that we ignore the Na because it's a spectator ion and we solve it as a weak base problem. But this is a good question. How do I know that NAF is basic and not acidic? Now, every now and then on the AP exam, they'll ask you questions about this. They'll give you an ionic compound, perhaps a random ionic compound, and they'll ask you, is this ionic compound or is this salt acidic or basic in its nature? And there's a little trick that we can do to figure that out. It's actually quite simple. All you have to do is take that ionic compound and add it to water. Now, as you write water, I would recommend that you write it as HOH. And what you do there is you just swap the ions just like you were writing a double replacement reaction back in first year chemistry. So the Na gets with the O to make NaOH and the H gets with F to make HF. And you realize NaOH is a strong base and HF is a weak acid. And so strong base versus weak acid, the base wins. And so that means that NAF is a basic compound. That's a little trick that we can use to kind of figure that out. Now, it's I'm not saying that NAF actually reacts with water like this. This is just a little trick to help us figure that out. Now, likewise, if we have ZNB Br2, zinc bromide, we could do the same thing. We could add it to water and just use this little trick here to see if ZNB Br2 is going to be acidic or basic. Well, in this case, the zinc gets with hydroxide to make zinc hydroxide, and then H gets with the Br. Now, we know that zinc hydroxide is not a strong base, so it has to be a weak base. And HBr is one of our big six strong acid. So, this time it's weak base strong acid. The acid wins. And so, that means that zinc broomemide is somewhat acidic in nature. And what if we have something like this, like KCl, potassium chloride. Well, same trick. You can add it to water. And K gets with O to make KOH. H gets with Cl to make HCl. And we know that KOH is a strong base. HCl is a strong acid. And guess what? Strong base, strong acid, they neutralize. That's a tie, which means that KCl is neutral. It's neither acidic nor basic. And so that's just a little trick that you can use to figure out if a salt is acidic or basic. Now, speaking of acids and bases, it's important to realize that in an acidbased titration, the goal of that titration is normally to determine the concentration of an acid or base. And very often on the AP chemistry exam, you'll be presented with an acidbased titration curve. And it often looks something like this. Now, let's say that we have a question like this. If 10 milliliters of a weak acid was used in this titration, determine the acid's concentration. I would recommend that you use the titration equation for something like this. The marity of the acid times the volume of the acid equals the marity of the base times the volume of the base. And so the marity of the acid in this problem is what we're trying to solve for. It says determine the acid's concentration. So we're solving for m sub a. Now V sub A the volume of the acid is given to us in the question. It says it's 10.00 milliliters. So that gets plugged in for V sub A. Now M subB is the marity of the base. Well the problem doesn't say what the titration curve does. It says it was.1 molar of base. So I'm going to plug that in there for the marity of the base. And the volume of the base is a little piece of information that we have to get out of the titration curve itself. We see that the inflection point, this equivalence point right here. It's where where it's labeled X is at 25 milliliters. So that means that we use 25 milliliters of the base to get to the equivalence point. So now we just have to solve using simple algebra and you find that the value for M sub A is about.25 molar of acid. So that's how you can solve a simple question like this using a titration curve. Now, how about if we're asked to estimate the Ka, the acid dissociation constant of the weak acid used in this titration? Well, a key little nugget of information that you need to know is that at the half equivalence point or halfway to the equivalence point, the pH at that moment is equal to the pKa or the negative log of the Ka of the acid. So 25 milit got us to the equivalence point. So halfway there is 12 1/2. So it looks like it's right here. It's denoted by point V on this curve. That seems to be it's kind of hard to eyeball that. Probably somewhere around 5 and a half. 5.4 is the pKa of that acid. So that means that the Ka is 10 to the 5.4 power which is around 4.0* 0* 10 -6. So we can estimate the Ka of an acid of a weak acid used in a titration just by looking at the titration curve. Now let's hone in on this titration curve a little bit more and let's estimate the primary reaction component of of course other than water present in the titration flask at each of the points labeled on the titration curve. Now at point V, notice that's the halfway point. At the halfway point, we are essentially at a point where the concentration of the acid is exactly equal to the concentration of the conjugate base. And so we can say that the acid and the conjugate base HA and A minus are equal at that point on the titration curve. Now, how about point W? Well, notice that we have a pH that's higher, that's more basic than the halfway point. So, it goes to reason that if we have a pH that's more basic than the halfway point, well, guess what's going to predominate? The conjugate base at this point. So, that means that as the pH is rising above that halfway point, the conjugate base is the primary component. Now there is still some HA some acid still there but it is decreasing as that titration continues at point X we are at the equivalence point all of the HA has been reacted all that's left is the conjugate base and so A minus the conjugate base is your primary component in the titration at the equivalence point and essentially only at the equivalence point. Now, if we go past that point, we've overshot the titration. Well, if you just keep adding hydroxide, just keep adding, keep adding, keep adding. Well, guess what's there? It's mainly hydroxide. So, it's hydroxide O minus that is the primary component in the reaction at those points, both Y and Z, because you've basically overshot the titration. Now, let's move on to unit nine, which is about the second law of thermodynamics and electrochemistry. So a very important concept in unit 9 especially about the second law of thermodynamics is this concept of entropy. Entropy is a measure of the total number of possible energy states in a material. Sometimes we talk about it in terms of how much disorder there is or how much chaos there is. Very often it's also described as a measure of how well dispersed a material is. All of these are are nice descriptions to help us understand and and conceptualize entropy, especially in AP chemistry. Essentially, we need to know that solids have very little entropy. They have the least entropy of all of your states of matter. Essentially, there's not much dispersal of the matter in a solid. In a liquid, the matter is somewhat more dispersed, and so we have more entropy. In an aquous solution, most of the time we have even more entropy than we had before. And then in a gas, of course, there's a whole lot of dispersal. Those molecules are moving around all over the place. So a gas has the most entropy of your states of matter. Now, it's important to realize that if you have a material at a higher temperature, that has more entropy than materials at lower temperatures. But what if you have perhaps uh two systems and they're both gases? Perhaps they're at the same temperature. Well, the tiebreaker in that case is to start counting molecules because a system with more molecules is going to have more entropy than a system with fewer molecules. So let's take a few examples here. Let's predict the sign of delta S, change in entropy for each of these processes. So in this first equation here we have a solid that is transitioning into a mixture of a solid and a gas. Now solids have the lowest entropy. Well solid and gas would have higher and so in this case we're increasing entropy so it's a positive sign for delta S. In this next equation here, it looks like we have a mixture of a solid and a gas transitioning into all solids. So once again, a solid and a gas changing to all solid. Well, that's a decrease in entropy. So our delta S should be negative for this one here. Now, how about this case? Here we have two ions. Those are aquous and they're transitioning to a solid. Well, aquous has a moderately high entropy. solid has the lowest. So in this case we are decreasing in entropy. So we have a negative sign for delta S. And then let's do one more. Here we have this combustion reaction. Now notice that everything is a gas. And so states of matter will not help us here. Like I said earlier, if there's a tie, we've got to count the molecules. So on the reactant side, we have five molecules of gas. On the product side, we have six molecules of gas. So 5 to six is an increase in the number of molecules. Hence that's an increase in entropy. So we have a positive sign for delta S in this case. Now as we talk about entropy it's also important to realize there's this concept of thermodynamic favorability and this answers the question is a reaction going to take place and if it does then we say it's thermodynamically favorable. Now, there are two driving forces that can determine if a reaction is going to be thermodynamically favored. The first one, which we learned about back in unit 6, is enthalpy. That's the delta H. You might know that most of the reactions that you're familiar with in the lab, tend to be exothermic because exothermic reactions tend to be favored by the universe. The universe likes exothermic processes. And so if you have a reaction that's exothermic, that exothermicity is a driving force for that reaction. Now the other driving force is entropy. And we know that the universe likes it when entropy increases. And so if you have a reaction that has a delta S that's positive, then that's going to be a driving force. The entropy is going to be a driving force for that reaction. Now, if you ever encounter a reaction that is both exothermic and has increasing entry, so that's a negative delta H and a positive delta S, then the reaction is really going to be favored basically at all temperatures. If it's the opposite though, if you have an endothermic reaction and it decreases in entropy, well, the universe doesn't like either of those. So, that process is never going to be favored no matter what the temperature is. Now if only one of those driving forces is favored by the universe the process is going to be thermodynamically favored only at certain temperatures. So for example if the delta H is positive endothermic and delta S is positive increasing entropy that process is going to be favored only at relatively high temperatures. Now likewise if the delta H is negative it's exothermic and delta S is negative decreasing entropy then that reaction is going to be favored only at relatively low temperatures and so we have to understand how that works. Now how can we measure thermodynamic favorability? Well there there is a numerical value for that. We call that delta G. That's the Gibbs free energy. And there are a couple ways to calculate delta G, Gibbs free energy. We can use this equation right here. Delta G equals delta H minus T * delta S. Uh we can also use this equation if we know the equilibrium constant. We can take delta G and that's equal to R which is 8.314 JW per mole per kelvin times the Kelvin temperature times the natural log of the equilibrium constant. We can also calculate delta G in a similar way to we did earlier with delta H. The sum of the Gibbs free energies of the products minus the sum of the Gibbs free energy of the reactants. That's another way to do that. Uh that could pop up on the exam as well. Remember that if delta G is a positive value, the reaction is not thermodynamically favored. On the other hand, if delta G is a negative value, then that reaction is going to be thermodynamically favored. So be aware of how that works for thermodynamic favorability and delta G. Now the last few parts of unit 9 have to do with electrochemistry. So let's take a look at a galvanic cell problem here. A galvvenic cell sometimes called a voltaic cell is basically a battery. It is a thermodynamically favored process where we have essentially an oxidation reduction reaction or a redux reaction taking place. Now this is different from an electrolytic cell. An electrolytic cell is electrolysis. You have to put energy into that electrolytic cell. It is a non thermodynamically favored process. It requires an external power source to run that electrolytic cell. In this example though, we're going to look at a galvanic cell. This is the one that basically runs on its own. It's a battery. So here we have a picture and I have silver and silver ions on one side of this and on the other side I have nickel with nickel ions. Of course I have some spectator ions. I have some nitrates and some chlorides thrown in there for good measure. And we want to take a look at the half reactions. Now each half reaction is normally going to be given to you on the test. And so in this case we have these half reactions. Notice that these half reactions are always written as reductions. That's just how we do it. Okay? I know that we have to have one reduction, one oxidation, but in the tables, they're always going to write them as reductions. And so in these, let's first of all calculate the total voltage or the cell potential here. Now, the cell potential, the EC cell, is going to be the standard reduction potential of the cathode minus the standard reduction potential of the anode. And the way we do this is we have to subtract these numbers in such a way that we get a positive value for EC cell. Since all galvanic cells are thermodynamically favored, the value of the voltage, the value of the EC cell has to be positive. So that means that the more positive value goes in the position of the cathode and the more negative number goes in the value of the anode. So we have to subtract it like this.80 volts minus a negative.25 volts. And so our EC cell is 1.05 volts. If we had tried to subtract it in the other direction, we would have gotten the negative value. And you can't have a negative voltage for a galvanic cell. Now this also tells us some information. It tells us that the 0.80 volts, that's the value for silver is the cathode. So that means that silver is the cathode and we know that reduction always takes place at the cathode and the.25 volts that was the value for nickel that is associated with the anode. So the nickel is the anode and oxidation always takes place at the anode. Couple different ways to remember that. Anode and oxidation both start with vowels. cathode and reduction both start with consonants or if you prefer red cat and an ox that's how uh I tell my students to remember that sometimes now we know that in the galvanic cell the electrons are going to travel through the wire from the anode to the cathode so like this think about AC as in like air conditioning anode to cathode AC so it always travels in that direction now let's take this one step further Let's write the overall balanced equation. Now we know that the cathode half reaction is written as it is. So that means that the silver half reaction stays just like it was in the in the chart here. Well, the anode is the one that you have to flip around. Whereas I say flip an ode. Okay? So the anode gets flipped around. So it's Ni yields Ni2+ plus two electrons. And whenever you add the two half reactions, you have to make it so that these electrons cancel out completely. So they're not canceling out completely right now. So I have to multiply the silver equation by two just like this. So that the two electrons will now cancel out on both sides. And so the overall balanced equation is 2 ag+ plus ni yields 2 ag plus ni2+. Whenever you have a galvanic cell with metallic electrodes like this, I always tell my students, the cat gets fat, which means that the cathode increases in mass. So the cathode is the silver. So as the reaction runs, you're going to see the mass of silver over here increasing and then the anode is going to decrease in mass. So this Ni, this nickel would be uh being corroded away and would actually decrease in mass. at the salt bridge. Don't forget that cations ions flow toward the cathode and annions flow toward the anode. So if you were to zoom in on this, if there were any cations ions like you know sodium or something like that, it would be flowing toward the cathode in this direction. The annions like maybe chloride or uh most likely nitrate or something like that that that would be flowing toward the the anode over here. Wow, that was quite a ride. I hope that was helpful to you. And if it was, please remember to like this video and don't forget to share it with your classmates. I hope you do a great job on the AP exam. Just remember what you've learned all year. You've spent this whole year working hard on this course. Now, it's time to shine. So, go out there and get that score. Thanks for watching and I'll see you soon.