AP Chemistry Sample Multiple Choice Lecture Notes

In this video we&#39;re going to work through a bunch of sample multiple choice questions for the AP chemistry test. Now before you go through this whole video, I really recommend you check out my website. Here you can download these problems first and do kind of a mini test. It only takes 30 minutes and then you can come back here for the answers, and also, I’m going to be adding all sorts of AP chemistry resources to help students with a test so definitely check out this website. I think you&#39;ll find some stuff that helps you. Anyway, before we start, a quick legal disclaimer: the AP chemistry is a trademark owned by the college board, that&#39;s not me and the college board is not affiliated with and does not endorse this video or website—and of course they don&#39;t endorse this video because we&#39;re going to be walking through the trips and the traps that they put in the AP chemistry test to mess up students. We’re going to figure out how you can outsmart them and how you can get a better score on the test. Okay, so let&#39;s get started. Here&#39;s our first question: 10.0 grams of iron and 10.0 grams of sulfur react to completion. Assuming a 100 percent yield, which of the following describes the contents of the reaction vessel. Now, this is a question about stoichiometry but remember the AP exam doesn&#39;t allow calculators on the multiple-choice section. So, the math here is either going to be really easy or it&#39;s going to be conceptual in nature. Let&#39;s start with the equation for the reaction. Now if you didn&#39;t know what it was right away, the answer choices tell you what the product&#39;s going to be. It’s going to be FeS, iron (II) sulfide. Now, the molar ratio of the reactants is one to one which makes things simple. One mole of iron reacts with one mole of sulfur to produce one mole of iron (II) sulfide. Now, we know that both reactants have the same mass, 10 grams each. Let&#39;s pull up their molar masses from the periodic table and we can see that iron is 55.85 grams per mole and sulfur is 32.06 grams per mole. Now to find the moles of each reactant you could divide 10 by the molar masses and that would give you the moles, so you could figure out the limiting reagent, but you don&#39;t even need to go that far. After all, the question doesn&#39;t ask how much is left in the vessel, just which species--what&#39;s left over. So, think it through. Iron has a higher molar mass, so an equal mass of iron contains fewer moles than an equal mass of sulfur, right? That means that iron with fewer moles is going to be the limiting reagent. Because the problem states that there&#39;s 100 percent completion, it means all the iron is consumed. All the limiting reagent is consumed. So, what&#39;s left behind in the reaction vessel is going to be choice C, iron (II) sulfide, that&#39;s the product, and the leftover sulfur which is the excess reagent. Okay, next one. Identify the Bronsted-Lowry acid conjugate base pair, and we have these choices. Now a little background. A Bronsted-Lowry acid is defined as a species that donates a proton or an H+ ion, just different names. A base is a species that accepts a proton. So, let&#39;s look here at a weak acid in equilibrium. Here&#39;s the acid, HA, and it donates a proton, and the proton donation leaves behind a negatively charged ion, right? It leaves behind A-. It&#39;s lost the proton and this minus can re-accept a proton in the reverse direction, and that makes HA again. We call this species A-, we call this the conjugate base. A- is the conjugate base of the acid HA. So, HA/A- represents the acid conjugate base pair. And sometimes we&#39;ll talk about bases and conjugate acid pairs. That’s where we have B and BH+. But look, here&#39;s the most important part. In all cases a conjugate pair differs by only one single proton. Okay, take away an H+ from HA and you get A-. Add an H+ to B and you get BH+. They differ by one proton. Okay, so keeping that in mind let&#39;s take a look at these answer choices. Only choice D makes sense. Okay let&#39;s start with this guy H2PO4-. It’s going to donate or lose one proton and it&#39;s going to turn into HPO4 2-. Okay, this in turn can accept or gain a proton to turn back into H2PO4-. They differ by one proton because they&#39;re an acid conjugate base pair. But let&#39;s look at some common mistakes here so they don&#39;t trip you up. Okay, choice C might be tempting because it shows nitric acid and the ion left behind when it deprotonates, and these differ by only one proton or one H+. However, nitric acid is a strong acid. There&#39;s 100 percent dissociation and no equilibrium, right? So, there&#39;s no double arrow here. This means that the nitrate ion over here never accepts a proton to reform nitric acid the reaction doesn&#39;t happen in this direction, so nitrate is not a conjugate base—another reason why it&#39;s always useful to have your strong acids and strong bases memorized for the AP exam. Okay, choice A might be tempting because it represents the conjugate acid and the conjugate base of water itself. You know we often sort of talk about these two ions together when we&#39;re talking about water and pH, but they themselves are not a pair because they differ by two protons. You&#39;ve got to lose two protons to get this you got to gain two protons to get this so they&#39;re not going be a pair. And finally here choice B. This is two oxyacids. They differ by one oxygen they don&#39;t even differ by the number of hydrogens, so that&#39;s definitely not going to be the right answer. So, D it is. Okay, here [a]the following is a photoelectron spectrum of an element, and its ground state now many recent AP exams have been featuring photoelectron spectra. So, these PES graphs are useful to understand, Now they&#39;re not as complicated as they seem. Here&#39;s how we&#39;re going to break it down. The y-axis is always going to represent the number of electrons. Sometimes it&#39;s going to be specific, but more often it&#39;s going to be something like the relative number of electrons. Okay, the x-axis is always going to represent energy. It might be labeled in a bunch of different ways. You&#39;ll see a variety of different units, but the important thing to notice is that the axis decreases moving this direction and increases in this direction. That&#39;s standard. Now, for most of these problems the labels the units aren&#39;t really going to matter. What matters are the peaks here. These are always going to be the same—we’re going to see spikes or peaks of differing heights. And basically, these spikes represent the electrons in electron orbitals. Most PES questions are going to ask you to translate the graph into a specific element like we&#39;re doing here, or translate it into an electron configuration, so here&#39;s how you do that. First look at the smaller spikes, these usually represent filled s orbitals and look for a pattern moving from left to right. These spikes represent 1s2 2s2 3s2. Electrons that are closer to the nucleus like 1s2 have a higher binding energy so that&#39;s why they&#39;re over here where the values are higher. Okay, now the spike on the far right over here is smaller than the others. It doesn&#39;t fit this pattern so we&#39;re going to ignore it for now but these here 1s2 2s2 3s2 pretty sure about that now the next spike here is a p orbital notice how this spike is three times higher than the s spikes. That&#39;s because an s orbital, a filled s orbital, holds two electrons but a p orbital holds six electrons which is three times as many. Okay, so let&#39;s see what we have so far this is going to be 1s2 2s2 and then our first p orbital is going to be 2p6 then we&#39;ve got 2s2. And what&#39;s going to come next? Our outlier, our little spike over here. What do you think this short spike means? It&#39;s half the size of a filled s orbital so it only has one electron. Well, what comes after 3s2? 3p right? And there&#39;s only one electron in it so it&#39;s got to be 3p1. So, our complete electron configuration is 1s2 2s2 2p6 3s2 3p1. We can add up the number of electrons that are in all these orbitals we get 13. So that is choice B, aluminum, which in its ground state has 13 electrons. Now, let&#39;s look at a couple of common mistakes some people may be tempted to choose boron, A, which has five electrons. Now why might you do that? Well because there&#39;s five spikes one two three four five. But remember, the x-axis is the relative number of electrons these spikes represent electrons in orbitals not individual electrons. If you picked D, potassium, you probably thought that this little spike at the end was 4s1. But remember, you have to go in order of a standard electron configuration so you before you get to 4s1 you have to have 3p6, and we don&#39;t have that here so that&#39;s why aluminum with 13 electrons a neutral aluminum atom is our correct choice. All right, a saturated solution of (NH4)3PO4 has a concentration of ammonium ions of 1.50 molar. What is the concentration of ammonium phosphate? Okay, ammonium phosphate is an ionic compound that&#39;s fully soluble in water, and we know that because all ammonium compounds are soluble ammonium phosphate dissociates according to this equation. This three subscript on the ammonium means we get three ammonium ions after dissociation, okay? And there is a one to three ratio of ammonium phosphate to ammonium ion so because of this ratio, if the concentration of ammonium ion is 1.50 molar—that’s given to us right here in the problem, right? The concentration of ammonium phosphate must be one third of that amount. That&#39;s 0.50 molar. So choice A is the correct answer, but some common mistakes. You may have been tempted by choice C because you see the 1.50 there, but that would only be true if the ratio were one to one like ammonium chloride. Here then, the concentration of each component would be 1.50 molar. And choice D here represents a 1 to 3 ratio, but it&#39;s reversed there&#39;s three times as much ammonium phosphate as ammonium and that&#39;s backward okay. So that&#39;s why choice A is correct. All right this is going to be a little bit of a conceptual math workout. An antacid tablet containing calcium hydroxide and we get the molar mass is titrated with a 0.100 molar solution of hydrochloric acid HCl. The end point is determined by using an indicator. Based on the information below that&#39;s going to be this here what was a mass of the calcium hydroxide in the tablet. And we&#39;re given the buret readings. The initial reading is 5.1 milliliters, and the final is 25.1 milliliters. Okay, where do we start? Well this question is about titration we have a base calcium hydroxide and an acid hydrochloric acid. Now, it also mentions the end point here and that&#39;s fundamental to any acid-base titration when the neutralization process is complete. Okay the end point is when the moles of H+ from the acid equal the moles of hydroxide from the base. It’s also called the equivalence point. So let&#39;s work through this problem. So first, how many moles of HCl do we need to neutralize the base? Okay the buret contained the HCl, the volume started at 5.1 milliliters and went to 25.1 milliliters so we can do some subtraction here giving us 20.0 milliliters. That&#39;s the volume of acid that we used, but that&#39;s not the number of moles, that&#39;s the volume, okay? So to find the number of moles, we have to use the molarity equation: molarity equals moles divided by liters. Now you probably have this memorized because you use it a lot, but it is one of the given equations on the AP exam. So rearranging this equation shows you that moles equals molarity times liters. So we can start with moles equals 0.100 molar. But now, remember that the equation needs you to use liters, and this number here is milliliters. Okay, so you need to convert milliliters to liters before you can use this equation. There’re a thousand milliliters in one liter so you can divide by a thousand or just shift the decimal three spots to the left. Okay so we had a 20.0 and we go one two three point zero two zero liters very important. And when you think about the math, you&#39;re multiplying by 0.1 which just means that you&#39;re going to be moving this decimal spot one more to the left, okay? This tells us how many moles of HCl are needed to neutralize the base. Okay, it&#39;s getting a little bit cluttered here so let&#39;s get rid of the stuff that we don&#39;t need right now, and we&#39;ll bring it back later. Also, I’m going to stop writing these zeros every time, just remember that there are two zeros here at the end for sig fig reasons, but usually sig figs aren&#39;t a big deal with calculation problems on the multiple choice anyway. Okay, that&#39;s better, a little cleaner here. So we know that 0.002 moles of hydrochloric acid neutralize the base. We can use this information to find the mass of calcium hydroxide. We need to look at the balanced neutralization equation for this reaction. You might even want to sketch it out on the test. Okay, here&#39;s our equation. It shows that two moles of HCl are required to neutralize one mole of calcium hydroxide. So if it took 0.002 moles of HCl to reach the end point there&#39;s only 0.001 moles of calcium hydroxide present. See, twice as much acid is needed. And look at the formula too, right? One mole of calcium hydroxide contains two moles of hydroxide ion so that makes sense, right? Anyway, we have 0.001 moles of calcium hydroxide. Now, how many grams is that? Well, we can use molar mass to calculate that; and we&#39;re even given the molar mass in the problem. So, 0.001 moles of calcium hydroxide. Do this math when you are multiplying by 0.001, you&#39;re moving the decimal place three spots to the left. So you&#39;re going to end up with 0.0741 grams. Okay, now let&#39;s take a look at our answer choices and there we go, number one or I should say A, but look, we&#39;re going to spend a second looking at some common mistakes so that you won&#39;t make them. There’s a reason why the AP test has these other choices they&#39;re traps designed to trick you. Okay, one thing we talked about it a minute ago that students often forget is to convert milliliters to liters when using the molarity equation. Okay, so this is what would have happened if the student forgot to convert to liters. And this gives the impression that there are two moles of acid. Carry that through, look at that, 74.1 grams it&#39;s waiting there to trick you. Okay, another error, this is very common, is to forget about the mole ratio here okay and to think that you need 0.002 moles of calcium hydroxide to go with 0.002 moles of hydrochloric acid. Okay, so if you carry this error through look at that you get C. And finally, D is an example where the student made both of those mistakes and that incorrect answer is waiting right there with open arms to welcome them in and to trap them. So make sure that you&#39;re cognizant of some of those common mistakes so that you won&#39;t make them on the test. Which of the following 0.10 molar solutions would experience the highest percent ionization? Now, the first thing to notice here is that all of these answer choices are acids. They all have a hydrogen cation attached to a polyatomic anion. Also, percent ionization questions on the AP exam are usually about weak acids. Okay, since these are all acids, they ionize. This is the same as deprotonation. Now these three acids are also a specific type of weak acid we call these oxyacids because the proton is attached to an oxygen. Here, we&#39;re going to show their structural formulas so you can see this a little bit more clearly. You don&#39;t need to know the structure of these acids to answer the question, but these structures are going to help us illustrate the main points. Okay, so each of these hydrogens is going to dissociate. We want to know which of these acids experiences the highest percent ionization, and the higher the percent ionization the stronger the acid. The thing to keep in mind for oxyacids is this: as you pull electrons away from the OH bond that, weakens it. And the hydrogen ionizes more easily. The red arrows here represent the forces pulling on the electrons. There’s two things that impact the pull and the AP exam might test either one or both. Okay, first the more electronegative the halogen the stronger the oxyacid that&#39;s because the electronegative halogen will be pulling electrons away from the OH bond. But in this case, our halogen is always bromine so that doesn&#39;t matter here, but keep in mind for other questions like this. The second thing to remember is that oxygens are electronegative too. So the more oxygens you have, the stronger the acid, because they pull electrons away from the OH bond. In this question choice C has the most oxygens, therefore it pulls hardest on the electrons in the OH bond, weakening it. This acid deprotonates most readily, meaning a higher percent ionization and a stronger acid. Okay, given a constant temperature which of the following reactions is most likely to proceed spontaneously. And then we have all these different reactions with different delta H values and different reactants, and different products. How are we going to break this down? Well, this question is about spontaneity, and when we&#39;re thinking about spontaneity there are two main factors to keep in mind: the first is enthalpy, and the second is entropy. If a reaction is both exothermic and increases entropy, it&#39;s guaranteed to be spontaneous. If a reaction is both endothermic and the entropy decreases, it&#39;s guaranteed to be non-spontaneous. Now, any combination in between these two options requires you to know the temperature and the exact values of entropy and enthalpy in order to determine spontaneity using the Gibbs free energy equation. So this question says that temperature is constant, so we don&#39;t need to worry about those in between cases. We’re just going to look at these two options they&#39;re either always spontaneous or always non-spontaneous. Okay, first let&#39;s look at enthalpy. Okay, we&#39;re given enthalpy values for each reaction, that is the delta H. Okay just remember a negative delta H indicates an exothermic reaction and a positive delta H indicates an endothermic reaction. That’s because we&#39;re looking at the change in enthalpy, and exothermic reactions lose heat, so that change is going to be negative. Remember that because it&#39;s commonly tested on the AP exam. Okay, so right off the bat okay we&#39;re looking for spontaneous reactions. So under these conditions they have to be exothermic. We can cross out A and B right away because these are endothermic. Okay, now it&#39;s not that endothermic reactions can&#39;t be spontaneous, but we&#39;re looking for a candidate that&#39;s going to meet both of these conditions. Okay, now to narrow it down between C and D we need to look at entropy, the measure of disorder. There are two things to keep in mind about entropy questions. The first is that entropy increases when you move from solids to liquids to gases. Okay, these phases of matter have increasingly more entropy. The second is that entropy increases when you produce more moles of gas. That’s because gas has a lot of entropy, so the more gas particles you make, the more entropy your reaction is going to be contributing. So let&#39;s look at these two changes for entropy in choice C. An aqueous reactant and a solid reactant produce two moles of gas, that&#39;s a definite increase in entropy. That’s good. And C is probably the correct answer, but let&#39;s look at choice D just to be sure. Here we see one two three, remember the two here, one two three moles of reactant gas produce two moles of product gas. That’s a decrease in the moles of gas; it&#39;s a decrease in entropy too, which is not favorable to spontaneity, so we can cross this out. And the correct answer is choice C. It’s the only one that meets both of these conditions; it’s exothermic and entropy increases.[b] All right, here’s a lab-based question. A student is performing a redox titration by oxidizing Fe2+ ions to Fe3+ ions using a solution of 0.0010 molar KMnO4. The target solution is iron (II) sulfate. And then we&#39;re given the balanced redox reaction. Okay, so that&#39;s the background. Now here&#39;s the actual question: assuming all glassware has been cleaned and thoroughly rinsed with distilled water, what is the proper laboratory procedure before adding the KMnO4 solution to the buret? And we have choices. Now as we said, this is an example of a lab procedure question and you can expect a couple of these on the multiple choice section of the test. And some are pretty common sense but some get a little bit particular. Okay, the best way to approach these questions is to focus on two main concerns. The first is lab safety and the second is preventing error. Almost all the lab questions the lab procedure questions you&#39;ll get on the test have to do with one of these two themes. So here we&#39;re not talking about safety; here it&#39;s about preventing error. Now your first impulse might be to select choice A because it&#39;s never a bad idea to make sure that your glassware is clean, but the problem has already stated that the buret has been thoroughly rinsed which means that it has been cleaned but there are still drops of water inside the buret, right? So what happens when you add KMnO4 solution? Well that extra water that&#39;s in the buret dilutes the solution which lowers the molarity. So you think you&#39;re working with a solution that has this concentration, but since there was water left over inside the buret that diluted the solution, what you&#39;re working with is actually slightly more dilute than this value here because of the leftover water. So the correct answer choice is the buret first should be washed with excess permanganate solution to clear out any pure water that has stayed behind so that it won&#39;t get diluted by the leftover water. Okay now there are some traps here that you might have picked and that&#39;s okay. Let’s talk about why if you picked choice D. Okay, you might have noticed that the balanced equation has some H+ ions in here as reactants and like many complicated redox reactions, you have to acidify the reaction to balance it properly, but here when you&#39;re doing it in the lab, the acid is actually added to the iron sulfate solution. So there&#39;s nothing in the problem that suggests that to get ready for this titration you would add acid to the permanganate; so avoid making that assumption. And finally choice C is very very wrong because you&#39;d be introducing traces of one reactant to the other reactant before you actually do the lab procedure, so that would contaminate the buret and it would be very poor procedure. So our only good choice here is choice B. Okay for this hypothetical reaction the following initial rate data were obtained. The temperature remains constant. And then we got this table here showing four experimental trials along with the initial concentrations of the reactants and the initial reactant rate the initial reaction rate. So, this is a problem about kinetics. And the AP exam often asks rate law questions a lot like this in the multiple-choice section. Usually, you&#39;re given a table of initial concentrations and rates and asks to find the rate law. To solve a problem like this you&#39;d use the method of initial rates. You find two rows where only one concentration changes and you look at the effect that that has on the rate. And there are only three possibilities. If you change the concentration and there&#39;s no effect on rate that species is zero order. If you change the concentration and the rate changes by the same proportion that species is first order. For example if you double the concentration and the rate doubles that&#39;s classic first order. And finally if you change the concentration and the rate increases by the square of the change, the species is second order. So if you double the concentration and the rate quadruples, times four, that would be second order. Okay, so let&#39;s use that method here between experiment one and experiment two you can see the only thing that changes is the concentration of Q. Okay, what happens to it? Well Q doubles 0.2 to 0.4. Okay, now let&#39;s take a look at what is going on with the rate. This quadruples, okay? We have double here, quadruple here, that means the reaction is second order with respect to Q because 2 squared is 4. We’re done with Q. Now before we move on though let&#39;s talk about rates and scientific notation. The AP exam loves giving reaction rates in scientific notation like this so when you&#39;re comparing rates pay close attention to the superscripts on ten. Okay, it might sound weird to say that 6.0 times 10 to the negative third quadruples to 2.4 times 10 to the negative second, but 10 to the negative 2 is a larger number than 10 to the negative third. If you expand these numbers out of scientific notation you can see it more clearly okay 0.0060 times 4 equals 0.024. Okay, got it? So keep an eye on the scientific notation. Let’s get back to the problem now. Experiment one and experiment for M and Q stay the same, but X doubles. What happens to the rate bam bam no effect so with respect to X this is zero order, all right? And the last thing to look at is the concentration of M. Experiment one and experiment three, Q and X stay the same, M doubles. What happens to the rate? The rate doubles M doubles rate doubles classic, boom, first order. Same proportion and once again if we are mindful of the scientific notation we can see that this times two equals 0.012 1.2 times ten to the negative second. Okay, so we know all these orders we figured out from the table; let&#39;s now put together our rate law. Rate laws always start with k. Zero order reactants drop out when we write the rate law so we&#39;re just not going to put them in there, so our rate equation, our rate law, is going to be rate equals k M that&#39;s first order, Q squared—the square because it&#39;s second order. Now let&#39;s bring back our answer choices, and right here D. Dis the correct answer ah-ha. But before we go to the next problem let&#39;s pause for a second, right? There’s an even quicker way to get to this answer. You remember how we immediately established that this equation was, or this reaction was second order with respect to Q? Well look at this look at the answer choices only one of them has Q squared as a term, so if you&#39;re feeling really confident you could have selected D immediately and moved on. Now look that&#39;s a gutsy move but if you really know what you&#39;re doing you could save yourself a lot of time. Okay, the above gaseous reaction occurs and comes to equilibrium in a closed container. Which of the following changes to the system would cause the reverse reaction to speed up? And there are four things that we could do to this system here. Now, AP questions asking about changes to a system at equilibrium are generally going to be about Le Chatelier&#39;s principle. And this states that if a system in equilibrium is disturbed, the system will shift to counteract that change. Okay, here the question is asking us to select the change that would cause the reverse reaction to speed up. So, let&#39;s look at our answer choices one by one to see which would lead to that result. Choice A is to remove some HCl. Now if you add more of a chemical the equilibrium shifts in the direction that consumes the addition. If you remove a chemical, the reaction shifts in the direction that replaces the missing chemical. Okay so in this system HCl is a product so if we remove some product Le Chatelier’s principle kicks in and the reaction has to make more product. Because we&#39;ve removed it, that would shift the reaction forward. That’s not the direction we&#39;re looking for so let&#39;s go ahead and cross off A. Choice B has us adding some hydrogen. Le Chatelier’s principle kicks in again, but this time the reaction has to consume the hydrogen to re-establish equilibrium. You add more, it&#39;s got to use it up, it&#39;s going to move in this direction. So again the reaction shifts right, boom, cross it off. Now the next two choices here are about temperature. Another way to offset equilibrium is to change the temperature of the reaction. To solve these you need to treat heat energy like a reactant or product. So does this reaction create heat or does it absorb heat? This reaction has a negative delta H, so the reaction is exothermic—it releases heat. So we can treat heat as a product that is made by the reaction. Heat can be considered a product. Choice C is to lower the temperature, which is the same thing as removing heat energy, okay? So if we were to remove a product, remember we&#39;re treating heat like a product, the reaction would have to shift right to produce more heat. We got rid of heat, the reaction produces heat, the reaction is going to move to the right to make more heat. So cross off choice C. Even though choice D is the only one left let&#39;s reason through it to be sure. If you increase the temperature you&#39;re adding heat, okay? In an exothermic reaction that&#39;s like adding product; it&#39;s like adding something to this side of the equation so to re-establish equilibrium, the reaction shifts left to consume that extra heat energy. Rasing the temperature is our correct answer. Which of the following diagrams provides the best representation of an aqueous solution of HNO3? This question is testing our understanding of the dissociation of ionic compounds and strong acids. Before we take a look at the diagram let&#39;s write out the dissociation equation for nitric acid in aqueous solution. Here it is, nitric acid dissociates into hydrogen cations and nitrate anions. There are two things you need to recognize here. First nitrate is a polyatomic ion. That means that nitrate remains a unit when nitric acid dissociates or ionizes. We get NO3 1- and so that means that right away we can get rid of choice B because polyatomic ions don&#39;t break into individual atoms when they dissociate. It&#39;s always a good idea to have your polyatomic ions memorized for the AP exam. Now the second thing to notice here is that nitric acid is a strong acid. That means that it undergoes complete dissociation or ionization, so we can eliminate choice A because this shows no dissociation at all, right? These hydrogens are still attached to the NO3 in A. We can also eliminate choice C because this shows partial dissociation, right? Here, two of the molecules have the hydrogens attached, and only one of them shows it being dissociated. Okay so this would be correct for a weak acid, but a strong acid undergoes complete dissociation—another good reason to have your strong acids and strong bases memorized for the exam. So that leaves us with choice D, which is the correct answer. It shows one hydrogen ion for every one nitrate ion. 100 percent complete dissociation. Okay, a 0.50 gram sample of zinc solid was placed in a solution of hydrochloric acid, aqueous. The zinc reacted completely with the acid. Which one of the following equations best represents the reacting species described above? This question describes the metal zinc reacting with hydrochloric acid, and then it asks you which equation best represents the reacting species. Note the wording here in red. We’re going to come back to it in a second. Now the first thing that you should remember is that the reaction of a metal in an acid is an example of a single replacement reaction. Now at first you might be tempted to select choice C. That’s pretty understandable because this is actually the full equation for the single displacement or single replacement reaction that would happen. However, remember that we said pay close attention to the wording. The question is asking you which equation best represents the reacting species. That’s a different question, so what we&#39;re looking for here is actually the net ionic equation. That’s a type of abbreviated equation that only shows the substances that are actually reacting. So let&#39;s take a closer look at this reaction. Because HCl is a strong acid, it completely dissociates. And because zinc chloride is a soluble chloride salt, it&#39;s also dissociated in solution. So if we focused on the ions in this equation, our full equation would look like this. Now, look at the chloride ion here. See how it doesn&#39;t react? It starts as chloride, Cl- and it ends up as chloride. We call an ion like this a spectator ion. it doesn&#39;t actually react. So it&#39;s not included in the net ionic equation. So the chloride drops out and we have our net ionic equation. We keep this and we keep this because these actually react, they actually change, during the course of the reaction. This is what we&#39;re left with this is the equation that best represents the reacting species. Look at this. There it is, choice A. the AP exam loves net ionic equations. Sometimes they&#39;re pretty clear about it, but sometimes they get really sneaky like with this question, so pay attention to the wording. Let’s look at the other two answer choices to double check our answer. Choice D looks like a single replacement reaction but notice the hydrogen—it remains as H+ ions instead of being reduced. Oh sorry there, to H2, tempting but wrong. And finally choice B is just wrong. It’s not even a single replacement reaction so we can just get rid of that. Okay, beaker A, there it is, contains 250 milliliters of pure deionized water and beaker B contains 500 milliliters of 1.0 molar KCl dissolved in water we would assume. Silver chloride crystals are added to each beaker. After stirring the contents of the beaker, some solid AgCl remains at the bottom of each as shown below. Which of the following is true? This question is about solubility and the common ion effect. The AP exam loves the common ion effect, so problems like this are pretty common. Let’s take a closer look. It’s more complicated than it seems; there’s a lot going on here. So to make some room we&#39;ll shrink our diagram. Okay, so we have two beakers. One contains 250 milliliters of pure water, the other 500 milliliters of 1.0 molar KCl. Silver chloride has been added to both beakers to saturate the solution. Which beaker has the highest concentration of silver ions? At first glance you might be tempted to select D because you probably memorized that silver chloride is an insoluble salt, okay? It doesn&#39;t dissolve well but if you&#39;ve studied solubility equilibrium you know that all salts are actually soluble even if just a tiny bit. Okay, so some AgCl is going to dissolve, so we can eliminate D right away. It’s a trap, don&#39;t step into it. Okay here&#39;s another kind of trap: you might be tempted by the fact that beaker B has twice as much solvent, twice as much water, so it should dissolve twice as much AgCl. Or you think it might dissolve twice as much HCl, but we&#39;re talking about the concentration of silver ions, which is a ratio of moles per liter. So the concentration of silver ions is not the same as the amount of silver ions. That doesn&#39;t mean that choice B is wrong, it just means don&#39;t pick it for the wrong reasons, okay? So we&#39;re going to put a question mark next to it because the AP exam can be tricky, and it plans for student mistakes. They might be anticipating the fact that some students are going to think, “oh well more should dissolve in beaker B” and forget about concentration. Okay, so we&#39;re just going to put this question mark there, maybe you know be wary of it we&#39;ve marked these two traps or potential traps. And now we&#39;ll solve the problem. This is a solubility equilibrium problem. Okay the salt HCl dissociates according to this equation. As we said it&#39;s a weakly soluble salt, insoluble essentially, so it&#39;s going to be in equilibrium here and we&#39;re going to have an arrow that goes in both directions. Alright, now for an insoluble salt like this you can write an equilibrium expression using Ksp, the solubility product constant. And every salt has a unique Ksp. The equation for silver chloride is Ksp equals Ag+, the concentration times the concentration of Cl-. Now you don&#39;t need to know the actual value for Ksp. Here you just need to know that Ksp is a constant, a specific number, so the concentration of silver ions times the concentration of chloride ions is always going to equal the constant number here. Now beaker A contains pure water so the only source of silver and chloride ions is the AgCl that gets dissolved. And because silver chloride dissociates with a 1:1 molar ratio, both values are the same we&#39;ll just call them X, okay? X for silver X for chloride. The actual value doesn&#39;t matter, okay? Now here&#39;s where things get interesting. Beaker B isn&#39;t pure water—it&#39;s a solution of KCl, and that means there are chloride ions already in solution. Okay, this is going to impact the solubility equilibrium. The solution doesn&#39;t care where the chloride ions came from. They can come from potassium chloride, silver chloride, doesn&#39;t make any difference. Now remember that Ksp has to be the same value for both beakers. It’s a constant. Over here in beaker B there&#39;s a lot more chloride ions, okay? Those came from the KCl. So if the chloride concentration is larger, which it is, that means that the silver concentration has to be smaller. That’s the only way that they can still equal Ksp when you multiply them together, okay? Algebraically speaking, X for silver is smaller in beaker B because this, the Cl- concentration, is larger, okay? Just once again Cl- concentration is larger in beaker B so Ag, silver, has to be smaller so they can still equal the same value. This means that the concentration of silver ions is greater in beaker A, which is choice A. Now this is an example of the common ion effect. When you have a solubility equilibrium, the addition of a soluble common ion, in this case chloride, depresses the solubility of the insoluble ions. Here, the chloride from the potassium chloride depresses the solubility of the silver chloride, which means fewer silver ions in this beaker. If you know the basic gist of the common ion effect, questions like this are easier to answer and you don&#39;t even need to go through doing math for the whole equilibrium expression. Which of the following is a nonpolar molecule that still contains polar bonds? The AP exam loves to ask questions about polarity, so going into the test it&#39;s helpful to know a couple things. First C-H bonds are essentially nonpolar. Now, know that fluorine is the most electronegative element, followed by oxygen, nitrogen, and chlorine. You should also know that the shape of a molecule affects its overall polarity. In this question we need to meet two criteria. First, the molecule has to be nonpolar. Second, the molecule has to feature a polar bond. So let&#39;s answer them one at a time. First, are any of the molecules here polar? If so we can eliminate them immediately. Okay, choice D is the only polar molecule. As you can see from the Lewis structure here, it would have a tetrahedral shape but two of the bonds are polar C-F bonds, right? That F is way more electronegative than the carbon, than the C, so electrons are going to be pulled in this direction, okay? Now these two bonds are polar. We can now consider the molecule as a whole and we can see that one side of it here has the electrons being pulled to one side. So it&#39;s unbalanced, okay? That means that this is a polar molecule. Now this could be a little bit tricky because this is a tetrahedral shape, which is one that we sometimes associate with nonpolar molecules like methane--methane there—but we want to keep in mind what the individual bonds are and how they&#39;re oriented, alright? So polar molecule, boom, cross it off. Now three nonpolar molecules left. So now we&#39;re looking for polar bonds. We just said that C-H bonds are nonpolar. So boom, that&#39;s why we can get rid of choice C, right? Here, choice A is diatomic, all right? We have a nitrogen bonded to a nitrogen which means that both atoms have the same electronegativity, so we can definitely cross out that choice. That leaves us with choice B, carbon dioxide. Okay the C-O bond is polar. Oxygen has a significantly greater electronegativity than carbon, so we have polar bond there, polar bond there, electrons getting pulled. But the shape of the molecule is linear so it cancels out. These two polar bonds going in this direction cancel each other out, so we have a non-polar molecule that still contains two polar bonds. Okay, which of the following statements is true regarding lithium and cesium? Okay, here we&#39;re comparing some of the properties of two elements, and specifically we&#39;re looking at ionization energy, atomic radius, and electronegativity. The first step here is to recognize that these three things are all periodic trends, okay? They are properties that vary in a predictable manner based on where an element is on the periodic table. The AP exam is never going to expect you to memorize the atomic radius or the ionization energy of every element, but it will expect you to compare the properties of elements based on where they fall on the periodic table. Bam! So to answer this question we&#39;ll want to use our periodic table. We’ll get rid of the answer choices now, we&#39;ll bring them back when we&#39;re ready. Okay, so let&#39;s find lithium and cesium on the periodic table. They’re in group one, the alkaline metals, and this means that the question is asking us to predict how periodic trends change from top to bottom of a group like this. Let’s review these trends. We’ll get rid of the rest of the periodic table and just focus on group one here. Okay, as we move down a group, atomic radius increases, and this makes sense because as we add energy levels the atom gets bigger. So we can see that lithium will have a smaller atomic radius than cesium, which has more energy levels with electrons. Okay, so lithium has a smaller atomic radius. However, ionization energy and electronegativity both decrease as you move down a group. They increase in this direction, but they decrease in this direction. This is also because of the extra energy levels, okay? As we move down a group, the atom&#39;s valence electrons are further from the nucleus. The core electrons shield these valence electrons from the attractive forces of the nucleus that are pulling them in so it&#39;s easier to remove valence electrons. The more core electrons you have shielding the nucleus. So, it&#39;s easier to remove valence electrons, that&#39;s ionization energy, and the nucleus has less attraction for the electrons in a bond and that&#39;s electronegativity, okay? So keep that in mind. So these trends mean that lithium will have a smaller atomic radius, but it will have a higher ionization energy and a higher electronegativity than cesium. Okay, now that we&#39;ve figured out these three things we can run through the answer choices to see which one fits. We’ll start with A. Well that&#39;s wrong because cesium has a lower ionization energy than lithium, so we can cross that. Okay, choice B states lithium has a higher ionization energy and the higher electronegativity. That fits with what we know so that&#39;s our correct answer. And choice D is wrong because lithium has a smaller atomic radius than cesium. So yep, B was our correct answer. Before we move on, one word about periodic trends. Well… a couple words about periodic trends. Sometimes the AP exam will ask about trends moving across a period instead of down a group, okay? So it&#39;s best to know all your trends. Just for the record, when moving left to right across a row, ionization energy and electronegativity both increase, but atomic radius increases moving right to left across a row. That&#39;s the one students usually forget, so be prepared for all the possibilities. A 1.0 liter sealed container contains equal masses of neon and argon. The total pressure of the system is 12 atm. The temperature remains constant. What is the partial pressure of argon? In order to solve this problem you need to understand Dalton’s law of partial pressures, and that can be expressed as a formula or it can be expressed simply in words. Okay, the total pressure of a system is the sum of each partial pressure. Now here, the total pressure of the system is the partial pressure of the neon plus the partial pressure of the argon. So how are we going to find these two pressures? Well we can use the ideal gas law PV = nRT. We have to rearrange a little bit. If we divide both sides by V, we can see that the pressure of the gas equals number of moles times the gas constant times temperature divided by the volume. But look carefully. We don&#39;t have to do that. We don&#39;t have to do that much math here. In this system the temperature and the volume remain constant, and R is always a constant. So this is going to make things easy for us because the variables T and V are the same for both gases; we don&#39;t have to worry about them. The only thing different for each gas will be the number of moles n, boom, so basically each partial pressure is directly related to n, the number of moles. The problem tells us that we have equal masses of neon and argon but n is all about moles, not mass. So let&#39;s compare their molar masses. Neon has a molar mass of about 20 and argon about 40 grams per mole. Now, argon is almost exactly twice as large as neon. Argon&#39;s molar mass is almost exactly twice as large as neon’s. So if the masses that we have are equal, the molar amounts have a two to one ratio. Okay, let&#39;s think through this a second, okay? For example, let&#39;s say you have 40 grams of each gas. It doesn&#39;t tell us how much we have but it says equal masses so we can just assume, I don&#39;t know, 40 grams of each. Okay, what would happen then? That would be one mole of argon and two moles of neon, right? This molar mass is half as much as this, so you need twice as much neon to have the same mass of argon. You need twice as many moles of neon to have the same mass as argon and that is a 2 to 1 molar ratio. Now, the total pressure of the system is 12 atmospheres. Remember that for this problem pressure is equivalent to the number of moles. So if 12 atmospheres represents a 2 to 1 molar ratio of neon to argon, the pressure of neon must be twice as much as the pressure of argon. So the partial pressure of neon must be eight atmospheres and the partial pressure of argon must be four atmospheres, choice B. Remember Dalton’s law, here 8 plus 4 equals 12 the total pressure of the system. Okay, which the following species has the electron configuration shown below? Now we could go through each one of these and we could write electron configurations for all of them, but that&#39;s going to take too much time. Let’s think of a quicker way to do this. This electron configuration tells us the number of electrons in each one of these orbitals so we can add this up and get 36 total electrons. Now which of these atoms or ions has 36 total electrons? Well a neutral krypton atom has 36 protons and 36 electrons, but check it out, that&#39;s not a choice here. That means that the electron configuration must be for an ion where protons and electrons are not equal, so right away since we know we&#39;re looking for an ion, we can cross out A and we can cross out D, since these are neutral. So it&#39;s going to be between B and C. Now let&#39;s take a look at choice B, the strontium ion. Its atomic number is 38, so it has 38 protons but the 2 plus charge means that there&#39;s two more protons than electrons, so it&#39;s missing two electrons which means it has 36 electrons. So we know that&#39;s right, but what about choice C, Br+? If you just glanced at it quickly you might be tempted to think that the plus one charge means plus one electron, which means you go from 35 protons to 36 electrons, right? No, it&#39;s a common error. To make a plus one charge would mean you have one more proton than electrons, so you&#39;d actually have 34 electrons. Now also halogens like bromine don&#39;t form plus one ions, okay? So it&#39;s important to keep in mind that sometimes the AP exam gives an answer that doesn&#39;t exist in real life. Those answers are never the correct ones like Br+, so don&#39;t be fooled by that. And there&#39;s an even faster way to do this problem using the periodic table, okay? Krypton is a noble gas and this is a noble gas electron configuration. Full s shell, full s and p shell, okay? So krypton is here. Abd strontium is here in the second column with two more electrons, right? Krypton is here, one, two, strontium. So its electron configuration with two more electrons is going to be krypton 5s2, with those two more electrons. Now strontium will often lose these two electrons to get a noble gas configuration, which gives us a plus two strontium ion with krypton&#39;s electron configuration. So again, choice B. Magnesium has a specific heat capacity of 1.0 joules over grams degrees celsius. A sample of magnesium is heated to 80 degrees celsius. It’s placed in a calorimeter filled with 100 grams of water at 19 degrees celsius. The magnesium cools and reaches thermal equilibrium with water at 20 degrees celsius. If the specific heat of water is 4.2 joules over grams times degrees celsius, what is the approximate mass of the magnesium? You can probably tell there&#39;s going to be a lot of calculation here but remember this is on the multiple-choice section so all the math you&#39;re going to be able to do in your head. There might be some tricks to keep in mind. Okay, now the first thing to remember is that the equations you&#39;re going to need are probably going to be on that equation sheet, and one of these is q equals m c delta T, which stands for heat transferred equals mass times specific heat times a change in temperature. So, this is the equation that we&#39;re going to be using here, and when we&#39;re approaching specific heat problems like this a great thing to do is to write down a little table, all of your knowns and unknowns. Specific heat problems often involve the transfer of heat from one substance to another, so you want to remember to keep track of both substances. Here, that&#39;s magnesium and water. Okay, we&#39;ll make a little bit of room write out a tiny little table here. Okay, so now we&#39;ll start writing down our knowns and unknowns. Our unknown is the mass of magnesium—we’re asked that in the question, okay? The specific heat of magnesium is this. Specific heat of water is this. The magnesium starts at 80 degrees and it&#39;s put in 100 grams of water. The water has an initial temperature of 19 degrees, and the problem says that thermal equilibrium happens at 20 degrees. That means the final temperature for both the magnesium and the water is going to be 20 degrees, okay? Now, the equation up here needs delta T, the change in temperature, so we can do a little bit of subtraction in our heads or on a piece of paper, and we can get the delta T for magnesium and the delta T for water. Now that we have our knowns and unknowns categorized here, we can start solving the problem. We&#39;ll take this, rearrange it for m, the mass of magnesium. So, the mass of magnesium equals the heat lost by the magnesium divided by the specific heat of magnesium and the change in temperature of magnesium. Now we know this, we know this, but we don&#39;t actually know the value of q here, right? We&#39;re given a change in temperature, but we don&#39;t know how much heat was actually lost by the magnesium. But we can figure that out. If you&#39;ve ever done a specific heat lab or specific heat problems like this, you probably know how we&#39;re going to do this. With problems like this there&#39;s usually a switch step based on heat transfer. To find q we make the assumption that the heat lost by magnesium was gained by the water in the calorimeter. So let&#39;s do a little bit more cleanup here. We&#39;ll add subscripts to the variables so we can keep track of things. So this is our equation for the mass of magnesium. Now we have to find q, the heat gained by the water. Then we get that we can swap it into this equation. Okay, so to find the heat gained by the water we&#39;ll go back to this heat equation and we&#39;ll make a version of it for water, okay? We put this little blue box here water water water, so you can keep track of the fact that this is for water. Okay, so we can fill in these variables we have the heat gained by water equals the mass of water, well, that was given to us. We have the specific heat of water, that was given to us. And the change in temperature, delta T, one degree celsius. Calculate this through math you can do in your head. You don&#39;t worry about the one, you&#39;re essentially just multiplying 4.2 times 100. It&#39;s going to give you 420 joules. Okay, so that is the heat gained by the water, 420 joules. So now we can plug that into the equation for magnesium. 420 joules there divided by the specific heat and the delta T. Okay. We do the math, it is 420 essentially divided by 60, because you don&#39;t have to worry about the one, and that gives us seven grams. Let&#39;s take a look at our answer choices. There it is, B. Now there are lots of common mistakes when you&#39;re doing specific heat problems, and most of them involve forgetting which the variables are for which substances. So that&#39;s why we kind of made that little table there at the beginning to get everything organized. Okay, so if you correctly calculated q to be 420 joules, but you used the final temperature of the magnesium instead of delta T, well that would give you 21 grams which is choice D. In fact, all these other choices represent wrong choices that are typical math error. So that&#39;s why it&#39;s important to write down and label your unknowns, and don&#39;t forget that delta T here equals the absolute value of T1 minus T2. Okay, now for our last two problems, the multiple-choice section of the AP exam often features a group of questions that are related to a single diagram, or a graph, or something like this. So we&#39;re going to look at two questions that use that pattern and they&#39;re focused on electrochemistry here. Here we have a diagram of a galvanic cell there&#39;s an illustration and we have some standard reduction potentials and it&#39;s important to know that when you see something like this on the multiple-choice question, the questions don&#39;t depend on each other. In other words, the correct answer for one isn&#39;t necessary to get the second one or the following questions correct, right? So that&#39;s a little bit different than the free response questions where you often need to get one right in order to get the next ones correct too. Okay, so this question takes up a whole screen so we&#39;re going to break it apart and deal with each section on its own. If you have a pdf of the test available, you might want to keep that handy to keep referring back to this stuff. Okay, so let&#39;s dive in. The first problem is which expression gives the standard cell potential in volts. We don&#39;t need the diagram to answer this, but we do need the redox reaction and the reduction potentials. The standard cell potential of a cell, sometimes called the EMF or the electromotive force, is calculated by this equation, okay? E cell equals or I should say E naught cell equals E naught cathode minus E naught anode. For whatever reason this equation isn&#39;t given on the AP exam, so it&#39;s useful to know. There are other equations that you can also use to correctly calculate the E naught of a cell, but we&#39;re going to be using this one. Now you should know that in an electrochemical cell reduction happens at the cathode and oxidation happens at the anode. A useful way to remember this is that oxidation, or the anode, and oxidation both start with vowels. Okay, now this means since we have to find the E naught of the cathode and the E naught of the anode, that we need to figure out which species is getting reduced, and which is being oxidized. So let&#39;s look at the reaction and start looking at what happens to the iron here. Okay, well iron goes from Fe 3+ plus to Fe 2+. That means that iron gains an electron, and gaining electrons means reduction. So iron is reduced at the cathode. Now, this chart gives us the standard reduction potential. So iron is being reduced so we can plug this value in, 0.77 volts, for the cathode. Alright. Now the lead, neutral lead Pb, goes to the lead two plus ion. And that means that the lead loses two electrons, and losing electrons means oxidation. This is happening at the anode, okay? So now we know what&#39;s happening at the anode. We&#39;re going to subtract this value. Don&#39;t forget that this reduction potential is a negative number, okay? So the equation is E naught cell equals 0.77 volts minus negative 0.13 volts. We&#39;re subtracting the anode, so now we look at that. Well that&#39;s not an answer choice! How are we going to solve it? Or is it right? Mathematically, subtracting a negative is the same as adding it, okay? So this is actually perfectly represented by answer choice A. Now you don&#39;t even have to do the math, right? As you can see, you don&#39;t even have to add these together. We did the math here, but this just gives you the answer choices before the numbers are combined. Okay let&#39;s look at some common mistakes. The most common error when doing EMF problems happens when students are confused by subtracting negative numbers. Remember standard reduction potentials here can have negative values, so don&#39;t forget to use the correct signs when doing the subtraction, right? If you don&#39;t remember to subtract the negative, well there you go, choice B is waiting to trick you. Now there&#39;s something else that&#39;s important when you&#39;re doing problems like this. A common mistake is thinking that voltage is affected by stoichiometry, okay? If you&#39;re tempted by choice C, you probably doubled the value of iron, right? You see this two here, you might have double the value of ion because the reaction is talking about two moles of iron but EMF of a cell is not affected by the coefficients in the balanced redox reaction, so you don&#39;t want to be multiplying this value by two. And choice D combines both mistakes. The student forgot the minus and they multiplied by the coefficient from the equation. Alright, the last problem reads: which of the following accurately describes the change in mass occurring at the electrodes during the operation of the cell? Now in order to answer this we&#39;ll bring back the illustration, there it is. This cell features two types of electrodes; electrode one, the anode, is made of solid lead. We can see the equation for what&#39;s happening to lead, the half reaction that&#39;s happening to lead, and you can see that Pb solid is turning into Pb 2+. That means that the electrode itself, which is made of lead, is oxidizing to produce Pb 2+ ions. That means that atoms of solid lead in the electrode are turning into ions and are floating off into solution. So as this solid, as the atoms in this solid lead are turning into ions and are floating off into solution, the mass of this electrode is going to decrease. So that&#39;s very important. The mass decreases here. That means that we can cross off choice B and choice C, because we know that the mass has to decrease. So now we have a choice A or D. What&#39;s going to happen to electrode 2? well here&#39;s the cathode. Now at the cathode this reaction is taking place okay Fe 3+ to Fe 2+. We don&#39;t have ions getting reduced and turning into solid metal, we have ions getting reduced to make ions. Both Fe3+ and the Fe 2+ are going to stay dissolved in solution, so the electrode is not part of the redox reaction. It just provides a way for electrons to flow into the cell reducing Fe 3+ to Fe 2+. Now because of that, this electrode wouldn&#39;t be made out of iron, it would be made from inert platinum. Its only job is to transfer electrons. It’s not gaining mass, it&#39;s not losing mass, so that means that choice A is correct, and platinum is not part of the redox equation. There’s no change in platinum mass, so choice A is our correct answer. [a]Check with audio. [b]29:12

In this video we&amp;#39;re going to work through a 
bunch of sample multiple choice questions for   the AP chemistry test. Now before you go through 
this whole video, I really recommend you check out   my website. Here you can download these problems 
first and do kind of a mini test. It only takes 30   minutes and then you can come back here for the 
answers, and also, I’m going to be adding all   sorts of AP chemistry resources to help students 
with a test so definitely check out this website.   I think you&amp;#39;ll find some stuff that helps you. 
Anyway, before we start, a quick legal disclaimer:   the AP chemistry is a trademark owned by the 
college board, that&amp;#39;s not me and the college board   is not affiliated with and does not endorse this 
video or website—and of course they don&amp;#39;t endorse   this video because we&amp;#39;re going to be walking 
through the trips and the traps that they put in   the AP chemistry test to mess up students. We’re 
going to figure out how you can outsmart them   and how you can get a better score on 
the test. Okay, so let&amp;#39;s get started.   Here&amp;#39;s our first question: 10.0 grams of iron and 
10.0 grams of sulfur react to completion. Assuming   a 100 percent yield, which of the following 
describes the contents of the reaction vessel.   Now, this is a question about stoichiometry but 
remember the AP exam doesn&amp;#39;t allow calculators   on the multiple-choice section. So, the math here 
is either going to be really easy or it&amp;#39;s going   to be conceptual in nature. Let&amp;#39;s start with the 
equation for the reaction. Now if you didn&amp;#39;t know   what it was right away, the answer choices tell 
you what the product&amp;#39;s going to be. It’s going   to be FeS, iron (II) sulfide. Now, the molar ratio 
of the reactants is one to one which makes things   simple. One mole of iron reacts with one mole of 
sulfur to produce one mole of iron (II) sulfide.   Now, we know that both reactants have the same 
mass, 10 grams each. Let&amp;#39;s pull up their molar   masses from the periodic table and we can see that 
iron is 55.85 grams per mole and sulfur is 32.06   grams per mole. Now to find the moles of each 
reactant you could divide 10 by the molar masses   and that would give you the moles, so you could 
figure out the limiting reagent, but you don&amp;#39;t   even need to go that far. After all, the question 
doesn&amp;#39;t ask how much is left in the vessel,   just which species--what&amp;#39;s left over. So, 
think it through. Iron has a higher molar mass,   so an equal mass of iron contains fewer 
moles than an equal mass of sulfur, right?   That means that iron with fewer moles 
is going to be the limiting reagent.   Because the problem states that 
there&amp;#39;s 100 percent completion,   it means all the iron is consumed. All the 
limiting reagent is consumed. So, what&amp;#39;s   left behind in the reaction vessel is going to be 
choice C, iron (II) sulfide, that&amp;#39;s the product,   and the leftover sulfur which 
is the excess reagent. Okay,   next one. Identify the Bronsted-Lowry acid 
conjugate base pair, and we have these choices.   Now a little background. A Bronsted-Lowry acid 
is defined as a species that donates a proton   or an H+ ion, just different names. A base is 
a species that accepts a proton. So, let&amp;#39;s look   here at a weak acid in equilibrium. Here&amp;#39;s the 
acid, HA, and it donates a proton, and the proton   donation leaves behind a negatively charged ion, 
right? It leaves behind A-. It&amp;#39;s lost the proton   and this minus can re-accept a proton in the 
reverse direction, and that makes HA again. We   call this species A-, we call this the conjugate 
base. A- is the conjugate base of the acid HA.   So, HA/A- represents the acid conjugate base pair. 
And sometimes we&amp;#39;ll talk about bases and conjugate   acid pairs. That’s where we have B and BH+. But 
look, here&amp;#39;s the most important part. In all cases   a conjugate pair differs by only one single 
proton. Okay, take away an H+ from HA and you get   A-. Add an H+ to B and you get BH+. They differ 
by one proton. Okay, so keeping that in mind   let&amp;#39;s take a look at these answer choices. Only 
choice D makes sense. Okay let&amp;#39;s start with this   guy H2PO4-. It’s going to donate or lose one 
proton and it&amp;#39;s going to turn into HPO4 2-.   Okay, this in turn can accept or gain 
a proton to turn back into H2PO4-.   They differ by one proton because 
they&amp;#39;re an acid conjugate base pair.   But let&amp;#39;s look at some common mistakes here so 
they don&amp;#39;t trip you up. Okay, choice C might be   tempting because it shows nitric acid and the ion 
left behind when it deprotonates, and these differ   by only one proton or one H+. However, nitric acid 
is a strong acid. There&amp;#39;s 100 percent dissociation   and no equilibrium, right? So, there&amp;#39;s no double 
arrow here. This means that the nitrate ion over   here never accepts a proton to reform nitric acid 
the reaction doesn&amp;#39;t happen in this direction,   so nitrate is not a conjugate base—another 
reason why it&amp;#39;s always useful to have your   strong acids and strong bases memorized for 
the AP exam. Okay, choice A might be tempting   because it represents the conjugate acid 
and the conjugate base of water itself. You   know we often sort of talk about these two ions 
together when we&amp;#39;re talking about water and pH,   but they themselves are not a pair because they 
differ by two protons. You&amp;#39;ve got to lose two   protons to get this you got to gain two protons 
to get this so they&amp;#39;re not going be a pair.   And finally here choice B. This is two oxyacids. 
They differ by one oxygen they don&amp;#39;t even differ   by the number of hydrogens, so that&amp;#39;s definitely 
not going to be the right answer. So, D it is.   Okay, here [a]the following is a photoelectron 
spectrum of an element, and its ground   state now many recent AP exams have been 
featuring photoelectron spectra. So, these   PES graphs are useful to understand, Now they&amp;#39;re 
not as complicated as they seem. Here&amp;#39;s how we&amp;#39;re   going to break it down. The y-axis is always 
going to represent the number of electrons.   Sometimes it&amp;#39;s going to be specific, but 
more often it&amp;#39;s going to be something like   the relative number of electrons. Okay, the 
x-axis is always going to represent energy.   It might be labeled in a bunch of different 
ways. You&amp;#39;ll see a variety of different units,   but the important thing to notice is that 
the axis decreases moving this direction and   increases in this direction. That&amp;#39;s standard. 
Now, for most of these problems the labels the   units aren&amp;#39;t really going to matter. 
What matters are the peaks here.   These are always going to be the same—we’re going 
to see spikes or peaks of differing heights.   And basically, these spikes represent the 
electrons in electron orbitals. Most PES questions   are going to ask you to translate the graph 
into a specific element like we&amp;#39;re doing here,   or translate it into an electron 
configuration, so here&amp;#39;s how you do that.   First look at the smaller spikes, these 
usually represent filled s orbitals and look   for a pattern moving from left to right. These 
spikes represent 1s2 2s2 3s2. Electrons that are   closer to the nucleus like 1s2 have a higher 
binding energy so that&amp;#39;s why they&amp;#39;re over here   where the values are higher. Okay, now the spike 
on the far right over here is smaller than the   others. It doesn&amp;#39;t fit this pattern so we&amp;#39;re going 
to ignore it for now but these here 1s2 2s2 3s2   pretty sure about that now the next spike here is 
a p orbital notice how this spike is three times   higher than the s spikes. That&amp;#39;s because an s 
orbital, a filled s orbital, holds two electrons   but a p orbital holds six electrons which is three 
times as many. Okay, so let&amp;#39;s see what we have   so far this is going to be 1s2 2s2 and then our 
first p orbital is going to be 2p6 then we&amp;#39;ve got   2s2. And what&amp;#39;s going to come next? Our outlier, 
our little spike over here. What do you think   this short spike means? It&amp;#39;s half the size of 
a filled s orbital so it only has one electron.   Well, what comes after 3s2? 3p right? And there&amp;#39;s 
only one electron in it so it&amp;#39;s got to be 3p1.   So, our complete electron configuration is 1s2 
2s2 2p6 3s2 3p1. We can add up the number of   electrons that are in all these orbitals 
we get 13. So that is choice B, aluminum,   which in its ground state has 13 electrons. 
Now, let&amp;#39;s look at a couple of common mistakes   some people may be tempted to choose 
boron, A, which has five electrons.   Now why might you do that? Well because there&amp;#39;s 
five spikes one two three four five. But remember,   the x-axis is the relative number of electrons 
these spikes represent electrons in orbitals   not individual electrons. If you picked 
D, potassium, you probably thought that   this little spike at the end was 4s1. But 
remember, you have to go in order of a standard   electron configuration so you before you get to 
4s1 you have to have 3p6, and we don&amp;#39;t have that   here so that&amp;#39;s why aluminum with 13 electrons 
a neutral aluminum atom is our correct choice.   All right, a saturated solution of (NH4)3PO4 has a 
concentration of ammonium ions of 1.50 molar. What   is the concentration of ammonium phosphate? Okay, 
ammonium phosphate is an ionic compound that&amp;#39;s   fully soluble in water, and we know that because 
all ammonium compounds are soluble ammonium   phosphate dissociates according to this equation. 
This three subscript on the ammonium means we get   three ammonium ions after dissociation, okay? 
And there is a one to three ratio of ammonium   phosphate to ammonium ion so because of this 
ratio, if the concentration of ammonium ion   is 1.50 molar—that’s given to us right here in 
the problem, right? The concentration of ammonium   phosphate must be one third of that amount. That&amp;#39;s 
0.50 molar. So choice A is the correct answer, but   some common mistakes. You may have been tempted 
by choice C because you see the 1.50 there,   but that would only be true if the ratio 
were one to one like ammonium chloride. Here   then, the concentration of each 
component would be 1.50 molar.   And choice D here represents a 1 to 3 
ratio, but it&amp;#39;s reversed there&amp;#39;s three times   as much ammonium phosphate as ammonium and that&amp;#39;s 
backward okay. So that&amp;#39;s why choice A is correct.   All right this is going to be a little bit of 
a conceptual math workout. An antacid tablet   containing calcium hydroxide and we get the molar 
mass is titrated with a 0.100 molar solution of   hydrochloric acid HCl. The end point is determined 
by using an indicator. Based on the information   below that&amp;#39;s going to be this here what was a 
mass of the calcium hydroxide in the tablet.   And we&amp;#39;re given the buret readings. The initial 
reading is 5.1 milliliters, and the final is   25.1 milliliters. Okay, where do we start? Well 
this question is about titration we have a base   calcium hydroxide and an acid hydrochloric 
acid. Now, it also mentions the end point here   and that&amp;#39;s fundamental to any acid-base titration 
when the neutralization process is complete. Okay   the end point is when the moles of H+ from the 
acid equal the moles of hydroxide from the base.   It’s also called the equivalence point. 
So let&amp;#39;s work through this problem.   So first, how many moles of HCl do we need to 
neutralize the base? Okay the buret contained   the HCl, the volume started at 5.1 milliliters 
and went to 25.1 milliliters so we can do some   subtraction here giving us 20.0 milliliters. 
That&amp;#39;s the volume of acid that we used,   but that&amp;#39;s not the number of moles, that&amp;#39;s the 
volume, okay? So to find the number of moles,   we have to use the molarity equation: 
molarity equals moles divided by liters.   Now you probably have this memorized because you 
use it a lot, but it is one of the given equations   on the AP exam. So rearranging this equation 
shows you that moles equals molarity times liters.   So we can start with moles equals 0.100 molar. But 
now, remember that the equation needs you to use   liters, and this number here is milliliters. 
Okay, so you need to convert milliliters   to liters before you can use this equation. 
There’re a thousand milliliters in one liter   so you can divide by a thousand or just shift 
the decimal three spots to the left. Okay   so we had a 20.0 and we go one two three 
point zero two zero liters very important.   And when you think about the math, you&amp;#39;re 
multiplying by 0.1 which just means that   you&amp;#39;re going to be moving this decimal spot 
one more to the left, okay? This tells us   how many moles of HCl are needed to neutralize the 
base. Okay, it&amp;#39;s getting a little bit cluttered   here so let&amp;#39;s get rid of the stuff that we don&amp;#39;t 
need right now, and we&amp;#39;ll bring it back later.   Also, I’m going to stop writing these zeros every 
time, just remember that there are two zeros here   at the end for sig fig reasons, but usually sig 
figs aren&amp;#39;t a big deal with calculation problems   on the multiple choice anyway. Okay, 
that&amp;#39;s better, a little cleaner here.   So we know that 0.002 moles of hydrochloric acid 
neutralize the base. We can use this information   to find the mass of calcium hydroxide. We need 
to look at the balanced neutralization equation   for this reaction. You might even want to sketch 
it out on the test. Okay, here&amp;#39;s our equation.   It shows that two moles of HCl are required 
to neutralize one mole of calcium hydroxide.   So if it took 0.002 moles of HCl to reach 
the end point there&amp;#39;s only 0.001 moles   of calcium hydroxide present. See, twice as much 
acid is needed. And look at the formula too,   right? One mole of calcium hydroxide contains 
two moles of hydroxide ion so that makes sense,   right? Anyway, we have 0.001 moles of calcium 
hydroxide. Now, how many grams is that?   Well, we can use molar mass to calculate that; and 
we&amp;#39;re even given the molar mass in the problem.   So, 0.001 moles of calcium hydroxide. Do this 
math when you are multiplying by 0.001, you&amp;#39;re   moving the decimal place three spots to the left. 
So you&amp;#39;re going to end up with 0.0741 grams. Okay,   now let&amp;#39;s take a look at our answer choices 
and there we go, number one or I should say A,   but look, we&amp;#39;re going to spend a 
second looking at some common mistakes   so that you won&amp;#39;t make them. There’s a reason why 
the AP test has these other choices they&amp;#39;re traps   designed to trick you. Okay, one thing we talked 
about it a minute ago that students often forget   is to convert milliliters to liters 
when using the molarity equation. Okay,   so this is what would have happened if 
the student forgot to convert to liters.   And this gives the impression that there are two 
moles of acid. Carry that through, look at that,   74.1 grams it&amp;#39;s waiting there to trick you. 
Okay, another error, this is very common, is to   forget about the mole ratio here okay and to think 
that you need 0.002 moles of calcium hydroxide to   go with 0.002 moles of hydrochloric acid. Okay, 
so if you carry this error through look at that   you get C. And finally, D is an example where 
the student made both of those mistakes and that   incorrect answer is waiting right there with open 
arms to welcome them in and to trap them. So make   sure that you&amp;#39;re cognizant of some of those common 
mistakes so that you won&amp;#39;t make them on the test.   Which of the following 0.10 molar solutions would 
experience the highest percent ionization? Now,   the first thing to notice here is that all of 
these answer choices are acids. They all have   a hydrogen cation attached to a polyatomic anion. 
Also, percent ionization questions on the AP exam   are usually about weak acids. Okay, since these 
are all acids, they ionize. This is the same as   deprotonation. Now these three acids are also a 
specific type of weak acid we call these oxyacids   because the proton is attached to an oxygen. Here, 
we&amp;#39;re going to show their structural formulas so   you can see this a little bit more clearly. You 
don&amp;#39;t need to know the structure of these acids   to answer the question, but these structures are 
going to help us illustrate the main points. Okay,   so each of these hydrogens is going to dissociate. 
We want to know which of these acids experiences   the highest percent ionization, and the higher 
the percent ionization the stronger the acid.   The thing to keep in mind for oxyacids is this:   as you pull electrons away from the OH 
bond that, weakens it. And the hydrogen   ionizes more easily. The red arrows here represent 
the forces pulling on the electrons. There’s two   things that impact the pull and the AP exam might 
test either one or both. Okay, first the more   electronegative the halogen the stronger the 
oxyacid that&amp;#39;s because the electronegative halogen   will be pulling electrons away from the OH bond. 
But in this case, our halogen is always bromine   so that doesn&amp;#39;t matter here, but keep in mind 
for other questions like this. The second thing   to remember is that oxygens are electronegative 
too. So the more oxygens you have, the stronger   the acid, because they pull electrons away from 
the OH bond. In this question choice C has the   most oxygens, therefore it pulls hardest on the 
electrons in the OH bond, weakening it. This acid   deprotonates most readily, meaning a higher 
percent ionization and a stronger acid.
  Okay, given a constant temperature which of the 
following reactions is most likely to proceed   spontaneously. And then we have all these 
different reactions with different delta H   values and different reactants, and different 
products. How are we going to break this down?   Well, this question is about spontaneity, and 
when we&amp;#39;re thinking about spontaneity there are   two main factors to keep in mind: the first 
is enthalpy, and the second is entropy. If   a reaction is both exothermic and increases 
entropy, it&amp;#39;s guaranteed to be spontaneous.   If a reaction is both endothermic and the entropy 
decreases, it&amp;#39;s guaranteed to be non-spontaneous.   Now, any combination in between these two 
options requires you to know the temperature   and the exact values of entropy and enthalpy in 
order to determine spontaneity using the Gibbs   free energy equation. So this question says 
that temperature is constant, so we don&amp;#39;t need   to worry about those in between cases. We’re just 
going to look at these two options they&amp;#39;re either   always spontaneous or always non-spontaneous. 
Okay, first let&amp;#39;s look at enthalpy. Okay,   we&amp;#39;re given enthalpy values for each reaction, 
that is the delta H. Okay just remember   a negative delta H indicates an exothermic 
reaction and a positive delta H indicates an   endothermic reaction. That’s because we&amp;#39;re 
looking at the change in enthalpy, and exothermic   reactions lose heat, so that change is going to 
be negative. Remember that because it&amp;#39;s commonly   tested on the AP exam. Okay, so right off the 
bat okay we&amp;#39;re looking for spontaneous reactions.   So under these conditions they have to be 
exothermic. We can cross out A and B right away   because these are endothermic. Okay, now it&amp;#39;s not 
that endothermic reactions can&amp;#39;t be spontaneous,   but we&amp;#39;re looking for a candidate that&amp;#39;s going 
to meet both of these conditions. Okay, now   to narrow it down between C and D we need 
to look at entropy, the measure of disorder.   There are two things to keep in mind about entropy 
questions. The first is that entropy increases   when you move from solids to liquids to 
gases. Okay, these phases of matter have   increasingly more entropy. The second is that 
entropy increases when you produce more moles   of gas. That’s because gas has a lot of 
entropy, so the more gas particles you make,   the more entropy your reaction is going to be 
contributing. So let&amp;#39;s look at these two changes   for entropy in choice C. An aqueous reactant 
and a solid reactant produce two moles of gas,   that&amp;#39;s a definite increase in entropy. That’s 
good. And C is probably the correct answer,   but let&amp;#39;s look at choice D just to be sure. Here 
we see one two three, remember the two here,   one two three moles of reactant gas 
produce two moles of product gas.   That’s a decrease in the moles of gas; it&amp;#39;s a 
decrease in entropy too, which is not favorable   to spontaneity, so we can cross this out. And 
the correct answer is choice C. It’s the only one   that meets both of these conditions; it’s 
exothermic and entropy increases.[b]
  All right, here’s a lab-based question. 
A student is performing a redox titration   by oxidizing Fe2+ ions to Fe3+ ions 
using a solution of 0.0010 molar KMnO4.   The target solution is iron (II) sulfate. And then 
we&amp;#39;re given the balanced redox reaction. Okay,   so that&amp;#39;s the background. Now here&amp;#39;s the 
actual question: assuming all glassware has   been cleaned and thoroughly rinsed with distilled 
water, what is the proper laboratory procedure   before adding the KMnO4 solution 
to the buret? And we have choices.   Now as we said, this is an example of a lab 
procedure question and you can expect a couple of   these on the multiple choice section of the test. 
And some are pretty common sense but some get   a little bit particular. Okay, the best way to 
approach these questions is to focus on two main   concerns. The first is lab safety and the second 
is preventing error. Almost all the lab questions   the lab procedure questions you&amp;#39;ll get on the test 
have to do with one of these two themes. So here   we&amp;#39;re not talking about safety; here it&amp;#39;s about 
preventing error. Now your first impulse might   be to select choice A because it&amp;#39;s never a bad 
idea to make sure that your glassware is clean,   but the problem has already stated that 
the buret has been thoroughly rinsed   which means that it has been cleaned but there 
are still drops of water inside the buret, right?   So what happens when you add KMnO4 solution? Well 
that extra water that&amp;#39;s in the buret dilutes the   solution which lowers the molarity. So you 
think you&amp;#39;re working with a solution that has   this concentration, but since there was water left 
over inside the buret that diluted the solution,   what you&amp;#39;re working with is 
actually slightly more dilute   than this value here because of the leftover 
water. So the correct answer choice is   the buret first should be washed with excess 
permanganate solution to clear out any pure water   that has stayed behind so that it won&amp;#39;t 
get diluted by the leftover water.   Okay now there are some traps here that you might 
have picked and that&amp;#39;s okay. Let’s talk about why   if you picked choice D. Okay, you might have 
noticed that the balanced equation has some H+   ions in here as reactants and like many 
complicated redox reactions, you have to acidify   the reaction to balance it properly, but here when 
you&amp;#39;re doing it in the lab, the acid is actually   added to the iron sulfate solution. So there&amp;#39;s 
nothing in the problem that suggests that to   get ready for this titration you would add acid to 
the permanganate; so avoid making that assumption.   And finally choice C is very very wrong because 
you&amp;#39;d be introducing traces of one reactant to   the other reactant before you actually do the 
lab procedure, so that would contaminate the   buret and it would be very poor procedure. So 
our only good choice here is choice B.
  Okay for this hypothetical reaction the following 
initial rate data were obtained. The temperature   remains constant. And then we got this table here 
showing four experimental trials along with the   initial concentrations of the reactants and the 
initial reactant rate the initial reaction rate.   So, this is a problem about kinetics. And the AP 
exam often asks rate law questions a lot like this   in the multiple-choice section. Usually, you&amp;#39;re 
given a table of initial concentrations and rates   and asks to find the rate law. To solve a problem 
like this you&amp;#39;d use the method of initial rates.   You find two rows where only one concentration 
changes and you look at the effect that that   has on the rate. And there are only three 
possibilities. If you change the concentration   and there&amp;#39;s no effect on rate that species is 
zero order. If you change the concentration   and the rate changes by the same proportion 
that species is first order. For example   if you double the concentration and the 
rate doubles that&amp;#39;s classic first order.   And finally if you change the concentration and 
the rate increases by the square of the change,   the species is second order. So if you double the 
concentration and the rate quadruples, times four,   that would be second order. Okay, so let&amp;#39;s 
use that method here between experiment one   and experiment two you can see the only thing 
that changes is the concentration of Q. Okay,   what happens to it? Well Q doubles 0.2 to 0.4. 
Okay, now let&amp;#39;s take a look at what is going   on with the rate. This quadruples, okay? We 
have double here, quadruple here, that means   the reaction is second order with respect to 
Q because 2 squared is 4. We’re done with Q.   Now before we move on though let&amp;#39;s talk about 
rates and scientific notation. The AP exam   loves giving reaction rates in scientific 
notation like this so when you&amp;#39;re comparing rates   pay close attention to the superscripts 
on ten. Okay, it might sound weird   to say that 6.0 times 10 to the negative third 
quadruples to 2.4 times 10 to the negative second,   but 10 to the negative 2 is a larger number than 
10 to the negative third. If you expand these   numbers out of scientific notation you can see 
it more clearly okay 0.0060 times 4 equals 0.024.   Okay, got it? So keep an eye on the scientific 
notation. Let’s get back to the problem now.   Experiment one and experiment for M and Q stay the 
same, but X doubles. What happens to the rate bam   bam no effect so with respect to X this is zero 
order, all right? And the last thing to look at   is the concentration of M. Experiment one 
and experiment three, Q and X stay the same,   M doubles. What happens to the rate? 
The rate doubles M doubles rate doubles   classic, boom, first order. Same proportion and 
once again if we are mindful of the scientific   notation we can see that this times two equals 
0.012 1.2 times ten to the negative second.   Okay, so we know all these orders 
we figured out from the table;   let&amp;#39;s now put together our rate 
law. Rate laws always start with k.   Zero order reactants drop out when we write the 
rate law so we&amp;#39;re just not going to put them in   there, so our rate equation, our rate law, is 
going to be rate equals k M that&amp;#39;s first order,   Q squared—the square because it&amp;#39;s second 
order. Now let&amp;#39;s bring back our answer choices,   and right here D. Dis the correct answer ah-ha. 
But before we go to the next problem let&amp;#39;s pause   for a second, right? There’s an even quicker 
way to get to this answer. You remember how we   immediately established that this equation was, or 
this reaction was second order with respect to Q?   Well look at this look at the answer choices only 
one of them has Q squared as a term, so if you&amp;#39;re   feeling really confident you could have selected D 
immediately and moved on. Now look that&amp;#39;s a gutsy   move but if you really know what you&amp;#39;re doing 
you could save yourself a lot of time.
  Okay, the above gaseous reaction occurs and comes 
to equilibrium in a closed container. Which of   the following changes to the system would cause 
the reverse reaction to speed up? And there are   four things that we could do to this system 
here. Now, AP questions asking about changes   to a system at equilibrium are generally going to 
be about Le Chatelier&amp;#39;s principle. And this states   that if a system in equilibrium is disturbed, the 
system will shift to counteract that change. Okay,   here the question is asking us to select 
the change that would cause the reverse   reaction to speed up. So, let&amp;#39;s look 
at our answer choices one by one to see   which would lead to that result. Choice A is to 
remove some HCl. Now if you add more of a chemical   the equilibrium shifts in the direction that 
consumes the addition. If you remove a chemical,   the reaction shifts in the direction that replaces 
the missing chemical. Okay so in this system   HCl is a product so if we remove some product Le 
Chatelier’s principle kicks in and the reaction   has to make more product. Because we&amp;#39;ve removed 
it, that would shift the reaction forward.   That’s not the direction we&amp;#39;re looking 
for so let&amp;#39;s go ahead and cross off A.   Choice B has us adding some hydrogen. 
Le Chatelier’s principle kicks in again,   but this time the reaction has to consume 
the hydrogen to re-establish equilibrium.   You add more, it&amp;#39;s got to use it up, it&amp;#39;s going 
to move in this direction. So again the reaction   shifts right, boom, cross it off. Now the 
next two choices here are about temperature.   Another way to offset equilibrium is to change 
the temperature of the reaction. To solve these   you need to treat heat energy like a reactant or 
product. So does this reaction create heat or does   it absorb heat? This reaction has a negative delta 
H, so the reaction is exothermic—it releases heat.   So we can treat heat as a product that is made by 
the reaction. Heat can be considered a product.   Choice C is to lower the temperature, which 
is the same thing as removing heat energy,   okay? So if we were to remove a product, remember 
we&amp;#39;re treating heat like a product, the reaction   would have to shift right to produce more 
heat. We got rid of heat, the reaction produces   heat, the reaction is going to move to the 
right to make more heat. So cross off choice C.   Even though choice D is the only one 
left let&amp;#39;s reason through it to be sure.   If you increase the temperature you&amp;#39;re adding 
heat, okay? In an exothermic reaction that&amp;#39;s like   adding product; it&amp;#39;s like adding 
something to this side of the equation   so to re-establish equilibrium, the reaction 
shifts left to consume that extra heat energy.   Rasing the temperature is our correct answer.   Which of the following diagrams provides the best 
representation of an aqueous solution of HNO3?   This question is testing our understanding 
of the dissociation of ionic compounds   and strong acids. Before we take a look at the 
diagram let&amp;#39;s write out the dissociation equation   for nitric acid in aqueous solution. Here it is, 
nitric acid dissociates into hydrogen cations and   nitrate anions. There are two things you need to 
recognize here. First nitrate is a polyatomic ion.   That means that nitrate remains a unit when nitric 
acid dissociates or ionizes. We get NO3 1- and   so that means that right away we can get rid of 
choice B because polyatomic ions don&amp;#39;t break into   individual atoms when they dissociate. It&amp;#39;s always 
a good idea to have your polyatomic ions memorized   for the AP exam. Now the second thing to notice 
here is that nitric acid is a strong acid. That   means that it undergoes complete dissociation or 
ionization, so we can eliminate choice A because   this shows no dissociation at all, right? 
These hydrogens are still attached to the NO3   in A. We can also eliminate choice C because 
this shows partial dissociation, right? Here,   two of the molecules have the hydrogens attached, 
and only one of them shows it being dissociated.   Okay so this would be correct for a weak 
acid, but a strong acid undergoes complete   dissociation—another good reason to have your 
strong acids and strong bases memorized for   the exam. So that leaves us with choice D, 
which is the correct answer. It shows one   hydrogen ion for every one nitrate ion. 
100 percent complete dissociation. Okay,   a 0.50 gram sample of zinc solid was placed in 
a solution of hydrochloric acid, aqueous. The   zinc reacted completely with the acid. Which 
one of the following equations best represents   the reacting species described above? 
This question describes the metal zinc   reacting with hydrochloric acid, and then it asks 
you which equation best represents the reacting   species. Note the wording here in red. 
We’re going to come back to it in a second.   Now the first thing that you should remember 
is that the reaction of a metal in an acid is   an example of a single replacement reaction. 
Now at first you might be tempted to select   choice C. That’s pretty understandable 
because this is actually the full equation for   the single displacement or single replacement 
reaction that would happen. However,   remember that we said pay 
close attention to the wording.   The question is asking you which equation 
best represents the reacting species. That’s   a different question, so what we&amp;#39;re looking for 
here is actually the net ionic equation. That’s   a type of abbreviated equation that only shows 
the substances that are actually reacting. So   let&amp;#39;s take a closer look at this reaction. Because 
HCl is a strong acid, it completely dissociates.   And because zinc chloride is a soluble chloride 
salt, it&amp;#39;s also dissociated in solution. So if   we focused on the ions in this equation, our 
full equation would look like this. 
  Now, look at the chloride ion here. See how 
it doesn&amp;#39;t react? It starts as chloride,   Cl- and it ends up as chloride. We 
call an ion like this a spectator ion.   it doesn&amp;#39;t actually react. So it&amp;#39;s not included in 
the net ionic equation. So the chloride drops out   and we have our net ionic equation. We keep this 
and we keep this because these actually react,   they actually change, during the course of 
the reaction. This is what we&amp;#39;re left with   this is the equation that best represents the 
reacting species. Look at this. There it is,   choice A. the AP exam loves net ionic equations. 
Sometimes they&amp;#39;re pretty clear about it,   but sometimes they get really sneaky 
like with this question, so pay attention   to the wording. Let’s look at the other two 
answer choices to double check our answer.   Choice D looks like a single replacement reaction 
but notice the hydrogen—it remains as H+ ions   instead of being reduced. Oh sorry there, to 
H2, tempting but wrong. And finally choice B   is just wrong. It’s not even a single replacement 
reaction so we can just get rid of that.   Okay, beaker A, there it is, contains 
250 milliliters of pure deionized water   and beaker B contains 500 milliliters of 1.0 
molar KCl dissolved in water we would assume.   Silver chloride crystals are added to each 
beaker. After stirring the contents of the beaker,   some solid AgCl remains at the bottom of each as 
shown below. Which of the following is true? This   question is about solubility and the common ion 
effect. The AP exam loves the common ion effect,   so problems like this are pretty common. Let’s 
take a closer look. It’s more complicated than   it seems; there’s a lot going on here. So to make 
some room we&amp;#39;ll shrink our diagram. Okay, so we   have two beakers. One contains 250 milliliters 
of pure water, the other 500 milliliters of 1.0   molar KCl. Silver chloride has been added 
to both beakers to saturate the solution.   Which beaker has the highest concentration 
of silver ions? At first glance you might be   tempted to select D because you probably memorized 
that silver chloride is an insoluble salt, okay?   It doesn&amp;#39;t dissolve well but if you&amp;#39;ve studied 
solubility equilibrium you know that all salts   are actually soluble even if just a tiny bit. 
Okay, so some AgCl is going to dissolve, so we   can eliminate D right away. It’s a trap, don&amp;#39;t 
step into it. Okay here&amp;#39;s another kind of trap:   you might be tempted by the fact that beaker B 
has twice as much solvent, twice as much water,   so it should dissolve twice as much AgCl. Or 
you think it might dissolve twice as much HCl,   but we&amp;#39;re talking about the concentration 
of silver ions, which is a ratio   of moles per liter. So the concentration of 
silver ions is not the same as the amount   of silver ions. That doesn&amp;#39;t mean that choice B is 
wrong, it just means don&amp;#39;t pick it for the wrong   reasons, okay? So we&amp;#39;re going to put a question 
mark next to it because the AP exam can be tricky,   and it plans for student mistakes. They might 
be anticipating the fact that some students are   going to think, “oh well more should dissolve 
in beaker B” and forget about concentration.   Okay, so we&amp;#39;re just going to put this question 
mark there, maybe you know be wary of it   we&amp;#39;ve marked these two traps or potential 
traps. And now we&amp;#39;ll solve the problem. This   is a solubility equilibrium problem. Okay the 
salt HCl dissociates according to this equation.   As we said it&amp;#39;s a weakly soluble salt, 
insoluble essentially, so it&amp;#39;s going to   be in equilibrium here and we&amp;#39;re going to 
have an arrow that goes in both directions.   Alright, now for an insoluble salt like this you 
can write an equilibrium expression using Ksp,   the solubility product constant. And every salt 
has a unique Ksp. The equation for silver chloride   is Ksp equals Ag+, the concentration times the 
concentration of Cl-. Now you don&amp;#39;t need to know   the actual value for Ksp. Here you just need to 
know that Ksp is a constant, a specific number,   so the concentration of silver ions times the 
concentration of chloride ions is always going   to equal the constant number here. Now beaker A 
contains pure water so the only source of silver   and chloride ions is the AgCl that gets dissolved. 
And because silver chloride dissociates with a 1:1   molar ratio, both values are the same we&amp;#39;ll 
just call them X, okay? X for silver X for   chloride. The actual value doesn&amp;#39;t matter, 
okay? Now here&amp;#39;s where things get interesting.   Beaker B isn&amp;#39;t pure water—it&amp;#39;s a solution of 
KCl, and that means there are chloride ions   already in solution. Okay, this is going to 
impact the solubility equilibrium. The solution   doesn&amp;#39;t care where the chloride ions came 
from. They can come from potassium chloride,   silver chloride, doesn&amp;#39;t make any difference. 
Now remember that Ksp has to be the same value   for both beakers. It’s a constant. Over here 
in beaker B there&amp;#39;s a lot more chloride ions,   okay? Those came from the KCl. So if the chloride 
concentration is larger, which it is, that means   that the silver concentration has to be 
smaller. That’s the only way that they can   still equal Ksp when you multiply them together, 
okay? Algebraically speaking, X for silver   is smaller in beaker B because this, 
the Cl- concentration, is larger,   okay? Just once again Cl- concentration is larger 
in beaker B so Ag, silver, has to be smaller   so they can still equal the same value. This means 
that the concentration of silver ions is greater   in beaker A, which is choice A. Now this is an 
example of the common ion effect. When you have   a solubility equilibrium, the addition of a 
soluble common ion, in this case chloride,   depresses the solubility of the insoluble ions. 
Here, the chloride from the potassium chloride   depresses the solubility of the silver chloride, 
which means fewer silver ions in this beaker.   If you know the basic gist of the common 
ion effect, questions like this are easier   to answer and you don&amp;#39;t even need to go through 
doing math for the whole equilibrium expression.   Which of the following is a nonpolar 
molecule that still contains   polar bonds? The AP exam loves to ask questions 
about polarity, so going into the test it&amp;#39;s   helpful to know a couple things. First C-H 
bonds are essentially nonpolar. Now, know that   fluorine is the most electronegative element, 
followed by oxygen, nitrogen, and chlorine.   You should also know that the shape of 
a molecule affects its overall polarity.   In this question we need to meet two criteria. 
First, the molecule has to be nonpolar.   Second, the molecule has to feature a polar 
bond. So let&amp;#39;s answer them one at a time. First,   are any of the molecules here polar? If so we can 
eliminate them immediately. Okay, choice D is the   only polar molecule. As you can see from the Lewis 
structure here, it would have a tetrahedral shape   but two of the bonds are polar C-F bonds, 
right? That F is way more electronegative   than the carbon, than the C, so electrons 
are going to be pulled in this direction,   okay? Now these two bonds are polar. We can now 
consider the molecule as a whole and we can see   that one side of it here has the electrons 
being pulled to one side. So it&amp;#39;s unbalanced,   okay? That means that this is a polar molecule. 
Now this could be a little bit tricky because   this is a tetrahedral shape, which is one that we 
sometimes associate with nonpolar molecules like   methane--methane there—but we want to keep in 
mind what the individual bonds are and how they&amp;#39;re   oriented, alright? So polar molecule, boom, 
cross it off. Now three nonpolar molecules left.   So now we&amp;#39;re looking for polar bonds. We just 
said that C-H bonds are nonpolar. So boom,   that&amp;#39;s why we can get rid of choice C, right? 
Here, choice A is diatomic, all right? We have a   nitrogen bonded to a nitrogen which means that 
both atoms have the same electronegativity,   so we can definitely cross out that 
choice. That leaves us with choice B,   carbon dioxide. Okay the C-O bond is polar. Oxygen 
has a significantly greater electronegativity   than carbon, so we have polar bond there, polar 
bond there, electrons getting pulled. But the   shape of the molecule is linear so it cancels 
out. These two polar bonds going in this direction   cancel each other out, so we have a non-polar 
molecule that still contains two polar bonds.   Okay, which of the following statements is true 
regarding lithium and cesium? Okay, here we&amp;#39;re   comparing some of the properties of two elements, 
and specifically we&amp;#39;re looking at ionization   energy, atomic radius, and electronegativity. The 
first step here is to recognize that these three   things are all periodic trends, okay? They are 
properties that vary in a predictable manner based   on where an element is on the periodic table. The 
AP exam is never going to expect you to memorize   the atomic radius or the ionization energy of 
every element, but it will expect you to compare   the properties of elements based on where they 
fall on the periodic table. Bam! So to answer this   question we&amp;#39;ll want to use our periodic table. 
We’ll get rid of the answer choices now, we&amp;#39;ll   bring them back when we&amp;#39;re ready. Okay, so let&amp;#39;s 
find lithium and cesium on the periodic table.   They’re in group one, the alkaline metals, and 
this means that the question is asking us to   predict how periodic trends change from top to 
bottom of a group like this. Let’s review these   trends. We’ll get rid of the rest of the periodic 
table and just focus on group one here. Okay, as   we move down a group, atomic radius increases, and 
this makes sense because as we add energy levels   the atom gets bigger. So we can see that lithium 
will have a smaller atomic radius than cesium,   which has more energy levels with electrons. 
Okay, so lithium has a smaller atomic radius.   However, ionization energy and electronegativity 
both decrease as you move down a group. They   increase in this direction, but 
they decrease in this direction.   This is also because of the extra energy 
levels, okay? As we move down a group,   the atom&amp;#39;s valence electrons are further from 
the nucleus. The core electrons shield these   valence electrons from the attractive forces 
of the nucleus that are pulling them in so   it&amp;#39;s easier to remove valence electrons. The more 
core electrons you have shielding the nucleus. So,   it&amp;#39;s easier to remove valence electrons, that&amp;#39;s 
ionization energy, and the nucleus has less   attraction for the electrons in a bond and that&amp;#39;s 
electronegativity, okay? So keep that in mind.   So these trends mean that lithium will 
have a smaller atomic radius, but it will   have a higher ionization energy and a higher 
electronegativity than cesium. Okay, now that   we&amp;#39;ve figured out these three things we can run 
through the answer choices to see which one fits.   We’ll start with A. Well that&amp;#39;s wrong because 
cesium has a lower ionization energy than lithium,   so we can cross that. Okay, choice B states 
lithium has a higher ionization energy and the   higher electronegativity. That fits with what we 
know so that&amp;#39;s our correct answer. And choice D is   wrong because lithium has a smaller atomic radius 
than cesium. So yep, B was our correct answer.   Before we move on, one word about periodic trends. 
Well… a couple words about periodic trends.   Sometimes the AP exam will ask about trends moving 
across a period instead of down a group, okay? So   it&amp;#39;s best to know all your trends. Just for the 
record, when moving left to right across a row,   ionization energy and electronegativity both 
increase, but atomic radius increases moving   right to left across a row. That&amp;#39;s the one 
students usually forget, so be prepared for   all the possibilities. A 1.0 liter sealed 
container contains equal masses of neon and argon.   The total pressure of the system is 12 atm. The 
temperature remains constant. What is the partial   pressure of argon? In order to solve this problem 
you need to understand Dalton’s law of partial   pressures, and that can be expressed as a formula 
or it can be expressed simply in words. Okay,   the total pressure of a system is the sum of 
each partial pressure. Now here, the total   pressure of the system is the partial pressure of 
the neon plus the partial pressure of the argon.   So how are we going to find these two 
pressures? Well we can use the ideal gas law   PV = nRT. We have to rearrange a little bit. If 
we divide both sides by V, we can see that the   pressure of the gas equals number of moles times 
the gas constant times temperature divided by the   volume. But look carefully. We don&amp;#39;t have to do 
that. We don&amp;#39;t have to do that much math here.   In this system the temperature and the volume 
remain constant, and R is always a constant. So   this is going to make things easy for us because 
the variables T and V are the same for both gases;   we don&amp;#39;t have to worry about them. The only thing 
different for each gas will be the number of moles   n, boom, so basically each partial pressure 
is directly related to n, the number of moles.   The problem tells us that we have equal masses   of neon and argon but n is all about moles, 
not mass. So let&amp;#39;s compare their molar masses.   Neon has a molar mass of about 20 and argon about 
40 grams per mole. Now, argon is almost exactly   twice as large as neon. Argon&amp;#39;s molar mass 
is almost exactly twice as large as neon’s.   So if the masses that we have are equal, the 
molar amounts have a two to one ratio. Okay,   let&amp;#39;s think through this a second, okay? For 
example, let&amp;#39;s say you have 40 grams of each   gas. It doesn&amp;#39;t tell us how much we have but 
it says equal masses so we can just assume,   I don&amp;#39;t know, 40 grams of each. Okay, what would 
happen then? That would be one mole of argon and   two moles of neon, right? This molar mass is half 
as much as this, so you need twice as much neon   to have the same mass of argon. You need twice 
as many moles of neon to have the same mass   as argon and that is a 2 to 1 molar ratio. 
Now, the total pressure of the system is 12   atmospheres. Remember that for this problem 
pressure is equivalent to the number of moles.   So if 12 atmospheres represents a 2 to 1 molar 
ratio of neon to argon, the pressure of neon   must be twice as much as the pressure of 
argon. So the partial pressure of neon   must be eight atmospheres and the partial 
pressure of argon must be four atmospheres,   choice B. Remember Dalton’s law, here 8 plus 4 
equals 12 the total pressure of the system.
  Okay, which the following species has the electron 
configuration shown below? Now we could go through   each one of these and we could write electron 
configurations for all of them, but that&amp;#39;s going   to take too much time. Let’s think of a quicker 
way to do this. This electron configuration   tells us the number of electrons in each one of 
these orbitals so we can add this up and get 36   total electrons. Now which of these atoms or 
ions has 36 total electrons? Well a neutral   krypton atom has 36 protons and 36 electrons, 
but check it out, that&amp;#39;s not a choice here.   That means that the electron configuration must 
be for an ion where protons and electrons are not   equal, so right away since we know 
we&amp;#39;re looking for an ion, we can cross   out A and we can cross out D, since these are 
neutral. So it&amp;#39;s going to be between B and C.   Now let&amp;#39;s take a look at choice B, the strontium 
ion. Its atomic number is 38, so it has 38 protons   but the 2 plus charge means that there&amp;#39;s 
two more protons than electrons, so it&amp;#39;s   missing two electrons which means it has 
36 electrons. So we know that&amp;#39;s right,   but what about choice C, Br+? If you just glanced 
at it quickly you might be tempted to think that   the plus one charge means plus one electron, which 
means you go from 35 protons to 36 electrons,   right? No, it&amp;#39;s a common error. To make a plus 
one charge would mean you have one more proton   than electrons, so you&amp;#39;d actually have 34 
electrons. Now also halogens like bromine   don&amp;#39;t form plus one ions, okay? So it&amp;#39;s important 
to keep in mind that sometimes the AP exam gives   an answer that doesn&amp;#39;t exist in real life. Those 
answers are never the correct ones like Br+,   so don&amp;#39;t be fooled by that. And there&amp;#39;s an 
even faster way to do this problem using   the periodic table, okay? Krypton is a noble gas 
and this is a noble gas electron configuration.   Full s shell, full s and p shell, okay? 
So krypton is here. Abd strontium is here   in the second column with two more electrons, 
right? Krypton is here, one, two, strontium. So   its electron configuration with two more electrons 
is going to be krypton 5s2, with those two more   electrons. Now strontium will often lose these 
two electrons to get a noble gas configuration,   which gives us a plus two strontium ion with 
krypton&amp;#39;s electron configuration. So again,   choice B. Magnesium has a specific heat capacity 
of 1.0 joules over grams degrees celsius.   A sample of magnesium is heated to 80 degrees 
celsius. It’s placed in a calorimeter filled   with 100 grams of water at 19 degrees celsius. The 
magnesium cools and reaches thermal equilibrium   with water at 20 degrees celsius. If the specific 
heat of water is 4.2 joules over grams times   degrees celsius, what is the approximate mass of 
the magnesium? You can probably tell there&amp;#39;s going   to be a lot of calculation here but remember this 
is on the multiple-choice section so all the math   you&amp;#39;re going to be able to do in your head. There 
might be some tricks to keep in mind. Okay, now   the first thing to remember is that the equations 
you&amp;#39;re going to need are probably going to be on   that equation sheet, and one of these is q equals 
m c delta T, which stands for heat transferred   equals mass times specific heat times a change in 
temperature. So, this is the equation that we&amp;#39;re   going to be using here, and when we&amp;#39;re approaching 
specific heat problems like this a great thing to   do is to write down a little table, all of your 
knowns and unknowns. Specific heat problems often   involve the transfer of heat from one substance 
to another, so you want to remember to keep track   of both substances. Here, that&amp;#39;s magnesium and 
water. Okay, we&amp;#39;ll make a little bit of room   write out a tiny little table here. Okay, so now 
we&amp;#39;ll start writing down our knowns and unknowns.   Our unknown is the mass of magnesium—we’re asked 
that in the question, okay? The specific heat of   magnesium is this. Specific heat of water 
is this. The magnesium starts at 80 degrees   and it&amp;#39;s put in 100 grams of water. The water 
has an initial temperature of 19 degrees, and   the problem says that thermal equilibrium happens 
at 20 degrees. That means the final temperature   for both the magnesium and the water is going to 
be 20 degrees, okay? Now, the equation up here   needs delta T, the change in temperature, so we 
can do a little bit of subtraction in our heads   or on a piece of paper, and we can get the delta 
T for magnesium and the delta T for water.
  Now that we have our knowns and 
unknowns categorized here, we can start   solving the problem. We&amp;#39;ll take this, 
rearrange it for m, the mass of magnesium.   So, the mass of magnesium equals the heat lost 
by the magnesium divided by the specific heat   of magnesium and the change in temperature of 
magnesium. Now we know this, we know this, but   we don&amp;#39;t actually know the value of q here, 
right? We&amp;#39;re given a change in temperature,   but we don&amp;#39;t know how much heat was 
actually lost by the magnesium. But   we can figure that out. If you&amp;#39;ve 
ever done a specific heat lab   or specific heat problems like this, you 
probably know how we&amp;#39;re going to do this.   With problems like this there&amp;#39;s usually a 
switch step based on heat transfer. To find q   we make the assumption that the heat 
lost by magnesium was gained by the water   in the calorimeter. So let&amp;#39;s do a little bit 
more cleanup here. We&amp;#39;ll add subscripts to   the variables so we can keep track of things. So 
this is our equation for the mass of magnesium.   Now we have to find q, the heat gained by the 
water. Then we get that we can swap it into this   equation. Okay, so to find the heat gained by 
the water we&amp;#39;ll go back to this heat equation   and we&amp;#39;ll make a version of it for water, okay? We 
put this little blue box here water water water,   so you can keep track of the fact that this 
is for water. Okay, so we can fill in these   variables we have the heat gained by water equals 
the mass of water, well, that was given to us.   We have the specific heat of water, that was given 
to us. And the change in temperature, delta T,   one degree celsius. Calculate this through math 
you can do in your head. You don&amp;#39;t worry about   the one, you&amp;#39;re essentially just multiplying 4.2 
times 100. It&amp;#39;s going to give you 420 joules.   Okay, so that is the heat gained by the water, 420 
joules. So now we can plug that into the equation   for magnesium. 
420 joules there divided by   the specific heat and the delta T. Okay. We do 
the math, it is 420 essentially divided by 60,   because you don&amp;#39;t have to worry about the one, and 
that gives us seven grams. Let&amp;#39;s take a look at   our answer choices. There it is, B. Now there are 
lots of common mistakes when you&amp;#39;re doing specific   heat problems, and most of them involve forgetting 
which the variables are for which substances. So   that&amp;#39;s why we kind of made that little table there 
at the beginning to get everything organized.   Okay, so if you correctly calculated q to be 420 
joules, but you used the final temperature of the   magnesium instead of delta T, well that would 
give you 21 grams which is choice D. In fact,   all these other choices represent wrong 
choices that are typical math error. So   that&amp;#39;s why it&amp;#39;s important to write down and label 
your unknowns, and don&amp;#39;t forget that delta T here   equals the absolute value of T1 minus T2. Okay, 
now for our last two problems, the multiple-choice   section of the AP exam often features a group of 
questions that are related to a single diagram,   or a graph, or something like this. So we&amp;#39;re going 
to look at two questions that use that pattern   and they&amp;#39;re focused on electrochemistry here. Here 
we have a diagram of a galvanic cell there&amp;#39;s an   illustration and we have some standard reduction 
potentials and it&amp;#39;s important to know that when   you see something like this on the multiple-choice 
question, the questions don&amp;#39;t depend on each   other. In other words, the correct answer for 
one isn&amp;#39;t necessary to get the second one or the   following questions correct, right? So that&amp;#39;s 
a little bit different than the free response   questions where you often need to get one right in 
order to get the next ones correct too. Okay, so   this question takes up a whole screen so we&amp;#39;re 
going to break it apart and deal with each section   on its own. If you have a pdf of the 
test available, you might want to keep   that handy to keep referring back to 
this stuff. Okay, so let&amp;#39;s dive in.   The first problem is which expression gives the 
standard cell potential in volts. We don&amp;#39;t need   the diagram to answer this, but we do need the 
redox reaction and the reduction potentials.   The standard cell potential of a cell, sometimes 
called the EMF or the electromotive force,   is calculated by this equation, okay? E 
cell equals or I should say E naught cell   equals E naught cathode minus E naught anode. 
For whatever reason this equation isn&amp;#39;t given   on the AP exam, so it&amp;#39;s useful to know. There are 
other equations that you can also use to correctly   calculate the E naught of a cell, 
but we&amp;#39;re going to be using this one.   Now you should know that in an electrochemical 
cell reduction happens at the cathode   and oxidation happens at the anode. A useful 
way to remember this is that oxidation, or the   anode, and oxidation both start with vowels. 
Okay, now this means since we have to find   the E naught of the cathode and the E naught 
of the anode, that we need to figure out   which species is getting reduced, and which is 
being oxidized. So let&amp;#39;s look at the reaction   and start looking at what happens to the iron 
here. Okay, well iron goes from Fe 3+ plus to Fe   2+. That means that iron gains an electron, 
and gaining electrons means reduction.   So iron is reduced at the cathode. Now, 
this chart gives us the standard reduction   potential. So iron is being reduced so 
we can plug this value in, 0.77 volts,   for the cathode. Alright. Now the lead, 
neutral lead Pb, goes to the lead two plus ion.   And that means that the lead loses two 
electrons, and losing electrons means oxidation.   This is happening at the anode, okay? So now 
we know what&amp;#39;s happening at the anode. We&amp;#39;re   going to subtract this value. Don&amp;#39;t forget that 
this reduction potential is a negative number,   okay? So the equation is E naught cell equals 0.77 
volts minus negative 0.13 volts. We&amp;#39;re subtracting   the anode, so now we look at that. Well that&amp;#39;s not 
an answer choice! How are we going to solve it?   Or is it right? Mathematically, subtracting a 
negative is the same as adding it, okay? So this   is actually perfectly represented by answer 
choice A. Now you don&amp;#39;t even have to do the math,   right? As you can see, you don&amp;#39;t even have to 
add these together. We did the math here, but   this just gives you the answer choices before 
the numbers are combined. Okay let&amp;#39;s look at some   common mistakes. The most common error when doing 
EMF problems happens when students are confused by   subtracting negative numbers. Remember standard 
reduction potentials here can have negative   values, so don&amp;#39;t forget to use the correct 
signs when doing the subtraction, right? If   you don&amp;#39;t remember to subtract the negative, well 
there you go, choice B is waiting to trick you.   Now there&amp;#39;s something else that&amp;#39;s important when 
you&amp;#39;re doing problems like this. A common mistake   is thinking that voltage is affected by 
stoichiometry, okay? If you&amp;#39;re tempted by choice   C, you probably doubled the value of iron, right? 
You see this two here, you might have double the   value of ion because the reaction is talking 
about two moles of iron but EMF of a cell is not   affected by the coefficients in the balanced redox 
reaction, so you don&amp;#39;t want to be multiplying   this value by two. And choice D combines both 
mistakes. The student forgot the minus and they   multiplied by the coefficient from the equation. 
Alright, the last problem reads: which of the   following accurately describes the change in 
mass occurring at the electrodes during the   operation of the cell? Now in order to answer this 
we&amp;#39;ll bring back the illustration, there it is.   This cell features two types of electrodes; 
electrode one, the anode, is made of solid lead.   We can see the equation for what&amp;#39;s happening to 
lead, the half reaction that&amp;#39;s happening to lead,   and you can see that Pb solid is turning into 
Pb 2+. That means that the electrode itself,   which is made of lead, is oxidizing to produce 
Pb 2+ ions. That means that atoms of solid lead   in the electrode are turning into ions and are 
floating off into solution. So as this solid,   as the atoms in this solid lead are turning 
into ions and are floating off into solution,   the mass of this electrode is going to decrease. 
So that&amp;#39;s very important. The mass decreases here.   That means that we can cross off choice B and 
choice C, because we know that the mass has   to decrease. So now we have a choice A or 
D. What&amp;#39;s going to happen to electrode 2?   well here&amp;#39;s the cathode. Now at the cathode this 
reaction is taking place okay Fe 3+ to Fe 2+.   We don&amp;#39;t have ions getting reduced and turning 
into solid metal, we have ions getting reduced   to make ions. Both Fe3+ and the Fe 2+ are going 
to stay dissolved in solution, so the electrode   is not part of the redox reaction. It just 
provides a way for electrons to flow into the cell   reducing Fe 3+ to Fe 2+. Now because of that, 
this electrode wouldn&amp;#39;t be made out of iron,   it would be made from inert platinum. Its only job 
is to transfer electrons. It’s not gaining mass,   it&amp;#39;s not losing mass, so that means that choice A 
is correct, and platinum is not part of the redox   equation. There’s no change in platinum mass, 
so choice A is our correct answer.
  [a]Check with audio.
[b]29:12

Transcript for:AP Chemistry Sample Multiple Choice Lecture Notes

Transcript for:
AP Chemistry Sample Multiple Choice Lecture Notes