Transcript for:
Comprehensive AP Chemistry Exam Review

the AP Chemistry exam is just days away and I know you're all working hard to make sure you get every single point you deserve on that exam hi there my name is Jeremy kug and we're going to have a little cram session we'll be working our way through all nine units of the course over the next hour and a half or so and if you need some more practice don't forget to check out my detailed videos right here on YouTube that cover all 91 topics in the course as well as full study guides practice exams and more over at ultimat review packet.com now let's get started so just so you know we're going to go through this entire course from the beginning all the way through to the end and we're going to hit as many high points as we can in this cram session so as you follow along uh try to work the problems with me and I think that's going to help you quite a bit as we get ready for the AP exam here in just a few days so the first item I'd like to share with you is mass spectrometry so often what you'll have in a question like this is you'll see a graph that looks a whole lot like this to be honest each bar represents an isotope and uh on the x axis we have the atomic mass of the isotope so for example this graph represents two isotopes because we have two bars there one of the Isotopes has a mass of 85 atomic mass units the other bar the other isotope is representing a mass of 87 atomic mass units the high represents the percentage abundance of each of those Isotopes so this first one here that's the tall one it looks like that's somewhere around probably probably around 75% abundant of all the atoms of that isotope this other short one seems to be somewhere around 25% you can take the atomic mass of the isotope which is 85 multiply it by the decimal equivalent of its percentage which is 75 here and you get 63.75 do do the same thing for the other isotope so the 87 times the 0.25 which is 25% and you have that product then you add those two together and you find that the overall average atomic mass is somewhere around 8550 now this is an estimate because it's kind of hard to tell exactly how tall these are but this is a really good estimate now we can match this up to the periodic table and we can see what element we're talking about so this seems to be rubidium because it's very very close to that atomic mass that we had estimated now very often you'll see these types of questions pop up in the multiple choice section of the test and so you might have a question like this where there seem to be five different heights of bars here that represents five Isotopes and so what we have to do is determine which element we have and so it's not so important necessarily that we calculate an answer if it's a multiple choice question but we can estimate the mass and we can probably just eyeball that you can see that this Mass over here with about 64 is you know about half of the atoms and we have several that are in the 67 68 range a few that are higher or lower but we could kind of eyeball this and say that the overall weighted average here is probably somewhere around uh 65 somewhere in that neighborhood so what we could do is we could match this up to the periodic table table and find an element that has the closest atomic mass to about 65 and we find that that would be zinc and so if they give you four choices one is zinc and one is hydrogen and one is uranium or something like that should be pretty obvious as to which element you're talking about so that's mass spectrometry now another important point that pops up in unit one is periodic trends and so there are these Trends or these patterns on the periodic table that you need to be very aware of as you work through the exam so the first one that I want to share with you is atomic radius now generally speaking atomic radius is the greatest at the left and the bottom of the periodic table so that means these elements down here like cesium and uh rubidium and all these over here those are going to be the largest elements on the periodic table and the reason that's the case is that atoms at the bottom of the periodic table have more electron shells they have more occupied energy levels and because of that that last shell is the farthest away from the nucleus and so it's important that you're aware of that so that if this is on an essay you're able to explain that now atoms toward the left side of the periodic periodic table have a less effective nuclear charge now what that means is you know all of these atoms in uh period six for example all have six electron shells and so you might think that their their atomic radius is about the same the thing is the atoms on the left side have fewer protons to pull in those six energy levels or those those six electron shells so that means that those outermost electrons can Veer farther away from the nucleus and so that's why these atoms on the left side are larger and the ones on the right side of the table are smaller now another Trend that you want to be aware of aware of is ionic radius and here is a nice graphical representation of that as you can see just generally speaking cat ions the positive ions tend to be smaller than the negative ions the the anion and if we look at just a few of these here let's look at the nitride oxide fluoride ions as well as the sodium magnesium and aluminum ions you'll notice that if you start looking at the number of uh protons electrons and those electron configurations there they all have the same number of electrons so in chemistry we say that they're ISO electronic they all have uh 10 electrons so why do they all have very different sizes in fact aluminum is absolutely tiny as far as an ion goes compared to nitride well the reason is the atom with the most protons if they're isoelectronic is going to be the smallest and that's because it has more protons right aluminum has what something like 13 protons I think so it can pull in those electrons a whole lot more tightly than something like nitrogen which only has you know about seven uh protons I believe so you know we see that the one that has fewer uh protons is going to be the largest for isoelectronic species now here's another Trend that you want to be aware of first ionization energy and generally speaking this is going to be greatest toward the right and the top of the periodic table so these elements like helium and neon and Florine up here they're going to have very high first ionization energies and the ones down here toward the bottom left like cium they're going to be very very low and the reason for that is that atoms at the top of the periodic table have fewer electron shells so that means that that last shell is a whole lot closer to the nucleus and so it has stronger attractions to its outermost electrons we say it has stronger kulic attractions to its electrons now on the right side it's kind of the same way as we had for uh atomic radius on the right side you have greater effective nuclear charge so that means that since you have more protons as you go to the right it's going going to be able to pull in those electron shells a whole lot more tightly than those atoms that have fewer protons and so that's the reasoning that you want to share with your reader if you have an essay question about this now if we look at another concept this is something called photo electron spectroscopy and this will appear on the exam uh you're going to have a photo electron spectroscopy graph here and what you want to realize is that each one of these Peaks represents a suel and so it's important that you're able to to write electron configurations the way this works is it goes from left to right in order of in increasing energy so that means you have 1 s is always the first one and then 2 s is your next Peak and then after 2 s is 2 p after 2 p is 3s and after 3s is 3 p so those are the sublevels that are represented in this PES diagram now the height of each Peak represents the relative number of electrons that you have there so notice that the 1s the 2s and the 3s Peaks all have the same height and if you've been studying chemistry for very long you know that those sublevels max out at two electrons don't they so these all are two two and two now notice that the 2p Peak here is exactly three times taller than those S Peak so that means it has three times as many electrons 2 * 3 is 6 so this is 2p 6 right here now this 3p electron it seems to be U you know I I suppose you could say 56 as high as the 2p Peak so that means that this peak has five electrons in it so that's 3p5 and so if you know your electron configurations you can you know look at the electron configuration that this uh graph represents and determine that we're talking about the chlorine atom here with this particular PES diagram now let's try another one let's say we have this PES diagram hopefully you can plot out those sublevels just like I did before it goes 1 s 2 s and 2 p and then we know that the s's have two a pce so it's 1 S2 and 2 S2 and the last Peak is about you know one and a half times as tall as the S Peak so this represents three electrons so it's 2p3 and if you look at the periodic table you can see that we're talking about nitrogen here something that has seven electrons and so that's how you can interpret a PES diagram there are some more involved things here as there are with lots of the topics that we're going to hit in this cram session and of course you can check out my full length videos uh of course here before the exam we just kind of want to hit the high points so that we get those things back in our memory now let's take a look at uh unit two and one of the very important parts of unit two is being able to draw a Lewis electron do diagram so if you have a molecule like sulfur di fluoride you need to realize that the central atom of course goes in the center of your structure so I'm going to put sulfur in the middle here and we have the two Florine atoms one on either side I always recommend to students that you start with the outside and you work your way toward the center so we have Florine and Florine if you look at the periodic table you know Group 17 it has seven veence electrons so I'm going to put seven dots around this one and seven dots around that one right there and sulfur is in group 16 so it has six veence electrons so I'm going to put six dots around the sulfur there trying to pair these up as much as I can and we have that so now the goal is we're trying to follow the octet rule if at all possible does everything have eight veence electrons and yes indeed everything does we have the Florine can lay claim to you know these here in the middle we have a a couple shared pairs here everything has eight so this is a good structure so you you can write your your Lewis electron do diagram like this a lot of students like to replace the shared pairs with a line so you can have something like this and that's good as well so there's the ls electron do diagram for sulfur di fluoride let's try this next one which is c l which uh is a chemical compound that has carbon as its Central atom so once again just like in the last example we're going to always start with the atoms on the outside and work our way toward the inside so this chlorine down here is in group 17 so seven veence electrons so I'll put those there and then the same thing with the chlorine atom on the right seven veence electrons right there and then the oxygen is in group 16 so it has six so six dots for that and the carbon is in group 14 so four dots 1 two 3 4 trying to pair these up as I can and now I am trying to get everything with eight so does everything have eight veence electrons the way it's drawn in this structure well no it doesn't does it the oxygens and the and and the chlorines there all have eight but the carbon only has six the way it's drawn here so what I have to do is move some electrons from the outside to the inside so I think I'll take these two dots on top of oxygen and move those down like that and now everything has eight and so what I have here is two single bonds with the the carbon and chlorines there but I have a double bond between carbon and oxygen so when I draw the structure it's going to look like this so we have a good example with a a double bond there now let's take a look at uh type of bonds and and hybridization now we know that this structure here has two single bonds now just so you know every single Bond every Bond that's a single bond is a sigma Bond so that means this molecule has two Sigma bonds and the way you find the hybridization is you take the number of Sigma bonds touching that Central atom and you add the number of unshared electron pairs on that Central atom as well so in this case we have two Sigma bonds plus two unshared electron pairs 2 + 2 is four that's called the steric number and that corresponds to a hybridization of sp3 now you have to know that they're not going to tell you that on the exam so in this case here we have two single bonds but there's a double bond here well a double bond is composed of a sigma and a pi Bond Pi bonds don't count toward hybridization or molecular geometry so this molecule has three sigma bonds plus zero unshared pairs that's a steric number of three and so that's going to be a hybridization of SP2 because it has three sigma bonds one Pi Bond no unshared pairs on that Central atom now we can also do this for somewhat more complex molecules that have multiple Central atoms so this uh organic molecule here has three carbon atoms that are all essentially functioning as a central atom so carbon number one has a single Bond and a triple bond touching it well don't forget a triple bond has one Sigma and two Pi bonds so if we want to add up the total number of bonds here we have 1 2 3 4 five Six Sigma bonds and because there's a triple bond there that's two Pi bonds so that's why it's six uh Sigma and two Pi bonds each Central atom each carbon atom here has its own hybridization so carbon number one has two Sigma bonds touching it no unshared pairs so that steric number of two corresponds to SP and carbon atom number two is the same way it's got two Sigma bonds no unshared pair so that's SP hybridization as well and then on carbon atom number three we have four Sigma bonds no unshared pairs so that's sp3 hybridization because its steric number is four now we can take this a step further and talk about the molecular geometry and the bond angle for each of these molecules so on this one we have a molecular geometry since it has two Sigma bonds two unshared pairs that's called a bent geometry now you might be thinking that doesn't look bent it has a bond angle of 105 degrees it doesn't look like that the way it's drawn as it turns out most chemists will not draw a bent molecule like this we actually draw it more like this to kind of reflect the bent structure that we actually have in the real world so just be aware of that whenever we draw these uh on this next molecule the CO2 we have three sigma bonds and zero unshared pairs well that's called trigonal planer and that has a bond angle of 120° so you have to know that now on this last molecule here since there are three Central atoms guess what there are three molecular geometries uh the first one is linear because it's two Sigma bonds no unshared pairs the same thing for carbon number two it's linear as well for carbon number three we have four Sigma bonds and no unshared pairs that's called tetrahedral now you need to know that linear is 180° tetrahedral is 109.5 degre so there isn't a whole lot of memorization on the AP exam but those molecular geometries and bond angles and hybridizations those are things that you do need to remember for the exam now another property that we can talk about in terms of these molecules is is it polar or is it nonpolar now if a molecule is a polar molecule that means that it has a lopsided electron distribution it's almost like it has an unbalanced region of negative charge on the molecule it doesn't really have anything to do with is the molecule uh symmetrical or not that is a common misconception it's not doesn't have anything to do with symmetry for example this molecule here sf2 it's symmetrical isn't it but it's a polar molecule because there's this lopsided electron distribution up here at the top we have these two unshared electron Pairs and that makes it a polar molecule now what kind of intermolecular forces does this have as we kind of ease our way into unit three well we know that every molecule no matter what it is is going to have London dispersion forces so if you see a question and it asks what kind of intermolecular forces does it have you'll never go wrong by saying London dispersion forces however polar molecules also have dipole dipole forces so this one here has both dipole dipole and London dispersion now this second one here this is also a little bit lopsided isn't it we have you know four uh dots those two unshared pairs whereas these have uh three pairs that's a polar molecule as well and so that means we have London dispersion forces and we also have dipole dipole forces now in this last structure notice that there is no lopsided electron distribution everything seems to be very even there are no um unshared electron pairs that aren't canceled out by something else so this is a non-polar structure and considering that it's non-polar it would only have London dispersion for forces so that's one of the ways that you can figure out the intermolecular forces you really need to draw the structure and see if it's polar or non-polar to see what kind of forces it's going to have now let's take a look at this molecule here we have the ammonia molecule NH3 and hopefully you can see that it's a very polar molecule we have this unshared electron pair that's just hanging out up there not canceled out by anything else so it's absolutely a polar molecule and so it's going to have intermolecular forces that are London dispersion it also has dipole dipole but this is a very special type of dipole dipole Force if you have a molecule that has a nitrogen hydrogen bond or an oxygen hydrogen bond or a Florine hydrogen bond it's also going to have this hydrogen bond so instead of calling it dipole dipole we just go ahead and normally call it hydrogen bonding so that's how you'd discuss the intermolecular forces for that molecule now what about something that's a little bit uh uh different here this is not really a molecule is it sodium fluoride is not really a molecule at all it's an ionic compound so the forces that it's going to have in between its uh its own structure would be ionic forces and as we think about ionic compounds these are structured differently than molecular or calent compounds they have this uh lattice structure now of course this would be a three-dimensional structure in the real world I'm just simplifying it to make it look like this on the screen here but you can see how every sodium and every fluoride these are kind of alternating each other and so that's the structure of an ionic bond uh these these ionic compounds don't exist in U individual molecules like like ammonia or uh the fos Gene molecule would they are in this nice repeating lattice here now as we talk about ionic compounds there's another Force going on here that you need to know about these are called ion dipole forces and colge board kind of calls this an intermolecular force it's not exactly an intermolecular force but it is this is a very important force that you need to know about and it is the force between individual ions in a compound and polar molecules like water is probably the best example if you take sodium fluoride and dissolve it into water well we know that water molecules are very polar they have a negative pole and they have a positive pole and in the case of water the side that has the hydrogens here that's going to be the the positive pole of water and the oxygen side over here that's the negative pole so what happens is these water molecules are going to surround that fluoride ion right there and they're literally going to drag that fluoride ion away and this is what an ion dipole force is it's the force between the individual ions and the polar molecules the the individual poles in the polar molecule and that's essentially how ionic compounds dissolve in water if a substance is soluble if it's ionic and it's soluble that means that this ion dipole force is essentially stronger than the ionic forces holding this thing together uh we could say the same thing for the sodium over here if we had water molecules uh surrounding that sodium over there the negative pole of the of the water molecules are going to be attracting the positive sodium ion so that very or that fairly strong ion dipole force is literally going to drag that sodium ion out of the structure uh and so that's how these ionic compounds dissolve in water you have these ion dipole forces that are actually taking place here very important Force to know about uh as we move on let's take a look at some gas laws here so this is also part of unit three knowing about the gas laws so here we have a gas mixture that is collected in a 2.000 l rigid container at a temperature of 295 Kelvin if the pressure inside the container is measured to be 1.02 atmospheres how many moles of gas are in the mixture so we have a perfect setup here for ideal gas we have a pressure in atmospheres we have a volume we have a temperature we want to know the moles so we just have to plug and chug into this equation here so the pressure is 1.02 atmospheres we just plug that in and the volume is 2 L right there we're trying to find moles so n is going to be our unknown here R is that universal gas constant it's 0821 ler atmospheres per mole Kelvin on the exam they actually give this to you with one extra significant figure 8206 so I'm going to go ahead and use that number here temperature is 295 kelv so we just do a little bit of algebra here and we can find that the number of moles of gas in the mixture is about 0.843 moles and so that's how we would solve a fairly straightforward ideal gas law problem now what if we have this what if it is determine that 0329 moles of that mixture is oxygen the other gas gas would compose how many moles well if you have two gases and they have an individual amount of moles they're going to add up to give you the the total moles aren't they so if the total mole value is 0843 and one of the gas is 0329 we just have to subtract and so the other gas is going to be 0.0514 moles and so that's how we'd solve something like that very simple uh just subtraction here now let's go step further if it is determined that this other gas has a mass of 2056 G what is the molar mass of this gas well all we have to do there is recognize that molar mass is measured in grams per mole and so that implies we have to take grams and divide it by the moles well the problem tells us that it has 2056 G and the number of moles well we just got that from the last problem it was 0514 moles so when you divide that out it's about 4.00 G per mole so if I had to guess by looking at the periodic table I'd say that we're talking about helium in that other gas there so let's take a look at another important concept that is going to show up on the exam and this is spectrometry and this is where we take an instrument called a spectrometer and we're trying to use it to measure the concentration of a certain ion or compound or something like that now normally the way this is presented on the AP exam is we have a graph and the graph is going to look something like this where we have on the x axis the concentration of the the substance whether it be the ion or molecule or something like that is presented on the x-axis and the absorbance that is read right off of the Spectra photometer is plotted on the Y AIS on the the vertical axis here so the first thing you want to do when you're doing an experiment like this is to run a blank and so I'm just going to do a quick check and make sure that there's a blank so we'll have a little dot there at the origin 0 comma 0 to show that we have a blank that when there's no uh ion that we're looking for in here that the absorbance read zero that's what you want to start with then we're going to run several known concentrations and we're going to record their absorbances and so as we do that we just plot those on the graph like this we have a bunch of dots now notice that these dots roughly give us a line so we're going to draw a best fit line it's not going to be perfect but we should have a pretty good idea as we draw that best fit line so it's probably going to look something like this as you draw that now this is sometimes called a calibration curve I know it's a line but we often call it a calibration curve in chemistry and now we can determine the absorbance of our unknown and match it up to its concentration so if we run the the unknown maybe it has an absorbance of right here about 0.20 so we just line that up to the best fit line right like this and we drop down and we can see that the concentration of the unknown seems to be right around 0.55 moles per liter so that would be our uh concentration of the unknown very simple basically just being able to read a graph as you do that now a couple of things that sometimes they do throw at you you on the multiple choice section watch out for outliers and so they like to ask these questions if you have a calibration curve like this and you see a random dot that's kind of off the line it's it's like above where it ought to be that's probably because the sample got contaminated with some solution that had a higher concentration than you expected so maybe you know according to this maybe you had um you know you you had a025 and its way up here maybe it got contaminated with some uh 60 mol or something like that or if you have an outlier that's like way down here it's way down below the line that's probably a sign of a sample that CAU contaminated with distilled water most likely as you were washing out the cuvette or possibly it was contaminated with something that had a concentration that was too low so maybe you thought it was. 35 and it actually got contaminated with something that was like a 0.1 or something like that so just be aware of those outliers they like to ask those questions on multiple choice so we have that now going on to unit four we're moving right on to the concept of net ionic equations and so when we write equations especially if it's in solution uh you need to understand that there are ions in solution so if we have this reaction here where a piece of magnesium metal is added to a solution of zinc chloride the first thing that I like to do is write out what you have on the reactant side so the magnesium metal of course is just mg so we write that and then the solution of zinc chloride well anytime you have a solution of an ionic compound it's existing in its ion form if it's soluble so zinc chloride will exist as zinc 2+ ions and chloride ions CL negative now we have to remember that generally speaking if we have ions and metals kind of mixed together metals are going to react with metal ions and likewise if you ever have a non-metal the non-metal is going to react with non-metal ions well in this problem we have a metal which is the mg and we have a metal ion which is the zn2+ so those are going to react with each other now what's going to what's that chloride going to do well the answer is not going to do anything it is a spectator ion do not include spectator ions in your net ionic equation that's a very common mistake that students make on the AP exam do not include those in your net ionic equation so the way this works is the Magnesium is going to have to turn into something else well the only other form of magnesium I can think of is mg2+ right that's the ion form of magnesium and then the zinc 2+ is going to have to turn into zinc metal so we'll have the ZN and so now we get to balance these on the first one here it looks like I can balance this half reaction by putting two electrons on the right side right here balancing out the charge and then on the second one I can balance the charge by adding two electrons to the left side just like this and just so you know uh whenever we have a case like this where we're losing electrons that half reaction is called oxidation or if you prefer the charge on the Magnesium is going up that's also oxidation going from a zero to a plus two that means that the second one is a reduction because we're gaining electrons and the charge on the zinc is going down from plus two down to a zero so now we can add these together and we have our overall balanced equation this is an oxidation reduction process we often call that Redux now let's try another one what about sodium metal is burned in air well sodium metal of course is na and when something is burned in air it's always being reacted with oxygen gas O2 so essentially we just have to put those two elements together to make sodium oxide which is na2o and we have to balance these equations uh probably want to balance the oxygens by putting a two right here and then I have four sodiums on the right side so I put a four over here and now I have a balanced equation on the next one here we have a sample of propane gas c3h8 that's burned in a so kind of the same thing except this time we have c3h8 being reacted with O2 but anytime we have a complete combustion reaction we know that the products are going to be carbon dioxide and water and so uh we can write those out we need to balance this I can balance my carbons by putting let's see three carbons over here so a three right here I can balance my hydrogens I have eight hydrogens on the left side and only two on the right side so if I put a four right there that balances the hydrogens and then the oxygens well I have uh six plus four more oxygen right there which is 10 so by placing a five in front of this oxygen now I have a balanced equation so we have that now let's try a precipitation reaction let's say we have Solutions of lead to nitrate and sodium carbonate that are mixed in a test tube once again realize that if you have an ionic compound and it's soluble it's going to be in its ionic form so lead to nitrate solution is actually pb2+ and NO3 with a negative charge we have those ions kind of swimming around then we have sodium carbonate same thing we have sodium ions and then we have carbonate ions and our job is to determine which of those two combinations is going to make the solid precipitate now in college chemistry or in general chemistry you actually do have to learn all those solubility rules on the AP exam they give you kind of a gimmick where they tell you that anything that has a group one ion in it or a nitrate or an ammonium ion is going to be soluble and the other substance is going to be your precipitate so that's kind of how they they write these questions that way you don't have to learn all of those solubility rules for this course so we notice that sodium and nitrate are in one of the combinations and so those are going to end up being our spectator ions so that means that the lead 2 ions and the carbonate ions are going to get together to make uh the the lead two carbonate so that's what we would have as our product which means that the sodium and the nitrate are spectator ions so if I clean this up I can write the net ionic equation like this lead 2+ aquous plus car carbonate aquous makes our PB CO3 solid precipitate so that's how you do a precipitation reaction so let's take a look at a little bit of stochiometry here so often we have to solve stochiometry problems this is kind of a big part of unit 4 we have an equation given to us here it says in a chemical analysis excess silver ions are added to a solution containing carbonate ions after the reaction the total amount of silver carbonate solid is weighed and found to have a mass of 582 G how many gr of carbonate ions were present in the original sample well what we have to do here is start out with the 582 G of silver carbonate and the question is asking how many grams of carbonate did we have starting out so way down here at the end I'm going to write grams of carbonate down here and whenever we have a stochiometry process I always tell my students we have a three-step process here Step One is convert to moles step two is do the mole ratio and step three is convert to whatever your your final unit is in this case it's looking like it's it's going to be grams so step one convert to moles so that means in our denominator we have to put grams down here so it'll cancel and moles on top now we can consult the periodic table and we can see that one mole of of of silver carbonate ag2 CO3 has a mass of about 275.50 grams we can cancel grams top and bottom and we're in moles of silver carbonate now we can go on to step two which is the mole ratio so the mole ratio is where we take the actual substances out of the equation here and put them into our stochiometry here so ag2 CO3 has to go on the bottom so it'll cancel out we're converting to carbonate ions so those will go on top and now it looks like we have one carbonate for every one silver carbonate those are just the coefficients of the balanced equation so now silver carbonate goes out top and bottom we in moles of carbonate we want to be in grams of carbonate so step three is convert to that final unit which is grams so moles on the bottom and grams on top so according to the periodic table it's about 60.0 G in one mole of CO3 to negative so moles are out and so on my calculator I just take 582 divide by 275.50 times 60.0 you know all these ones kind of cancel out and I find that the answer is about1 127 G of the carbonate ions right there so that's how you would do a fairly simple stochiometry problem now let's try another one let's try one where we have some solutions stochiometry because they do like to ask these on the exam and sometimes this throws off a lot of students because of the because of these units that we use so in this problem it's the same equation but this time it says a student adds 100 Mill of .100 molar silver nitrate to a solution of excess sodium carbonate assuming the reaction goes to completion how many moles of silver carbonate should be produced so the first thing I want to do is I want to figure out how many moles of the silver ions did we use because the problem gives me enough information to figure out uh I have 0.1 lers and I have 0.1 moles per liter so to find the moles you just take the marity times the liter so 0.1 molar time .1 L is 01 moles of silver ions so that's what I'm going to start with the 0010 moles of silver ions and the question is how many moles of silver carbonate should be produced so that's what I'm going to put down here at the end of my uh stochiometry process so essentially right now I just have to do a mole ratio because you know Step One is convert to moles and I'm in moles right now so I can put ag+ on the bottom and ag2 CO3 goes on the top and looking at the equation uh to do the mole ratio it's one silver carbonate for every two silver ions and so now I just have to divide 0.0010 / 2 and the answer is about 5 * 10 the -3rd moles of silver carbonate or 00500 if you prefer that so those are some fairly straightforward stochiometry problems make sure that you're able to do those now there are others that you might want to learn how to do or perhaps I should say relearn or remember how to do like uh limiting reactant problems percent yield I have videos for all of those as well let's go on to unit five which is kinetics and so the concept of rate probably the most important thing that you need to know how to do as far as kinetics goes is know how to write a rate law and understand how to find the order of a reaction with respect to a specific reactant you're going to see some sort of a table like this on your exam somewhere where you'll have a balanced equation and you'll have these multiple trials and we're trying to figure out the orders and the rate laws and things like that so the first part of this says find the order of the reaction with respect to nitrogen dioxide what we have to do here is find two trials where nitrogen dioxide is the only thing that's changing in the experiment so it looks like experiments one and two are going to fit the bill here because FL chlorine is being held constant from trials 1 to two and looks like NO2 is being doubled so it goes from 01 to 02 and notice that while that's happening the rate is doubling as well the 2.37 * 10us 3r is doubling to 4.73 * 10us 3r so 2 equals 2 and what's the power that makes this a true statement well it's a power of one so that means we say that this is a firstorder process with respect to nitrogen dioxide if we had doubled the concentration but the rate had quadrupled well that would have been a second order process or if we had doubled the concentration and the rate hadn't changed at all that would have been a zero Thor process now let's do the same thing but for F2 this time and this time I'm going to choose trials one and three because it looks like from trials 1 to three NO2 is being held constant this time and Florine is the one that's changing except this time it's not being doubled it's being tripled 01 to 03 it's being tripled and looks like the rate is being tripled as well from trials 1 to three so that's a first order process as well so it's first order with respect to both of the reactants and if the question ever asked asks you what is the overall order well it's just the sum of the individual order so 1 + 1 is 2 so that means it's overall second order in this process now let's take this a step further let's write the rate law so if we write the rate law for this reaction it's always written in the format rate equals K time the concentration of the first reactant which is NO2 raised to its order which is one so that's why there's no uh visible exponent here times the next reactant which is F2 raised to its power which is a one as well now I want you to notice that in order to write a rate law you have to have some sort of data have to have some sort of data like experiments here perhaps a mechanism you can't just look at an overall balanced equation and guess what the rate law is we can't say you know some students might say well there's a two right here in front of NO2 so why isn't it second order well it doesn't work that way okay you have to have some sort of experimental data in order to write a rate loss so that's why it's first order in both of these now let's take it a step further let's determine the rate constant for the reaction at this temperature well what we have to do here is take any of these trials it doesn't matter which trial you use one two or three but take the data from that trial and plug and chug into the rate law now I'm going to use trial one just because it has a bunch of ones in it and it's kind of convenient to use but I'm going to plug those numbers into the rate law here so the rate in trial 1 is 2.37 * 10 -3rd m per second now the K is what we're solving for so that's my unknown the concentration of NO2 in trial one seems to be 010 moles per liter and the concentration of F2 in that Tri is also .010 moles per liter so now all I have to do is solve for K now the calculating part of this is pretty easy we find that K is about 23.7 on these questions they really want you to make sure that you have the correct unit and the units are a little tricky here so that's why I want to lead you through this so to get the units I want you to notice that we have to isolate the unknown which is K so we have a marity times a marity left over here so in order to get rid of that I'm going to multiply that by a marity to the -2 okay that's going to get rid of your two moities there in order to make that work if I do something to one side of the equation I have to to do the same thing to the other side of the equation so over here I'm going to also multiply by marity to the minus 2 now this molarity that's here time marity to the minus 2 gives me marity to the minus1 so that's why my units are marity to the minus1 time seconds to the minus1 when I drag the seconds back up to the numerator okay so that's my appropriate Ray constant with units when you have this you got to make sure you have the right unit in there it's a little tricky so make sure that you do that okay now another way to determine the order with respect to a reactant is to do this graphically so let's say we have a reaction so here's a uh have 2 ni yields H2 and I2 and we've decided to graph these values here as a function of time on one graph we have the concentration versus time on another graph we have the natural log of concentration versus time and on yet another graph we have one over the concentration versus time so how do we find the order of the reaction with respect to hi well here's how it works if if the order is zero order then the concentration versus time graph would be a straight line well it's not this is a curve so it's not zero order if it's a first order process the natural log of concentration versus time is going to be a straight line it's a little straighter but it's still not straight is it so it's not first order if it's a second order process the graph of one over concentration versus time is GNA be a straight line and that is a straight line so that's why I would say it's a second order process now as it turns out this is the only reactant in the reaction so writing the rate law is pretty easy at this point it's just rate equals K * hi to the second power because it's second order so that's how you can determine this graphically they're probably going to have that on the exam at some point as well so make sure that you can solve something like that now how could you determine the rate constant well to determine the rate constant you just look at the at those three graphs and look at the one or focus on the one that's a straight line so that would be the 1 over hi graph in this case so you want to find the slope of that line okay the absolute value of the slope is equal to the rate constant okay that's all you got to do and I say absolute value because notice that the hi and the natural log of hi those are sling downward and so if you have a uh if you have a slope of those lines it's going to be a negative value and rate constants are always positive so that's why I say absolute value that works for this as well of course it's a positive slope anyway but that that still works so to find the rate constant it's just the it's just the absolute value of the slope of the straight line okay now another important thing to know how to do in uh kinetics is work with reaction mechanism so if we have a mechanism for a chemical reaction and this is what you have here um we need to realize that there are often reaction intermediates notice that there's a little substance here this bf2 that's produced in an early step and then it's uh used up in a later step so when you add these equations or these uh steps together you end up canceling them out okay that's the reaction intermediate the BF 2 is the reaction intermediate it's essentially uh used up about as fast as it's produced and so it this is kind of hard to uh to to to isolate and to measure uh very easily now what's the overall Balan equation this is the point where you just have to add those two together so we have BF3 plus 2 o2s get us Boron plus O2 F2 plus o2f and so there's the overall balanced equation now often we're given a little bit of extra information and notice that I just put on the the board here that the first step is a slow step and the second step is a faster step perhaps there's an equilibrium here I don't have that drawn in but it's often how it works so which one is the rate determining step well the slow step is always the rate determining step so in this case it's the first step and why well because it's the slow step and that's how that works now what is the overall rate law for the reaction well the rate law for the overall reaction is the same as the rate law for the slow step and so the way you write that is just rate equals K times the concentration of the first reactant of that slow step times the concentration of that other reactant in that slow step as well so it's going to be rate equals K * BF3 concentration O2 concentration and that's basically all you have to do for that one right there okay so that's a simple uh uh review of reaction mechanisms and how to write rate laws there there are some more things you might need to be able to do but I have a video for that if you want to go more in depth into mechanisms now let's go on to unit six which is about thermodynamics and specifically the first law of thermodynamics and let's say we have a question like this a student adds 50 g of warm copper metal to a 52.77 g sample of water after stirring the temperature of the copper metal drops from 3 56.0 de C down to 23.0 De C the specific heat capacity of copper metal is 0.385 Jew per gam degre C and the specific heat capacity for water is 4.18 Jew per G de C assume that no heat is lost to the surrounding ings so the first question I have for you here is how many jewels of heat were lost by the copper well anytime we have a problem that talks about specific heat and we're dropping something in and there's a mass and something like that we're going to use qal MC delta T So once again this is just a plug-and chug we're trying to solve for Q which is jws so we're going to solve for that the mass of the copper is 50 G so that goes in for m c is specific heat capacity so the specific heat capacity of copper is the 385 Jew per gam degre Cel and notice that the temperature of the copper drops from 356 de down to 23.0 De so when you subtract that I think that's about 333° C notice it's a drop in temperature so that's why I have a negative value here now when you use your calculator and multiply these together you find that it's about 64.1 Jew and that negative sign just implies that heat has been lost so that means we can safely say that about 64.1 Jews of heat were lost by the copper now if that's the case how many jewels of heat are gained by the water well we're assuming that no heat is lost to the surroundings so that means that whatever heat was uh lost by the copper is going to have to be the same as the amount of heat that's picked up by the water so that means that the heat gain by the water is 64.1 Jew so it's just the same amount now what was the original temperature of the water well we're going to use the same equation qal MC delta T now this time we know how many jewels were picked up by the water it was 64.1 Jew right out of the problem here the mass of the water the m is it's up here it's about 52.77 G so that goes in there for m c is your specific heat capacity which is 4.18 Jew per gam degre C and delta T is your change of temperature and that's what we're solving for here it says what was that original temperature we're going to solve for delta T so when you uh key this into your calculator you find that delta T is about. 29° C now that means that the water temp temperature went up by about. 29° C now let's reason through this if the temperature of the water ended up at 23° C and we know that because when you take something that's hot and you drop it into something that's cold the hot object is going to drop in temperature the cold object is going to rise in temperature and their final temperatures will both be the same won't they so that means that if Copper's final temperature was 23° water's final temperature had to be 23° as well right so if it ended up at 23° and it Rose by 0. 29 what was its initial temperature well it has to be about you know 0. 29 less than that so that's why we can safely say that the original temperature of the water was about 22.7 de um so that's your initial temperature I'm trying to use a proper number of sig figs as much as I can because we didn't have the original temperature out to the hundreds place on that one so that's why I said 22.7 de now as we heat something up speaking of heating things we can take a material or a substance and we can we can heat it up and if it starts out as a solid its temperature is going to rise just like we have here as we add heat its temperature rises and anytime you have just a just an object that's in a specific state of matter it its temperature goes up whenever you add heat to it however once it gets to a phase change when you add heat its temperature stays level so notice that as we get to the melting point of whatever this substance is it just stays constant and the value that goes along with that is the heat of fusion or the enthalpy of fusion okay once it's all melted and it's all a liquid and you add heat well then it keeps going up again okay so it temperature goes up until it hits the boiling point and then you add heat and it levels out until all the substance is boiled away you keep adding heat then it goes up uh as a as a gas and its temperature goes up again now you can do this in Reverse it would be called a cooling curve if you uh start out at a high temperature and you're getting you know cooler that would be called a cooling curve now speaking of enthalpies and heats let's say we have this reaction here and let's work with enthalpies of formation a little bit because they often like to ask you something like this on the exam and let's say that we know that the change in enthalpy for this reaction is 16.38 K per mole of reaction and the question asks us to determine the enthalpy of reaction of n2o okay and somewhere they're going to have to tell us the enthalpy of reaction of n and N2 in order for us to figure this out well if that's the case we don't know what n2o is so we're going to call it X and we have two moles of this so that means we have to multiply it by two 2 * X of course is just 2x they're going to have to tell us somewhere on the exam what the enthalpy of formation of nitrogen monoxide is so it's about 9.29 and we have two moles of that as well so we have to times that by two so that's where we get 180.5 K and nitrogen gas is a zero and just in case they don't tell you this uh any element any free standing element in its most natural state has an enthalpy of formation of zero okay so they may tell you this they may not but just be aware it's zero there's only one mole of that so that's still 0 kles and the way this works is the Delta h of the reaction is equal to the sum of the products minus the sum of the reactants okay so the sum of the reactants well we just have a 2X there so we'll deal with that in a second the sum of the products you know 18.58% 0 is 180.5 Kil and the Delta h of a reaction is the sum of the products sum of the products minus the sum of the reactants and so that means I have to take the Delta H which is 16.38 given to us in the problem equals the 180 .58 - 2x and if I can solve for x I'm going to find the enthalpy of formation of the N2 so just use some algebra here subtract 18.58% formation of all of these substances and you just have to plug them in and find the Delta H for the whole reaction but on the exam on the AP exam sometimes they like to throw you a curveball and do something like this so that's why I wanted to work a problem like this for you let's say we have equilibrium let's go on to Unit Seven here which is equilibrium and a very important skill very fundamental skill actually is being able to write equilibrium constant expressions for pretty much any reaction and remember remember it's written by taking products over reactants raised to the power of the coefficients and you're going to want to Omit any liquids or solids that are given to you in the problem you only have gases and aquous substances in the expression so in this one here notice we have a couple of solids and so we have to ignore those or at least omit them from the equilibrium constant expression if if we're going to write this in terms of concentration that's called KC and you would write these in Brackets and so uh the brackets imply the marity of the of the solution or the substance so KC equals the concentration of the aluminum ions over the concentration of the silver ions cubed and once again that cubed is because there's a little three as a coefficient right there now sometimes we're given a a reaction that has all gases and in that case it's a a lot more convenient to measure this in terms of pressures or partial pressures and we call that KP in that case you'd write it more like this where KP equals the partial pressure of N squared and notice that squared comes from the fact that there's a coefficient of two times the partial pressure of nitrogen all over the partial pressure of dinitrogen of monoxide squared and once again the squared comes from that coefficient of two right there so be able to write an equilibrium constant expression now let's solve a problem here we have that very same problem that I think we just looked at or that same equation it says at a certain temperature the reaction below occurs and a chemist investigates the process by filling a container with dinitrogen monoxide gas to a pressure of one atmosphere after the mixture rains equilibrium the pressure of D nitrogen monoxide gas has dropped to 080 atmospheres calculate the equilibrium constant KP at this temperature so all we have to do here is set up an ice box I strongly recommend that if you have an initial uh pressure or concentration and final there's a lot going on here so use an ice box and so the initial pressure of n2o is one atmosphere so that goes in right here in that position and it says that's all that's put in there so we can safely assume that the initial pressures of your products are both going to be zero unless it says otherwise now it also tells us that once you get to equilibrium the pressure of n2o is down to 080 atmosphere so for equilibrium n2o I'm going to put in 080 atmospheres now notice that's all that's given to us but guess what that's enough to solve the puzzle we can figure out what every other slot in that ice box is going to be so let's start by realizing that the 1 to8 that's that's a drop of .2 atmospheres now on the other side of the reaction if n2o is going down then the products have to be going up that's how that's how chemical reactions work and since this is a 2:2 ratio well the nitrogen monoxide is going up by the same amount that the n2o went down so that's why the n o is a plus2 and since it's a 2:1 ratio N2 is going up by half as much so this is a0 one0 so our equilibrium pressures are 20 and10 and so now I just have to take these equilibrium values and plug them into the equilibrium constant expression now I have to write that uh it's you the same as it was earlier whenever we wrote that on the last slide here it's it's the pressure of n^ 2 * N2 all over n22 and I just plug those numbers into each appropriate slot like this. 2^ 2ar * .1 all over8 2 and I find that the KP value is equal to 6.3 * 10 -3rd remember equilibrium constants have no units now one thing I might notice is this KP value value is kind of a small number isn't it anytime you have a small value is in like way less than one okay that means you're going to have a whole lot of reactants and not very many products okay that's just how that works if you have a very large value of K like a million or a billion or something like that something much greater than one it's going to be the opposite you'll have a whole lot of products and not very much reactant left over so be able to interpret uh the meaning of an equilibrium constant as well now as we move on let's say we have another problem at the same temperature as the previous example a mixture of uh these three gases is what that should say that these three gases were was added to a closed container the partial pressure of n2o gas was 1.35 atmospheres the partial pressure of n gas was2 atmospheres and the partial pressure of N2 gas gas was 3 atmospheres what is the total pressure in the container well we have to understand that the total pressure is just the sum of the individual partial pressure so we have to take that uh you know that 1.35 the 20 and the0 30 and add them together and you'll find that the total pressure in the container is 1.85 atmospheres now the next part of the question says as the reaction proceeds which gases will increase in pressure and which gases will decrease in pressure now this is a very typical example of a reaction quotient question what we call Q versus K we don't know if this mixture here is an equilibrium or not and so we have to plug this into the reaction quotient expression and it's written exactly the same as the K expression we just can't call it k because we don't know if it's an equilibrium or not so we call it Q so we have the same expression I wrote before except I'm calling it q and I'm going to plug those same pressures into the expression so I plug in you know the 20 is the n o so that gets plugged in there the N2 is30 so that's plugged in there and the n2o is 1.35 so it gets plugged in there and I find that the value for Q once I mathematically calculate that is about 6.6 * 10 -3rd and once I get the Q I compare that to the equilibrium constant now I'm at the same temperature as I was in the previous example so the equilibrium constants the same so the Q is 6.6 * 10 -3 the K is 6.3 * 103 so it looks like Q is a little bit greater than K now what does that mean students get this confused sometimes so what I like to do is I draw in little a pacman here so if I do this I can see that the Pac-Man is eating toward the left okay if it's eating toward the left that means that you know as it goes toward the left there it's going to be going toward the side of the left which is the reactant side so that means the reactant side is going to predominate it's going to increase so I would expect that the n2o is going to go up and it's going to do that at the expense of the other side which is n o and N2 so the n o and the N2 are going to go down okay so that's how you would solve these Q versus K problems and sometimes they're in a little bit of a disguise kind of like you have here it it doesn't say uh toward which side does it go it just says which gases are going to go up which side is going to go down so this is a good Q versus K problem now how about this one here after the reaction has returned to equilibrium how would each of the following changes affect the partial pressure of nitrogen gas in the reaction vessel so this is a lattia principle question if we add some n right here anytime you add something it's going to go toward the other side so in this case we're adding a product so it's going to go toward the reactant side so it's going to the n2o is going to go up and and the N2 will have to go down in response so that's why the N2 decreases how about if we take away some n2o okay well if we take away something the reaction is going to shift in such a way to replenish whatever was taken away so whatever we took away is actually going to go back up so if we take away N2 then the n2o is going to increase right and it's going to do that at the expense of the other side so guess what the N2 is also going to decrease okay what about if we add some helium gas well helium gas has nothing to do with this problem does it it's it is an irrelevant inert gas and so uh there's going to be no change if we add some helium they like to ask you that sometimes and some students get tripped up on that but don't let that trick you it's no change if it's an inert gas that doesn't have anything to do with the reaction what about if the pressure is decreased by increasing the volume okay well whenever you increase the volume of a container the equilibrium is going to shift toward whichever side has more moles of gas because that's the side that takes up more space well in this case we have two moles of gas on the reactant side and we have three moles of gas you know 2+ 1 on the product side so whenever we increase the volume in this case it's going to go toward the toward the product side and N2 is a product so we're going to have N2 increasing so that's uh the result for that one how about what if the temperature of the container is increased well this is why we use our Delta H in here notice that the Delta H is positive so that means this is an endothermic process right so that means that heat is acting as a reactant as it is in every endothermic process so when you increase the temperature we're adding heat aren't we so that's like adding a reactant so it's going to shift toward the other side so when you add heat it's going to shift toward the toward the product side isn't it so N2 is going to increase as well so be aware of how these different changes these shifts are going to affect things uh using later's principle now let's take a look at Unit 8 which is uh acids and bases and let's think about the relationships between hydron ions and hydroxide ions as well as pH and P now this is probably the fundamental equation here the hydronium ion concentration times the hydroxide ion concentration for any solution is always equal to KW okay now at 25° C it's kind of nice that KW works out to this nice round number it's 1.0 * 10 -14 so that means that as long as you're at 25° C the hydronium ion concentration times the hydroxide ion concentration equals 1.0 * 10 -14 so if you know one of these you can calculate the other and then to add on to that we have these other equations pH is equal to the negative log of the hydronium ion concentration and likewise p is the negative log of the hydroxide ion concentration ation I don't have it on the screen but pH plus P equal 14 okay that's kind of a different way of stating this this one here but pH plus P equals 14 so let's try this problem here at 25° C A solution has a pH of 9.58 determine its P its hydronium ion concentration and its hydroxide ion concentration well we'll do the easy one first first let's find the PO if pH plus P equal 14 well all we have to do to find the p is subtract the pH from 14 so 14 minus 9.58 gets us about 4.42 so that's the P of this solution let's do the next one the hydronium ion concentration well if we know the pH we can figure out the hydronium ion concentration by taking the negative anti-log of the pH that just means we take 10 to the Negative PH power so 10 to the 9.58 is a concentration of 2.6 * 10 -10th so that's the concentration of hydronium and if you want to find the hydroxide concentration well there are a couple different ways to do that uh we could take you know uh we could use this equation here take h3o+ * o minus yields KW or we could also do this we could take hydroxide equals 10 to the P power and so hydroxide is 10^ the 4.42 we got back in the first part of this and so the hydroxide is equal to 3.8 * 105 moles per liter okay so moles per liter are the units on both of those there now um just to check your work if you want to if you take the hydronium ion concent conentration time hydroxide ion concentration you should get a number that's very very close to 1 * 10 -14 so we have that now let's think about strong acids and strong bases for a few minutes here um there are six strong acids you need to know those and for those six strong acids the concentration of that acid is equal to its hydronium ion concentration so if you want to find the pH of a strong acid it's really easy just take the negative log of its concentration so for example if we have .25 moles per liter of nitric acid all we have to do uh to find the pH is just take negative log of 0.25 and you find that that pH is uh 60 pretty simple okay for strong bases it's it's fairly straightforward as well there's a little extra step in there there are eight strong bases essentially the group one and two hydroxides it's done the same way the concentration of the compound is equal to the hydroxide concentration just be aware that on the group twos we have a 2:1 ratio so so you have to multiply by two so if we have 0.25 molar calcium hydroxide well the hydroxide concentration is50 and so to find P this time it's negative log of50 and that means our p is30 and the pH you know 14 minus that is is 13.7 so that's how you find the pH of a strong acid or a strong base what if it's a weak acid or a weak base well we have to consider that this is an equilibrium problem essentially because weak acids and weak bases don't dissociate all the way we have to work these out as equilibrium so let's try a fairly straightforward weak acid problem here it says determine the pH of 1.25 moles per liter acetic acid and it gives us the KA that Ka is 1.8 * 10 -5 and so the first thing you want to do here is write the dissociation for this write the equation for it and so we have our weak acid the acetic acid it's being added to water now anytime we have an acid base reaction like this an acid is always reacting with a base right so if this is a weak acid that means that water is acting as a weak base here okay now as we think about the products well the acid is a proton donor and the base is a proton acceptor right well once that water has accepted the proton or accepted the H+ it becomes h3o+ hydronium which makes sense because this is an acid problem right and likewise that acetic acid once it's donated that proton Don donated that H+ all that's left is the acetate ion okay so notice we have two conjugate acid base pairs the acetic acid and the acetate ion are a conjugate acidbase pair here and likewise the hydronium and the water make a conjugate acid base pair as well so you may be asked that in fact I'm pretty sure you will be asked that at some point on the exam to identify conjugate acid base pairs now let's go back to the problem here let's set up the ice box and so our concentration of the acid is 1.25 moles per liter so that goes here and then the hydronium and acetate ions are essentially zero we don't know the change so we know that this acetic acid has to go down by X and the products have to go up by the same amount which is X and so at equilibrium we have 1.25 - x x and x so we're going to plug these values into the equilibrium constant expression so we write that like this Ka equals hydronium ion concentration times acetate concentration all over the acetic acid notice that liquids like water are not in this so we just plug and chug the KA is 1.8 * 10-5 = x * X over 1.25 - x now a little mathematical trick that we do often in these problems is when we realize that this is a small equilibrium constant we're going to go ahead and ignore the minus X there to make our math a little bit easier and we can cross multiply and you find that x^2 equal 1.25 * 1.8 * 10-5 which is x^2 = 2.25 * 105 take the square root and X which is equal to the hydronium ion concentration is 4.74 * 10 -3rd moles per liter and to find the pH we just take the negative log of that number so the pH here is 2324 and so makes sense it's an acidic pH um it should have a pH much less than seven shouldn't it now let's do one little calculation here let's calculate the percent dissociation so the percent dissociation is where you take the hydronium ion concentration in the case of a weak base it would be the hydroxide concentration but it's hydronium here divided by the initial concentration of the of the substance which is 1.25 of course times 100 I ran out of room here so that's why I didn't put the times 100 but when you find the percent dissociation it's about 379 per dissociated and that makes sense because this is a weak acid you would expect it to have a fairly or very low percent dissociation now let's talk about acidbase titrations for a couple minutes we know that the goal of a titration is normally to determine how concentrated an acid or a base is so whenever we take a titration if we uh plot the volume of a substance that's added usually the base on the x- axis and we plot the pH on the Y AIS we can get this curve this this graph here this is called a titration curve and there's some interesting data that we can get out of this titration curve now let's take a look at the first question here let's say that if we have 10.00 milliliters of a weak acid that was used in this titration determine the acid's concentration well we're going to plug this into the titration equation marity of the acid time volume of the acid equals marity of the base time the volume of the base so marity of the acid is what we're trying to solve for so ma is the unknown the volume of the acid what says right here we use 10.00 milliliters so that goes in for VA now the marity of the base is uh given to us in the graph it says that the base is .10 molar so that goes in for M subb and then the volume of the base well notice that there's this inflection point right here right around the 25 Mill Point that tells me that was the point where we got to the equivalence point so that means that VB is 25 milliliters so I put that in right here I can solve for M Suba using simple algebra and I find that it was .25 moles per liter of the acid so using that titration equation in this curve we can get some pretty good information there now let's estimate the KA of the weak a used in this titration well the way we do that is we have to find the half equivalence point so notice that it took us 25 milliliters to get to the equivalence point well the halfway point would be 12.5 so I go to the 12.5 Mill Point which is right here looks like it's marked V and at that point pH equals PKA and so the pka let's see that pH is about uh somewhere around 5.4 if we just kind of eyeball that so that means that the KA is 10 to the 5.4 which is very close to 4.0 * 10 -6 and so that's how you can use a titration curve to estimate the KA or the pka also of the acid that you're using now let's go back to this titration curve and let's try to analyze the prim primary reaction component other than water of course that's present in the titration flask at each of the following points well at Point V that's the halfway point you need to realize that at that halfway point you have the same amount of acid as you have of the conjugate base now we don't know what the acid is so I'm just going to use the the uh generic ha here but whatever it is the ha and the conjugate B base of that acid a negative are going to have equal concentration so it's actually a tie at Point V now at point w we've added as you notice we've continued to add the base and so the base the conjugate base anyway is going to predominate at point w so as the pH is rising faster we have more of the conjugate base than we have of the weak acid so the a negative is our primary component at that point even though we do still have a fairly significant amount of the weak acid now once you get to point x that is the equivalence point and your acid is all gone at that point and so all you have left is just a negative so all you have uh other than water of course in the flask is the conjugate base at the equivalence point now once you go past that and you have essentially overshot your titration you have an excess of hydroxide as you keep just you know pouring in the hydroxide and that's why the pH is keep shooting up so fast after that equivalence point so that's why hydroxide is your primary component once you've overshot the titration at Point at both points y and z now let's go on to unit 9 and talk about thermodynamics and specifically the second law of Thermodynamics in the first part of this now when we talk about the concept of entropy we need to understand that there are a couple different ways of talking about this entropy if you look it up in the dictionary it says it's a measure of the total number of possible energy states in a material uh and that is a very good definition we sometimes say it's also how well something is dispersed that's that's also a very good definition I found that students sometimes find it easier to think of it in terms of disorder or chaos entropy is a measure of the disorder in a material so if you uh think about it in terms of that you're GNA it might be easier for us to kind of Reason about it in that uh in that way as well so if you have a solid not much disorder there everything is Crystal and it's it is in this nice orderly fashion so you have very little uh entropy in a solid so solid has the least entropy As you move up an entr in entropy and have a liquid you have molecules that are able to move around freely they're slipping and slid in around each other so liquids have more entropy uh if you have an aquous solution you have that liquid but now there's even more disorder because we have uh we have these ions or perhaps these other molecules that are swimming around in the liquid so that has even more entropy and once you get to a gas well every molecule is independent of each other so gases have the most entropy if you're comparing uh solid liquid a lious and gas and so that's the hierarchy of entropy if everything is at the same state then we can say that materials at a higher temperature are going to have more entropy than than substances that are at lower temperatures and that makes sense because your molecules are moving around faster at higher temperatures so that sounds like more disorder doesn't it uh if it's a tie and you're at the same temperature well a container that has more molecules is going to have more entropy than those with fewer molecules so if everything else is the same you have the same temperature the same state of matter and you have a a container with 10 molecules or 10 moles it's going to have more entropy than a container with three molecules or three moles and so more molecules implies more entropy so let's predict the sign for the change in entropy Delta s for each of these processes well in this first one we have a gas actually two gases that are changing into a solid so a gas down to a solid that's a drop in entropy so that's why it's a negative Delta H for that one what about this one here we have a solid that's changing to a gas well solids have a low entropy gases have a higher entropy so that's an increase in entropy increase in Delta H how about this one here we have two ions that are aquous they're swimming around in solution and they end up turning into a solid well aquous has a fairly High entropy and solids have the lowest don't they so that's a drop in entropy negative sign of Delta s how about this one well here we don't have any change in state do we they're they're all gases here so the tiebreaker is how many molecules do we have so notice that in this one we have U two molecules of gas on the left and 2 plus one so that's three molecules of gas on the right so if it goes from two to three that's an increase in entropy isn't it so that's why the sign for that reaction is a positive Delta s so you should be able to predict the signs of Delta s just by looking at a chemical reaction for pretty much every uh reaction that you could could encounter now will a reaction take place well when you answer that question you're saying if a re reaction is thermodynamically favorable now as we think about this there are basically two forces that are going to determine if a reaction is thermodynamically favored the first one is enthalpy now we talked about that a little bit earlier we said that enthalpy is Delta H that's the heat right and we find that exothermic reactions tend to be favored by the universe it's almost like the universe likes exothermic reactions so if something is exothermic that's something in its favor there's a good chance that at least at some Condition it's going to be thermodynamically favorable and the other process or or the other Force I should say is entropy now we find that processes that have an increase in entropy tend to be favored by the universe so if you look at a reaction and it's sign for Delta s is positive then there's a good chance it's going to be thermodynamically favored at least under some conditions if you ever find a reaction that is both exothermic and has increasing entropy it's going to be favored at all temperatures okay it's just at all temperatures always thermodynamically favorable however if it's the opposite of that if it's an endothermic reaction and it has a decreasing entropy then that reaction is never going to be thermodynamically favored under any condition so we have that now if only one of the factors is favorable the process will be thermodynamically favored only at certain temperatures so that means that if you have a reaction that is endothermic Delta H positive and it's increasing an entropy Delta s positive it's going to be favorable at high temperatures know they're both positive at very positive temperatures it'll be favorable if it's the other direction if they're both the negative you know it's exothermic and the entropy is decreasing it's going to be favorable at lower temperatures now as we talk about this business of thermodynamic favorability we know that that is Gibbs free energy Delta G is kind of a numerical way of calculating or determining or talking about thermodynamic favorability and there are a couple different ways to calculate thermodynamic favorability the two probably most common ones are shown here we can calculate it by saying Delta g equals the Delta h of a reaction minus its temperature in Kelvin times its change in entropy you can calculate Delta Delta G if you know the uh equilibrium constant you can figure it out as well Delta G equal R which is 8.314 Jew per mole per Kelvin times the temperature in Kelvin times the natural log of the equilibrium constant so you can find the Delta G in a couple different ways here and there's another one that we talk about in our daily lessons as well if you have the Delta G Gibbs free energy that is a positive number it's not thermodynamically favored if it's negative it is thermodynamically favored so that's just a little quick review of thermodynamic favorability and the second law of Thermodynamics as we wrap things up in unit 9 we want to talk about galvanic cells galvanic cells or sometimes called the voltaic cells are basically just batteries these are thermodynamically favored processes where we have an oxidation reduction reaction taking place and we're harnessing it to power some sort of a load like maybe a a light bulb or a cell phone or something like that and whenever we have this we have to have two different electrodes we have a wire here we have ions in solution that are corresponding to what's in the electrodes there and we have a salt bridge now every half rea ction as we look it up in our data tables or in our chemistry textbook or what's given to you on the AP exam it's going to be given to you as a reduction half reaction so as you think about the silver here we would write it as The Silver ions are being reduced into silver metal by that half reaction right there and its uh cell or half potential I should say is positive. 80 volts and we can say the same thing for aluminum ions aluminum + three being reduced down to aluminum by means of three electrons would be 1.66 volts now I guess the the main question here is which one is the cathode which one is the anode well the way we do this is we use this reaction or this equation right here e cell equals e cathode minus E of the anode and this reaction or this equation is not given to you on the AP exam you have to remember this one and E cell the overall cell potential or voltage has to be positive which means we have to arrange these two numbers in the equation in such a way that we get a positive number and the only way to do that is to arrange it like this with the +80 in the cathode position and the - 1.66 in the anode position so when you calculate this you find that the E cell is 2.4 4 6 volts so that's the the voltage that's going through this wire here as long as we're at standard conditions which is 1 Mo per liter and 25° C and all that now that also tells us that 08 which is the silver well that's the cathode and the cathode is where reduction is taking place so yes the silver is being reduced into silver metal here the silver ions I should say reduced into silver metal that means that the aluminum is where the oxidation is taking place that's called the anode which means whenever we write the overall balanced equation we have to flip this okay because the aluminum is actually turning into aluminum ions now the electron flow goes like this it always goes through the wire from anode to cathode okay that's how electrons always flow Through the Wire if we think about the salt bridge and I don't have much drawn in here but if you have inert ions in that salt bridge well your cations I are going to be attracted toward the cathode which I think kind of makes sense and the anode well your anion are going to be attracted toward the anode through the salt bridge I think that kind of makes sense as well okay now as we think about this let's write the overall balanced equation we know that the the cathode half reaction is written as it is but the anode you know think about like anti you that's the one that you flip so that's why it's written as Al yields al3+ plus three electrons and to balance this this reaction we have to multiply the first half reaction by three because you want those electrons to cancel out and now when you add this up we can clean it up and there is our balanced equation for this galvanic cell okay and generally speaking when you have metals like this the cathode is going to increase in Mass you know so the silver is going to get more massive the cat gets fat and the anode the aluminum here is going to decrease in Mass so there we have our entire course in a relatively quick cram session I hope you've been able to learn something from this uh video from this cram session if you have please hit that Thumbs Up Button if you uh stayed with me this long I really appreciate it and I know you're all going to do great on your AP Chemistry exam uh I have lots of review videos the ones that are uh you know have each unit summarized in 10 minutes I've got the longer videos for all 91 of the uh topics in AP Chemistry so don't forget about those if you need more review and if you haven't subscribed go ahead and subscribe so that after the AP exam you can come back and watch my uh problem walkthroughs for this year's AP questions I'm going to try to get those online as as soon as I can and I wish you all the best for your AP exam I hope you spread this video around to all of your friends and your colleagues and everyone in your AP Chemistry Community thanks for watching and I hope to see you very soon