 Yeah, I think everything is set now. I need to give you the link, right? Hello? Yeah, so you need to start the link. Yeah. I've shared the link as well, OK? Yeah. Yeah, so they can access. How many of you are here on the Google Hangout link? Kevin is there. Vihan is there. And Sundar, are you there? Kevin? Yes? I cannot hear you, Kevin. There's some problem with your mic, I guess. Yeah, no problem. OK, but I know that you are there because you tried to unmute. So I know that you are able to listen to me. OK, let's continue. Let's continue. We are losing time. Yeah, so the idea is just to have the proportions balanced, OK? Now, the next thing that we need to understand in chemical equilibrium is our law of mass action, OK? Now, what is law of mass action? Law of mass action is basically it is one thing that relates rate to the concentrations. The underlying principle of law of mass action is the collision theory, OK? So what does the collision theory say? The collision theory says, look, when any reaction is happening, it is basically an interaction between two molecules. And how do these molecules interact? They basically bump into each other, OK, with high velocities. Whenever the bumping velocity is greater than the threshold velocity, bonds break, OK? So there is a threshold energy, bonds break. As the bonds break, the molecules are searching for newer bonds. And as they search for newer bonds, new products are formed. And newer products also are formed because the new bonds that these molecules form are lesser in energy. That means they're more stable. And hence, they retain that structure and move on. Does that mean that once the bonds are broken, reactants are not formed again? Yes, they are absolutely formed. The problem with reactants being formed is that these reactants do not have a lesser energy or more stability. And therefore, they cannot stay for long. Eventually, they have to get converted to products in subsequent bumps. Or subsequent is the next collisions that will happen, subsequent bumps, right? OK, now, so what is the law of mass action? If we understand this, we know that the number of collisions is basically proportional to the number of particles. And it is also inversely proportional to the volume in which it is. This actually gives, this is the root where the collisions are proportional to the concentration of the substance. Now, there is one more catcher. Let's say we have a equation like H2 plus O2 giving twice H2O, OK? Now, in this reaction, if we balance this reaction out, we get 2 and 2, which means unless two molecules of H2O react with one molecule of oxygen, we don't end up getting even a single molecule of water, OK? But when the reaction is completed, we end up getting two molecules of H2O. Now, here there are two molecules of H2O required. And therefore, the concentration of H2O, we will have to take it twice because unless the collisions are happening with both the molecules, the rate will not be proportional. And now, therefore, since the rate in turn is proportional to collisions, we actually end up getting what we call as law of mass action, which is R is proportional to the concentration of the reactions. Race to the power in. Is that Vihan calling? Hello? Vihan, are you there? Sir, you left a handoff call and on YouTube, I don't think the screen is not being broadcasted. Oh, is it? OK. Yeah. I think now the screen will be visible on the YouTube. Sir, after you spoke about collision theory, you just stopped hearing you. OK. So I'm going to revise that one more time. So we were saying that in law. Let's see. Yeah, the screen is shared now. You can see on YouTube, I guess. Sir, I can't see on Hangouts because the YouTube is large. OK, so I'm sharing on Hangout as well. OK. You can see on YouTube, I guess. Sir, I can't see on Hangout because the YouTube is large. Yeah, I'm sharing on Hangout as well. Don't worry. Yes, I'm sharing on Hangout now. You will see on Hangout. Because the YouTube is large, it has a large screen. Yeah, I am. Oh, OK. Yes, Vyan, is everything visible now? OK, OK, cool. So now let's begin. So what I was saying is that underlying principle of loss marks action is collision theory, where they say that basically reactions happen because two molecules collide with each other. And whenever they collide, the energy, it actually goes from beyond certain energy which we call as the threshold energy. Only then the reaction is possible. That threshold energy means that the kinetic energies of both the colliding molecules should be enough to break the bonds. Now, once the collision happens, obviously the number of collisions will be larger if you have more particles. And secondly, the collisions will also be larger if these particles are in lesser volume. So collisions are directly proportional to number of particles and inversely proportional to volume. This is where actually we come to the point where collisions are proportional to the concentration. That's a very important point. Now, since collisions are proportional to concentration, collisions are nothing but how much reaction is happening. And therefore the rate is proportional to collisions is the underlying principle of loss of mass action from where we get R proportional to concentration. But there is one more additional point that we need to remember here. Sometimes we require more than one particle colliding with each other at the same point of time. For example, we require two of hydrogens to collide with each other. Now, if the concentration of hydrogens is doubled, obviously the collisions will also get doubled. And therefore, the rate of reaction is proportional to the concentration raised to the power coefficient of that particular element or that particular compound in the reaction. This brings us to a very common reaction. For example, if we have AA plus BB giving CC plus DD, this is a general representation of any reaction, then we know that the rate of forward reaction is going to be K forward into A to the power A and B to the power B. And the rate of backward reaction will be equal to KB into C to the power C and D to the power D. So these are the two different rates that are possible. One is this is the forward reaction and this is the backward reaction derived from law of mass action, nothing different. Now, when equilibrium is achieved, equilibrium is said to be achieved when the concentrations no more change. The concentrations do not change. Now, please note, in equilibrium, new substances are always formed. In equilibrium, the same substances do not remain as reactants or products. In fact, the substances that were or the molecules that were reactants produce products. They get converted into products. But at the same time, whatever was products also get converted to reactants. And therefore, these two conditions all the time are happening. Having said this, the concentration as a whole remains constant. This is something that we have to understand very well. This is a dynamic equilibrium where up this circle, this cycle of product to reactants is constantly happening. The only thing is at a macro scale, the concentration remains same. So the collisions are happening, reactions are happening, new products are formed, new reactants are getting consumed, new reactants are also formed and new products are also getting consumed. All of this thing is happening all the time. Only that the concentration remains fixed. So this one condition of concentration becoming fixed at a certain temperature is what we call equilibrium. Now, once equilibrium is achieved, the rate of forward reaction becomes equal to the rate of backward reaction because only when the rate becomes equal will the concentrations get frozen. So the capacity of A and B getting formed is equal to the capacity of A and B getting consumed. If both of them match, then the overall concentration of A and B will never change. And that's the point where equilibrium gets achieved. So when RF is equal to RB, that is where we call the equilibrium achieved Now, let's see what happens when we do this. When we do this, we basically end up getting KF into A to the power A and B to the power B is equal to KB into C to the power C and D to the power D. Now, this actually gives us KF by KB is equal to C to the power C into D to the power D divided by A to the power A and B to the power B. Now, we know from these two reactions is that KF and KB are constants. Constants only dependent on temperature, nature of the products, nature of the reactants and if it is a gaseous substance, also the pressure and other conditions that we'll be keeping, the larger conditions, number of moles, we take all of that. Now, having said these, if these two are constant, obviously the ratio is also going to be a constant and this is where the genesis of equilibrium constant comes up. So equilibrium constant is not something that we define. Equilibrium constant is something that we find to be happening in the reaction because of the collusion theory and at equilibrium we get an entity which looks like a constant. This is what we end up calling KC. So KC is called as equilibrium constant at in the moles per liter format and it is written as concentrations of C to the power C, D to the power DA, to the power N, B to the power B. Now, sometimes you might not have to use this part. Most of the times, we have solved problems where we are using the concentrations of the reactants and products but there are problems where you will simply be given the equilibrium constant of forward reaction, equilibrium constant of backward reaction and you will need to find the equilibrium constant of the reaction and you can do that. Now, a good thing to see is that let's imagine that all of these substances were gaseous. If these substances are gaseous, only when all of them are gaseous, please note this. When all of them are gaseous, we can write PV equal to NRT for any individual gas and for all the collection of the gas also. For example, PA equal to NART. Please note that the volumes of ABC and D, all the gases, now I'm going to write here for gases so that it remains with you. Please note that the volume for ABC and D, all of these will remain the same. The temperature also going to remain the same. So there is nothing called as VA, VB, BC or VD but there is something called as PAPB, PC and PD. Now, if we know this, we would understand that PA is nothing but NA by V into RT and NA by V is actually our concentration term because we define concentration as moles per liter and therefore, instead of C, I can simply write it as RT divided by PC, comma RT divided by PD, so on and so forth where PC, PD are partial pressures. What do you mean by partial pressure? If the total number of moles in this product is eight and there are two, two, two, two moles of each of them, then if the total pressure of this container is eight atmosphere, each individual gas will have two atmosphere pressure. So this PA, PD, everything will be two to two, okay? So it is the partial pressure. Remember, pressure gets divided into number of gases. So if the total pressure is 10, each gas can have its individual pressure as one atmosphere, two, three, four, depending on what number of moles it has. So pressure gets divided. Volume never gets divided into gases. Temperature never gets divided into gases. A lot of people make mistakes saying that if there are 10 moles of three different gases, then each gas occupies 3.3 liters of the 10 liters. No, wrong. Each gas will occupy 10 liters, okay? Fundamental stuff, but people forget to realize this. Okay, once we understand this, we can replace all of these concentrations, which we had by PA by RT, in fact, so the concentration is going to be PA by RT the other way around, okay? Not RT, so I can write the concentration as PA by RT, PB by RT, so on and so forth. Now you will realize if I do that, my KC actually ends up becoming PC to the power C, PD to the power D divided by PA to the power A and PB to the power B, okay? And I will get RT to the power, here it will become A plus B minus C plus D, okay? A plus B minus C plus D, where what are ABCDs? Here ABCDs are basically number of moles. I should write NA, ND, NC and ND, but just for, just to have, or I will write this as delta N, where delta N is nothing, but sum of, sum of products, this is minus delta N actually, okay? Sum of products, minus sum of moles, okay? Sum of moles of products, minus sum of moles of reactants, okay? Now, if that's the scenario, I can simply write KC is equal to KP. This entire thing can now be defined as KP into RT to the power minus delta N. Please note that we are done here, A plus B minus C plus D. This is the product side and this is the reactant side. And delta N we are defining as products minus reactants. So this becomes minus delta N. Or in other cases, we can simply write KP is equal to KC into power, RT to the power delta N. So I took- So why do we need delta again? Delta is the difference in total number of moles that are produced minus total number of reactants. Now, please note, these are not the actual difference of moles when the reaction is happening. This is the difference of moles when equilibrium is achieved, or simply to say difference of moles as coefficients in a standard equation, in a reversible standard equation. Getting me? Thank you. Sir? Sorry? Even- So only T raised to the power delta N, right? Not R or- No, no. See, if you look at this reaction, this equation, P, see concentration is entirely raised to the power C, right? So P by RT, this entire thing is raised to the power C, correct? Because NA by V is concentration. So here, your entire RT is raised to the power delta N. Okay, so I should put a bracket here. Good point that you- Yeah, so I should put a bracket so that it is more clear. Got it? Yeah. So both R and T are raised to the power delta N. Okay. Now, so this is the relationship between KP and KC, where KP is this portion, I'm gonna take a different color maybe. KP is defined as the ratio of just pressures raised to their coefficients, and KC is defined as the ratio of concentrations raised to their coefficients. Now, I'm going to give you one very important example here. If KC is going to have some, so firstly, understand that if we have a heterogeneous system, okay, which is the one that creates trouble. Now, in this scenario, we all considered that everything was gaseous. Let me actually write an example for you. Let's take a heterogeneous system where we have AA plus BB giving CC plus DD. Okay. Now, in this scenario, we are going to take D as gaseous. We are going to make CS solid. B, we will make as liquid, and A, we will make as gaseous. Okay. Now, in this scenario, first thing is I will define my KC as only D to the power D divided by A to the power A. And I gave you the reason a few minutes ago. Why? Because the change in concentrations of D and A are going to be so significant that changes in concentrations of B and C are going to be minuscule with the comparison to DEM. And therefore, I will get a workable number to do the calculations, although this still is an approximation. So this is one kind of problem. But we have also encountered problems where you actually will get a definition of KC, which is equal to D to the power D and C to the power C. I'm taking C also just to keep maximum variation. Here you will get A to the power A and B to the power B. It's not that you cannot define KC like this. You can define. Your numbers are going to be very fluctuating. But if you have defined KC like this, you will not be able to relate it to KP at all. Please note that there will be no relation with KP. You cannot connect this KC to any KP. But this KC can definitely be connected to KP and it will have delta N that you take as equal to only ND minus NA. Please note, it is not ND plus NC. It is ND minus NA. So you are only considering. I'm going to make a huge big note here with the arrow. Only consider gaseous products. Only consider a gaseous product. Where are we considering gaseous products? Kevin, you're there with me. Kevin, Sundar, Whom sir are there on the hangout? I'm there also. Yeah. Okay. Good. Prithvi and Akshath who are joining on the YouTube. If I would recommend that you actually also join us on this thing so that we can talk. Else it will just be a passive. If it's not possible, that's okay. But if possible, you can do that. Now, Kevin and Sundar Vihan who are there. Please note that this delta N is the delta N we are talking here. And we are talking this as the difference of the gaseous products only. Here we are taking difference of all of them because all of them were gaseous. And the second condition we spoke about was that this delta N, everything is possible only when you define KC as only the ratio of the gaseous product divided by the gaseous product. That is exactly, we should always use this. If we are given a choice, we should use this as the definition and do all the KPKC calculations. So that is, okay. So this is where we finish understanding KPKCs. Now, what are the things that equilibrium constant depends on, let's say KC, what does it depend on? Firstly, definitely it depends on the temperature, which is the very evident thing. A lot of students actually confuse whether it is dependent on catalyst or not. Please note that the equilibrium constant is independent of catalyst. Why? Because catalyst as much as it increases the forward reaction, it also increases the backward reaction. But please note, KF and KB are dependent on catalyst. Why? Because the presence of catalyst actually increases the forward reaction and also increases the backward reaction. So independently, both RF and RB are affected, but the ratio of RF and RB remains unaffected, okay? So this is a very important aspect. Also, the presence of catalyst actually will help us achieve the equilibrium very fast. Please note, equilibrium constant does not tell you anything about the amount of time that is necessary to achieve equilibrium, then how slow or fast the reaction is. It simply says that this is the concentration that we expect when your equilibrium will be achieved. Now that equilibrium might be achieved after two years, God knows, right? But KC does not tell us anything about it. The only difference that happens with catalyst is the moment you add catalyst, your KC is achieved faster. As in the equilibrium concentrations are achieved faster, not KC, that's the wrong word. The concentrations that are necessary for equilibrium are achieved much faster. So that is one very important thing. The third factor that actually affects KC is this stoichiometric coefficient or just to say the reaction itself. For example, if I am talking about H2 plus O2 giving H2O, I can write this reaction in two formats. I can put two and two here, or I can write H2 plus half O2 giving H2O, okay? Now in this scenario, my KC is defined as KC is equal to H2O. I'm assuming that all of them are in gaseous form, guys. So please note that every time I'm going to speak further, by default it will be gaseous form, unless and until I say so, otherwise. It's going to be H2O square divided by O2 divided by H2 square. And the KC, let me call this KC1, the KC2 for this reaction is going to be H2O divided by O2 to the power half and H2. So you will quickly realize that KC1 is nothing but KC2 square, okay? The thing that I did here, the division that I did here, so I divided this by two, okay? This thing has turned as a power here. So remember stoichiometric coefficients completely affect the equilibrium constant, right? So that's the third factor. The fourth factor that affects the equilibrium constant is actually your concentration units, okay? So what units are we measuring the concentration? So if you're measuring the concentration in moles per liter, obviously I'm going to get a different KC. I can also use that in normality, okay? Moles per liter is nothing but molarity. I can also use that as equivalence per liter, which is basically nothing but normality. I can also use molality. So the moment you change the concentration units, your KC value is going to change and the units of the K are also going to change. There are no defined units of K fixed. Every reaction has a different unit of K because it depends on what are the gaseous or other substances that you're defining your K with. So that's a very important fundamental thing, okay? So that's the last thing that K depends on. Now that closes equilibrium discussions. Now there is one trick I would say that we use or in fact property that is very important or good to understand in equilibrium constant, which is what we call as the reaction quotient, okay? So what is the reaction quotient? Reaction quotient is nothing but the definitely, it is as Q. It is the ratio of concentration of products for any reaction upon ratio of concentration of reactant in the same reaction at any point in time, okay? So only difference that Q we bring in is, we bring in the time factor. So we are saying, look equilibrium constant, I know, it tells me that these are the concentrations at equilibrium, but let's define a Q which will give me the same ratio at any point of time or stage of the reaction. Now, obviously the question comes, why are we defining another Q? Let's look at it, why? Let's say the value of Q actually comes out to be equal to the value of K. This means that the concentrations of individual reactants and products are actually equal to the concentrations that I expect at equilibrium, which means that the reaction has achieved equilibrium. So the time that we are talking right now is a time of equilibrium already happened. Let's take another example. Let's say Q is less than in value than K. Then this ratio means that the concentrations of the products are lesser than what concentrations I am expecting at equilibrium, which means that more of the products must be formed in time to come. Now this means that our time that we are talking about is where the reaction is proceeding towards equilibrium in the forward direction. Why forward direction? Because more products are forming and less of reactants should happen so that the ratio of Q will increase and become equal to K. Let's take the third and the final case where Q is greater than K. When Q is greater than K, the products and the numerator are in very large in concentration than what I expect at equilibrium. This means that the reaction must be going in the reverse direction more to give more of reactants than products and therefore the backward reaction must be much more strong, right? So the Q suggests that backward reaction is happening much faster and here the forward reaction is happening much faster. So Q is actually a tool which will give us an understanding as to what point in time is the reaction heading to, right? Yeah, tell me. Can you tell the difference between Q and K again? So K is actually equilibrium constant where K simply tells us whenever equilibrium is achieved, these are the concentrations of reactants and products. It does not say when would the reaction we achieve equilibrium is, does not say how is it going to achieve the equilibrium? Nothing, it does not talk about anything. It simply says this is the ratio that you should have, period. Now Q is a ratio that I'm taking at any point of time. I don't know the reaction when you do start. Let's say Sundar started the reaction yesterday night and I want to check whether equilibrium is achieved or not. What I'm simply taking is taking a sample of the reaction mixture and I'm measuring how much is the concentration of ABCD. Once I measure, I take this ratio and I call that ratio as Q. If that ratio is equal to K, I know that equilibrium is achieved. If that ratio is greater than K, then I know that Sundar came and actually added some products in the, in midnight so that the reaction could achieve equilibrium but then he lost track of time saying, because I know that because of addition of products, what happened is that the reaction started going back. So therefore, my Q will be greater than K. Are you getting my point? Yeah, so Q is just like at any instant, at a snap of time, what is the reaction mixtures ratio? Simple, right? So that's your Q, right? Now, okay. Now let's talk about some estimates of what are the values of K? How large K can be, okay? So some estimates of K. Now let's say estimates of K, okay? Now let's say your K is a very large value than one. For example, the value is about 10 to the power seven to 10 to the power 15 odd, okay? Then we can definitely say that this reaction is almost proceeding to completion, which means that if you actually take the reaction mixture at equilibrium, I'm not talking at any instant now, I'm talking only about equilibrium. If you take the reaction mixture at any point after equilibrium is reached, you'll find that the reactants are barely present, their concentrations are minuscule. Now, if the K value is greater than one, but not so high, not too large, say up to about 10 to the power three, or even up to 10 to the power seven, or maybe in some unit digits or so, then we know that the forward reaction is more favorable, which means that the concentration of products is going to be more than the concentration of reactants. Having said that, the difference is not very high. So all we are saying is that products are more than reactants, but difference is not very high. Here we are saying only products are formed. There is no reactants formed, where very minuscule amount of reactants will be remaining. If K is a very rare case, the concentration of products, as in the product of the concentration of products, that's actually how we should all the time say, but just to keep it simple, products will mostly be approximately equal to the reactants. And obviously, as we go below K less than one, the reaction and the backward direction will be much more favored. So these are a couple of things that we can speak about, the values of K. Now, the last important topic of this chapter is actually what we call as Leche-Atelier's principle. Now, what is this principle? A lot of you generally tend to forget this, but it's a very simple principle. What this principle says is, look, if you are at equilibrium, this principle is applicable only at equilibrium. If you do in between, then its effect is not very visible, but you can actually predict, but we'll talk about that in some minutes. Here we are saying, look, at equilibrium, when everything is very happy and going very well, if you come and do a disturbance, if you come and do a disturbance, the reaction is going to consume this disturbance. It's just going to eat it up. So these are in very colloquial or simple terms. Technically, what he says is that, if you're going to bring a change in the equilibrium, then the reaction will proceed in such manner that the change is nullified. Now, for example, what type of changes can you do? I can add products. I can simply add products to the reaction. Second is, I can increase its temperature. Third is, I can change its pressure. Fourth is, I can change its volume. So all of that you can do. Now, let's talk one by one. If you change the products, which means that you are changing the concentration of the products, then the reaction will proceed to nullify that. So if you add products, the reaction will go in the direction to consume the products. If you remove products, the reaction will go in the direction of making more products. And therefore, when ammonia is formed in Haber's process, when nitrogen and hydrogen are mixed to give ammonia, the more amount of ammonia you take out, the more amount of ammonia gets created. So as soon as ammonia is formed, you take out that ammonia either through HCl vapors and then reform ammonia of sorts, or you simply try and condense ammonia and take it out as liquid. Because nitrogen and hydrogen are very difficult to condense, ammonia condenses faster. It dissolves in water also faster. Nitrogen and hydrogen do not dissolve in water faster. So as soon as ammonia is being taken out by dissolving it in water or condensing it or through HCl or whatever mechanism, you'll find that more amount of ammonia will get started creating. So the equilibrium is never left to get achieved and settled. You know, all the time you keep on disturbing the equilibrium to get more and more product. So that's with product, not temperature. If you have exothermic process, then increasing the temperature will make the reaction go in opposite direction. Why? Because it was the habit of the reaction to increase the temperature by giving out heat. But you yourself are increasing the temperature. So now the reaction is compelled to consume the external heat and to decrease the temperature. You're getting my point. That's a very important thing. Now, so how would the reaction, exothermic reaction behave when we increase the temperature? The exothermic reaction will go towards the reactants more than the products. Let me write this down for you because it might be slightly confusing. This is reactants and they are in equilibrium with products. Okay, and heat is given out. So I'm going to write delta at the top and given out just as a symbol so that it is explicitly clear. So if I increase the temperature opposite, which is where heat will be taken in now. Why? Because if I go from products to reactants, it becomes an endothermic reaction. So heat is taken in. So if I increase the temperature, the products will go more towards reactants and backward reaction will be favored. So this is the temperature factor. Now volume, it depends on the units of K. For example, let's say the my unit of K is equal to moles per liter. Which part? So if you increase the temperature, if the forward reaction is exothermic, then increasing the temperature means that the reaction now wants to consume the temperature back, consume the heat that is there in the surrounding. So the backward reaction will be favored because the backward reaction is going to be endothermic. So increasing the temperature does not help exothermic reactions. Okay, it is a lower temperature that helps exothermic reactions a lot. Okay, yeah. Now let's talk about volume. Let's say the unit of K is moles per liter. This is K1 and let's say in another case, the unit of K2 is moles per liter, liter to the power minus one. Okay, the unit of K can be anything. So now in this scenario, if I increase the volume, we know that our K value is going to go down. Okay, so our K value is going to go down means that the amount of products are going to be lesser and the amount of reactants are going to be higher. So increasing the volume for this kind of a unit of K is going to favor the backward reaction. But increasing the volume here is going to increase the K value, which means that the concentration of products is going to go higher and the concentration of reactants is going to go lesser. So all of these factors can be figured out from the units of K as well, when it comes to volume. Pressure, you can go to the units of Kp. Instead of Kc, you try and compare the units of Kp and it will be very visible what will be the effect on that. Okay, yeah. So give me just a quick minute. I'm just going to put on light here and give me a minute. Yeah. So now, so this is late at least principle. Now the last thing that we need to understand is how is Gibbs free energy related? And I had promised you that, you know, we'll be doing all Gibbs free energy thing. So now that we are doing chemical equilibrium, we'll also understand how Gibbs free energy is related to the equilibrium constant. Now for any standard reaction, generally Gibbs free energy is related to its Kp value, okay? And the expression is given as delta G, which is the standard free energy is equal to delta G naught, which is in the STP form in the standard form, right? Plus RT into ln Kp, okay? Now, please note that whenever equilibrium is there, delta G is equal to zero, okay? When there is no equilibrium, delta G is equal to delta G naught plus RT ln Kp. Now, when it is Kp, we get, by this rule, we get delta G naught is equal to minus RT ln Kp, okay? So these are the two expressions that you should remember. This is when delta G becomes zero, because at equilibrium, delta G is equal to zero at equilibrium, okay? And since delta G is zero, so what are the factors here? Delta G naught is at the standard forms, okay? The standard change in free energy, and delta G is the net change in free energy at any position, right? So net change in free energy is equal to the standard plus RT ln Kp. Now, at equilibrium, delta G itself is zero. So delta G naught ends up being minus RT ln Kp, right? So delta G is equal to delta G naught RT ln Kp. This is for Qp, when Qpy, when it is not at equilibrium, when not at equilibrium, okay? So this is a quick relationship between reaction quotient and Gibbs free energy. There is one more parameter, I think, is also what we can look at, which is called as the Vanthoff factor. Now, let's understand what is this Vanthoff factor? So let's say you, Vanthoff factor, okay? So let's say we have a general chemical reaction, which is AG giving BG. Remember, both of them are gaseous still. And let's say it gives Y. So it is basically dissociating. So a gas is dissociating into its components. Then at equilibrium, per mole, per mole of gas, if X have dissociated, I'm going to get YX moles of B gas, okay? Where X actually is the degree of dissociation. Degree of dissociation means how many moles were dissociated per unit mole. So if I had one mole taken at initial, X moles dissociated and YX will be formed. Because for one mole, Y of B are formed. For X, YX will be formed. Where X is degree of dissociation. Degree of dissociation means for one mole of the substance, how many moles are getting dissociated. Now, if we see this, we understand that the total number of moles at equilibrium, the total moles at equilibrium. We have done this in solutions, I guess. I don't know if you guys remember. It turns out to be is one minus X plus YX. Or in other words, it is simply one plus X into Y minus one, okay? Now, if the initial volume... Now, what is the density after dissociation? If we take that, then we realize that the density before dissociation, it was D. So this is before. And let's say density after dissociation is small D after dissociation. Then I'm going to take to the next page, you know? So D is going to be equal to M by V, the density before dissociation. The mass remains the same, mass is never going to change. The volume itself also is not going to change. But what happens to the density after dissociation is that, which is small D, the density after dissociation is, mass will remain the same, but the volume of gas has now increased. For one mole, it was V volume. Now, the total number of moles has increased. So the volume has increased by so much factor. So these are the total number of moles after the gas has dissociated. So the volume has increased by so much of times into V. Now, if we take this density ratio, if we take the density ratio, which is nothing but our event of factor, if you remember density ratio or I, what we used to call in that is nothing but D by D. Then we realize that this D by D turns out to be one plus X into Y minus one. So this at the denominator will go to the numerator and it will come out to be this. Or in other words, the degree of dissociation in terms of the densities turns out to be, now we are just rearranging this. So we'll get X is equal to D into Y minus one. Please note that the degree of dissociation is actually dependent on the number of component that the gas dissociates into. So if it dissociates in three or four components, that is an integral part of what the degree of dissociation is. So this is the last factor, just a small addition to our understanding of equilibrium. This is where we actually end up, end chemical equilibrium. I just want to pause here for a minute and connect to all those who are at least there on the hangout. Yeah. Are you connecting guys? Vihan, Sundar, Kevin, your voice is still not audible. But I think you're getting it, yeah? Yeah, let's do a quick few problems. I'm going to give you a problem and I'm going to pause for a few seconds. So let's say, let me give you a problem. Okay, now let's say we have a closed system where A solid is giving two B gaseous plus three C gaseous, okay? This is a closed system, please note A is solid. Now the partial pressure of C is PC. So PC is the partial pressure of C. We are going to double this. So partial pressure of C, C final is equal to twice of partial pressure of C initial. So we've just doubled the pressure. Then the partial pressure of B will be. So the question is, what will happen to the partial pressure of B? Yeah, Sundar, Kevin and Vihan. It's on to you guys too. Even the guys who are there on YouTube, you can actually attempt, you can just write the answers. Yeah, you can just write the answers in your comments. Yes, so what will happen if partial pressure of C is doubled? What will happen? Yes, guys. It will also be doubled. It's gonna be halved or doubled. It's gonna be halved or doubled. It's gonna be halved or doubled. It will be halved or doubled, okay? A very important question that you guys need to ask me is, sir, if we are doubling the partial pressure of C, are we talking about partial pressure of B at equilibrium or are we talking about partial pressure of B immediately after the instant that I've doubled the partial pressure of C? Okay, Sundar, what are you considering it at? Then I thought it's final. Okay, and how are you arriving at this inference, Vihan and Sundar? Kevin, you also trying? Yes, guys. Okay. So let's write... The V value will remain the same, right? Yes, so let's write Kp. Kp is equal to Pc cube into Pb square, okay? Okay, I didn't consider the partial pressure. Then that's going to be an important thing. Yeah, yeah. Okay, so this is Pci and Pci. Are you guys trying the ones who are on YouTube? Are you guys also trying? Yes, Pithvi and Akshath on YouTube also you can attempt. Write the answer there as a comment or something. One by three times, okay? Sundar, what's your answer? Do you have an answer, Sundar? Okay, now let's see. C, Kpf is now equal to, it will remain the same as Pcf cube, okay? Into Pcb cube, Pcb square, right? Now, Kp and Kpf are going to be the same. Therefore, Pci cube into Pb i square is going to be Pcf cube into Pcb square. But we know that Pcf is equal to twice of Pci. So this is going to be two to the power of three, that is eight, okay? Eight times Pci cube into Pcb, sorry, Pcf, this should be Pc. What am I saying? Yeah, this should be Pb. Yes, sir, that's how many we take as P, one by eight, sorry. Yeah, let's see. So, okay, Pbf square, right? Now, Pci cube and Pci cube will get canceled out. So Pb i square is nothing but eight times Pbf square. Therefore, Pbf is nothing but one by two root two times Pb i, which means that the pressure, the partial pressure of B by one by two root two times of Pb i. Okay, it will have to decrease by one by two root two times of Pb i once you have doubled. Do you get the point? Do you get this? So, work out on equations directly, you know, instead of really, you know, jumping. So sometimes, you know, you might just get baffled with what is happening around, okay? Let's do one more expression, one more question, okay? Okay, let's do this. There are two equations, Pcl5 gaseous is in equilibrium with Pcl3 gaseous plus Cl2 gaseous and CoCl2 gaseous is in equilibrium with Co gaseous plus Cl2 gaseous, okay? Now, they both are simultaneously in the equilibrium in a vessel at constant volume, okay? So the volume is constant and they both are at equilibrium. That's the given information. Now, some amount of Co is introduced, okay? So we introduce some amount of Co in the vessel. Then the new equilibrium, when this is introduced, then the question asked is, what will happen with the new equilibrium? New equilibrium, what is the condition, okay? Now, there are four options, multiple options. I think the same thing there. Sorry? Both Pcl5 and CoCl, you are in the same container. Yes, they are in the same container and both of them are at equilibrium, okay? They are in the same container and both of them are in equilibrium. Now, the question is, when they were at equilibrium, Co2 was introduced, okay? Co2 was introduced and we are asked what will happen to the new equilibrium and there are four options. First, in the new equilibrium, Co2 is greater, is greater in amount. The second option is, Pcl5 is less in amount, Pcl5 is less. The third equation is, Pcl3 will remain unchanged. Pcl3 remain, okay? And the fourth equation is, D is, Pcl5 will be greater, Pcl5 is greater. Yeah. Can you guys attempt and see what do you think is? You tell me, how many of them are true? Okay, Pcl5 is less, okay. What do you, what does Sundar, Kevin, you think? Prithvi and Akshar, try an attempt and the YouTube as well. Pcl3 remains unchanged. Okay, so Sundar says Pcl3 remains unchanged. What, and you say Pcl5 is greater? Pcl5 is less. Is less, is your answer? Okay, any other attempts on the YouTube? Guys, do you want to attempt on the YouTube? Then a little bit of a lap on the YouTube. Yeah, I know, I know. Yeah, I know, but I've already said that a couple of times earlier. So therefore, I hope they must have heard it. Okay, cool. So, I'm attempting this because we have two answers. Kevin, do you want to say something? Your voice is also not audible at all. Kevin, I'm not able to hear you. Okay, you can type if required. Kevin, if you feel that you need to say something, you can type sometimes, okay? Okay, so I'll attempt this for you. Now see, the moment CO is introduced, I'm going to use blue ink to do all new calculations. As soon as CO is introduced, this reaction is going to go backwards. It's going to go backwards. The moment it goes backwards, what happens is that Cl2 amount here will get decreased. Now please note, whether it is the Cl2 here or it is the Cl2 there, this reaction for the equilibrium of this reaction, the concentration of Cl2 is one and the same. For example, I'm going to write these equations. The K1 for PCL5 reactions will be concentration of Cl2 into concentration of PCL3 divided by concentration of PCL5. In this expression, this Cl2 is not something that is only coming from PCL5. Whether it comes from COCl2 or PCL5, it does not matter. In K2 reaction, the equilibrium is going to be Cl2 again of the container into CO divided by concentration of COCl2. Again, which Cl2 it is, it does not matter. The moment I am adding more CO, this reaction is going to go backwards, which means that the concentration of Cl2 in the entire container is going to get decreased. The moment the concentration of Cl2 is getting decreased, PCL5 would want to add to Cl2. This reaction will go forward. Why? Because for this reaction, the equilibrium is disturbed by decreasing Cl2. So therefore, this is going to go forward. Therefore, PCL5 in equilibrium is going to be lesser is the correct answer. The second point also will be Cl2 will not be greater. We cannot comment on how much Cl2 will happen. Why? Because we are not sure whether the Cl2 that is getting consumed here and here, how they both, their interdependence is not very clear. PCL5 will definitely be lesser. Why? Because Cl2 here as soon as it goes backward to compensate Cl2, there is only one way that it can happen is where PCL5 gets forward. So this is a definite step that will happen. PCL5 is greater, this is definitely not going to happen. PCL3 remains unchanged. If PCL5 is going forward, PCL3 will also increase. So this, we can very clearly say that PCL3 will be larger in quantity. So this statement is definitely wrong. Cl2 is greater, this we cannot determine. This we are ambiguous about, we are not sure. But PCL5 being larger and PCL3 being larger, these two things we definitely can conquer. Okay, so do you understand? Is everyone clear? How to think around such problems? Yeah. Okay, good. Okay, now one more question. There are four reactions that are given. H2 gaseous plus Cl2 gaseous giving twice of HCl gaseous. Okay, this is one. Then twice of CO gaseous plus O2 gaseous is being given to twice of CO2 gaseous. Then N2 gaseous plus thrice of H2 gaseous. And we can have twice of NH3 again in gaseous format. And the last is PCL5 gaseous plus, sorry, it gives PCL3, PCL3 gaseous plus Cl2 gaseous. Okay, now the question is this. In these four systems of which are at equilibrium, at constant temperature, if I double the volume, then I am going to get more products or the equilibrium will shift to right. Okay, the question is temperature is constant, equilibrium is achieved. Now I am going to simply double the volume, volume is doubled. Okay, the question that is asked is, in which of these four reactions will the equilibrium shift towards the products? Will the reaction shift towards the products? Yes. Sundar, Rihanna. Sundar, Rihanna. Yeah, try. Yeah. ATL when you increase the volume, ATL will shift backwards. ATL. So, we don't have to go on. Go on. So, I think ATL, I think NH3 will shift backwards and the ATL will be increased in the system. So, the ATL will increase. Okay. What do you, what do other guys think? I am going to give you a hint here. Try and write the equilibrium expressions. From there, you will get some idea. No, the volume is doubled of the container. You're kidding me? There is no product and reactants, right? Volume is just doubled of the container. Everything is there in the same container. You cannot just eliminate product volumes and reactant volumes. Differentiate that. Is PCL5 right? Yes, PCL5. You're telling me the answer or you're asking me? Answer, answer. Okay. I'll say that in a minute. I'm just waiting for the others to complete. Why? See, it was an easy one. Okay. So, it requires greater volume. Well, I'll tell you why it, how to look at this. Always calculate delta N for all of them. The delta N for this is equal to zero. Delta N for this is equal to two minus three is minus one. The delta N for this is two minus four, which is minus two. And delta N for this is equal to two minus one, that is one. I hope all of you understand how do I calculate delta N? It is simply the number of moles of products, minus number of total number of moles of reactants in gaseous form. Now, please note that your Kp is equal to Kc into RT to the power delta N. Okay. Now, okay. Instead of that, actually we can do in a more simpler way is that we can simply write. If delta N is that, you know that the K is always equal to mole per liter to the power delta N. Okay. Now, why is that? Think about it. When you have products, let's say C to the power C and D to the power D divided by A to the power A and B to the power B, then when you're writing the units, it is simply moles per liter divided to the power C, moles per liter to the power D, moles per liter to the power A, moles per liter to the power B, which is nothing but the units of K will turn out to be moles per liter to the power delta N because it will be C plus D minus A plus V. Now, if your delta N is positive, then your K is inversely proportional to volume. And if your delta N is negative, then your K will be directly proportional to volume because this liter will come on the top. Okay. So in simple words, what I'm saying is, if you're doubling the volume and you want your equilibrium to shift towards right, your K should become smaller because your Q, when you double the volume, you basically at that instant, you are going to get a Q which should be less than K. When Q is less than K, the reaction will go forward and that is worse, right? When will Q be less than K? When you're doubling the volume, your Q should be inversely proportional to volume. So your Q will become K by two or earlier stuff by two. Only when your K is inversely proportional to volume, the moment you double the volume, it will no more remain K, it will become Q and Q will be half of the earlier K. It's value will be half of K. When it says half of K, it will want to achieve K and will go forward. Yeah. Is everybody getting this? I don't know. Maybe I'm just using too many words. Are you getting this? Sir, I didn't answer the Q part. Okay, listen. The moment I double the volume, the reaction is not at equilibrium. So I cannot say K anymore. I will have to say Q. But having said Q, the expression of moles per liter to the power delta N, this remains the same. Now, only if K is inversely proportional to V, if I increase the volume, my Q is going to get smaller. And I want my Q to be smaller because if Q is smaller, then it goes forward towards K. I want it to go forward, so I have to make Q smaller. Now, since K is inversely proportional to volume, the Q will become smaller if I increase the volume, not just doubling, but any increase in volume. My Q, Q will become smaller. Now, for that, my delta N has to be positive. For that, my delta N has to be positive. Only then K will be inversely proportional to volume. If my delta N is zero or negative, my K will be proportional to V. My K is going to be, because this will, this lead minus, if it is negative, liter will come to the top. Okay, because one by X to the power two is nothing but X to the, minus two is nothing but X to the power two. So it is directly proportional to volume. Are you kidding, Vian, are you kidding this? Right, so I have to only look at the option where my delta N is positive because only in this scenario, my equilibrium constant will be inversely proportional to volume and therefore increasing the volume, my Q is suddenly going to drop below equilibrium constant and then it will make an effort to achieve equilibrium constant by going forward. Okay, so another good question. Now, do you want, guys want to do some more questions or should I go to Ionic equilibrium with V, Vian? Huh? Yeah, what do you think, Sundar, with V? Are you there? Okay, Sundar is now lost. Okay, okay. No problem, yeah, let's go to, let's go to Ionic equilibrium, okay. Now, one of the important things in understanding in Ionic equilibrium is why are we talking about Ionic, okay? So far, chemical, et cetera, was very good, but is Ionic equilibrium also equally true? Do everything that we spoke about chemical equilibrium valid for Ionic equilibrium? Now, let's understand the similarities and differences. Now, basically when we talk about reactions, we are talking about two products, okay? So for example, N2 plus H2 is actually giving me NH3. So a compound is reacting with another compound to give me a new molecule, okay? Having said this, this is possible only if you're talking about pure compounds, not solutions, not dissolved gases or any mixtures. You're talking about an actual nitrogen taken to hydrogen and ammonia getting formed, but most of your real life situations will actually be solutions where either gases are dissolved or any substances are dissolved. And the moment substances are dissolved, the ability to form a fixed compound is very little. For example, when you have HCl plus NaOH, the most simplest example that we can have given out, you actually do not get out NaCl precipitated immediately out of the water and what a new amount of water form, no. In fact, you will find that in the solution, neither NaCl is present nor NaOH is present nor HCl is present. By the way, in the solution, what we actually find is H plus Cl minus Na plus OH minus and H2O and H2O, okay? Maybe some amount of, but therefore I'm going to write this in bracket, maybe some amount of NaCl, HCl and NaOH, but very, very little. Why? Because it is the nature of these ions to remain stabilized in the ionic form itself, okay? So if that is the scenario, I no more can talk in terms of simple compounds. I have to go to a much deeper level and therefore I talk about ionic reactions. Now, as these compounds are getting formed and they get rebounded back to form reactants from products, similarly, these ions also can get formed and get rebounded back to form the core reactants. And hence, and if you remember, when we spoke about law of mass action, we were actually talking about collusion theory and in collusion theory, we said that look if two ions or two atoms are reacting, only then we will be able to do all of these reactions, okay? Guys, you're gonna have to spare me for 30 seconds, you know, just one, few seconds here. So, you know, all these ions are formed, okay? Now, yeah. Now, let's say we take a weak substance, okay? For example, something like NH4 OH, okay? Now, we will find that the moment I put NH4 OH in water, I actually end up getting ions which are NH4 plus, plus OH minus. And I would also have H plus ions in what are in the water plus I'll have H2O, okay? Now, in this kind of a scenario, and I know that all the time as NH4 and OH are formed, they also bump into each other and give me back NH4 OH. Now, according to our collusion theory, this is an exact case of threshold energies where collusions happen and the product is formed, all of that is present. Now, if that is the scenario, then I can apply our law of mass action to actually write an expression even for NH4 OH. Now, our law of mass action says products raise to its concentrations divided by reactants raise to its coefficients, right? So, the same thing we apply here and we can write the dissociation of NH4 OH as if it dissociates into NH4 plus, plus OH minus as nothing but K is equal to OH minus into NH4 plus divided by NH4 OH. Okay? Now, okay, one second. Okay, now, right. Now, let's understand before going this that we all know what are the acid and base theories, right? There were three definitions, Arrhenius, then Laurie-Bronstead and the final is the Lewis theory. Now, whatever we wrote, this is nothing but our equilibrium constant expression. We should also understand that if I'm writing this expression for a base, what I call this equilibrium constant is nothing but an acid constant, a base constant, okay? And it is written as KB, KB, okay? Similarly, if I would have written this for an acid which is CS3COOH and I would have written an expression which is CS3COO minus plus H plus, I will get this as KA, okay? And I will call that as an acid constant, okay? Acid constant. So this is one part. So we are just writing like we said equilibrium constant and we said KP, which is a pressure equilibrium constant or equilibrium constant in terms of partial pressures. Here we call this as base constant or acid constant, right? There is another habit of writing pKa and pKb values. The reason for pKa and pKb is because we will deal with a lot of ions and most of the ionic reactions would be generally an acid-base reactions that we'll have to deal with. And we have a familiarity with pH. We also try and define pKa and pKb, which is nothing but minus log of kA's and minus log of kB. Now you have to understand here very importantly is that pKa values will be inversely proportional. So which means that lower the pKa values, stronger the acid, but higher the kA values, stronger the acid. So kA and pKa, they are inversely proportional. Strength is inversely proportional to pKa values and strength is directly proportional to kA and kB. This has to be remembered, right? Now we all also know conjugate acid-base theory, right? In acids and bases, we have seen. Just to quickly revise the conjugate acid base is nothing but two pairs with a difference of either a H or OH minus. So for example, HCl and Cl minus are basically, if you HCl gives H plus and Cl minus takes H plus to become HCl, okay? So these are conjugate acid-base pair. Of course, since this gives H plus, this is an acid, whereas this takes H plus, it is an base. A quick thing that I'm going to say here is that stable the conjugate of the substance, stronger will be the reactivity of the substance, which means if Cl minus is more stable, HCl will be a very reactive acid. Why? Because of course HCl will want to be in the Cl minus state as much as possible and therefore it will leave HCl state and go to Cl minus state, okay? And therefore it will be very reactive. Now, now, you know, Ionic equilibrium chapter also deals with a lot of understanding of the strengths between different acids and bases, right? Now remember that the strength of acids is generally on four different factors. I'm just going to see this quickly as it is our basis and move on because Ionic equilibrium has further more topics to deal with. The first thing that the strength of acid-base depends on is on the stability of conjugate pair, most important factor, stability of conjugate pair, okay? Or just the conjugate in simple terms. The second factor that the strength depends on, the charge and on what I atom the charges, the charge on what atom, okay? For example, let's say I have CH3SH, which is a high, you know, hydrogen sulfide of methane or, and I have CH3OH, when I break this bond I end up getting CH3S minus, I end up getting CH3O minus. Now between S minus and O minus, the negative charge is more stable on the O and therefore this is going to be more acidic than the upper one, okay? So that's the second property of the charge. The third is resonance. More the resonance structures, stronger the acidity. More resonance means more stability. If you're having more resonance, obviously your, your, you know, stability will be larger and therefore you would want to get into the conjugate. So I'm talking resonance of the conjugate, okay? So if you have more resonance of the conjugate, you will have more strength and you will want to get into the conjugate, okay? Okay, now, and the, and the final factor is what we call as, you know, the inductive effects or just called as induction. So for example, you know, in this scenario, if I have to compare just the oxygens, if I have CH3 thrice CO minus, this is less stable than CH3O minus, why? Because you're the inductive effect will be very large and O minus, which was stable with the, with this negative sign is now one, is now being burdened with the negative charge. And since it is being burdened with the negative charge, this is less acidic than the upper one, okay? So, yeah, so, so that is the last factor, okay? Now, now this is acidity-basicity aspects of the, you know, in, in any, now let's talk about what are the major rules or what is the real crux of this entire chapter? The first major rule of this entire, you know, chapter on what everything is based on is called as Oswald's dilution law, okay? You might have heard this earlier, Oswald's dilution law, okay? So what is Oswald's dilution law? It says, look, if you have a binary compound, like AB, okay? And if it goes to A plus, plus, B minus, okay? Then you can calculate the expression of concentration and degree of dissociations for this compound. So for that, first, let's understand what is degree of dissociation? As we had defined earlier, alpha, which is degree of dissociation, degree of dis is sometimes written, is, is nothing but the amount of moles, the number of moles that get dissociated per unit mole of the original compound or the parent compound, per unit mole, okay? So number of moles, alpha is number of moles are dissociated, okay? Per one mole of parent, okay? Now, if alpha moles are dissociated, let's say at time t equal to zero, I had taken only compound A. So I took, let's say, C concentration of the A. C concentration of AB, and this will be obviously zero, zero. Then at time t equal to t, when actually equilibrium is achieved, if for one mole, alpha was dissociated, for C moles, one minus alpha gets dissociated. And C alpha and C alpha gets formed, okay? Guys, give me two seconds, okay? Just give me a second. Now, if this happens, then the, you know, you can write your equilibrium constant as K equal to concentration of A plus into concentration of B minus divided by concentration of AB. If you put in all the values, you'll realize that you get an expression which is C alpha squared divided by one minus alpha. Now, this is the root Oswald's law of dilution, where it relates three things, okay? It is relating degree of dissociation. It is relating the initial concentration of AB initial. Please note not, it is not intermediate. It is the initial concentration and it relates the equilibrium constant. Now, there are multiple assumptions that we can do here. Let's say the degree of dissociation is very small. When I say small, it should be less than 5% or so. You know, 5 to 7% is maximum that you can allow. When it is very small, this one minus alpha is very insignificant for us. One minus alpha would be 0.97, which we can approximate it as one. That brings on when, so I will write here when alpha is very less than one. We write this K is equal to C alpha squared or your alpha becomes nothing root of K by C, okay? And if alpha is root of K by C, then concentration of A and B at any point of time is the concentration of A plus is equal to concentration of B minus is equal to C into alpha, okay? C into alpha. I hope that you understood this. So I'm going to just mark this into a lot of bold. This is the most important step to understand. If you understood this step, everything else is very easy. How does this step happen? It is basically C into one minus alpha. For one mole, it is alpha dissociating. So for C moles, it will be one minus alpha that will get dissociated, right? You can also write this as C minus C alpha, right? So then C alpha, C alpha are getting formed. Now, concentration at equilibrium of A plus and B minus is C alpha. This is nothing but if I substitute the value of alpha here, then I get this as equal to root of KC, okay? Because C into one by root C, they'll get canceled out. So this is what is the key. Now, these two expressions are something that are very important. Please note they are for very less alpha, which means very weak dilutions. And alpha is given by root K by C and C alpha is given by, or which is the concentration of A, B as root KC. This is the only theorem that is applicable over and over again at multiple places in this chapter. Now, let's see its application. How does it really affect our understandings? Now, there's one more quick thing that I will go before that is what we call as water product, okay? So what is water product? Water product is nothing, but the product of concentrations of, denoted as KW is a product of concentrations of H plus and OH minus at 25 degrees Celsius and one atmosphere pressure. This atmospheric pressure is to maintain the vapor pressure, okay? Now, please understand that this product is, yeah, water product is nothing but the product of concentration of H plus and OH minus ions at 25 degrees Celsius inside water, okay? So if I take one liter of water, how much of, in fact, it does not matter whether you take one or other liter because we're talking in terms of concentrations. So whenever H plus and OH minus are present in water, their product, it turns out to be 10 to the power minus 14, okay? This is a fixed quantity and because their product is 10 to the power minus 14, our pH scale becomes 14 and I'll demonstrate it how. So let's say what is pH? Your pH is nothing but minus log of H plus ion concentration. If I take minus log on both the sides, I get minus log of KW, which is nothing but 10 to the power minus 14 is equal to pH plus POH, right? Which means that pH can range from a maximum entity because pH cannot be negative. So pH can be zero. So this value turns, this value firstly turns out to be 14. So 14 is equal to pH plus POH and therefore pH can range from 14 less than equal to 14 or less than equal to zero, right? Because we do not talk about negative pH. So if pH is zero, it is highly acidic and pH is 14, it is highly basic. This is the reason why our pH scale is constrained from zero to 14. But please note, this is only at 25 degrees Celsius. If you change the temperature, your pH scale will differ. So here's a question for me to you. If you take it at let's say 50 degrees Celsius, will my pH scale be lesser than 14 or more than 14? If you take at 50 degrees Celsius instead of 25, will I have a pH scale from zero to, for example, 10? Or will I have a pH scale from zero to, let's say 20? Or will it remain same, zero to 14? What are these three? What do you guys think? Three degrees. Okay. What do others think? Guys. Vihan, Sundar. Okay. Now let's analyze this. How will you do that? Say, KW is equal to 10 to the power minus 14. In a perfectly neutral solution, my H plus and OH minus are going to be the same. So each of them is going to be H plus square. You know, I'm just doing to, because they are same, so I'm just going to make it as H plus square, which means that my H plus concentration is 10 to the power minus seven. And therefore, when I take pH, my pH turns out to be seven when H plus and OH minus concentrations are same. Now, important point. When pH is seven, it is a neutral and my H plus is 10 to the power minus seven. If I increase the temperature, let's say I take it to 50 degrees Celsius, my H plus concentration is going to increase. So it is going to go from 10 to the power minus seven to 10 to the power minus two, for example. Now, please note, 10 to the power minus two is a higher entity than 10 to the power minus seven. It is not smaller, it is bigger. This is 0.01, this is 0.0006 times one. So when this is more, then my KW product is not going to remain 10 to the power minus 14. It's going to become 10 to the power minus seven or minus eight, because 10 to the power minus two into 10 to the power, let's say minus three is 10 to the power minus five, which means my pH scale will now become from zero to five. So if I increase the temperature, my pH scale actually shrinks, okay? So it decreases, okay? So that's a quick answer. Now, from here, you can also understand that when H plus ion concentrations are more, you will get pH as lesser than, when it is more, for example, it is minus two, your pH will be two, and therefore solution becomes acidic. So when pH is less than seven, the solution is acidic. When pH is more than seven, the solution is basic, okay? Now, let's talk about, there are a lot of stuff in ionic, which we, for example, weak acid, weak bases, and all of this. So whenever we determine pH of a weak acid, okay, of a weak acid, and mostly we are talking about binary compounds, okay? So CH3COOH, I'm writing it purposely like this to understand that this is nothing but AB, okay? So this is a binary compound. It will also dissociate as CH3COO minus and H plus, which is nothing but something like B minus and H, A plus. Now, if you apply Oswald's law of dilution, we know that from Oswald's law of dilution, A plus ion concentration was C alpha, which is also equal to root of KC, okay? So if I take initial concentration of C, and if this equilibrium is a weak acid equilibrium, so this K will be KA, so H plus will remain equal to under root of KA into C, which means that my pH value will be nothing but minus log of square root of KA into C, and I can also write that as minus half of log of, into bracket log of KA plus log of C, right? So this is just a derivation of Oswald's law of dilution for weak acids. Now, as I have found the expression for H plus ion concentration in terms of KA and C, similarly, I can also write the expression for alpha, which is the degree of dissociation, which is nothing but root of KA divided by C, okay? And you can put in the values of K and C and get the degree of dissociation of how much CH3COH actually dissociate, right? So that is one. Now, there is one more thing that we need to understand here, which is the common ion effect, very similar to the previous problem that we did. Let's say I have two compounds, HA, which dissociates into H plus plus A plus A minus, and I have HB, which dissociates into H plus plus B minus. This is the most important thing that I would like you to really understand. A lot of problems are just framed around this. Please understand, because HA dissociates, it gives H plus, and this also dissociates, it gives H plus. We know from Leigh-Chatteldis principle that if I simply take HB, and I'm already having an extra product, which came from HA, this reaction is going to get backwards, it is going to get suppressed. But as it is going to get suppressed, this reaction will be pressed forward to give more H plus. Having said this, they both will come to an agreement saying that, look, whatever H plus is coming from HB and whatever H plus is coming from HA, they both will have to be balanced, both the equilibrium's one for HA and one for HB also. And therefore, the dissociation of both HA and HB will get suppressed, not one, both. Please remember, because of a common ion that comes out, the dissociation of both the compounds get suppressed. Now let's see how it will really work out. Let's say initially I had taken a concentration, C1 of this, and C2 of this. At equilibrium, this is going to be C1 into one minus alpha, okay? And now I'm going to take this as a very strong acid, okay? And this as a very weak acid. So this is a strong, just to keep some simplicity else, I will have to deal with too many alphas. If I'm doing this as a strong acid, at equilibrium, entire thing will get dissociated into giving C1 and C1 here. This, in fact, being weak, will get dissociated with some alpha, one minus alpha, and I will get this as C2 alpha and C2 alpha here. Now, because the concentrations are same, technically, we have to add both of them, okay? This will be C2 alpha, and I will have to add C1. Why? Because they are in the same container. So I cannot keep one concentration of H plus here and second concentration of H plus. Also, I want you to realize that adding simply the concentration of C2 is a very wrong step to do. But having said that, this is the most applicable step that, and the only step that we can do, why? Because how the exact negotiation happens between HA and HB, given that their strengths were similar, is very difficult to ascertain. What we are doing is we are taking actually a shortcut, which is a very good approximation also, saying that look from HA, C1 is going to come out. There is no denial about it. It's strong, so entire C1 is going to come out. From HB, entire C2 alpha is going to come out, but because of C1, it is going to get suppressed. So I don't know how much this association technically is happening, but let's say the final dissociation was alpha. So if alpha would have happened, let's say the contribution of HB was C2 alpha, is what we are assuming and going forward. If we do this, now I do not need to consider or worry about the K here. I only need to worry about the K here because it is a weak acid and let me call that as KA. So KAHB will be nothing but C2 alpha into, sorry, yeah, correct, into C1 plus C2 alpha divided by C2 into one minus alpha, okay? Now this is the expression for the KA of this guy, okay, of this mixture. Now I've just given you a way how to do that. You can do for multiple variations and find any kind of equation around it. Now if you want to find pH, if you want to find pH, then pH is nothing but the concentration of C1 plus C2 alpha, C1 plus C2 alpha, right? Now to really get concentration of, sorry, this will be minus log, not this concentration. It would be minus log of C1 plus C2 alpha, okay? Minus log of C1 plus C2 alpha. Now you can, if KA value of this guy is given, you can find the value of alpha from this expression, substitute here and you will get the pH of this kind of a solution. So this is one way, okay? Now what I'm going to do is I'm going to show you just two more variations. One variation I'm going to show you for, for, you know, not a binary compound but maybe a di-basic compound. For example, when you have H2A, it firstly dissociates into HA minus plus A minus, okay? Now there is a different dissociation constant for this. Let's call that as KA1. And now HA minus is not going to remain silent but it will also dissociate further to give H plus, plus A minus. Now, please note that A minus is, is the A minus minus actually, sorry, A minus. Oh, this will also be H plus, what am I doing? Yeah, sorry, this is going to be H plus, okay? So HA minus and H plus and H plus and A minus, right? Now you will realize that H plus is the common ion that will get into these two equations, okay? So now let's say I have, I have two of these. I have two, you know, this is KA1 and this is KA2. This equilibrium constant is KA2. Now let's say I had taken C as my initial concentrations and this degree of dissociation for this is alpha one. So this will become one minus alpha one at equilibrium. This will become C alpha one and this will be again C alpha one. Now after the second dissociation, this is the concentration of HA minus that will dissociate. So I will write this as C alpha one and my degree of dissociation for this reaction is going to be different. So this will be into one minus alpha two. I'm just gonna pause here for a minute and really ask you if you have understood this. Please note how am I writing all of these C and this will be C alpha one into alpha two because this C alpha one itself is my initial concentration of my reactant now. And my A2 minus is again going to be C alpha one, alpha two. I want to ask you guys here. Did you really understand how am I writing these concentrations? This is one of the most important things for you to understand. Are you getting it everyone? I'm in this area. Okay. Yeah. Sundar, did you understand? Okay. I'm going to write this as even a more simpler step. Initially I had concentration of alpha as C and I had zero and zero of this, okay? So this is initial, okay? Now alpha is a degree of dissociation which means for one mole, alpha are going to dissociate. For C moles, how much will dissociate? C alpha. So C into one minus alpha dissociated and it got first into H A minus and H plus. See this H2A is going to dissociate in two stages. It's not going to dissociate in one stage. Firstly, it will give out one H plus. Second time it will give out another H plus. For giving out one H plus, the equilibrium constant is K1. For giving out second H plus, the equilibrium constant is K2, okay? So what can we do? So what we have to do is we will have to have the equilibrium constant now match to each other. Now the common ion here is H plus, H plus, okay? If common ion is H plus, then after the first dissociation, the H A minus is my reactant now and H plus and A minus are my products. So with this as my concentration, now alpha two is the degree of dissociation for this reaction. This was alpha one, alpha one dissociated in this thing. Alpha two are dissociating in this thing. So with this concentration, one minus alpha two will dissociate and this concentration into alpha two and this concentration into alpha two will be repeated. For example, we wrote C into alpha one here. Now it will be C alpha one into alpha two, C alpha one into alpha two. Are you getting this? Is there any other query? Yeah, just ask me if you have a query, okay? Now please note that since there is a common ion effect, I cannot just simply keep this as C alpha one. I will have to add a C alpha one, alpha two from this point here and here I will have to add C alpha one. See how complex it starts getting from here, okay? Now, my K A one expression will be simply this one, you know the upper first reaction. Let's call that as first and second reaction. So in the first reaction, it is H plus into H A minus. Yeah. Why do we have to add the alpha one? See, because in one container, you cannot have two concentrations of H plus, right? Then H plus concentration is going to be the same. So if C alpha one came out of this reaction and C alpha one, alpha two came out of the second, I have to add both of them to keep the concentration same. Okay, yeah. Are you getting my point? So now you will find that in this entire, even if there would have been five reactions, all the H plus in them has to have the same concentration. I cannot have five different concentrations of H plus because H plus is the same in the container. So therefore we add up. So now I'm doing K A one as H plus concentration into H A minus concentration divided by H two, which is C alpha one into C alpha one plus C alpha one, alpha two divided by C into one minus alpha one. And my K A two expression will be C alpha one, alpha two into C alpha one plus C alpha one, alpha two divided by C alpha one into one minus alpha two. Okay. All these expressions look pretty serious, but you will realize that a lot of factors actually cancel out. Okay. A lot of factors cancel out. Now, I'm going to stop here, you know, because stop here for this reaction because no point of classifying it further. But what I really want to make you understand is the things that the examiner sometimes or what you have is you have K alpha, K A one, you have K A two. You are definitely going to have C. So you basically have only two variables. What are those two variables? The one in pink alpha one and alpha two. Okay. Now you have two expressions and two variables. These two expressions can easily be used to find what is the degree of dissociation at both of these stages. Okay. And that is our purpose. The moment we find this degree of dissociation, my H plus iron concentration is very visible, which is nothing but C alpha one plus C alpha one into alpha two. Another important point that I want to tell you or as a part of strategy is most of the times you might have to simply neglect C alpha one, alpha two because this value becomes very small. You are absolutely free to neglect that. Remember that whenever you are omitting or neglecting this, you have to understand that what is the comparison that you're doing? You cannot compare one with 10. If you're neglecting one out of 10, that is a huge number. Can you neglect one out of thousand, one out of hundred? Yes. So generally remember within the range of 5%, 5 to 3%, you're allowed to neglect. So anything that is three out of hundred, you're allowed to neglect. One out of thousand, definitely you can neglect. Okay. What do I mean by neglection? See, the moment I simply cancel out C alpha one, alpha two to be zero, because as compared to hundred, if this is only three, then I do not need to take 103. I can write this as a hundred. If I just do that, my expression becomes very easy. It simply becomes C alpha one, alpha one square divided by one minus alpha one. And this expression becomes C alpha one, alpha two divided by one minus alpha two. So my expressions become very easy to solve and manage. So remember, Ionic equilibrium is all about probability and all about approximation. So you have to understand what are the probabilities that can possible and what are the approximations that I'm implying? Okay. So that's the crux that's I think the most complex that you can do, but I'm going to give you one more, you know? So I, since I promised, I would show you two different variations. I'm going to show you one more variation where I'm going to take both weak assets and both different weak assets, okay? So that was a, this was an expression for, remember, only one asset, but a die basic asset, okay? Which means it has two H plus. Now I'm going to give you an expression where we have HA and HB, okay? So these are two weak assets. So HA and HB will dissociate as H plus plus A minus and this is going to dissociate as H plus plus B minus. Now please note that the concentrations this time are going to be different because there are two weak assets being put in the same container. So if this is C1, there will be one minus alpha one dissociation, this will be C1 alpha one and this will be C1 alpha one. Since it is a weak base, it will be one minus alpha two dissociation. This will be C2 alpha two and C2 alpha two here. The common ion being C1 alpha one, I will add your C2 alpha two and I will add your C1 alpha one, okay? Now this is going to be KA1, KA1 and this is going to be KA2, okay? Now we have two expressions, KA1 is equal to C1 alpha one into C1 alpha one plus C2 alpha two divided by C1 into one minus alpha one and I have KA2 is equal to C2 alpha two into C2 alpha two plus C1 alpha one divided by C2 into one minus alpha two. So I've done nothing but return H plus into B minus divided by HB, H plus into A minus divided by HA, right? So these are the two expressions I have. Again, what will be the entities that should be given to you? You should be given C1, you should be given C2, you should be given KA1 and you should be given KA2, okay? What are the entities that you will have to find? You'll have to find alpha one and alpha two. You have two equations and two variables. You will be easily able to find it, okay? Now, once you have this structure in your head, all the constraints can be applied to it. For example, I can give you a question saying that, look, I want to suppress the dilution of HB to 90%, which means that only 0.1% of HB has to come out, which means that alpha two will become 0.1, right? For every one mole, 0.1 is alpha one. So if 0.1 is alpha one with that constraint, how can I squeeze these expressions to give me a concentration of HA can easily be found out? So you simply put alpha one, alpha one here, you already know KA1, KA2 now, and you know how much of C2 was being put in, right? So you'll easily be able to find out C1. So multiple of such questions can be designed around it. Now what I want to do is instead of just going ahead, I want to do a couple of problems on this so that we at least do one or two problems and see what is possible. Okay. Maybe I want to, I'll just do two more things which are important before doing the problem because if you don't get time, then it will be an issue. There is something called as a buffer also, okay? Now what is a buffer? A buffer is a solution which has a constant pH. How does the pH become constant? Because it has two components. If you add acid to it, one component will eat up the acid part. If you add base to it, the second component eats up the base part. Let's look at how this happens. Let's take a salt of weak acid and strong base, okay? So weak acid is HA and a mixture of, a weak acid and it's salt with strong base. So this is, let's say NA, okay? So this is the salt of this weak acid with a strong base. For example, weak acid could be CS3-COOH. This is CS3-COONA, okay? So A minus is CS3-COOH, right? So now this is a mixture of a weak acid, a weak acid. What do you mean by definition of buffer again? Buffer is a mixture which resists the change in pH. How does it resists the change in pH? What do you mean by resisting change in pH? Which means that my acidity or basicity, whatever it is, remains fixed. Which means if I add some acid to it, the equilibrium will eat up all extra H plus. And if I add some OH minus to it, the equilibrium will eat up that extra OH minus. When they eat that thing up, they will form the compounds and ions will be reduced or maintained constant inside the solution. And if H plus and OH minus ions are maintained constant, then my pH will remain constant and therefore I will have a more stable acidity for it. Why is this important? This is important for a lot of plants. This is important for a lot of food items. This is important for a lot of medicines, right? If the pH of the soil differs, then the plants die. So you want to add something to the soil which will keep on resisting any addition of acidity or basicity to the soil by any means of nature. How is that achieved? That is achieved by multiple means. You can take a weak acid and it's salt. You can take a weak base and it's salt. You can take two different salts. So all of that is possible, right? Now, how does this happen? For example, I'm going to give you one example. So weak acid and let's say this is the salt. It's salt with a strong base, okay? Now, what happens in this solution is that this is going to get dissociated with equilibrium constant into H plus plus A minus and this will be dissociated with a KA. This dissociation is going to be complete. Why? Because it is a salt. Remember, salts always dissociate completely, right? This is going to get dissociated like this, right? Now, if I have taken C1 as its concentration and C2 as its concentration, this will be one minus alpha and this will be C alpha and C alpha. This is going to be C2, simply zero C2 and C2 here, right? So my A minus will become C alpha plus, sorry, C1, C1, I should write it, C1 alpha plus C2. Now, please note, if I add a lot of H plus to this solution, okay? Now, let's firstly calculate pH. So what is my pH? My pH is nothing but C1 alpha and how do I find my alpha from? I get my alpha from the expression of KA which is C1 alpha into C1 alpha plus C2 divided by C1 into one minus alpha. pH is going to be minus log of this, huh? Minus log of this, right? Now, at the moment I've found, so I can get alpha from this equation because here I know C1 and C2 and I know KA. So I can find alpha from here and I put in here, I'll get the concentration of H plus. Now, let's say to this solution, I add a lot of H plus, okay, some amount of, not lot. The buffer generally works within a range. If you break the range, then it's no more going to keep the acidity. So it has a tolerance, what do you call as tolerance of buffer. But let's say if I add a lot of H plus to this solution, obviously what's going to happen is all of these reaction is going to go backwards, okay? The moment it goes backwards, A minus is going to get consumed. As soon as A minus is going to get consumed, this equilibrium is also, if there is no equilibrium because the salt is very strong, but this A minus also gets affected and it will push the solution forward, okay? Now, if I add a lot more OH minus, then more of H plus will get created from this HA. So the effect of addition of H plus is compensated by A minus. Effect of addition of OH minus is compensated by HA. So this is one reaction. In fact, I will tell you one more buffer, which is a much better example. Let's take, I take a buffer of CS3CONA and HCL. Yeah, so let's take CS3CONA. So this is going to dissociate as CH3COO minus plus NA plus, okay? And this actually is also in conjugation with CS3COOH, okay? This is a very good example, okay? Now say, if I add H plus to this solution, okay? What's going to happen? This equilibrium is going to get hit. Why? Because CS3COOH is in equilibrium with CS3COO minus and H plus. As soon as I add H plus, this is going to get the reaction towards CS3COOH and H plus will get reduced and pH will get maintained. Let's say I add OH minus. The moment I add OH minus, the OH minus will react with any of the H plus inside of the solution. And more of H plus will come from CS3COOH. Also, all of these OH minus is going to have a neutrality effect with NA plus as well, right? Why? Because NA and NAOH will have all the solutions in ionic form. So if CS3COO is formed, you are having NA plus stabilized in the solution only because of, also because of OH minus. So any addition of OH minus is going to demand more of H plus coming from CS3COOH. And this equilibrium, again, K will be hit and our pH is getting balanced. So you might ask, what is the need of taking a salt here? Why we are taking salt is we are having an additional supply of CS3COO minus available in the solution because of the salt. CS3COOH is only responsible for giving me H plus, okay? If I need H plus to be consumed, I'm going to use the supply from CS3COO minus to go back to CS3COOH. If I need H plus to be supplied, I'm going to use CS3COOH to get me H plus. So CS3COO minus is to consume H plus, whereas CS3COOH is to supply H plus, okay? This CS3COO minus is coming from salt, and this CS3COOH is coming from acid. So basically buffer is consuming H plus if more are put and it's supplying H plus if less are available. Now there are some very important equations. I don't think so that we'll be able to finish all of this in today's class, but I want to talk about salt hydrolysis and a lot of these topics. But I'm just going to mention a few important ones. The pH of a buffer is given by pKa of the acid plus log of salt's concentration divided by the concentration of acid, okay? I will urge you guys to go through these derivations. I would have done the derivations, but because of positive time, I'm just writing this, okay? Similarly, now this is pH for a pH value for a salt and acid mixture, okay? Similarly, a pH for a salt and base mixture is nothing but pKb plus log of salt concentration divided by base concentration, okay? So these are the two equations. Now please note that this pH is actually coming out to be a factor of pKb plus this. Now if we take equal concentrations of salt and base, I am going to get log of this as zero, okay? This is also going to be one. If I take equal concentration of this, again it is going to be, it is going to be, this is going to be zero, and therefore pH is going to be constant. Why? Because k and kb is constant. Now let's say even if we take this as a ratio of one by 10, please note, log of one by 10 is minus one. So to an extent that I changed this ratio by a factor of 10. So you imagine that one was 10 moles per liter, the second is now becoming 100 moles per liter. So you have changed an entire multiple. My pH is simply changing by a factor of one, okay? It is still getting fixated at the pKb value but changing by a factor of one. So therefore pH is very difficult to change the pH when we have such buffer solutions mixed with each other, okay? All right, so that's a quick view on buffer. I'm just again going to pause here. I actually want to talk about hydrolysis of salt but we'll not finish that within the next few minutes. But I just want to stop here and I want to ask you, ionic equilibrium is something that we should spend more time on. What do you guys think? Yes, sir. Do you, are you relating to what we just said, what we are doing? Yes, sir. Prithvi, Kevin, yeah, one at a time. Vihanna is saying something, yeah. Go on, Vihanna. So what are the questions that they ask on this topic? Multiple questions can be formed, yeah. This topic is one of the most notorious topics to form questions and the worst is when you actually mix these questions with let's say Gibbs free energy or thermodynamics. But having said this, I don't want to scare you up but I'm going to give you a couple of questions that they can ask. For example, you can have two different solutions mixed together, okay? Two different ionic solutions mixed together and their concentrations asked. You might have questions of finding out pH when a change to an ionic equilibrium is done. You might be asked questions to find pH of some buffer solutions. So all of this are possible. All of this can be very well taken up. Okay. Your voice is not very audible, Vihanna. Can you speak? Sir, this is important to talk about the CD. Look, there is a lot of weightage of chemical and ionic definitely, okay? So as much as the weightage of thermodynamics or chemical bonding is there, ionic also has so much of weightage. In CET, I don't expect you to solve complex problems. In CET, most of the times you will be given formula-based applications. Common ion effect problems also are not very rigorously used. But in GE mains and GE advance, they definitely exploit this chapter. They also definitely test your ability to do approximations and probability. For example, are you able to neglect a certain term and simplify the problem to get the answer fast? If you go through the rigorous methods of solving the tedious equations that we have just written, you are going to be in a mess within the time that you are going to have. So having said this, a good idea is always to do substitution. I always tell students is that if you're solving ionic equilibrium problems, simply substitute the answers to begin with and see if you are able to get somewhere. The second thing is at least try to get a range. So you can always estimate the range by common sense without getting into the math of it. If you're able to estimate the range, then one of the few options can be really gotten to faster. Okay, so that is, that are some of the ways to really handle this topic. You know, go on. Yeah, see, titration is, I mean, the most basic thing is, I mean, again, titration where different pHs are used and where di-basic titrations are also involved. That topic is again a large one, but to basic titration is where we have N1, V1 equal to N2, V2. We know this, right? This is the basic equation where normalities are used. There are multiple problems. I'm just going to classify titration for the time being and we'll look at titration again probably in the next class because we'll have to understand for that what are pH curves. So there is something called as pH curve. For example, let's say this is the concentration of the titrant, titrate that you're adding to the titrant. So titrant means whatever is there in the flask and titrate means whatever is there in the, in your view rate, okay? Now, and if you see the pH, your pH scale is somewhere like this, okay? So this is pH. So it is somewhere like this, which means that as you are pH would change rapidly with concentration is the other way around. It's, it should go top, okay? It's going to go like this. The concentration usually, yeah. So it is going to be like this. So you start from here, yeah, and then you go here. Yeah, something like this, okay? So you see that as you are increasing the concentration at certain point of time, the pH suddenly shifts to a very high value. And as you keep on increasing the concentration, the pH again stabilizes majorly, okay? So what happens is that you will find that, yeah, this is the right way to draw the curve, okay? Yeah, right way to draw the curve. Now, so this is the range where pH actually is hit. So the pH suddenly changes from let's say four to eight, okay? Four is a very big range, but maybe from six to eight is a good idea to say, right? So this is the way that pH changes when your titration happens. Your indicator is generally somewhere around this median point. So what happens is the indicator does not get affected so long that your titrate is getting added. Suddenly as soon as titrate is added, the pH shift being very large, you are in the basic zone more than the acidic zone or vice versa. And your indicator color shoots up giving you indication that your pH has changed heavily, okay? Now, what we have to do is there are multiple ways of these titration that can go on. So one is this way, but there are sometimes die basic titration. For example, when you are titrating, for example, NA2CO3, okay? So one is that when you get to NAHCO3 and then NAHCO3 further reacts with an acid to give you H2CO3, okay? Now, when this kind of a titration happens, there is a equilibrium point achieved twice. So what happens to your pH curve is that you will have a two pH curve, something like this. So this is when you're actually getting a pH curve for NAHCO3. And the next is when you get pH curve for H2CO3, right? So these are also another type of problems that are certain. How to deal with these problems? We'll just look at the basic understanding us as to what we do. To tackle these problems, we need to find out what is the H plus ion concentration for different stages? How to find H plus ion concentration if you saw die basic problem that we did a few minutes ago? What did we do? We wrote both the equations. We wrote the first equilibrium which will give me first H plus out and then we wrote the second equilibrium which will give me the second H plus out. When both of these equilibriums are happening, the first equilibrium depending on the K value, we have to see that which K value is getting hit first. If the first equilibrium value is getting hit first, I'm going to achieve a neutralization at that point and applying all the H plus common ion effect, Oswald's law of dilution, C alpha ones, I will be able to get the concentration of the products. I will also be able to understand what is the pH that I'm getting into. Then the second equilibrium is it, then the same pH is found by using the KA2 value which is the second dissociation values. Are you guys connecting to what I'm saying? Yes sir. Yeah? I'm going to bring this up one more time to show you. We look, whatever we were just saying about twice titrating is nothing but a die basic, it's like a die basic acid example. In this thing, one H plus is given out, second thing in second H plus is given out. There are two Ks also for that. If at the first H plus from this KA1, if I'm able to find the H plus, what was the H plus where first KA1 is reached, the moment I know KA1, I know what is alpha one. From alpha one, KA1 and C, I'm able to find out the H plus ion concentration that is reached. Now, how do I know what is the C? This is your Buret value. What is the acid that you're adding into the base or what is the base that you're adding into the acid? From that, you know what is a concentration. So as soon as the first pH change happens, your C is known. As you know your C, you know your pH. And as you know your pH, you know what is the indicator that you have used and where the equilibrium point has reached. Similarly, when the second equilibrium is achieved, you'll get the next C2 value. As soon as you know the C2 value, you can use KA2 and find the new pH that is achieved. So dealing with these problems is three things. I would say. First, writing the right reactions. You know, write the reactions that where equilibrium is happening. The second thing is writing the right, sorry, writing the right amount of concentration and degree of dissociation points, okay? So how these degrees of dissociation affect? This is the second point. If you are in doubt, forget everything and simply write the C alpha expressions to begin with. And once they are written, then observe their equations and the third thing that is very important is the common ion. So if you have a common ion, you simply add their concentrations and write expressions for KA2 and KA1, right? So there are three things I repeat. First, writing the right equation. Second, writing the right concentrations and equilibrium values of concentrations. And the third thing, writing the KA expressions and H plus expressions and finding the pH value. Yeah, are you getting it? Guys, yeah, okay. Now, I think we, you know, it's, you know, these online sessions are pretty concentrated, you know, as against our physical ones. So I think there is too much that we are doing on one single day. I want you to pause. But having said this, look guys, I want you to solve problems. Unfortunately, I would love to give you too many problems. You know, even if you do your Cengage problems, it is very good, but do problems. You know, we had discussed that these sessions of revisions would also be sessions where we should work on some problems. I want you to look at, you know, to do these problems and send me doubts or question up or take the available time on WhatsApp that I have. I do that for a lot of students and please make use of that because on these sessions, you know, all the problems we cannot solve. You can solve a couple of variety of problems. I will give you a set of problems that I feel are good enough. You know, that also I can help you with. Maybe a PDF or something, I can drop you on the group and we can work on that. Does that make sense? Yes, sir. Yeah. Please solve problems. Problem solving practice is a must. You cannot, you know, do without it. Especially, INEK and Chemical Equilibrium. Chemical equilibrium is a relatively easier topic. INEK equilibrium is kind of pinching. Yeah? Does that... Yeah, go on. We want customers to understand from now because we have exams coming up and every Friday we have an exam. Okay. We have a certain quality. Mm-hmm. So, when you are on Wednesday, typically on Wednesday we have a Friday with that. Yeah, so, I don't have a problem, but this has to be decided on case-to-case basis because Wednesdays I have a 315 to 515 class on some days, not always. Now, if I don't have, we can have or if there is a special... So, why don't you do one... You know, one of you guys, homesayer, are having an exam. Why don't you send me your schedule? Okay. Because what I will do is, according to the schedule, if I am able to shift my Wednesday class in this one of the schools, I will try and shift it to Thursday and probably give you a slot. Or, if you have a holiday, we can always shift the class to one of those holidays and have a revision class or, you know, a working class. Yeah, on the holidays we'll be studying for the exam because we have three finals going on right now for the board exam. Okay. So, week, day three and day four, like, every two days and Thursday is a holiday, so that day we'll be studying for the next day, the exam. So, we have Monday evening, Wednesday evening and Friday evening free. Monday evenings are all free of mind. I can have all Monday evenings if you guys are okay. Because Wednesday, there definitely is a clash, but as I said, this clash is not always some of the days. I'm okay with the Monday evening or on Wednesdays, a late night class, maybe after six or so. But, I will talk to everyone and get the news right. Sounds good, perfect. Yeah, all good. Yes. Okay, so I'll see you again and we'll meet up in the next session sometime. Yeah, please solve problems. I'm saying this again and again. Take any book, yeah, it does not matter. You know, take any type of problems, just go through all the problems serially and finish it off, okay? Yes, sir. Okay, take care guys, bye. Bye.