 Hi, Shant. Okay. Yeah. You have joined recently or what? She will continue with. I need to give him. Okay. We were doing last class. Okay. So like last class we have done. About the calculation of alpha, right? How and when we can neglect alpha? Few questions, one or two questions we have already done on that. So the next thing we have to discuss here is common ion effect. Have we discussed this common ion effect? Tell me common ion effect. Write down the heading common ion effect. Okay. So write down first of all, what is common ion effect? In this, we are talking about weak electrolyte. Okay. Electrolyte we have already discussed. Okay. And this thing is the degree of dissociation. Write down the degree of dissociation. Dissociation of a weak electrolyte. A weak electrolyte is suppressed by suppressed by the addition of, suppressed by the addition of another electrolyte. Another electrolyte. Okay. So we are talking about two electrolytes here. Okay. Two electrolytes. And we're talking about the dissociation of those electrolytes. What is the meaning of this particular concept is suppose we have two different electrolytes, right? One is suppose I am taking, suppose I will take the example of a weak electrolyte, which is CS3 COOH. When it dissociate, it forms what? CS3 COO minus and H plus. And another electrolyte, suppose I am taking, which is, suppose I'm taking a strong electrolyte here at CL. So this will form CL minus plus H plus. Okay. So both electrolyte if you compare, they contains one ion common. Okay. And what is that common ion? The common ion is H plus here. Right. So when the two electrolyte contains one common ion. Okay. So dissociation of the weak electrolyte is suppressed by the another one, which has a common ion. Okay. Okay. Okay. So since both the electrolyte contains one common ion, the dissociation of weak electrolyte is suppressed. Means the alpha value of CS3 COOH will be very, very less or minimum here. And suppose if you take these two electrolytes in a container. Right. So in this container, suppose I have added CS3 COOH and HCL. Right. CS3 COOH and HCL. Then what all ions will be there in this container? The ions will be what? We will have CL minus. We will have CS3 COO minus and H plus. Three ions will be there. Right. And what we consider because of the common ion, what we say this H plus is because of HCL, not because of CS3 COOH, if it is not mentioned. Understood. That the H plus, which is present in the solution is because of the dissociation of only HCL, not CS3 COOH. It doesn't mean that CS3 COOH will not dissociate at all. The point is, since this is strong and this is weak, and both contains one common ion, so the dissociation of weak electrolyte will be suppressed by the dissociation of strong electrolyte here. And hence, most of the H plus, 99% you can say, or more than that also even, can say most of the S plus in this solution or mixture, most of the H plus is because of the dissociation of HCL, not because of CS3 COOH. Okay. That's the meaning of common ion effect. Okay. Application of this also you will see in few questions. Okay. They'll ask this theoretical questions directly. Okay. Understood. Now suppose in this, if I take one more example, suppose I'm taking an electrolyte we have, and suppose that electrolyte is, which on dissociation gives B minus and A plus. Okay. And suppose it is a strong one. So when the dissociation takes place, this becomes zero and this becomes whatever the molarity of this compound we have, and this is also M, whatever the concentration we have. Another weak electrolyte, weak electrolyte, suppose I'm assuming that is HA, which on dissociation gives H plus and A minus, and A minus, here also it should be B plus and A minus, not A plus, with an ion we have. If this is C, zero to zero, C minus CL5 after dissociation, this becomes CL5 and this becomes CL5. Okay. So if I write down the KA value for this acid, that will be what? Concentration of H plus, concentration of A minus, divided by concentration of HA. Now you see these two are there in a mixture, right? Both are present in a mixture. Then only we can apply the concept of common ion effect. Okay. Now H plus concentration we know it is CL5, as I have written over here, CL5, and this A minus concentration is what this plus this, because both equilibrium is being affected by each other, because they have one common ion. So the concentration of A minus will be CL5 for H plus, for A minus it is CL5, plus whatever the molarity we have, divided by the concentration of HA, which is C into one minus alpha, right? Now the concept of common ion if you apply, because of this is strong electrolyte, the dissociation of this is suppressed, right? So alpha value is very less than, is very small or very less than one, we can say. Or what we can say, one minus alpha, which is here, one minus alpha, we can almost write it as one. We can neglect alpha since the dissociation of HA is very small. So we can neglect alpha with respect to one, right? So next thing here, if you write it is C alpha, now we are just solving this, C alpha into C alpha plus M divided by C. This C and C will get cancelled and when you multiply alpha, it becomes C alpha square plus M alpha, M alpha. Now since alpha is very small, so alpha square is even small, right? So we can also neglect this alpha square. Why? Since alpha is small, alpha is very small. So we can neglect alpha square also, okay? Hence, C alpha square can be neglected, can be neglected, right? So this expression becomes, K A is equals to M alpha, which gives alpha is equals to K A by M, okay? Alpha is equals to K A by M. This is the value of alpha we have when common and effect will apply. If you try to recall the same kind of expression we have already got in Ostwald dilution law, but little bit difference we have here, what in Ostwald dilution law, alpha is found to be K A by C root over of it, okay? But this root term is not there, right? This is the value of alpha when common and effect is there. Is it clear? Okay, now one question you write down and you try to solve this one. Calculate the degree of ionization. I'll just write down the question. Calculate the degree of ionization of 0.02 molar acetic acid, acetic acid, if it's K A value, if it's K A value is 1.8 into 10 to the power minus 5, what would be the degree of dissociation, degree of ionization, degree of ionization if the solution contains 0.01 molar sodium acetate. Sodium acetate. Yes, yes, yes, I'm sure. Because you see in the solution, you don't try to memorize this. You see this, the previous slide I'll show you. There's nothing to memorize here. Okay, whenever the solution contains 0, whenever the thing is there to memorize, I told you already that you have to memorize this. You see in the solution, what all ions we have? We have Cl minus CS3CO minus and H plus. So this H plus is coming from HCl and CS3COH. However, the dissociation or the ionization of this is suppressed from this HCl, which we are neglecting here also. 1 minus alpha is equals to 1. But since here also you see A minus and A minus. In the solution if you see, we have B plus, H plus and A minus. And we do not know from where this A minus is coming from. But we can say since these two has 1 comma ion, so most of the A plus, A minus is coming from H A only, which is a strong electrolyte. So whatever the A minus we have total, we are assuming it is both from this one and this one. But hence, the sense of this 2 contains 1 comma ion. So the degree of dissociation or ionization is this is very small and we can neglect this with respect to 1 and so on. That's why it is, right? Then 3 into 10 to the power minus 2. Is it 3 into 10 to the power minus 2? All of you are getting 3 into 10 to the power minus 2. Okay, I see. First of all, suppose if you have only acetic acid, CH3, COOH, only this one. If it dissociates, it forms what? CH3, COO minus and H plus. And it's alpha value if you calculate, which is not the question. If this alpha value you calculate from post-vault dilution law, that will be what? KE divided by its concentration root over of it. Anyone of you, if you have any doubt in this, you can ask me. Since I have done this in the last class, I have not known. Writing down the steps. How do we get this alpha value here? And when you write down the value here, 1.8 into 10 to the power minus 5 divided by 0.02 root over of it. And when you solve this, you will get 3 into 10 to the power minus 2, which is most of you are getting. Yes or no, this is what you are getting. But the question is not this. You'll get this as alpha when we have only CH3, COOH. This is not required to calculate. This is not the question. I have just calculated this to make you understand one point here. So first of all, this formula we can use when we have only CH3, COOH present. But here in this question, we have CH3, COOH and sodium acetate. The mixture we have is CH3, COOH and CH3, COONA. This is a strong electrolyte and this is a weak electrolyte. So here we have one common ion. And what is the common ion we have here? Can you tell me? What is the common ion? CH3, CO minus, yes. So don't get confused that common ion is always, you know, H plus or something like that, some cation. We can have any common ion. We are talking about ion, not cation or anion. So this will dissociate and forms CH3, CO minus and H plus. And when this dissociate, it forms CH3, CO minus and NA plus. So both contains one common ion here, which is CH3, CO minus. So obviously because of this common ion effect, the alpha value here for this weak electrolyte I'm talking about, because what is the question? Calculate the degree of ionization. If the solution contains this, means the degree of ionization of acetic acid we have to find out when this contains sodium acetate. So obviously alpha value is this when we don't have any common ion present. But when we have common ion like this in this situation, according to the logic, which is common ion effect, alpha value should be lesser than this, correct? Yes or no? The alpha value should be lesser than this, 3 into 2 minus 2. That is what our question is, right? Now that value must be less than this, otherwise the answer is wrong. Okay? So how do we calculate alpha here? Here you see, if only CH3, CO H is present, then it will dissociate according to the alpha value of this. But now since we have one common ion, so this CH3, CO H will not dissociate with the alpha value, which we have calculated already, which is this. So when this dissociate, it forms CH3, CO minus and H plus and its concentration it is given, which is 0.02. And suppose it dissociates X mole, which should be less than this, right? X and this will form X and this will form X. Right? Another electrolyte that we have, CS3, CO and A, when it dissociates, it forms CH3, COO minus and Na plus. Since this is a strong electrolyte whose concentration we know already, which is 0.01, so when it dissociates, it forms 0.01 molar of CS3, COO minus and 0.01 molar of Na plus. Right? K A value is given. So we have to write the expression of K A because we have to use this. So K A we can write for this one because this is an acid, this is a salt and this is an acid. So K A we can write only for acid. So K A value if I write here, see, K A is equals to what we can write, CS3, COO minus according to the first equation into H plus divided by CH3, COOH. Right? What is the concentration of CS3, COO minus? CS3, COO minus concentration is what? It is X from here and 0.01 from here. So concentration of CS3, COO minus is X plus 0.01 into concentration of H plus is X divided by concentration of CS3, COOH is 0.02 minus X and this K A value is given, which is 1.8 into 10 to the power minus 5. One more thing here you see, we have to calculate alpha, right? When CS3, COO Na is present, sodium acetate. So alpha value is what? It is a degree of ionization, which we can consider same as degree of dissociation. And what is that? Number of moles, number of moles reacted divided by, divided by total moles initially taken. We have done this in chemical equilibrium. Moles reacted divided by the number of moles, which we have taken initially. So what is the moles reacted in this case? X mole, here you see. So we can write X by initially what we take since we are talking about of CS3, COOH. So we will take 0.02, 0.02. This is the alpha dash I should write here because this is a new degree of ionization in presence of the common ion. So this alpha dash we have to find out, which is nothing but X by 0.02, right? So if you get X, you can find out this alpha dash, which should be less than this alpha here, 3 into 10 to the power minus 2. Now here you see since this X is very small because of the common ion effect. So we can neglect this X with respect to 0.01, right? So we'll write here 1.8 into 10 to the power minus 5. If you neglect X with respect to this, you'll get 0.01 into X. And this denominator also this X we can neglect with respect to 0.02, right? And when you solve this, X value you'll get is 3.6 into 10 to the power minus 5, which is obviously lesser than this value, right? Now X value is 3.6 10 to the power minus 5, which we can substitute here, 3.6 into 10 to the power minus 5 divided by 0.02, right? And when you solve this, you'll get 1.8 into 10, 1.8 into 10 to the power minus 4, right? 1.8 into 10 to the power minus 4. So this is the actual value of alpha dash when we have this common ion effect. You see this value is lesser than this, right? This value is lesser than this, which is our logic. 10 to the power minus 3, okay, 10 to the power minus 3. So this is the value of alpha dash here. Is it correct? So this is the answer, 1.8 10 to the power minus 3. So you remember most of you have calculated this. This we can use when we have only CS3 COH present. In presence of a common ion, we have to calculate like this. Next to write down, pH also we were discussing, right? Next one is H of the mixture of weak acid and strong acid. We have a mixture of weak and strong acid. We have to calculate the pH of that. Suppose a strong acid here is I am taking HB and this weak acid here is HA. So if you write down the equation here, HB being a strong acid will dissociate like H plus and B minus. Sorry, H plus and B minus, okay? And if the concentration of this suppose I am assuming it is C1 and this is C2. So since it is a strong acid, so it is initially C1, 00. And since it is a strong, it dissociates completely C1 and C1. This is for a strong. Now for weak acid suppose, which is HA, which also gives H plus and A minus, which is C2, 00 initially. Now C2 minus C2 alpha, C2 alpha and C2 alpha. So we have to calculate what? We have to calculate pH. And the formula of pH is what? Minus log of H plus minus log of H plus. So what is the concentration of H plus here? The concentration of H plus is C1 from the strong acid and C2 alpha from the weak acid. So if you know this C1 and C2 will be given in the question, right? With K value will also be given for the weak electrolyte. See, when I write down the K expression for this, the weak electrolyte, what is the K expression of this? It is equals to concentration of H plus, concentration of A minus divided by the concentration of HA. This expression, which is nothing but the Ostwald deletion law, which is only applicable for weak electrolyte, not for strong electrolyte. So in this type of question, the K value of weak electrolyte will be given. And this will be equals to C2 alpha plus C1, the concentration of H plus because we have the mixture of these two. Sorry, A minus concentration will be C2 alpha divided by the concentration of HA will be C2 into 1 minus alpha, right? K value will be given in the question. C1, C2 will be given. From this expression, you can calculate alpha, right? So from this relation, you can calculate alpha. And when you get alpha substituted here, you'll get H plus concentration and then you can find out pH, which is nothing but minus log of C1 plus C2 alpha, right? Substitute this alpha value here. C1, C2 will be given. You can find out the answer. Is it clear? Understood this? Make sure of weak acid. The only thing you have to keep in mind, the only thing you have to keep in mind, is the calculation of H plus. Okay, thank you. Thanks all. When you have to calculate pH, you have to find out H plus concentration, right? And H plus concentration, depending on whether you have weak acid, strong acid, and mixture of acid and base, which will neutralize each other. Accordingly, we'll calculate H plus concentration and we'll use this formula of pH. Okay? That's the only thing you have to do. Anyways, next question. Thank you all of you. Thanks. Okay. Now you see the next question. What concentration of HCOO minus is present in a solution, present in a solution of 0.015 molar HCOOH, 0.02 molar HCL. K value is given. That is 1.8 into 10 to the power minus 4. Solve this one. 1.35 into 10 to the power minus 4. Concentration of CS3 HCOO minus is 1.35 into 10 to the power minus 4. Yeah, it's correct. Yeah, answer is correct. What we'll do into this one? You tell me one more thing. Like some of you may have this confusion that there are two acids given. One is HCOOH and other one is HCL. This K value is given for which one of these? For which one of these acid? K value is given for which one of these acids? Yeah, it's for V. K value is always given for V acid. In the question, it is given that K is for this HCOOH. It is given, but I did not write it down here. Because you should understand this K, whenever it is written, it is always for V acid. And HCL, obviously it is a strong one. So this is given for HCOOH. Here also we have 1,9, which is H plus. You see, I'll just quickly write down the solution for all of you, if you haven't got it. Yeah, HCL is a strong acid. You see the concentration of HCOOH is given 0.015. HCL is 0.02. K value, if you write for the weaker one, which will be H plus into HCOO minus divided by HCOOH. H plus concentration is given, which is 0.02. See, one is weak, other one is strong. H plus, what we assume because of the common ion effect, it comes only from HCL. So that will be the concentration of H plus here will be 0.02 into concentration of HCOO minus, which we have to find out. And HCOOH concentration is 0.015. K value is 1.8 into 10 to the power minus 4. When you solve this, you'll get concentration of HCOO minus will be equals to 1.35 into 10 to the power minus 4 molar. You can also do this by assuming alpha, x and all. Eventually, what do we have to do? We have to write down this expression. And finally, we have to neglect alpha with respect to this, with respect to this concentration value. So ultimately, you'll get this only. Like this also you can do. This also you can do directly can substitute. And this you'll get the answer. Both ways are correct. One more question. Calculate H plus concentration. Calculate H plus concentration in a 0.20 molar solution of dichloroacetic acid. Dichloroacetic acid. I want you to write down the formula of this also. Dichloroacetic acid. It's K value is given, which is 5 into 10 to the power minus 2. That also contains 0.1 molar sodium dichloroacetate. Sodium dichloroacetate neglect hydrolysis of sodium salt. 0.1. 0.18. 0.1. Okay. What about others? Now you see. See, if you calculate alpha for this dichloroacetic acid, what is the formula of dichloroacetic acid? Tell me. Formula of dichloroacetic acid. First of all, we'll calculate alpha for this from host world dilution law. I have discussed this last class. And I guess you forgot. Alpha is equals to K a by the concentration C root over of it. Right? And this K a value will be 5 into 10 to the power minus 2 divided by 0.20 root over of it. When you solve this, you'll get 2.5 root over of it, which is obviously more than alpha. This value is more than 0.05. Right? And if you remember, I have discussed this when alpha value is more than 0.05, then we'll do the what? Exact calculation. We will not neglect. If you remember this, is it root 0.25? 5 by 2, we are getting now. 5 by 2, 0.5 into 10 to the power minus 2 divided by 0.20. It's 0.25. Correct. It's 0.25. Then also it is greater than 0.05. Correct. Yeah, my calculation mistake. Sorry. Right? Then also alpha value is greater than 0.05. Correct. Okay. So we hence we have to do the exact calculation. The first question you see, the first question you see, I know you'll miss this here. This alpha value is what? It is 0.03, which is less than 0.05. Yes. Hence we have neglected alpha here. You see, hence we have neglected x or alpha here and we get the answer. That's why here in this question, in this question, you cannot neglect alpha, but you have to do the exact calculation. Did you understand? Exact calculation we have to do. We cannot neglect alpha here. Now you see when you write down the equation here, this is dichloroacetic acid. C-S-Cl2-C-O-H and it gives C-H-Cl2-C-O-O minus and H plus. And for this, it is 0.0200 initially. Now it becomes 0.02 minus 0.02 alpha, 0.02 alpha, 0.02 alpha. Next we have dichloroacetate, sodium acetate. That will be C-H-Cl2-C-O-O-NA and it gives C-H-Cl2-C-O-O minus and H plus. Its concentration is given 0.1 molar and it's a strong electrolyte 0.1 and 0.1. So if I write down the K-A expression, again K is given for weak electrolyte, which is this one. So K-A will be concentration of C-H-Cl2-C-O-O minus H plus divided by C-H-Cl2-C-O-O-H. C-H-Cl2-C-O minus concentration will be what? 0.1 plus this? 0.1 plus 0.02 alpha into H plus concentration is 0.02 alpha. This is what we have to find out. First we'll get alpha from this equation. Then 0.02 into that will be the answer. Okay, divided by C-H-Cl2-C-O-O-H, which is 0.02 into 1 minus alpha. Now this quadratic you have to solve. 5 into 10 to the power minus 2, which is 5 into 1 minus 2, which is 0.05. Or we can say 5 divided by 100. Now you solve this equation and you find out the value of alpha. Quadratic you have to solve. What is the value of alpha you are getting? 0.02, 5 by 20 root over 8. 2.5 by 10 is 0.25 root under. That will be 5 by 10, 0.5 I think we are getting. Check once, is there any calculation mistake? What is K value? K value is 5 into 10 to the power minus 2. 5 into 10 to the power minus 2 K value, right? So 5 by 100 is just a calculation. Tell me the value of alpha you are getting here. What is the value of alpha you are getting? Concentration is 0.2, correct. Okay, I have taken it wrong. Anyways, you take 0.2 only. Whatever the concentration you take that only. Okay, I have written here it is 0.02. You take 0.2. Here also it is 0.2, 0.2 and 0.2. Find out alpha and then the answer will be 0.2 into alpha. This is the value we have to find out. H plus concentration. Yes, but alpha is 0.25, correct. And hence H plus will be what? 0.5, 0.25 into 2, 0.5. The alpha value you get here is 0.25. Hence H plus concentration 0.25 that gives 0.5. By mistake I have taken 0.02. Understood, you must take care of this thing. Here only usually we make mistake. Calculation of alpha. You must check alpha whether it is less than 0.05 or greater than 0.05. Okay, you must do this calculation. Next you write down when we have di-basic acid. Next, like I told you already that in this chapter we have calculation of pH. There are a lot of questions of pH calculation. So write down pH of di-basic acid, di-basic acids and polyprotic and polyprotic acid. Which means where we have more than one H plus ion present. Suppose the acid general expression I will write down. Suppose the acid we have H2A, di-basic acid 2H plus ion it can give. Concentration is C and when it dissociates it forms HA minus plus H plus, right. So when it dissociates initially it is 0 and 0. So C minus C alpha 1, this is C alpha 1 and this is C alpha 1. Next what happens since this HA minus can also produce H plus. So when this dissociates it forms H plus plus A minus 2, right. 2H plus ion it can give. That's why it is di-basic acid. So this concentration is already we have that is C alpha 1 and when it is 0 and 0 next when it dissociates it forms C alpha 1 minus C alpha 1 alpha 2 where alpha 2 is the degree of ionization of this. Alpha 1 is the degree of ionization of this. The H plus ion that you get here will be C alpha 1 alpha 2 and A minus 2 will be C alpha 1 alpha 2. Now here one thing you must take care of that this is we have a reaction vessel or we have a beaker in which this di-basic acid H2A is present and all these reactions 1 and 2 is taking place in this vessel only, in the same vessel. So in this in one reaction HA minus is forming and in other reaction HA minus is dissociating, right. So concentration of HA minus will be same here, here and here it will be same. Here the concentration is this but here the concentration is C alpha 1 minus C alpha 1 already. So concentration of HA minus will be here also C alpha 1 minus C alpha 1 alpha 2. This is the concentration of HA minus because both reactions are in equilibrium with each other, right. From H2A first of all HA is forming and simultaneously this is also dissociating into A minus and A minus 2 and H plus. So when this equilibrium is established the concentration of HA minus is C alpha 1 minus C alpha 1 alpha 2. The same concentration of HA minus should be here also. Is it clear? So for this one for this suppose it is KA1 and for this reaction suppose the equilibrium constant is KA2. So if I write down KA1 that will be equals to concentration of HA minus, concentration of H plus divided by the concentration of H2A, right. Concentration of HA minus will be C alpha 1 if I take common 1 minus alpha 2, H plus will be C alpha 1 alpha 2. Again you see H plus is present in both reaction. So here also the concentration of H plus will be C alpha 1 alpha 2, correct. So this is again H plus in this, okay one thing, one thing. This is again, this is not there, correct, correct. Okay, so now you see in both reaction we are getting H plus concentration. Here H plus concentration is this and already C alpha 1 is present. So total concentration of H plus is what? C alpha 1 alpha 2 we have to add into this one and here also we have to add C alpha 1. So the total concentration of H plus will be C alpha 1 plus C alpha 1 alpha 2. So H plus concentration is this, I'll write down this here. This into C alpha 1 if I take common 1 plus alpha 2 bracket close divided by H2A concentration is what? C into 1 minus alpha 1. Like this we'll write down the expression of KA1, okay. Similarly KA2 will be what? Will be H plus A minus 2 divided by HA minus. H plus is again same C alpha 1 into 1 plus alpha 2. A minus 2 will be C alpha 1 alpha 2 divided by C alpha 1 into 1 minus alpha 2. Now you can also solve this. H plus the important point is what? Since we have to find out pH so we need H plus concentration. So the important thing here it is what? The H plus concentration is C alpha 1 plus C alpha 1 alpha 2. Which is nothing but nothing but C alpha 1 1 plus alpha 2. And then we can solve this and we'll get pH. Both these value will be given in the question. With the help of these two value we can calculate alpha 1 and alpha 2. And then we'll find out H plus concentration. C A1 A2 plus C A1 which one? That's what I have written, correct? No. Okay can we move on? Write down the question based on this. Find the concentration of H plus HCO3 minus and CO32 minus in A 0.01 molar solution of carbonic acid. What is the formula of carbonic acid? If the pH of the solution is 4.18 you see pH is given. K1 is given 4.45 into 10 to the power minus 7 and K2 is 4.69 into 10 to the power minus 11. Solve this one. Yes, H2CO3 is carbonic acid. Solve this one. 16 to 6 10 to the power minus 5 minus 5 minus concentration is 4.78. Strong circuit. Check your calculation. So one thing I'll give you. pH is given so you can directly find out H plus concentration. pH is 4.18. So that we can write 4 plus 0.18. And that we can write 4 plus 6 log 2 because log 2 is 3. And this will be log of 10 to the power minus 4 plus log of 2 to the power 6. Is it correct? See, excuse me. We can write this 18 as 6 into 0.3. 0.03, right? Okay, it is 0.03. Yeah, that's a mistake I made. The derivative calculation we have in this. 4 into 0.6 into 0.3. Can we write this? Now 0.3 is log 2. 4 plus 2.6 log 2. So log of 10 to the power minus 4 plus log of 2 to the power 0.6. So this becomes 2 to the power 0.6 into 10 to the power minus 4 log of the whole this. So this concentration is this 2 to the power 0.6 into 10 to the power minus 4. What is the value of this one? 2 to the power 0.6. Can you tell me? Minus log, correct. So minus log, so we'll have, let me correct this. Minus log of h plus is equals to log of 0.6 into 10 to the power minus 4. So h plus concentration will be 2 to the power minus of 0.6 into 10 to the power 4. Is it correct? Yeah, correct. When you solve this, you'll get h plus concentration is 6.61 into 10 to the power minus 5 mole per liter. However, this solution is not given, right? They have only given the answer. Ph is this, so h plus is this. We can cross check your calculation. H plus concentration should be this. Now when you have this h2SO3, h2CO3, sorry. When it dissociate, it forms hCO3 minus and h plus. And kA1 for this will be concentration of hCO3 minus into h plus divided by concentration of h2CO3. So h plus concentration you have already, h2CO3 is given, right? You can also find out hCO3 minus kA value is given. So I'll just write down the answer here. hCO3 minus answer will be 6.73 you are getting, 6.73 into 10 to the power minus 5 mole per liter. And when you have this similarly with the help of kA2 value, you can find out the concentration of CO3 minus 2. For the second reaction, CO3 minus 2 concentration, the answer you will get is 4.78 into 10 to the power minus 11 mole per liter. This is the answer. Next I write down Ph of the mixture. Ph of the mixture. I'll just explain this one with the help of one question. And it is basically, you know, when you are talking about mixture, so mixture can be anything. We can have the mixture of anything possible. We can have acid plus acid mixture. We can have acid plus base mixture. And the last we can have base plus base mixture. Anything is possible into this. Now, when you have only acid present, so both this acid will give what? H plus ion. Here acid will give H plus and the base will give OH minus. So this gives you the neutralization process, neutralization reaction. Both will neutralize each other. Base if you have only base, then it gives OH minus. So in all the three cases, you may have the question to find out the Ph of the mixture. And again, for Ph, we need H plus concentration. So if you have acid directly, you can find out H plus from there and Ph we can calculate. So what we can say when we have acid plus acid mixture, acid is mixing with acid. So there are few data will be given in this kind of question. Calculation is mad. So let it be. Remember in the exam, you will not get those kind of calculation. It's only because of like you should have the practice of all kind of questions. So it is important for you to solve those kind of questions. Understood? Yeah, actually it depends upon just a second. Depending upon the option you can solve. And in the exam, you will not get those kind of tough calculations. Don't worry with that. Okay, but you have to do some questions like that. So if you have acid plus acid, right? So two, three things are there. Either normality will be given or molarity will be given for the acid, even for base also. This thing, what I'm discussing now, it is true for all these three cases, acid base, acid, acid and base base. Normality or molarity will be given. Volume will also be given into this. Okay. So if you have molarity given for any acid, so if the acid is also given, so we should also know the n factor of whatever acid and base we have. So if molarity is given, then we can multiply this by n factor and we'll get normality, right? And we know normality into volume that gives you the number of what? Normality into volume gives the number of, what is normality into volume? Number of equivalence, okay? And we also know in moral concept two, we have discussed that when two molecules are reacting acid base, acid, acid, then equal equivalence react, right? Equal equivalence react. So suppose you have acid one and acid two, we are mixing here. So whatever the equivalence of these two that we have to add to get the resultant equivalent of the mixture or resultant normality of the mixture, right? So what we can write here in this case, both acid and base will give you H plus, right? Acid one is mixing with another acid two. So since both are acid, so both will give H plus ion. So concentration of H plus will write here. That will be N1V1 plus N2V2 divided by V1 plus V2. Volume of acid one, volume of acid two, right? This gives you H plus and then we can find out pH if you have H plus, right? Case two, if you have acid plus base, acid A and base B. So in this case, when you have acid and base reaction, then it is a neutralization process, right? And in this case, there are three, in this particular section, there are three cases possible. What we'll do? We'll just calculate the number of equivalent of acid, which is NAVA and number of equivalent of base, which is NBVB. So there are three possibilities. One is when the equivalence of acid is more than to that of base, when both are equal, when the number of equivalent of base is more than to that of acid, right? So three cases possible, excuse me, three cases possible here. When we have NAVA greater than NBVB, right? So acid is more than that of base. So resultant mixture is acidic mixture. And since the mixture is acidic, so we'll get H plus concentration here. And this H plus concentration will be what? The number of equivalence of acid that has been neutralized that will be equals to the number of equivalent of base, right? This divided by VA plus PhoB. We are mixing acid and base, right? So whatever is more that will left behind after the neutralization reaction is complete, right? And whatever is less that will completely consume or indirectly we can also say whatever is less that will be the limiting reagent, okay? So the number of equivalence of acid that is left that gives the solution acidic property and since the solution is acidic, so we'll calculate H plus and by this, okay? Similarly, third, the second case or third one I'll discuss here, suppose, when the number of equivalence of base is more than to that of acid. So in this case, acid will get consumed completely. This is the limiting reagent. Here the limiting reagent is base. And hence, since base is more, solution will be what? Solution will be basic. For basic solution, we calculate OH minus. OH minus concentration will be what? The number of equivalent of base that is left. VA plus PhoB. So when you have OH minus, from here we can calculate what? PoH. And if you have PoH, right? And then we can calculate pH because we know pH plus PoH is equals to 14 at 25 degrees Celsius, right? These kind of questions usually they ask at 25 degrees Celsius only. If they change the temperature, they'll give you KW value for that, okay? So accordingly, you can add these two and put the value here and then we can find out pH. One more case possible here it is what? When third case possible is what? When the number of equivalence of acid is equals to the number of equivalence of base. In this case, the solution will be pH will be neutral and pH is 7 in this case. At obviously 25 degrees Celsius. Okay, here we are not considering the hydrolysis of salt. Neglecting the hydrolysis of salt. We are neglecting here the hydrolysis of salt. One more very important thing here it is what? In case one and two, in these two cases, in these two cases, right? Buffer solution may form. Buffer solution forms. What is buffer solution we'll discuss, okay? And in this case, if hydrolysis of salt we are considering, then pH will calculate according to the formula of hydrolysis of salt that also we'll discuss. But if hydrolysis is not considering hydrolysis of salt, then this is the case of, okay, sorry, then pH value will be different if you're considering the hydrolysis of salt. But hydrolysis of salt is only possible when the complete neutralization process is there. So in this case, hydrolysis of salt is possible, right? In this case, buffer solution forms. So in case of hydrolysis of salt, right, there is a different formula of pH that we'll discuss. We'll see the derivation of that also. Buffer solution, we have different formula of pH, different buffer solution, how it works and all, everything is there. We'll discuss that. So all these things comes under, these hydrolysis of salt and buffer solution comes under the reaction of acid and base. That too, what is the equivalent of acid and base you are taking? So all these things, in this kind of question, you have to, first of all, you have to check what is the equivalent of acid you are taking and base you are taking. Depending upon the equivalent of acid and base, we will calculate H plus, H minus concentration according to this. And if solution is forming a buffer solution, then we'll see what is the pH calculation over there in case of buffer solution. We'll discuss that suitcase into hydrolysis. Actually, you see what happens. A little bit I'll discuss now. I thought I should discuss this in the class. See, actually what happens, this acid and base, there are again four cases possible here. Acid can be strong and weak and base can be strong and weak. So if you take the, you know, four different possibilities, strong acid is strong base, strong acid weak base, weak acid is strong base, weak acid weak acid weak base. Like this four different cases are possible. Correct? So when you see, first of all, suppose, a little bit I'll discuss now because we are discussing something else. Since you have asked, I'll just discuss a little bit here. Suppose I am taking NaOH and HCl. Equal equivalents, NaOH and HCl. This forms what? NaCl and H2O. So this salt which is forming over here, this salt may also react with H2O. However, in this case it is not possible. But if you take, because both are strong, this is a strong base, this is strong acid. And there is one more thing that the hydrolysis of a strong acid and hydrolysis of the salt of a strong acid and a strong base is not possible. We'll discuss this later on. But if you take this weak acid, you have CS3, COOH and NaOH. So this gives you CS3, COO and Na plus H2. Now this is a salt we have and this gives the iron which is CS3, COO minus and Na+. Now this is the base. This reacts from salts plus water. This salt also has tendency to dissolve into the water. That is nothing but the hydrolysis. So what kind of salt can be dissolved? What kind of iron we should have here where the dissolution is possible that we'll discuss in the physical class when we do the hydrolysis of salt. So since this is forming, reaction is forming, we are getting salt over here. This salt, depending on its lattice energy, this salt will also have tendency to get dissolved into water. That is why hydrolysis of salt we are discussing. The point is these two are having equal equivalence but when the product forms, this product when we get whatever the salt we have, this salt may dissolve in this water depending on the nature of the salt. It has nothing to do with whether the number of equivalence of acid and base is same or not. That's why we said that acid and acid forms salt plus water. Usually what we say, acid base when reactant forms, the answer is salt plus water but it is not always true. Acid and base forms salt plus water whenever the acid and base is strong. If they are not strong, any one of these is weak. In this case, any one of these is weak. The acid is weak here. In this case, the reaction may go in the reverse direction. That's why this reaction is reversible reaction but this one is not reversible reaction. This is irreversible reaction. Why? Because the cation of base and an ion of acid which are strong is not hydrolyzed. There are many things to discuss into this. We will discuss this in the hydrolysis of salt. Let's not waste our time over here. The point is why it gets dissolved? Why it gets hydrolyzed? The point is whatever the product you are getting, the reaction of these two products has nothing to do with the number of equivalents of these two. Once it has been formed, depending on its nature, it will either dissolve into water or don't dissolve. That's the reason. We were discussing these four different cases here. Buffer solution will get when we have these two cases possible and hydrolysis of salt is possible whenever we have complete mutualization. These two you must remember. Now you see the question onto this. In the mixture, see we have sample 1, sample 2. pH is given. pH is 2. pH is 3. Volume is 1 liter. And here the volume is 2 liter. You have to find out the pH of the mixture. What is this pH of the mixture? Bharat is right. It is 2.3980. Approximately 2.4. And then it's also closed. 2.39, 2.4 you can have. Yes, 2.398. So basically we have pH, so we can get H plus concentration from there. And then concentration into volume, like this we write simply, the molarity of the resultant solution will be M1V1 plus M2V2 divided by V1 plus V2. M1 is nothing but 10 to the power minus 2 H plus concentration into 1 plus 10 to the power minus 3 into 2 divided by 3. This is what you have to solve. The answer will be 2.398 exactly. Next one you see. You have to find out the pH of the resultant mixture. First one. Volume is given of 0.2 molar, this CaOH whole twice. And this reacts with 25 ml of 0.1 molar HCl. H2SO4 plus CaOH whole twice. 0.01 molar. 0.01 molar. Third one. H2SO4 plus KoH. 0.01 molar. 10 ml. 0.1 molar. First one is acidic or basic. Check your answer. The first one, solution is basic pH value 12.63. Second one, the solution is neutral and hence pH value is 7. Third one, the solution is again basic and pH is 12.6. Is it? Do I need to solve this? Okay. Next one you see this is also important. Write down relation between pKa and pKb pKa and pKb for a conjugate acid base pair. Conjugate acid base have we discussed? I think we have discussed. Yes. Okay. So you see for an acid HX suppose acid we are taking HX this acid dissociates as H plus plus X minus and the K value for this will be H plus X minus divided by HX. This is the acid and this is its conjugate base. Okay. So this conjugate base for this if I write down the reaction for the conjugate base X minus dissolved in water and it forms HX again plus OH minus releases OH minus that's why it is base. Right? Acid are those compounds which releases H plus iron in water, RNA acid releases OH minus iron in water RNA space. Right? So for this Kb will be Kb will be HX OH minus divided by I will not consider into this. Okay. This is equation one and this is equation two when we multiply one and two left hand side will get KA into Kb and right hand side will have X minus X minus will get cancelled. HX HX will get cancelled. So we'll get what? H plus into OH minus which is nothing but the ionic product of water KW and the value of KW at 25 degrees Celsius we know it is 10 to the power minus 14 and when you solve this log, if you take both side, it becomes pKa sorry pKa plus pKb is equals to 14. This is the relation of pKa and pKb. Right? Now from this relation we can conclude one very important point here. As I think we have already discussed that acidity for any acid of any acid is directly proportional to its K value and inversely proportional to pKa value. Why it is directly proportional? Because you see here, acidity is more means H plus concentration is more when H plus is more, K should be more and K is equals to and pKa is equals to what? Minus log of K so when K is more then K will be less. That's why acidity is directly proportional to K inversely proportional to pKa. This is the relation we have. And the same thing is true for base also. Basisity for any base is directly proportional to Kb and inversely proportional to pKb. Same logic. Okay? So now when the acid is strong stronger acid will have lower value of K and higher value of pKb. When pKb value is more say strong acid, lower value of pKa lower value of pKa means higher value of pKb higher value of pKb means what? Low is the basisity. And this acid base here it is this is for conjugate acid base pair. So note down one point here first point you must keep this in mind it is important when we discuss acidic and basic nature in organic chemistry also. So write down first point here stronger acid stronger acid will have weaker conjugate base means the conjugate base of strong acid is weak. Second point you write down second point higher the value of pKa of an acid higher the value of pKa of an acid higher the value of pKa of an acid lower its acidic strength higher the value of pKa of an acid lower the value of lower its sorry higher the value of pKa of an acid what I said just a second higher the value of pKa of an acid lower will be its acidic strength higher will be the basic strength of its conjugate base higher the value of pKa of an acid lower will be its acidic strength and higher will be the basic strength of its conjugate base we are talking about. Higher will be the strength of its conjugate base. Is it clear? Have you written this? Yes. All of you have written. Can we move on? Okay. Write down one more question. After this we will take a break and then we will see the buffer solution. Calculate the H plus ion. Calculate concentration of H plus ion in a solution containing 0.1 molar of form 0.1 molar Cn. For this the Ka value 1.8 into 10 to the power minus 4 and for this the Ka value is 3.3 into 10 to the power minus 4. You have to calculate the concentration of H plus ion. Solve this. Both are weak acid. 7.19.9. Most of you are getting 9.9. The answer is 7.13 into 10 to the power minus 3. 7.13 into 10 to the power minus 3. Can you see the screen? Not visible. Just a second. I think some network issues there. The video has been paused. Audio is clear. That's why you are replying. Okay. I will tell you how to do this. We will take a break and we will fix it up. Okay. And then we will come back after 15-20 minutes. See, actually just you listen to me. Okay. HCOH and HOCN. Both are weak acid. Okay. So as you know, X is the degree of ionization of HCOH and Y is the degree of ionization of HOCN. Okay. Both will give H plus ion. Correct? So from HCOH, if X is the degree of ionization, so we will get X mole of H plus. For HOCN, if Y is the degree of ionization, we will get Y mole of H plus from there. So total H plus concentration will be X plus Y in the solution. We have to find out X plus Y, means X and Y value we have to find out. Okay. Then you write down the Ka expression for both acid. Okay. Solve those equations for X and Y. Okay. X Y, add it on. And then you'll get the concentration of H plus. Have you done this one only? Yes or no? Just tell me. The answer is 7.13 into 10 to the power minus 3. Check your calculation. Have you done this only? Yeah. Check your calculation then. 7.13 into 10 to the power minus 3 is the answer. Assume the degree of ionization of HCOH is X. If degree of ionization is X for HCOH, then X mole of H plus will get from that. Okay. Similarly, if Y is the degree of ionization of HOCN, then we'll get Y mole of H plus from that. Okay. Total H plus concentration will be X plus Y. Okay. Just on the expression of Ka for both acid. Solve those equations for X and Y. Best way is to take the ratio of equation one and two. You'll get X by Y directly. Find X plus Y, that is the answer. No, no, not Ka by C. Yeah, it's true. It's right. But 7.1, 10 to the power minus 3 is correct. Yes. H plus concentration is common. That's why we are adding from HCOH and Y from HCOCN. To try this, okay, I'll just, I will try to fix it because I'm at my hometown, I mean, in my hometown. So I think there is some network problem. That's why the video is not going through. Audio you are getting, right? So we'll take a break of 15 minutes. I'll try to fix it up. Otherwise, we'll see what can be done. Okay. By the way, what is going on in the school? Chemistry. It's okay. 7.2, it's close to correct hydrocarbons. Have they finished isomerism? Isomerism also, they have finished. What about stereo isomerism? Optical, they have finished optical isomerism. We have the finished two lines on optical. Anyways, anyways, anyways. So we'll take a break now. It's 11.20. We'll start at 11.35. If the video is, I'll try to fix it up. Otherwise, I don't think we can continue. Okay. So next week, do you have any classes? Is it holiday? And if possible, can we take like one session, one class next week online? Compensate this because we are not, if video is not going through. Fine. So let it be. We'll wind up the class here only. I'll take two hours next week. Okay. Any day I'll tell you the timing, but it will be online because the school is closed and the center is also closed, right? Okay. So fine. We'll wind up the class here only. Okay. I will, yeah, I'm coming. I'll be there on Monday. Monday by evening. I'll be there. Okay. So the schedule I'll let you know next week, what time. Okay. So you all enjoy. Have a great time. Enjoy Christmas.