 Sir, we're not able to chat with you. It says Gayathri. Yeah, I know because of the network issue, I left the meeting accidentally. So I joined now. I will be the host now. So I think you can not just ask me. Just let me check one just a second. No, I'm fine. I am the host now so you can chat with me. Sir, can you just send me a message? Just a second. Can you see this now? Yeah, it's fine with all of you. Yes, so last class we had discussed the levelling effect of water. Okay, and I asked you to find out the H plus concentration if pH is given as 2.1. Okay, so almost all of you have tried. So the last question that we were discussing is, suppose the pH value is given, right, I've given you this value 2.1. What is the concentration of H plus? Okay, concentration of H plus you need to find out. This said last class also that few know log values you have to memorize like log 2, log 3, log 5, at least these three. I'll give you all these value here. The value of log 2 is 0.301. 0.301. Log 3 value is 0.477 log 5 is 0.6999. And log 7 is 0.845. This is the exact value we have. For calculation, sometimes we also assume this to be 0.48 or this is almost equals to 0.704 calculation. We can do that assumption. Okay, now based on this, how can we get the value of H plus you see. We have this pH value 2.1, which we can write 3 minus 0.9. 3 minus 0.9. 0.9 we can write as this a second. Hello. Yeah, sorry, you got a call. Yeah, so 0.9 we can write 3 into 0.30. Why this I have written because 0.30 the value we have it is log 2. Okay, so we can further write this as 3 minus 3 log 2, which again we can write 3 as in minus of log 10 to the power minus 3 is 3 only minus 3 log 2 we can write log 2 to the power 3, which is 8. If you take this minus out, this becomes minus of log 8 into 10 to the power minus 3. Okay, so pH, we know this pH is equals to minus of log 8 into 10 to the power minus 3 and pH we know it is minus log of H plus. So if you compare the two, obviously the H plus concentration we are getting here is 8 into 10 to the power minus 3. This is the answer we have. Clear any doubt in this? Yes. So like this you can solve. You can also write this as 2 plus 0.1, but we are not writing it as 2 plus 0.1 because we do not have 0.1 as any log value here. 7 into, how it is 7 you are getting? Yeah, there are many different ways we can write. Just one of the method is this, the number 2.1 can be represented in many different ways. Yeah, that's why 3 minus this if you do that it is 2.155 since we have 2.1 over here. So approximation we can take like this 2.10 will have over there. Actually in this way you know you should understand that how to proceed. Correct. So we have 10 to the power minus 3 here in the option it is given 10 to the power minus 3. It means you need to write 3 plus or 3 minus something. One of the way is this. It's not like you cannot solve this by other means. You can do that. One of the way is this. So we were discussing calculation of pH. Right. So there are different, different, you know, conditions under which we need to calculate the pH of the solution of the mixture. Correct. What you need to do, first of all, for the calculation of pH you just require the concentration of H plus, but how to get this concentration of H plus in different different conditions that is important. The basic formula is this only you need to have the concentration of H plus substituted here you'll get the answer then. So the first or the next condition we have right down calculation of pH or pH of pH of the mixture of strong acid and strong base strong acid and strong base. So there are three different conditions we have over here. Right. So what is happening here, acid we take base will mix the two. Obviously acid and base will neutralize each other. Right. So it is a neutralization reaction. The solution that you get from this when you mix acid and base. The solution could be acidic, could be basic, could be neutral, all three possibilities are there. So under what condition the solution is acidic that depends upon what amount of acid or base we are taking. If acid is in excess, right, then the solution would be acidic. If base in excess solution would be basic. If both are exactly equal, then the solution would be neutral. Okay. So the case one what we have. We are assuming case one here in which the number of equivalents of acid and a VA is the number of equivalents normality into volume. If it is greater than the number of equivalent of base. If this is given, they will give you the number of equivalents like normality or molarity will be given like normality of acid and volume would be given if normality is not given, then molarity will be given. And we know normality is equals to molarity into n factor. Okay, so accordingly you can calculate and the VB is the normality and volume of the base we have. So when we have this, if you mix the two, depending upon the number of equivalents of acid and base we can have the property of the solution. I am assuming here that the case one is the number of equivalents of acid is greater than the number of equivalents of base. And under this condition, this base will be the limiting reagent. This is the base will be the limiting reagent and when base is the limiting reagent, then the solution is what? The solution is acidic in nature. Mixer will be acidic because we have acid left. And when the mixture is acidic, then we'll get H plus concentration. We'll get H plus concentration as the number of equivalents of acid that is NA VA and some part of it gets neutralized. We know the number of equivalents which gets neutralized is equals to the number of equivalents of base because we know equal equivalents react. The equivalents of base that we have equal equivalents of acid will react. This amount of H plus you'll get divided by the total volume VA plus VB. So this is the formula of H plus and that is what we need to know. We need to have the formula of H plus the value of H plus concentration. And then we can substitute this in pH formula you'll get the pH of the solution. Similarly, we can have the second case, I'm sorry, case two. Case two, I am assuming the number of equivalents of acid is less than the number of equivalents of base. So here the condition is exactly opposite. Hence the result is also exactly opposite. This NA VA acid is the limiting reagent here. And when acid is the limiting reagent, then solution is basic in nature. And when the solution is basic, then we'll get OH minus concentration, not H plus OH minus equals to the base which is left. That is the number of equivalents of base minus some part of it will get neutralized by acid by acid divided by the total volume VA plus VB. Once you know OH minus concentration, you can further calculate POH, right, is equals to minus log of OH minus concentration. Once you get POH, you can find out pH is equals to 14 minus POH provided the condition or the temperature is 25 degree Celsius. This is what you need to do in this one. Clear? Now, the third case is, yeah, that also you can do. You can directly divide by 10 to the power minus 14. You'll get H plus concentration and then pH. That also you can do. No problem. Okay. Next case three is when the number of equivalents of acid equals to the number of equivalent of base. In this case, it is the condition of complete neutralization case of complete neutralization. And in this case, pH would be seven. No need to calculate this neutral solution, pH is seven. Got it? Try this question. Calculate the pH of solution. And the solution is we have of H2SO4 and NaOH. H2SO4, it is given 0.01 molar and volume is also given over here. Volume is 200 ml of this. For NaOH, it is 0.05 molar and volume is 300 ml. This is mixed. You need to find out the pH of the solution. Try this. Yes. Done? Yes. So, when you mix this acid and base, what you will get, you see, first of all, you find out the number of equivalents of acid and base. So, molarity is given and volume is given. So, the number of milli equivalents are right. Since the volume is given in ml. So, number of milli equivalents also you can calculate. Not a problem. That is molarity. Number of milli equivalents, what do you want? We can write down here. Normality into volume. Normality is what molarity into n factor of H2SO4 is what? Could you tell me? What is the n factor of H2SO4? 2. So, 0.01 into 2 into volume is 200. So, this is the number of milli equivalent of acid. Number of milli equivalent of base would be its molarity into n factor is 1 only NaOH into 300. I am writing down here as milli equivalent because volume is in ml. So, if you don't write milli equivalent, also the answer won't change. But technically, number of equivalent we cannot write here. That would be wrong. This is the milli equivalent of base. If you solve this, you will get 15 here. 15 milli equivalent of base we have. If you solve this, you will get 4 milli equivalent of acid. So, obviously the one which is less is the limiting reagent. So, acid here is the limiting reagent and this milli equivalent will get neutralized. So, solution would be basic. OH minus concentration would be what? Is equals to 15 minus 4 divided by 300 plus 200 500. It is 11 divided by 500 OH minus concentration. So, when you solve this, you will get 11 by 5 is 2.2 into 10 to the power minus 2. Or we can also write this as 22 into 10 to the power minus 3. So, this is what? OH minus. If you find out TOH here minus log of 22 into 10 to the power minus 3, that would be 3 minus log 22. Log 22 value is 1.3 approximately. Hence, POH is 1.7. PH is 14 minus POH. That is 12.3. This is the answer we have for this question. Done. All of you understood this? This calculation you need to be a bit careful. You have to be in calculation part. Usually what happens? We calculate POH and we answer like this only. PH is this in hurry. I am getting 12.3. No, it is not 11.3. So, for this kind of question, you first calculate the number of equivalents of acid and base, find out which one is more, which one is less. Accordingly, you can get the property of solution, acidic and basic and then H plus or OH minus concentration you can find. Okay. Now, what happens if you have a mixture of acid? PH of mixture of acid. Suppose we are mixing two acid here. You can have more than two also, right? You can have more than two. I am assuming two here. So, suppose we have two acid HA1 and HA2 we are mixing. So, HA1 will have its own normality and volume. If not, then molarity will be given. Accordingly, you can find out normality. Similarly, this one will have its own normality into volume. So, when you mix these two into a container, right? Then in this container, the total number of equivalents would be what? In this container, the total number of equivalents would be the number of equivalents of these two acids only, number of equivalents of acid one, number of equivalents of acid two. And that would be N1V1 plus N2V2. This is the number of equivalents. So, further we can write, since we are mixing only acid, the total number of equivalents H plus concentration is N1V1 plus N2V2 divided by V1 plus V2, total volume. This is the formula of H plus and then further we can find out pH from this formula. Look at this question on this. You need to calculate the pH of solution. pH of solution. We are mixing three acids here. H2SO4 we are mixing. We are mixing H3PO4 and we are mixing HCl. Three acid we are mixing. For H2SO4, we have 200 ml, 10 to the power minus two molar solution, H2SO4. H3PO4, we have 200 ml and 1.5 into 10 to the power minus two molar. For HCl it is 200 ml into 2 into 10 to the power minus two molar. This is the data given. You need to find out the pH of the solution. Try this once. Then what is the answer? 1.52. Yeah, that's 1.52 is correct. Absolutely correct. Okay. So, first of all, the number of equivalents, we'll find out the number of milli-equivalents of this. Number of milli-equivalents would be volume into molarity into N factor. And for this one again, volume into molarity into N factor is 3. Here we have volume into molarity into N factor. So, number of equivalents here, it would be 4 milli-equivalent. Here it is 3 into 3, 9 milli-equivalents and here it is 2 into 2, 4 milli-equivalents. Okay. So, total number of equivalents would be this plus this plus this divided by total volume. So, H plus concentration equals to 17 divided by total volume, that is 200 plus 200 plus 200. We have a 600 ml here. 0.83 into 10 to the power minus two. I think slight change in the answer, actually 100 problem method is this only. H would be 2 minus log of, yes, what is the value you got? Yes, 1.55 approximately you are getting, that's fine. So, 1.55 is the pH value we have. Simple one, for acid, we just need to add the number of equivalents divided by total volume. Next is mixture of acid we got, mixture of bases, third one we write down, pH of the mixture of base. What is the value of log 2.83? Could you tell me? Log of 2.83, 0.45. So, it is 2 minus 0.45, correct. 1.55 approximately the answer is, okay, 1.55 is the answer. pH of the mixture of base again is very simple. Suppose we have two bases, BoH1 and BoH2. If you are mixing the two base, then obviously we will get the concentration of OH minus here, right. So, in this case we will have the total number of equivalents here equals to N1V1 plus N2V2 assuming the normality and volume of base 1 is N1V1 and this is N2V2. So, OH minus concentration would be N1V1 plus N2V2 divided by V1 plus V2, okay. Simple, like we did one question on base acid, similarly you can do it on base. Remember we need to find out pH, OH minus concentration we get, then you can find out H plus concentration and then you can find out pH. Okay, now the next one is pH of solution of weak acid, of weak acid, okay. So, weak acid we have, suppose I'm assuming an acid CS3 COOH, a weak acid we have, we know weak acid are the weak electrolytes, it won't dissociate completely, so it will be CS3 COO minus and H plus, okay. So, I'm assuming the initial concentration of acid is C, this is 0, this is 0, where it dissociates, if the degree of dissociation I'm assuming alpha to attain equilibrium state, so it is C minus C alpha, this is C alpha and this is C alpha. Ka is equals to concentration of product that is CS3 COO minus H plus CH3 COOH, this is the formula we have, which further we can write C alpha square by 1 minus alpha, this is what we can do. Now, if you're assuming alpha is very small, so we can write 1 minus alpha almost equals to 1, okay, hence Ka is equals to C alpha square, again you focus on this, we need to find out H plus concentration, that is C alpha we need to find out. Once you know this H plus, we can find out pH, okay, to get C alpha what we do, we'll multiply both side by C here, so C times Ka is equals to C square alpha square, so C alpha is equals to root under C Ka, which is nothing but the H plus concentration, copy this down, okay. So H plus concentration we got, now when you substitute this in order to find out pH, pH is equals to minus log of H plus, so H plus concentration is minus log of C Ka to the power 1 by 2, which is further equals to minus half log C plus log Ka. Minus sign if you take in, then it becomes half minus log Ka is P Ka minus log C, this is pH, formula of pH under this condition, yeah, this is again I'll go back, make sure of base, okay, okay. Now one more very important thing here, copy this down first, done this all of you, tell me, finished now, okay, now this is the formula of pH, but you see here, listen to me very carefully all of you, the concentration of H plus we get, once we have this assumption isn't it, we are assuming this that we do in most of the cases, but this formula is based upon this assumption, which we cannot always make. If you can make this assumption, then the formula of pH is correct, if you cannot make this assumption, then the only way is what we know this H plus concentration C alpha you need to find out C alpha and then you can get the answer. But the formula that I have written pH is equals to half of P Ka minus log C, this we can only use when we can make this assumption, if 1 minus alpha is almost equals to 1, this is the condition we have then we can use this. If we cannot, if you do not have 1 minus alpha equals to 1, then we know the concentration of H plus, I'll write down here, concentration of H plus equals to C alpha, and this is true in all condition. This is true in all condition, this we need to use this particular formula in order to find out pH, correct. The difference is what you see here, we have this expression, the expression is what? See the expression is Ka is equals to C alpha square by 1 minus alpha, we are having this assumption so that we can solve this expression easily. And to a certain extent, this assumption if you take, you will get a certain answer. If you don't take this assumption, then you have to solve the quadratic for alpha, and then you will get C alpha again. So in that way, the two answer that you are getting the difference is not that great. Hence we can always take this assumption, but like I said, we have a condition under which we can take this assumption, otherwise we cannot take this assumption. What is that condition I'll tell you? What you need to do? You need to find out alpha from this formula, alpha is equals to Ka by C, you just calculate the formula of alpha with this only, Ka is equals to C alpha square. So whatever question is given, you find out alpha with this formula. If the value of alpha you are getting is less than 0.1, is less than 0.1, then under this condition we can have this assumption 1 minus alpha equals to 1, right. But if the value of alpha you are getting greater than equal to 0.1, right, then we can say 1 minus alpha, we cannot assume it to be 1. In this case, you have to do the exact calculation. Exact calculation means what? You have to solve this quadratic, get the value of alpha, find C alpha, and then H plus concentration and then Ph, okay. In some book, this value they have written 0.05 as well. That also you must take care of. So this is the condition of alpha we have, under which we can take the approximation. Yes, clear understood all of you. Now we'll see this question you'll understand. First of all, all of you copy down this. Usually we use 0.1 also, 0.05 also you can take. Like, assume 0.05 only that's fine. You'll get questions in which the value won't lie in this range, because anyone you can take in mind. In some book they write 0.1, in some book they write 0.05. 0.05 you can take, not a problem. But keep this in mind, okay. Next, look at this question. Question is to calculate, calculate the Ph of 10 to the power minus 1 molar CH3 COOH solution if K value is given, K for CH3 COOH is given 2 into 10 to the power minus 5. Also calculate Ph of 10 to the power minus 5 molar CH3 COOH solution. Try this. Find out alpha first, you check whether you can neglect alpha or not depending upon its value and then use the formula accordingly. 2.85 is correct. The second one is not correct. Second one you cross check your calculation. The first one the answer is 2.85. Okay. See the first question is the first part of this question. We have concentration given that is 10 to the power minus 1 molar. K is same for both, right. So alpha is equals to what? We'll find out alpha, K by C root under. That would be 2 into 10 to the power minus 5 divided by 10 to the power minus 1 root under of it. So it is we are getting 10 to the power minus 2 into 1.414. So obviously this value is lesser than 0.05 we can say or 0.1 we can say, right. If this is there, then we can assume 1 minus alpha is almost equals to 1. And for this, the concentration of H plus we have, it is C into alpha, alpha value we have calculated. Right. So it is, or directly we can use the formula here. Both way you can solve that sort of problem. So pH is equals to half of pKa minus log C, half of pKa value is what? pKa of this is 5 minus log 2, 4.7 minus minus plus 1, 5.7 by 2. It is 2.85 the pH of the first part of this question. Okay. Obviously in the second one, if it is concentration is 10 to the power minus 5 here, this and this will get cancelled. Alpha we are getting more than one. Right. That value is to use. It means we cannot neglect alpha with respect to one. Correct. So in the second case, what we can write concentration is 10 to the power minus 5 and hence alpha value is greater than 0.1 or 0.05. Okay. So we can write H plus concentration here is C into alpha. So we need to find out the value of alpha because concentration we know 10 to the power minus 5. That formula we cannot use over here. Keep that in mind. To find out alpha what we can write kA is equals to C alpha square by 1 minus alpha. We need to solve the quadratic like I said. Yeah, 5.14 is correct I guess. Okay. So C alpha square is 10 to the power minus 5 alpha square divided by 1 minus alpha is equals to 2 into 10 to the power minus 5. So when you solve this quadratic, you will get alpha square plus 2 alpha minus 2 is equals to 0 alpha is equals to minus 2 plus minus 4 minus minus plus 4 root over divided by 2. So it is 12, 12 is 2 root 3, 2 and 2 will get cancelled. So minus 1 plus root 3 and we are getting 0.732, the alpha value. This alpha will substitute here and hence the H plus concentration would be 0.732 into 10 to the power minus 5. Ph is equals to minus log of H plus. So 5 minus log of 0.732. 0.732, the value is approximately 0.14 or 0.135. So it is minus of this, it is 5.135 or 5.14. Approximately the answer is because log of 0.732 is minus of 0.135 log value is minus 0.3135 log less than one is negative log value will be given. If you cannot find out with those 34 values of log, other values will be given in the question. You don't have to worry about it. Because you cannot memorize all the log values. So if it is required, it will be given in the question. If not, then you can solve it by using some log property. Clear any doubt?