 Hello, good evening students, can you hear me? Yes, so where were we in the last class? I think we were doing pH calculation, right? Common eye and effect we discussed. Mixer of acid and base, have we discussed that? Mixer of strong acid and strong base. We'll wait for a couple of minutes more, they haven't joined yet, let me drop a message. Okay, so last class we had discussed, the last thing we're discussing was the pH of calculation, like calculation of pH, okay? And I've given you one question in the last, that is the calculation of pH or H plus concentration of 2.1, what is the answer for that? All of you have done, like it was this, if pH is equals to 2.1 we have, what is the H plus concentration here? No, 80 is not correct, the correct answer is 79, 80 minus one, what volume is added? If you look at the question, Anusha, what volume is added? Final volume is 80, but the answer would be V2 minus V1. Yeah, so you see if the question is given like this, suppose pH is given, and you need to find out H plus concentration. This you can find out without any anti-log value, okay? How you see, just need to know three values of log, like we know log two is equals to 0.301, log two is 0.301, log three is 0.477, log five is 0.6999, log seven is 0.84, approximation we take here, and sometimes log three is also written as 0.48 for calculation. This is also written as 0.704 calculation, okay? So you must know these values, at least the first three values must, you have to memorize them, okay? So you must know these values, at least the first three values must, you have to memorize this first three value, two, three and five, okay? Now you see this pH is given, pH is equals to 2.1, which we can easily write three minus 0.9. Why not, why not this two plus 0.1? Right, that you will understand. Why I'm writing down this you see here, if I write 3.3 minus 0.9, further we can write it as three minus three into 0.30. We know the log term of this, right? Log expression of this, three minus three into log 30, 0.30 is log two, which further we can write three minus log two to the power eight, sorry, two to the power three, okay? So this further we can write log of 10 to the power minus three. I'll take one more minus sign here. This is nothing but three minus log eight. So further we can write log of eight into minus, if I take common, eight into 10 to the power minus three. This is minus of log this. This is pH on the left hand side. So pH is minus log of H plus. So if you compare the H plus concentration equals to eight into 10 to the power minus three, okay? If I write two plus 0.1, then 0.1 we don't have the log expression for this. Hence I write here three minus 0.9, which we can further simplify this. Did you get this all of you? And then you can use M1V1 equals to M2V2 to solve that. Seven into 0.3 if you write, then it would be seven into log two. Yes, that way also you can do. That way also you can do, you can find it, okay? So basically you should know these three four values. And when you get this kind of question in the exam, you can easily do this kind of calculation by these values. If it is not possible, then log anti-log value will be given in the question. No, this two, three log value you should know, Shiddish. Log two, log three, log five, at least you have to memorize. Okay, next. So we'll calculate now pH in different, different condition. Next write down pH of the mixture of strong acid, strong base, strong acid and strong base, okay? See when you have acid and base, so obviously both will neutralize each other, right? We know the reaction of acid and base, they'll give salt plus water, okay? So equal equivalence of acid and base reacts and it forms salt plus water. So this kind of mixture that you get or solution that you get, okay? The property of that solution, whether it is acidic or basic, it depends upon, it depends upon that what is the equivalent of acid or base you are taking. Suppose I am taking the acid of normality NA and volume BA, for base it is NB and volume BB. So there are three cases possible here. Case one, if the number of equivalents of acid NA VA is greater than the number of equivalents of base. Since acid is more over here, so obviously the base is, base will be the limiting reagent in this case and we'll get consumed completely and it gives acidic solution because we have some amount of acid left, acidic solution. Okay? If you have acidic solution, then concentration of H plus would be what? It depends upon the amount of acid left, which is NA VA minus NBVB divided by, divided by VA plus VB. This is the concentration of H plus and once you know the concentration of H plus, you can find out pH as well. Any doubt on this? Okay. Case two in this, that the number of equivalent of acid is less than the number of equivalent of base. Exactly opposite here. So this acid would be the limiting reagent and the solution is basic. And when the solution is basic, we'll get OH minus concentration. It is NBVB minus NAVA divided by VA plus VB. Okay? So once you get OH concentration, you can find out pH from this, sorry, POH from this. And then we know pH plus POH equals to 14. So we can find out pH further. Third and the last case we have in this, number of equivalence of acid equals to the number of equivalence of base. So in this case, we have complete neutralization, complete neutralization, okay? And the pH of the mixture would be seven. Obviously we are assuming 25 degrees Celsius, pH is seven. Yeah. Okay. You want me to go back? Which phase? So one second. This one. Now look at this question. Calculate the pH of solution, pH of solution obtained by mixing, by mixing, excuse me, 200 ml of 0.0 molar, 01 molar, H2SO4, 300 ml of 0.05 molar NaOH solution. Try this question. What is the answer? See, we have acid in base. That is H2SO4 is mixing with NaOH. Anyone got the answer? No. Okay. Try. Is it pH or I'm getting two different answers. Okay, see what we'll do here. We have acid H2SO4, correct? It's a, it's a molarity is given. Molarity is 0.01 and volume is 200 ml. So volume, you don't have to convert into litre. Okay, you'll get the same answer. For NaOH, the volume is given. Molarity is 0.05 molar and volume is 300 ml. Okay. So since we need to find out a number of equivalents, so we need to find out normality. Normality of acid is molarity into N factor. So molarity is 0.01. N factor for H2SO4 is two. It is 0.02 normal solution. Similarly, for base, molarity into volume again, N factor here, N factor for the base is one only. 0.05 into one to 0.05 normal. Then we calculate NAVA. NAVA is equals to 0.02 into 200. It is the number of milli equivalents. So we have four milli equivalent of acid. If you talk about base NBVB, we have 0.05 into 300, that is 15 milli equivalent of base. So obviously, which one is the limiting reagent here? Limiting reagent is the acid because it is lesser, right? So it is the limiting reagent, LR. Hence the solution is basic. OH minus concentration would be 15 minus 411 divided by 200 plus 300, 500. So it is 2.2 divided by this, 0.022. OH minus concentration. What is POH for this? What is POH? You can take the help of calculator and tell me. Yeah, yeah, yeah, tell me. 1.657, okay. Is it 1.657? Yeah, 1.657 you are getting. And then pH would be what? This is the answer, any doubt? Okay, next you see. Acid base mixture we had discussed. We can also have, we can also have the mixture of acid only, right? On pH of mixture of strong acids. So only acid we are mixing here, okay? I'm taking the example of two acids here, right? HA1, HA2, two acids we are mixing, HA1, HA2. This one will have some normality, some volume, right? This one will also have some normality and some volume. So the total, because this two, suppose we have separately. Now, when you mix this two in a separate V curve, right? So here the total number of equivalents would be what? N1V1 plus N2V2. So the normality would be or H plus concentration here is N1V1 plus N2V2 divided by V1 plus V2. If you have more than two acid present, more than two acid present, then we can have one more component of equivalents, N3V3. That is the only difference. So once you get H plus concentration, you'll get pH again. Okay, now do this question. The question is to calculate pH. And what happens here? We are mixing H2SO4, we are mixing H3PO4, we are mixing HCL. The data are given over here. H2SO4 is 100 ml, 10 to the power minus 2 molar. H3O4 is 200 ml, 1.5 into 10 to the power minus 2 ml. HCL is 200 ml and 2 into 10 to the power minus 2 molar. Try this one. If required, you can take the help of calculator. Okay, I got one answer. It is less than two, I guess. Could you check your calculation once? Okay, yes. See, here also what we do. I'll write down the number of milli equivalents directly here. Number of milli equivalent. If you talk about H2SO4, the number of milli equivalents would be its molarity, and the number of milli equivalents would be the number of milli equivalents. If you talk about H2SO4, the number of milli equivalents would be its molarity into n factor, this is normality, into volume. If you talk about H3PO4, what is the n factor for H3PO4? Tell me the n factor for H3PO4. Yes, n factor is 3. Okay. If you talk about HCL, we know n factor is 1. So it is 200 into 2 into 10 to the power minus 2 into 1. This is what I got, H plus concentration. Now pH would be what? pH is minus log of H plus. So that would be 2 minus log 3. Okay. log 3 value is 0.48. 2 minus 0.48 is 1.52. I'm getting the answer. Tell me what mistake you made. What approximation, Anusha? It's that calculation. No, you see. You're getting simply 9, 2 and 4, and then we add it. Direct one. Okay, anyways, the answer is 1.52 in this. Similarly, if you mix strong base, then the solution will be basic. You'll get OH minus concentration. See next. pH of the mixture of strong base. Suppose we have two bases, I'm assuming. Exactly same. We have BOH1 and BOH2. Right. If it's normality is again the same, N1B1. Normality in volume. Here it is N2V2. When you mix the two, right? So here the number of equivalents would be N1V1 plus N2V2. Okay, N1V1 plus N2V2. Total OH minus concentration equals to N1V1 plus N2V2 by V1 plus V2. This is the formula of OH minus concentration. Then we can find out POH and next pH. Yes, you can find out. If they ask you morality, you can find out from this. We know normality. We know volume. Obviously you can find out morality easily. Okay. Fine. Now next you see, these are the very direct questions we have. Okay. The next one is more important. And for here on words, you have to analyze a bit. Like what is the condition we have that you need to understand. The condition is pH of the solution of weak acid. pH of the solution of weak acid. Okay. Suppose we have a weak acid, say CH3COOH. When it dissociates, it forms CH3COO minus and H plus. So it is what? It is the partial dissociation. It is a weak electrolyte. So it won't dissociate completely. Okay. So at time T is equals to zero. If we assume C0 and zero. So at time T is equals to T equilibrium suppose when it is maintained. It is C minus C alpha. This is C alpha. This is C alpha. Any doubt in this? Any doubt here? Tell me. Yes or no? Okay. So KA would be what? Since it is an acid. So KA stands for acid. It is a concentration of product that is CS3COO minus into H plus divided by the concentration of reactant CH3COOH. Correct. CH3COO minus. Concentrate C alpha, C alpha for H plus. So it is C square alpha square C into one minus alpha. And we get KA is equals to C alpha square. Assuming what? Assuming one minus alpha is almost equals to one. The dissociation of acid is very weak. Okay. This is what we are assuming. So from here you see because you see we need to find out as pH, right? So for pH we require H plus concentration. H plus concentration is C into alpha. This is what you need to find out. C alpha. So we have C alpha square equals to KA. So what is C square alpha square? It is C times KA. So C alpha is equals to the concentration of H plus equals to C times KA. Concentration of H plus is this. What is pH minus log of H plus minus log of C into KA into log of A into B is log A plus log B. So pH is equals to minus log C minus log KA. Right. So the other it becomes pH is equals to one second at a CL5 is equals to this. So it should be root under of I missed a root over here. Right. We should have root under of it. Okay. We'll do this again. So this would be equals to minus of half log of C into KA. Okay. So what you solve this you will get pH is equals to half of this is a formula of pH of solution of weak acid we have. Yeah, I'm going back one second. Okay. See the condition you must take care of condition is pH of the solution of weak acid. Okay. So weak acid K is the equilibrium constant for weak acid. So that K will be given. Okay. And C is the concentration of acid that you are taking. Sometimes they also ask you to find out the degree of dissociation that is alpha alpha is equals to the formula we have already seen alpha is equals to KA by C root under of it. Right. So sometimes they ask you to calculate alpha as well. Look at this question now. Calculate the pH of pH of 10 to the power minus one molar CS3 COOH weak acid whose K value is given two into 10 to the power minus five. Also calculate also calculate the pH of 10 to the power minus five molar CH3 COOH solution. Done. Okay. See one thing here when you have this expression correct. We have done one assumption over here we have taken all of you must take care of it. This is the assumption we have here. This assumption we can take under only one condition when alpha is very small in comparison to one very small but what like what should be the minimum value of alpha. Beyond which we can consider this particular assumption. Right. So you need to first of all understand this if we can have this assumption if we can take this assumption then only H plus concentration is this and this formula is correct. Right. Which means this formula this two formula is correct. Only when only when we have this assumption which is one minus alpha is almost equals to one. If this assumption is not true then this formula and this formula also not valid. Okay. So in that case what we need to do if that assumption is not true then what we need to do if assumption is not true then also we have this expression correct. No. Like we can write always k a is equals to c alpha square divided by one minus alpha. So this relation is the true always whether the assumption is there like after this only will take the assumption. After this only we assume one minus alpha equals to one so k is equals to c alpha square and we get this an entire number. So till here it is correct without any assumption. So if we are not able to take this assumption then we need to solve this quadratic to find out alpha. So we solve this quadratic will get alpha and H plus concentration is c alpha always that you see here H plus concentration is c alpha. So we get alpha and H plus concentration is c alpha always that you see here H plus concentration is c alpha. This gives you exact calculation actually in order to calculate alpha we take this approximation. If you can't take this solve this quadratic get the value of alpha substitute here find H plus and then ph this is what we need to do. Did you understand what I said just now. Yes we are assuming that we are assuming that because if you do the exact calculation it won't affect the difference affect the answer much difference is not that great no alpha is already small to get the approximate answer we can do that kind of assumption. But if you want to calculate the exact value you can you have to solve the quadratic there's no other way we are doing this assumption because when you solve quadratic and when you do this assumption the difference we are not getting that much very small difference we have there. Hence we can do this assumption but like I said assumption we can also do under a given condition otherwise we cannot take this assumption in that case we cannot use the formula of ph that we derived just now then we have to go with this formula directly c alpha you find out H plus and then ph. So what is that condition you see how to how to how do we get to know this fact that we need to take that assumption or not what you do here it's a method that we use to solve this first of all you calculate alpha which is k by c root under of it and we get this value by assuming 1 minus alpha equals to 1 only but in this fact expression if you substitute the value of k and c if alpha you are getting less than 0.1 if alpha is less than 0.1 then we can try it 1 minus alpha is almost equals to 1 we can say alpha is very small in some book they have written here 0.05 as well right that also you can keep in mind if it is less than 0.1 1 minus alpha equals to 1 means we can take the approximation then if alpha is greater than 0.01 so 0.1 in this case we can't take the assumption we can't take the assumption and we need to do the exact calculation yes understood this is very important okay when to take alpha and when when to take that assumption and when not to take the assumption now coming back to the question the question is we have CS3 COOH and it's k value is given 2 into 10 to the power minus 5 k value of CS3 COOH and first case the concentration is given 10 to the power minus 1 molar so what we'll do we'll find out alpha first alpha is equals to k by c root under so k is 2 into 10 to the minus 5 what we get here 1.414 into 10 to the power minus 2 which is 0.01 we are getting this is less than 0.1 hence we can assume 1 minus alpha almost equals to 1 so under this case all the conditions are valid we can directly use pH formula pH is equals to half of pKa minus log c okay pKa value is 4.7 minus log c is log of 10 to the power minus 1 that is 1 1 plus 4.7 is 5.7 5.7 is 2.85 by 2 okay so pH for this one would be 2.85 this is the answer of the first part of this question tell me any doubt in this pKa we haven't discussed I told you know like pH is minus log of h plus so pKa is minus log of k this is the formula we have any doubt in this no okay now the second set of data we have k is same only k a is 2 into 10 to the power minus 5 but concentration is 10 to the power minus 5 it is given okay so k alpha is equals to root under k by c once again root under k by c which we can easily find out 2 into 10 to the power minus 5 so obviously when you solve this you will see that the value of alpha is greater than 0.1 and hence we cannot neglect alpha we cannot write 1 minus alpha equals to 1 this we cannot write then what to do in this process in this case we need to write down the expression of k a which is c alpha square by 1 minus alpha then we substitute the value of k a c and find out alpha 2 into 10 to the power minus 5 c is 10 to the power minus 5 alpha square by 1 minus alpha 10 to the power minus 5 and minus 5 will get cancelled so alpha square plus 2 alpha minus 2 equals to 0 solve this value and solve this quality and tell me the value of alpha here yes what is the value of alpha you got 0.414 2 will get cancelled and alpha is equals to 1.732 minus 1.732 I am getting isn't it 0.732 I am getting okay 0.732 we have then h plus concentration because for ph we require h plus concentration it is c into alpha c value is 10 to the power minus 5 into 0.732 so ph is equals to minus log of h plus this would be minus 5 minus log of 0.732 tell me the value of ph in this yes what is the answer yes tell me what is the value of log of 0.732 log of 0.732 is less than 1 it is negative of 0.135 and hence the ph value is 5.135 this is the answer we have understood so you understood what we need to do first of all you calculate alpha by k by c root under if the alpha value is coming out to be lesser than 0.1 then we can assume that 1 minus alpha equals to 1 and the formula of ph is correct in that case alpha is degree of dissociation situation like chemical equilibrium we discussed no alpha is the degree of dissociation out of one mole how many moles reacts ph cannot be negative Anusha negative ph if you take the log of h plus is positive negative concentration does not mean anything that's why you see when we solve this quadratic we are taking this positive value negative we are not taking we are ignoring this because negative concentration does not mean anything there is no any significance for that okay so you won't get any negative concentration condition you must take care of this is what we discussed about weak acid solution okay weak acid solution okay next sorry okay next condition you see we are considering dissociation of water we are considering