 Next condition you see pH of pH of a weak acid solution. If dissociation of water is considered all these formula you must keep this condition in mind. Okay, weak acid solution if dissociation of water is considered. Okay, suppose we have a weak acid HA and this gives. H plus and a minus water also dissociate gives H plus and OH minus we need to find out pH. Okay, logically also you can think see we have two source of H plus here one is this acid other one is water. So you need to find out total H plus concentration then only we can use then we can use the formula of pH. So how do we find out the total H plus concentration we can find out this by charge balance by charge balance but we can write total positive charge equals to the total negative charge a minus and OH minus. Total charge we are taking H plus and minus minus. Now for acid for acid k equals to H plus a minus HA. Remember we need to find out H plus what we are trying to do we are trying to substitute this a minus in terms of H plus because the concentration of this acid will be given. K value will be given in the question. Okay, correct. A minus we are trying to substitute in terms of H plus. Similarly, OH minus also will substitute in terms of H plus and we'll find out H plus there. This acid concentration will be given K value will be given in the question. So this if you find out the concentration of a minus that would be K a into H a divided by H plus concentration. This is one copy this down I'll go to the next slide. I have written this by charge balance total positive charge equals to total negative charge. I'm explaining this we have two source of H plus HA and H to acid and water total concentration of H plus equals to total concentration of negative charge a minus and O H minus because overall it should be neutral. A minus we calculated in terms of H plus. Now we'll find out OH minus in terms of H plus so for that. We have the expression of KW KW is equals to concentration of H plus OH minus. So concentration of OH minus would be KW by H plus concentration. So we have OH minus we have a minus now we substitute this two in the charge balance equation. What we get here we get H plus concentration equals to a minus would be K a into H a divided by H plus plus OH minus would be KW by H plus. So H plus concentration is square equals to K a into H a plus KW. Okay. This H a like I said, its concentration will be given in the question. See, KW value we know it is 10 to the power minus 14. Okay. But this value we won't substitute. I just I have just written it over here. So H plus concentration would be root over of K a into C plus KW. This is the formula we get all the value will be given. You just need to substitute all the value in this formula. You will get H plus concentration and then we can find out pH. Yeah, done. Okay. Now we have only one acid in water in this case. You can also have more than one acid multiple acids present. Okay. So suppose if you have, if two acids are present in the solution, then the formula is exactly similar. H plus concentration would be the root number of the first acid K a value K a one into concentration of that C one plus for the second one K two C two and KW. If you have three then one more component of this will have over there. Nothing much. Look at this question. Suppose we are mixing two acid in water. We are mixing two acid in water and dissociation of water we are considering here the K a one value of acid one is given 10 to the power minus eight C one is 10 to the power minus two molar. A two is 10 to the power minus three minus seven and C two is 10 to the power minus three molar. You need to find out the pH of this mixture. Try this. Done. Is it 17. It will be given on it in the question it will be mentioned that water is also dissociating like this it is mentioned. So like in thermodynamics, we are trying to discuss all possible conditions. And in a given condition what could be the possibility to find out H plus concentration and then pH. Okay, so in the question it will be mentioned that H plus concentration or water dissociation you have to consider or not. You're getting 17 guy three 17. 10 to the minus 14 only you will take but 17 is not possible is acid though. So it's a acid solution the pH value should be less than seven correct. Okay, so just we need to use the formula here pH equals to K a one into C one that is 10 to the power minus 10 K a two into C to 10 to the power minus 10 plus 10 to the power minus 14 root over of it. So minus 10 if you take common, then outside the root, it would be 10 to the power minus five divided by two plus 10 to the power minus four. So obviously 10 to the power minus four is too less in comparison to two. So we'll just ignore this at approximation also very important in this chapter okay you should know that how we can approximate the given expression. So it is. Yeah, it is the concentration of H plus my bad. Yes, it is H plus concentration. Okay, so it is root two into this. I'll write down as it is root two into 10 to the power minus five. Okay, so pH would be what pH is equals to minus log of H plus root two into 10 to the minus five. So five minus log of root two is half of log two log two is point three zero point one point point one five so five minus 0.15 it is 4.85 the answer. Okay, similar kind of formula we have when we have base and dissociation of water. So in this, we have a mixture of base. We're considering the dissociation of water here also. So in this case the concentration of OH matter I'm just giving the formula similar way we can derive this. We have KB1 C1 plus KB2 C2 plus KW root over of it. Now, so we are done first part of this, you know, chapter in which we have a mixture, which we forms by the mixture of acid base, water, everything we're done. Now we have to understand. Again, like I said, this chapter is all about the calculation of pH, so different different types of solution we may get like two very important concept we have here that is salt hydrolysis and the buffer solution. Okay. So first we are going to discuss next concept here that is salt hydrolysis heading all of you write down. Salt hydrolysis. What do you understand by this term salt hydrolysis. What is salt hydrolysis salt hydrolysis means. Yes, water and salt dissolution of salt in water. That is nothing but hydrolysis hydrolysis nothing but dissolution of any compound into water that is hydrolysis. Salt hydrolysis is reaction of salt and water. So basically if you look at this reaction acid reacts with base forms salt and water. What is this reaction, could you tell me reaction of acid and base is what it is neutralization reaction. It is neutralizing each other. Right, so it is neutralization reaction, but says that reaction is reversible. This reaction can also go into backward direction. Once it goes into backward direction. If it goes into backward direction, then this reaction we call it as what this is hydrolysis of salt, salt is reacting with water. So this is hydrolysis of salt. Tell me, is it clear to all of you, right. Clear. Okay, one very important point I'm going to tell you here which you will understand this clearly why this point I'm giving you, once we start the buffer solution. Okay, so for here you just keep this thing as it is in your mind. Okay, logic you will understand why we have this condition, once we discuss the buffer solution. Here you hear that for salt hydrolysis, the number of equivalents acid and base must be equal, must be equal, right. Why I'm giving you this condition that you will understand once we discuss buffer solution. Here you just keep this in mind. The number of equivalents must be same if it is not same, then that case will be a different case. We cannot use the formula of salt hydrolysis. Okay, just to keep this in mind. So whenever you get this question, right, I'll tell you one thing here suppose you have acid and base previously if you see we are discussing acid base only acid when mixed with base. We have acid base reaction. But the concept is different terms are also different here salt hydrolysis we are using. So when you are going to use acid base mixture or pH formula, or when you are going to use salt hydrolysis that you must understand. Okay, when you do not have equal equivalents of this. Correct. Then the solution, maybe what maybe acidic or basic if acid is more than the solution is acidic basis more than the solution is basic. Okay, if it is a basic solution, then it will be OH minus concentration equals to what number of equivalents of base minus number of equivalents of acid divided by total volume. Similarly for acid also we can take. Right, but acid and base when they are reacting. Okay, they forms saltless water. If you have a few amount of salt present with a given acid, it forms a different types of solution it has a different behavior. So the solution what is that we will discuss that later. So to avoid that situation salt plus acid situation if you want to avoid for that, you know, condition, you must have this acid gets, you know, acid reacts completely into the reaction. So once the reaction proceeds or completes, then there is no asset or base present into that solution, then only we can use salt plus water the reaction which is salt hydrolysis. If any one of these two present, then the reaction, you know, the solution has different property mixture of different property, and it has it is different condition we need to apply. The hydrolysis of salt, we apply only when either this or this not present in the solution, then only we can apply. So complete neutralization we need to think number of equivalents of acid and base must be first thing is that. Now, what is salt hydrolysis coming back to this point salt hydrolysis first point in this you write down it is the, it is the opposite of, or the reverse of neutralization process. It is the opposite of neutralization process or reverse of neutralization process. Okay. Next thing you see, when acid and base reacts it for salts and water, but we have four different conditions possible here. Like, we can take strong acid is strong base, we can take weak acid is strong base, we can take a strong acid week base, we can take weak acid week base. So there are four possibilities we have depending upon the property or the nature of acid and base that you are taking. For all these four property, we need to understand the salt hydrolysis and the formula of pH in each cases. So, one by one will discuss all the four over here first condition all of you write down. Hydrolysis of hydrolysis of the salt of a strong acid SA stands for a strong acid SB stands for a strong base hydrolysis of the salt of strong acid and strong base. I'll take one example. If you get all these things in the exam, you can also think of like this only assume one is strong acid is strong base for example I'm assuming at CL strong acid, NaOH strong base, what it forms. It forms salt and water. It forms NaCl and H2O. Okay, one second, one second, I forgot. Did you copy this all of you? Did you copy this all of you? Okay, let it be that. Just one more thing I just discussed over here and then we'll again come back to this. Suppose you have in general I am discussing this then again we'll come back to this. In general, we have an acid suppose HA and base is B over H when this two reacts it forms BA the salt and water H2O. In this BA you see there are two ions present that is B plus and A minus. So hydrolysis of salt means what? The reaction of these ion with water. The reaction of these ion with water. Suppose if B plus reacts with water then what happens? Okay, you see here, the salt has this an ion and this cation part. Okay guys, there's some construction some drilling work is going on. Can we take a break now? I think it will get over in 10-15 minutes. Can you hear me? Yes, okay, fine. So we'll, I think it's, we'll continue. If it happens again then we'll take a break. Just a second, we'll continue. If it happens again then we'll take a break. Okay, so what I was talking about, I was talking about the salt. It has two parts. One is cation part, other one is anion part. Okay, cation and anion part. So both this part get hydrolyzed in water depending upon some condition that condition we'll discuss. If the hydrolysis of cation is taking place then it releases H plus ion in the solution and this gives acidic property to the solution, isn't it? Is it clear? What I'm telling you, cation and anion part may get hydrolyzed like this. If the cation part is getting hydrolyzed, it gives acidic, acidic property to the solution. One second, Praku. If the cation part is getting hydrolyzed, it gives acidic property to the solution. If anion part is getting hydrolyzed, we'll get basic property to the solution. Correct? So point I'm trying to make that whenever you have this hydrolysis of salt, the solution could be either acidic or basic or neutral also possible. Correct? Because it releases either H plus or OH minus ion in the solution, correct? Just one thing we'll discuss here first. When you have this a strong acid and a strong base, right? One property you have to keep in mind. We'll finish this just a second. HCl and NaOH, this NaCl gives two ions like I said, one is cation or the one is anion, Na plus and Cl minus, correct? Now this cation is the cation of strong base, strong base, and this is the anion of strong acid, cation of strong base and anion of strong acid. Correct? Since it is the strong base and strong acid, so it exists in the form of ions only. Like you see, if I'm assuming this reaction Na plus dissolved in water, it gives NaOH and H plus. But still, this exists as Na plus and OH minus only since it is a strong electrolyte. So this reaction has no significance actually and it goes in backward direction only because it exists as Na plus. Hence we say the hydrolysis of cation of strong base is not possible, right? Similarly, if you talk about Cl minus plus H2O, it gives you the same kind of reaction HCl plus OH minus. HCl also a strong acid, so here also it exists as H plus and Cl minus. Hence we say what we conclude this and we say what that hydrolysis of cation of strong base of a strong base and anion of strong acid is not possible, is not possible. The solution is neutral, right? The solution is neutral and its pH value is 7. Did you understand this? Hydrolysis of salt of strong acid in strong base is neutral in nature and its pH value is 7. You must keep this in mind that hydrolysis of cation of strong base and anion of strong acid is not possible. Because it exists in the form of ions only because these two are strong electrolyte. So Na plus and Cl minus, weak acid and weak base, we generally don't define, but yes, if you look at this reaction, Na plus has very least tendency to release H plus ion. Hence it is a very, very, very weak base acid, sorry. Right? These two, a condition you must keep in mind hydrolysis of cation of strong base and anion of strong acid is not possible. Correct? Got it. Fine. We'll take a break now. Okay. We'll resume at 6. Okay. 6 o'clock we'll resume. Take a break. We'll continue from this. Yes, can you hear me guys? So we were talking about hydrolysis of salt. Salt contains cations and anions. So depending upon the nature of cations and anions, either like both can get hydrolyzed or none of them will get hydrolyzed. So all four possibilities are there. Right? If hydrolysis of both is not possible, that is there in the case of strong acid and strong base, and the solution will be neutral pH value is 7. Okay. And since you see this hydrolysis is not possible here, means backward reaction is not possible. And that's why I have given this irreversible sign. It's not reversible, right? Irreversible sign. Anyways, second case you see hydrolysis of the salt of weak acid and strong base. Okay. Any weak acid you can assume. I'm assuming CS3 COOH, CS3 COOH, and a strong base suppose NaOH we have, and it forms CS3 COOH Na and water H2O. This is the reaction. One of the ions we have here in the salt is CS3 COOH minus Na+. You see Na+, again it is the cation of strong base, so it's hydrolysis is not possible. Okay. So we can say for Na+, no hydrolysis. This one will go under hydrolysis, reverse reaction is possible here. The reaction if you see hydrolyzed reaction here, CS3 COOH minus dissolved in water forms CS3 COOH Na, sorry OH, plus OH minus. Concentration of this salt here, we're assuming it as C initially. Like when the reaction proceeds, you'll get this concentration of the salt, which gives the same concentration of CS3 COOH minus goes under hydrolysis forms this. So initially it is a zero and zero. And when the reaction takes place, it forms C minus CH. This is CH and this is CH. Here H is similar to alpha that we had. For hydrolysis, we are considering H over here. It is called as degree of hydrolysis. Alpha is degree of dissociation. Since the process is hydrolysis, we are assuming H here. Clear? Copy? Okay. Now, since it is the hydrolysis reaction, so it's constant for this reaction like we have Ka, Kb previously, Kc. Similarly, for hydrolysis, we define it as Kh. So Kh equals to CS3 COOH OH minus divided by CH3 COOH minus. You can write this. Okay. Now, the concentration of this will substitute here. CS3 COOH is CH. This is also CH. It is C into one minus H. Once again, I'll go back. So when you solve this, C and C will get cancelled. Kh equals to CH square divided by one minus H. And again, if H is very small, so one minus H is almost equals to one. And hence, H equals to Kh by C root over of it. This is one thing. This is mathematical derivation we have here. It's not difficult to understand. Now, this relation, you let it be at one place. This one, you just let it be here. Now, consider this one here. We have Kh equals to concentration of CH3 COOH divided by OH minus divided by CS3 COOH minus concentration. What I am doing, this here I'm multiplying with H plus and here also I'm multiplying with H plus. Just look at this. What is happening? Okay. Now, this entire term, you see here, this you can write as Kw. And this expression, if you see, this expression that you have, it is one by K. Isn't it? One by K. Okay. So this we can write, Kh equals to Kw by K. Any doubt in this? Tell me all of you. No doubt. So you have to memorize all this formula actually. And it's not like we have only one. Like this also we have another case, the third one and fourth one also we have. So we'll have three different types of relation over here. So you have to be very careful with all these, you know, expression in what condition we get what. I'll tell you some trick over here so that you can memorize. Before that, you see one more thing here. Further, this H what we can write in instead of this Kh, I'll write down Kw by K. So it is Kw by Ka into concentration C. Right. Now, what is the condition we have here? We are taking weak acid and strong base. Isn't it? We are taking weak acid and strong base. Correct. So whatever is weak for that only will write down the dissociation constant here. Means Kh equals to Kw by Ka. Why Ka here? Because acid is weak here. Right. So you can keep this in mind like this. The expression of Kh. No, in Kh we don't have an, here we have, I'm sorry. Here we have under root. I missed it. This is fine. What is Kh here? Kh is the, Kh is the hydrolysis constant, like K is the dissociation constant. It is hydrolysis constant. Okay. So you need to memorize the expression of H, which is degree of hydrolysis, expression of Kh, which is hydrolysis constant. How do you memorize? We always have Kw in the numerator. In that denominator, we'll write down the constant of the component, which is weak. We component will write down. So since it is for weak acid and strong base, so acid is weak. So for acid, I'll write down the constant over here, Ka. And then we'll have this only. Is it clear? Right. Here. So next our objective is to find out the concentration of H plus. And like I told you that the concentration of H plus we need to find out in order to calculate Ph. And here we have OH minus concentration, OH minus is CH. Correct. OH minus is CH. So as you have, C will be given in the question. So you can find out easily the concentration of OH minus, then POH and then Ph. So I'll write down here you see. Like you see this expression. First of all, the concentration of the H value is this Kw by Ka into C. So concentration of OH would be what? Will be CH. So C, you let it be as it is. H is Kw by Ka into C root over of it. So C and C will get cancelled. So it is Kw into C by Ka root over of it. This is OH minus concentration. Now when you solve this, you'll get POH. I'll write down the expression here directly. POH equals to PKW divided by 2 minus half of log C plus PKA. Okay. This is the expression at 25 degrees Celsius. What happens at 25 degrees Celsius PKW equals to 14. Hence POH equals to 14 by 2 that is 7 minus half of PKA plus log C. Okay. If you try to find out Ph, Ph would be 14 minus POH that is 7 plus half of PKA plus log C. This is a Ph and POH expression. Tell me, any doubt in this? You see here, since Ph we are getting greater than 7, 7 plus some value here. Hence the solution is what? Basic in nature. So you need to memorize the formula of Ph. You need to memorize the formula of H. You need to memorize the formula of Kh. All these three things they ask. Terms also you must keep in mind. H is the degree of hydrolysis. Kh is the hydrolysis constant. Okay. We are getting basic solution here, which is quite understandable because you see the condition is what we are considering here is strong base. Obviously this strong base will dominate this acid because it is weak. That's one thing. Second point is what? When it get hydrolyzed, it releases OH minus, which gives basic property to the solution. So ideally we should get Ph greater than 7, isn't it? Yes. Tell me guys. So all these things you can correlate. Since we are having a strong base, so solution must be basic Ph value must be greater than 7. So you won't get confused over here that 7 plus we should write or 7 minus we should write because the base is strong. So it is basic. So we must have 7 plus the expression. This expression is same for both, right? So plus and minus you won't get confused over here. Is it clear? Any doubt? All of your response please. Okay. But you need to understand again the condition of salt hydrolysis you can apply when the number of equivalents of what? Number of equivalents of acid and base must be equal. If it is not, then it is not the case of salt hydrolysis. The solution is different. We'll get different answer according to this. Okay. So condition you must remember. Now the third case we have here hydrolysis of hydrolysis of salt of strong acid and weak base. Strong acid and weak base. Okay. So strong acid for example I am taking HCl and weak is weak base is NH4OH. NH4OH and HCl suppose. Strong acid weak base. It forms NH4Cl and water H2O. So NH4Cl. The ions it produce is NH4 plus and Cl minus. Okay. So obviously the Cl minus is the anion of strong acid. So for this we don't have any hydrolysis. So no hydrolysis for this. Okay. No hydrolysis for this. This one will go under hydrolysis and the reaction would be we have NH4 plus plus H2O. It forms what? It forms NH4OH and H plus. Now two, three things logically you can understand here the one that we just discussed in the previous case. First of all, since this is a strong acid we have, hence the solution would be what? Solution would be acidic. It dominates the weak base. Second point, we are getting H plus here. H plus releases into the solution. This makes the solution acidic. So it's quite good. If you write and we'll derive this also but I'll just write down how to write down the, I'll just write down the formula here. How we can write down the formula directly you see. See the previous one KH is equals to we had written KW by KA and I said what in the denominator you will get the component which is, you will have the component which is weak here. Right. So since in this case we have weak base here. So instead of KA you will get KB. This is the formula of KH. I will do the derivation also but first you see because you cannot derive these things in the exam. Right. You should not derive this thing like that. So how to write down the formula. So KH is equals to KW by KB. If you write down the formula of H degree of hydrolysis that would be what could you tell me it is again KW by instead of KA will write down KB into C. Right. This is the formula of H. Yes, it will be a root under I'll write down that. So H is this root over of it. Right. Just you need to substitute the weak part over here like weak base we have so instead of KA will have KB. PH would be since it is acidic here. Since it is acidic here the previous one you see I'll just show you over here. The previous one you see the PH formula since the base acid was weak. So we'll get here the weak component KA. Right. Now we have here the base is weak plus the solution is acidic. So for acidic solution we know PH is less than seven. So seven minus you should write down because it is acidic seven minus everything is same half of P K we have the weak base so PKB plus log C. This is the formula of PH similarly POH is 14 minus PH. Did you understand how to write down the formula. Tell me how many of you understood this because you cannot derive this. Okay. You cannot derive this in the exam. Let me tell you. Okay. I have done the derivation so that you can understand how do we get it but then again you need to relate and then write on the formula. Okay. Easily you can write always weak component you should write here. Okay. Now how do we get this formula same kind of derivation we have quickly I'll write down consider this reaction I'll write down KH for this. So what I'll do here KH equals to a concentration of NH4 OH divided by NH4 plus we have H plus here. Now just to express this in terms of KW I'll multiply OH minus here and divide OH minus here same exactly same thing we have over here. So further KH is equals to what we get. We get KW by KB that's what I said. You get KW by KB and again if you write down the concentration term in terms of concentration then KH would be it is CH into CH divided by C into 1 minus H. All those approximation we make and we get concentrations H is equals to KH by C root over of it. Okay. Further this H equals to KH we know KW by KB. So we have KW by KB into C root over of it. This is the formula of H. We need to find out pH, pH concentration we know H plus concentration is C into H concentration of H plus is CH. H is KW by KB into C, C and root C cancel out so it is KW into C by KB root over of it and then you find out pH from this. So pH is equals to 7 minus half of PKB plus log C. And always remember this, this 7 that we get here it is because of the temperature that we are taking 25 degree Celsius. I haven't seen any question where they have no changes the temperature over here. So you need to consider this only but keep this in mind. It is PKW by 2 over there. Here it was PKW by 2. Now PKW is 14 at 25 degree Celsius hence we get 7 over here. If it is not 25 degree Celsius then we can have anything else. PoH would be 7 plus half of PKB plus log C. This is the expression. Yeah done. Okay. Now the last one we have when both are weak acid and base. Once again guys. Okay. So then the fourth part we have here. Hydrolysis of salt of weak acid and weak base. Now here both are weak. Correct. So in this case, yeah once again I will go back. Done. This one. Okay. Now weak acid and weak base we have. So for example I can take a weak base that is NH4 OH and weak acid is CS3 COOH. So it is CS3 COOH plus NH4 OH gives CS3 COOH, NH4 plus H2O. The ions if you see here it is CS3 COO minus and NH4 plus. Both are the ions of weak acid and weak base. So hydrolysis of both possible over here. So the reaction would be CH3 COO minus plus NH4 plus dissolved in water forms a weak acid CS3 COOH and the weak base NH4 OH. So if it is C, C, 0, 0 initially it is C minus CH, C minus CH, CH and CH. So KH would be here, CH into CH divided by C into 1 minus H, C into 1 minus H. Which is nothing but H square by 1 minus H. Copy this down. Okay. Done. Right. Now you see from this we can write down H by 1 minus H is equals to root over of KH. So expression of H equals to root over of KH by 1 plus root over of KH. This is one thing. Okay. This is the expression of H. Now, if you see the expression further of KH, we have CS3 COOH on the product side divided by CH3 COO minus. Then we have NH4 OH on the product side divided by NH4 plus. And to just get the expression of KW, I'll multiply here H plus, I'll multiply here OH minus. And then in the numerator also, we have to multiply by H plus and OH minus. This expression is one by KA, another one is one by KB and this one is KW. Hence the expression of KH is equals to KW by K into KB. Since both are weak acid and base, so we are getting both here in the denominator. So once you know this KH, you can find out degree of hydrolysis H also. Okay. We need to find out pH and for pH we require H plus concentration and for that, we'll write down the expression of KA, which is CH3 COO minus H plus CS3 COOH. So H plus concentration would be KA into CH3 COOH. Okay. So CS3 COOH is CHC into one minus H. C and C will get cancelled. So KA is into H by one minus H. What is H by one minus H? H by one minus H is nothing but root over of KH, root over of KH. KH expression just now we have, you know, find out, we find out this KH expression, which is equals to KW by KA into KB, root over of it. Which further we can write, KW into KA divided by KB root over of it. And once you solve this for pH, you will get the expression of pH 7 plus half of PKA minus PKB. Copy this down. Once again, this one done. Okay. See, here we cannot say whether this solution is acidic or basic because it depends upon PKA and PKB value. So we have three conditions over here. All three kinds of solutions possible here. So first thing is what? Case one, if KA, dissociation constant of an acid is greater than KB. If K is greater than KB, then we can write PKA is less than PKB. And then we can write PKA minus PKB is less than zero. Okay. Where it is less than zero. So under this condition, pH is what? Is less than seven? pH is less than seven, which makes the solution acidic. Case two, if KA is or KB, I'll write on KB is greater than KA. So PKB is less than PKA. Hence PKA minus PKB is greater than zero, which further means pH is greater than seven. And hence it is basic. Done? Okay. Third case, if KA is equals to KB, which further means PKA equals to PKB, which means pH equals to seven. And the solution is neutral in this case, neutral solution. Okay. So all three conditions you must take care of. Yeah, one second. I'll go back. How do you memorize this acidic or basic solution? See, whatever is more, if KA is more acidic, if KB is more basic. So whatever is more, that would be the property of the mixture. Done, all of you. Now you look at this question here. Salt hydrolysis is done over here. You can get question from this. So the question is calculate the pH of the solution. The first question we have, we are mixing CH3 COOH and we are mixing NaOH. The data is given over here. We have 0.1 molar and 100 ml of acid. NaOH, 0.1 molar and 100 ml of it. The first one is this. The second one is we have a weak base NH4 OH with HCl. For NH4 OH, it is again 0.1 molar and 100 ml. HCl, it is 0.1 molar, 100 ml. The KA value for acid is given. CS3 COOH, KB value for base is also given. Both are equal, NH4 OH. And this is equals to 2 into 10 to the power minus 5. Two more questions we have, I'll write down first and then I'll show you the question. The third one we have, we have HCl we are mixing with NaOH. 0.1 molar HCl and 100 ml volume. For this one it is 0.1 molar NaOH, 100 ml volume. And the last one is we have NH4 OH with CS3 COOH, both are weak. 0.1 and 100 here also, 0.1 and 100. Let's try this question. Done? Okay, the first one you see, first of all, you tell me the pKa and pKb value. If you calculate this pKa and pKb value, it will be same 4.7 you will get. It is a case of weak acid and a strong base. See the number of millimoles of this and millimoles of this is equal. So complete neutralization is there. Okay, it is 10 millimoles, 10 millimoles equal. So pH we know for weak acid, we know the solution is basic. So pH should be more than 7, 7 plus half of whatever week we have pKa plus log C. What is the value of C over here? 7 plus half of pKa is 4.7. Log C is 10 millimoles. So log 10 into 10 to the power minus 3 moles it is. So when you solve this, you will get pH equals to 7 plus half of 4.7 minus 2. Right, so 2.7, 2.7 by 2, 1.35, 1.35 plus 7. So it is 8.35, isn't it? One second, one correction I need to do. Yeah, one second, one second. We have log C here. What is the value of C? The value of C if you calculate, we have 0.1 molar, right? Or the number of millimoles one correction I will do over here. The acid the salt that forms, it forms what? It forms 10 millimoles, right? So 10 into 10 to the power minus 3 moles of salt we get, salt, right? Because this hydrolysis is happening with the salt only, but this is moles. It is not the concentration. Concentration would be 10 into 10 to the power minus 3 divided by the volume. Volume is 100 plus 100, 200, right? So this would be 2 into 10 to the power minus 4, sorry, 0.5 into 10 to the power minus 4. So log of this value you need to calculate. 0.05 concentration you'll get if you consider the millimoles over there, right? So millimoles we have here, if you take this as millimoles 10 only, you'll get this. 0.05, yeah, that's right. So log of here we get 0.05, which you will substitute here, 7 plus half of log of 0.05 is minus 1.3 is less than 1. So it is 5, 4.75. Could you check the log of 0.05? What is the value we have? Log of 0.05, what is the value? Tell me 0.7 minus 2. 0.7 minus 2 is again 1.3 minus 1.3, correct? So this would be 4.7 minus 1.3. Tell me what is this value you are getting? 0.7 is 2 and then 0.4, 3.4, 3.4 by 2. So 7 plus 1.7, 8.7 we are getting, basic. It should be basic only because base is strong over here, okay? Similarly, we have a strong acid. So directly I'll write down pH equals to 7 minus half of PKB plus log C. So that would be 7 minus half of PKB is 4.7 and log C would be what? Log C would be the same value, minus 1.3. So it is 7 minus 1.7, which is equals to 5.3. It's an acidic solution, 5.3. Minus we have here, so you see the minus sign, not plus. Yes, got it? The other two you try, these two. Yes, it is neutral because we have strong acid and strong base, but do check the number of millimoles here. Number of millimoles is 10. Number of millimoles is 10. Since we have equal millimoles, complete neutralization, pH is 7. If it is not equal, then the answer would be different. This one, weak acid and weak base. Yeah, I'll go once again. Last one, what is the answer? Yes. So in this case also you see, PKA is equals to PKB. Equal value we have, right? PKA is equals to PKB. This is also neutral. The pH value is 7 for this. Okay. Now the next concept we are going to understand, that is buffer solution.