Transcription
You are welcome to today's video lesson on chemistry Made Easy with Bryce Edo. In today's lesson, I'll be discussing the pH concept. So, the first question we have to ask ourselves is, what is the pH concept actually talking about?
It must be noted that the pH describes the degree of acidity or alkalinity of a solution. This must be noted. The pH describes the degree of acidity or alkalinity of a solution. This must be noted. And there's a scale that actually is used to measure the degree of acidity or alkalinity of a solution, and that scale is simply called the pH scale. This must be noted. The scale used to measure the degree of acidity or alkalinity of a solution is simply called the pH scale, and this pH scale ranges from zero to 14.
Now, you can see that I have one arrow moving towards zero and another arrow moving towards 14. This must be noted. Now, at the center here, we have seven. So, when we have a value that ranges from zero to 6.9 and 7.1 to 14. Now, it must be noted that whenever we have the pH scale to range from 0 to 6.9, we simply say that that particular solution in the pH scale is simply acidic. But if the value ranges from 7.1 to 14, we say that particular solution in the pH scale is simply alkaline or basic. But when we have the pH scale to be seven, we simply say that particular solution is neutral. It's at the neutrality stage. So, this must be noted. It must be noted that the pH scale ranges from 0 to 14, and from 0 to 6.9 is simply acidic, 7.1 to 14 is alkaline or basic, and seven itself is neutral. So, all of these must be noted.
Now, it must be noted that the concept of pH was brought about by a man called Søren Peder Lauritz Sørensen. So, this was the man that brought about the pH concept in the year 1909. So, this man called Søren Peder Lauritz Sørensen brought about the pH concept.
Now, let us assume two solutions. This is solution A and this is solution B. Now, let us say solution A has a pH to be 3, and solution B has its pH to be 5. Now, what is the question looking at these two solutions? Which of them is more acidic? Now, you can say that the arrow is going towards zero. As you are going towards zero, your acidity will increase. Definitely, pH of 3 will be more acidic than pH of 5. Though this is acidic, but this is more acidic. So, solution A here is more acidic.
But let us assume that this solution A has a pH of, let's say, 8, and solution B has a pH of 13. Which of these two solutions is more alkaline? We simply say solution B is more alkaline because you can see the arrow is going towards 14. So, all of these concepts must be noted.
Now, let us quickly move over to the quantitative aspects under the pH concept, that is, the calculative aspect on the pH concept. Now, moving over to the quantitative aspect on the pH concept, that is, the calculations on that pH concept. So, to make it very, very easy, I will classify these aspects into three parts. Now, the first of them is solving questions on pH of strong solutions. Okay, I'll be I'll be I'll be classifying them into three parts. The first is solving pH of strong solutions. After this, we simply go over to the aspects that talk about pH of weak solutions, okay, pH of weak solutions, and lastly, we simply go over to pH of buffers. Now, all of these must be noted. You can see I classified this part or this quantitative aspect into three cases, into three parts. The first of them was solving questions under the pH of strong solutions. After that, we simply move over to the part that has to do with the pH of weak solutions, and lastly, pH of buffers.
Now, it must be noted that anything that has to do with pH also has to do with pOH, and this must be noted. It must be noted that pH basically has to do with acid, and acid has to do with what we call H+ and what is H+? High hydrogen ion. This symbol here is hydrogen, so it's simply called hydrogen ion. Now, anything happening to acid must happen to base. So, here becomes pOH. Now, it must be noted that pOH has to do with this, and base has to do with the OH- which is called hydroxide ion. OH is called hydroxide, and there's a minus here, so I call it hydroxide ion. Now, all of this must be noted.
So, let us quickly talk about the first aspect that has to do with pH of strong solutions. So, let us quickly move over to the first aspect of this class, which is solving questions on pH of strong solutions, and this strong solution can be of two: strong acids and strong bases, okay, strong acids and strong bases. Same applies to this part, which is solving pH of weak solutions, which is weak acids and weak bases. Now, all of these are what we'll be discussing in this class, and after that, we simply move over to the pH of buffers.
Now, in the course of this class, I'll be talking about these two aspects simultaneously, whereby I'll be giving us important formulas that must be noted under these two aspects. Now, what is the formula used to solve pH of basically strong solutions? Now, the formula is simply this, which is pH is equal to negative, which is minus logarithm or minus log into brackets H+ close brackets. Now, this must be noted. Whenever I see H+ standing alone, I simply call it hydrogen ion, as we saw earlier here. Now, you can see that H+ was now inside the bracket, so the name will definitely change, and what will I call it? Hydrogen ion concentration. So, this must be noted. So, H+ is actually different from H+ in a bracket, and we already know that the S signs for concentration should be in moles. So, all of this must be noted.
Now, this is the formula used to solve pH of strong solutions. Recall we said anything happening to acid must happen to base. So, what is the formula for solving pOH? It's simply minus log into brackets OH-. whereby OH itself is called hydroxide ion. So, this is what we call OH hydroxide ion. But if OH is in a bracket, it's simply called hydroxide ion concentration.
So, let us quickly move over to the other formulas that are very, very important for solving questions on pH of strong solutions. Now, it must be noted that whenever I add pH and pOH together, I must have a value, and that value is 14. So, pH plus pOH will give me 14. And multiplication of the hydrogen ion concentration times or we simply say a multiplication of the hydroxide ion concentration is equal to 1 times 10 raised to the power of minus 14.
Now, you can see that I have deduced just four formulas that are very, very important for solving pH of strong solutions. This must be noted. The first of them is pH is equal to minus log into bracket H+, whereby I told you guys that H+ is called high hydrogen ion. But if H+ is in a bracket, we simply add concentration to the name, so it becomes hydrogen ion concentration. And I said anything happening to acid, which is pH, must happen to base. So, pOH becomes minus log into bracket OH-. That's basically called hydroxide ion concentration. And additional pH of pOH gives us 14. And multiplication of the hydrogen ion concentration, called H+ in a bracket, times the OH-, will give us 1 times 10 raised to the power of minus 14.
Now, you can see all the formulas that is used to solve questions under pH and pOH of strong solutions, either strong acids or strong bases. Now, what are the formulas that are very, very important in this aspect? That is what I'll be showing us right now. That is why I told us in the in the in the initial part of this class that both concepts will be discussed simultaneously. Now, before coming over to this aspect, I'm giving you guys the formula that is important under this aspect. It we have to be familiar with one concept, with the concept of Ka and KB. Now, what we call Ka? Ka is called acid dissociation constant. This must be noted. Ka is called acid dissociation constant. And recall, anything happening to acid must happen to base. So, what we call KB is simply called base dissociation constant. This must be noted.
Now, moving further. Now, what am I trying to say here? It must be noted that as we are having pH, so we are having what we call pKa. This must be noted. As we are having pH, so we have what we call pKa. And in chemistry, or in science, or in chemistry precise, the symbol p simply means minus log. So, what are we having here? pKa. We want to express it. It becomes minus log Ka. Now, anything happened to acid must happen to base. Oh, so what are we having here? pKb equals minus log Kb. Okay, this must be noted. pKa minus log Ka, pKb minus log Kb.
Now, it must be noted, I will follow our equation three, which is pKa plus pKb is equal to 14. I just said we are just inter, we are just changing all of the symbols because we have to be familiar with this concept right now before moving over to the formula for solving pH of a weak solution, either weak acids or weak bases. So, that is for that. And lastly, multiplication of Ka. Okay, mothers Ka times Kb is going to equal to 1 times 10 raised by minus 14. Okay, all of this must be noted. So, you can see we just interchange the formula for this to this because this is very, very important for come to this aspect.
So, what are the formulas used for solving the pH and pOH of weak solutions? The first of the formula here is simply pH of a weak solution is equal to minus log into root of KaC. This must be noted. This dot here means times, and what does C mean? C means concentration. Okay, that is what that C means. And concentration should be in what unit? Moles per DM cubed. We know we have various ways concentration can be expressed. You have expressing concentration in moles per DM cubed, that's molarity. Expressing concentration in moles per kg of solvent, that is molality. And also for normality. Okay, but this is basically expressing concentration in moles by DM cubed, which is also regarded to be called molarity. So, that is for that.
So, what is the formula for solving pOH? You know, anything happened to acid must happen to base. So, pOH formula becomes minus log into root of KbC. We are talking about this PO, it becomes KbC. So, you can see the two formulas that are very, very important when solving questions on the pH and pOH of weak solutions. So, with all what I've said, let us start solving practice questions under this aspect first before move over to this concept. Okay, and lastly, we'll go over to solve problems on the pH of buffers.
Now, before coming here, this particular concept must be noted. That is why I wrote it on the board. We can memorize and jot them down, but let us start solving questions on pH of strong solutions. So, you can see the first question written on the board. How do you know that this particular question is is on pH of strong solutions? Now, this particular acid here is very common. It is a strong acid called hydrochloric acid. So, when I see questions on this aspect, I simply say, "Oh, it is under pH of strong solutions." And also, they did not give us Ka or KB, so we can't apply the concept of pH of weak solution. This question is very straightforward, and what will be the solution? It is simply easy. They are asking us to get pH, and they gave us the hydrogen ion concentration to be 2.0 times 10 raised to the power of minus 5 moles per DM cubed. So, what formula should I apply? I simply apply the first formula that says pH is equal to minus log into bracket H+. You know, this H+ value should be imputed here. So, what becomes the pH? It becomes minus log into bracket 2.0 times 10 raised by minus 5. You know, I told you should be expressed in moles per DM cubed or molarity. So, when we do that, what becomes pH? It is equal to 4.7. So, you can see how questions under this aspect are being solved without stress. So, the next question here, you'll be solving it, and we solve a lot of practice questions. So, here is the question you'll be solving, and you've already answered in the comment section below. Meanwhile, let us quickly go over to the next practice question.
Now, let us quickly move over to question three, which is, calculate the pH of a solution of H2SO4. Wait, H+ that the hydrogen ion concentration of this value. Now, you know you've seen this compound now, which is interesting. For some persons, maybe like say, "Okay, because it's H2SO4, I want to times this part value by two." No, it doesn't work that way. Why? Because the question is straightforward, because they've already given us the H+ already. We don't need to times anything. When that particular aspect comes, I will let us know. So, look at how this question will be solved. Very straightforward. They are asking us to get pH, and they've given us the H+ that the hydrogen ion concentration already to be 3.0 times 10 raised to the power of minus 8 moles per DM cubed. So, what will I do? I simply recall my formula, which is pH is equal to minus log into bracket H+ or OH-. It is H+, not OH-. If it was OH-, here becomes pOH. So, what becomes the formula? The answer, rather, because minus log, let's impute. We are having 3.0 times 10 raised to the power of minus 8, close the bracket. So, we are having 7.523 as the pH of this particular solution. So, you can see how questions under this aspect are being tackled. So, let us quickly move over to the next practice question.
So, let us quickly move over to the next practice question. Now, let us quickly move over to question four. Now, let us quickly move over to the next practice question. Now, if it's look at this question carefully, the question is asking us to get pH of a 0.002 molar per DM cubed solution of HCl. Now, you can see that in this particular particular question, the configuration of the question doesn't changed. Why? You can see that they did not give us the the concentration of the hydrogen ion, but rather they gave us the concentration of the full compound, which is HCl. That's why they said 0.002 molar per DM cubed solution of HCl. So, we cannot just use the normal concept by saying pH is equal to minus log into H+. If you infuse this value as H+, you will not get the answer correctly. Why? Because they did not give us the H+ specifically. So, right now, we have to know the H+ data. I mean, the H+ and how do we do this? To do this, we have to know the concept of two terms. First of them is basicity of an acid, and the other is acidity of a base. Now, what do we call the basicity of an acid? Now, what's the concept actually talking about? Basicity of an acid. Now, it must be noted that the basicity of an acid is the number of replaceable hydrogen ions that is present in one molecule of an acid. Now, what am I trying to see? Now, let's take an example. This is one acid, which is HCl. Now, I want to dissociate it by breaking HCl. I'm having H+ plus Cl-. In the look of things, is this reaction balanced? Yes, because I'm having one atom of hydrogen. Same applies, a one, okay, one, and same applies here, which is one chlorine atom, or in this context, it's an ion, and same applies here, we have one. So, in the look of things, at the end of the dissociation, how many H+ did I have? I had just one H+. So, I will simply say H+ in this context is one. So, this particular compound is monobasic. This must be noted.
Now, let's go over to another example using another acid, which is H2SO4, because we have to be familiar with this aspect before we solve questions under this concept. So, I want to dissociate H2SO4. When I do that, I'm definitely going to have H+ from the from the definition that says the existence of an acid is the number of replaceable hydrogen ion in one molecule of an acid. This is one molecule of an acid, one molecule of an acid, plus the radical here, which is SO4 2-. Now, the reaction is not balanced, but I think if I put two moles of H+, it will be balanced. I'm having two hydrogen here. So, H+ in this context was two. Definitely, what becomes the basicity of this acid? It becomes dibasic.
Now, let's go over to the last example, okay, or the second and last example, H3PO4. Now, let's dissociate this. We are having H+ definitely, plus uh PO4 3-, is a radical actually called the phosphate radical. So, in this context, how many moles of H+ should I add? Three, because here was three. So, what becomes my basicity of this particular acid? It becomes three. And here, here becomes tribasic. So, all of this must be noted.
Now, let's take the last example using an organic acid. And all organic acids most at times are weak acids. So, this is one example, H3, uh, CH3COOH. This is called ethanoic acid. Okay, this is called ethanoic acid. Now, when we dissociate ethanoic acid, we are having CH3COO- and we are now left with just one hydrogen ion. So, in this case, the reaction is balanced because I'm having how many carbon? Two carbon, two carbon. How many hydrogen? Four. How many hydrogen? Four. How many, how many oxygen? Two and two. So, in this context, I'm having one H+. So, in this context, or or um, ethanoic acid has just one hydrogen or one replaceable hydrogen ion. So, in this context, we leave it. Tribasic? No. Will it be dibasic? No. Bora, that is what monobasic. So, all of this must be noted about the basicity of an acid because we have to understand this principle properly before we quickly move over to solve questions under this aspect. So, all of these concepts explain the basicity of an acid. As earlier said, as we are having basicity of an acid, so we are having acidity of a base. So, right now, if I talk about acid base, we'll be talking about all of, we'll be dissociating various bases.
Now, it must be noted that the the basis of this common acid must be noted because if you check this question now, the acid here is HCl. We are coming to this question, but before coming to this question, let's quickly talk about acidity of a base. So, let us quickly talk about the acidity of some common bases like the likes of NaOH, which is popularly called sodium hydroxide. So, let us dissociate this compound. I'm having sodium metal, which is Na+, which becomes sodium ion, plus I'm interested with the OH. Now, it has to do with base, it has to do with the OH- called the hydroxide ion concentration. It is not called hydroxyl. Hydroxyl is seen in organic chemistry. So, what are we having to be the acidity of this base? You can see here that the reaction is balanced. I'm having one mole of this and one mole of sodium. Same applies here, one sodium and, you know, one OH here. So, it tells us that the OH is simply one, and we say it is acidic. Yes, the compound is monoacidic. But I'm having just one OH-.
So, taking another example like the calcium hydroxide, which is represented to be this as this chemical formula. So, let us dissociate. We are having calcium ion, which is Ca2+, plus the OH-, which is the hydroxide ion. Now, in this context, how many OH have I have in here? Two. So, we have to pull two here. So, OH- here becomes two. So, we're having this particular compound called calcium hydroxide to be diacidic. So, this compound is diacidic. So, this must be noted.
Now, taking the last example, which is seen in aluminum hydroxide. Okay, aluminum hydroxide. So, let us dissociate it. So, when we dissociate it, we have in Al3+ plus OH-. So, we're having three moles here because here was three. So, it tells us here that this particular compound is triacidic. So, this must be noted. This must be noted about base. This is an acidity of a base.
So, with all what I've said, let us quickly move over to this particular question to solve questions under this aspect. Now, here's the question. The question says, calculate the pH of a 0.002 molar per DM cubed solution of HCl. Recall, this particular question we are seeing is quite different from the ones we've been solving. The ones we were solving, they gave us is straightforward. They gave us the H+. Now, here they did not give us H+. This we are seeing is the concentration of the full compound. So, how do we solve questions under this aspect? We simply remember the formula I gave, which is pH is equal to, that's not to get pH, so which is pH is equal to minus log into bracket H+. And we were not giving H+ in the question. Now, but we have an alternative way to get H+. H+ can be solved by formula, which is concentration of the acid, that's the concentration of the full compound, which is HCl in this context, times basicity. This must be noted. You know, acid has basicity, and bases have acidity. So, here, what's recorded? Let us record the basicity of this acid. We said initially, early from the breakdown, from the reaction, we say that this particular acid here is monobasic. So, the basicity is one. So, what do we do? This H+ I'm putting in place of bracket, the formula for H+ I put it here. So, we have been pH formula to be minus log into bracket concentration, I'm shorting concentration, conch, with dot, a, uh, concentration of acid times the basicity, which is one. Okay, so let us do this. Let's solve it. So, what becomes pH? Minus log into, what's the concentration of the full acid, which is this? So, 0.002 times 1. So, what becomes our pH? It becomes minus log into, um, 0.002. So, minus log into, um, 0.002, we have the pH of this particular solution to be 2.7. So, you can see how questions like this have been solved without stress. So, let us quickly move over to the next practice question.
So, let us quickly move over to the next practice question, which is question five, and it says, calculate the pH of of a 0.0025 DM cubed H2SO4. Now, it's still like the previous question. This I am seeing is not the H+ value, but rather the concentration of the full compound, which is H2SO4. So, how do we solve questions under this aspect? We simply remember our formula for solving pH, which is minus log into bracket H+. But we were not giving H+ in the question, and we already know an alternative formula for H+. So, pH here becomes minus log into, what's the alternative formula? I said initially, it is concentration of acid, concentration of acid times basicity. So, how do we solve questions under this aspect? We simply remember all what we've said initially. So, it becomes minus log into, what's the concentration of the acid in this context? It is simply 0.0025 times. Now, what's the basicity of this acid? Remember, we said it is dibasic, so the basicity is two. So, times two. So, what becomes pH? So, 0.0025 times 2 does, um, minus log into 0.005. So, uh, minus log into 0.005, we have the pH of this solution to be 2.30. So, that 2.301 as the case may be. So, you can see how questions on this aspect are being solved without stress. So, let us say, for instance, they are asking us to get pOH again. How do we solve it? Remember equation three I gave in the formula that says pH plus pOH is equal to 14. So, the actors will get pH in this context. So, pH becomes making subject, whether pH goes and becomes negative. So, how do we solve it? So, pOH will not become 14 minus pH. So, what becomes pOH? It becomes 14 minus what's our pH indication, which is 2. um, 2.301. So, what becomes pOH? It is equal to, uh, 14 minus 2.301. So, pOH becomes 11.7. So, so you can see how questions under this aspect are being solved without stress. So, that is how it is done.
Now, let us say the compound they gave to us was a base. Please get this straight. The compound they gave to us was a base. Let's say here was sodium hydroxide, NaOH, and definitely, and they are not asking us to get pOH. Okay, and the compound was not changed. So, a base, which is sodium hydroxide. Okay, so how do we solve questions like this? We simply remember from like assist or since the acid is for a base, and they actually get pOH, we simply just say pOH changes to OH-. Okay, and also, how do we get OH-? It must be noted that OH- might not has an other alternative formula as H+ is simply very, very familiar with this. So, it's simply concentration of base times acidity. Remember, base has to do with acidity, and acid has to do with what? Basicity. So, when we see questions relating to bases, we have to become and take these right steps and get the answer without stress. So, you can see how questions all that is expected are being solved without stress.
So, let us quickly move over to the next concept, which is solving pH and pOH of weak solutions. All of the examples we've been solving initially is for pH and pOH of strong solutions. So, let's quickly move over to solving pH and pH of weak solutions. So, here is the first question that explains pH and pOH of a weak solution. First of all, it must be noted that most or all organic acids are weak. This is a good example of an organic acid. So, when I see questions like this, I'll say, "Oh, it is talking about pH and pOH of a weak solution." And definitely, they give us Ka. You know, I gave you guys the formula initially. So, how do we solve questions under this aspect? We simply remember the formula for solving pH of a weak solution, which is pH is equal to minus log into root of KaC. So, this is the formula. So, let us impute parameters now. What is the pH? First of all, minus log into root. What was the Ka value for ethanoic acid? There is 1.8 times 10 raised to the power of minus 5 times. Now, what's the concentration giving the question? You know, anything has to do with moles per DM cubed or molar is concentration. You know, this is basically molarity. But even the small end, this is molar, and this has to do with molality. All of them are all concentration units, but in the concept of pH, we are interested with molarity, moles per DM cubed, or molar, not molal. So, how do we solve questions? What's our C, which is 0.0020? Okay, so let's impute them in our calculator and get our answer. So, minus log into root of 1.8 times 10 raised to the power of minus 5 times 0.0020. So, we're having the pH so be 3.722. Now, for instance, you asked us to get pOH, how do we do it? Remember our equation three that says pH plus pOH is equal to 14. And, uh, what do we do next? We simply make a pOH subject. Comes from 40 minus pH. When it comes here, so what are we having here? 14 minus our pH, which is, uh, 3.722. So, what becomes pOH? It becomes 14 minus 3.722. So, what becomes the answer? 10.2 or 10.3. Okay, so you can see how questions like this are being tackled without stress. So, let us quickly move over to another practice question under solving pH and pOH of weak solutions.
So, let us quickly move over to the second practice question under solving pH and pOH of weak solutions. Now, this question says, determine the pKa. We have to be careful here. Determine the pKa for a, they've given us concentration already, that's C, solution of ethanoic acid. Ethanoic acid is basically an organic acid, and all organic acids are weak, whose measured pH is 4.0. How do we solve this question? First of all, we must remember the formula, and what is the formula? It is simply pH is equal to minus log into root of KaC. Now, after this, what do we do next? We are looking for pKa, but first of all, I think we have to look for Ka. Yes, it's very, very important because they've given us pH already. The pH value was 4. Equal to minus log. This was not given, but they gave us concentration. So, let us work mathematically so as to get this answer without stress. So, the first of the step here, so the first step here is to basically take log inverse or antilog. So, what becomes the, uh, solution to this? We simply see, uh, this minus log comes to this direction, so it becomes log inverse, that shift log in your calculator, that's minus log into. Now, your pH changes to negative, minus pH, and this only remains equals the root of KaC. So, what becomes the next step? We simply say log, uh, inverse of what's our pH for, but in this context, it's minus four. Into root of KaC. So, first of all, let's get the value for log inverse of this. So, shift log into minus four, that is 1 times 10 raised to the power of minus four. So, bringing this here, we have in one, uh, 1 times 10 raised to the power of minus four is equal to, uh, root of KaC. So, the next step here is to bring the square roots here to make, uh, bring this query set to, so square this value. So, we have a 1 times 10 but minus four, or squared is equal to KaC. Okay, so the next step here is to make Ka subject because we are looking for pKa, but we have to make Ka subject. Divide by the coefficient of Ka, which is C, and C. So, you cancel. So, Ka becomes equals to, uh, or 1 times a minus four, um, r squared, or power, what's our, sorry, divided by DC, which is 0.005. So, what becomes Ka here? We are having, um, 1 times 35 minus 4, close brackets, or squared, over 0.005. So, we are having Ka here to be 2 times 10 raised to the power of minus, um, six, to be the Ka. And how do we get pKa? That's what the question is asking us. So, remember the initial part of this class, I said for we to get pKa, remember it's equal to minus log into Ka. You know, I said in chemistry, p means minus log. So, it's just a minus log into Ka. So, what becomes pKa value? Minus log into 2 times 10 minus six. So, when we hit our calculator here, minus log into 2 times 10 over minus six, uh, we are having 5.7 going to be the value for pKa. So, you can see how questions like this have been solved. If they are asking us to get pKb, for example, we simply say pKa plus pKb is equal to 14. No, I gave all of these initial equations initially in this video. So, you can see how questions on pH and pOH of weak solutions are being tackled without stress. Follow the step-by-step process, remember the formulas, and you are going to go.
So, with all what we've said, let us quickly go over to the last part of this video, pH. So, with all what I've said, let us quickly go over to the last part of this video, which is solving. So, so let us quickly move over to the last part of this video, which is solving questions on pH of buffers. Now, first of all, we have to understand what a buffer is. Now, a buffer is a type of solution that can resist change when the pH of that particular solution is altered. Now, what am I trying to say? Let's say we take an a quantity of acid, a quarter base added to a buffer solution, the pH of that particular buffer will resist change. It will not change. So, that is what we are trying to say here. But most importantly, is for us to be able to solve questions, calculative questions under this aspect, because both of us, we have two types. We have acidic and also we have basic buffer to be generalistic. So, what am I trying to say? You know, we have formula for solving a pH of a buffer. So, what's the formula? It's simply pH. You know, anything happening to acid must happen to base. So, the formula is simply pH is equal to pKa plus log concentration of salt. You know, anything that has to do with brackets must be called concentration over concentration of acid. So, this formula for solving pH of a buffer. So, what's the formula for solving pOH of a buffer? It's simply pOH is equal to NC, that's of new pH must have to do with pKb. So, here becomes pKb plus log, um, you know, we are talking about pH here should be base over salt. Now, what am I trying to say? Let's work with this first of all. Let's say we're having an acid, and also both of us has to do with weak acid and weak bases. So, let's talk about this guy. This guy is a weak acid, H2CO3, which is popularly called carbonic acid. Now, how do we get the salt of this weak acid? What do we do? We simply remove the hydrogen here and replace it with sodium, for example. You know, when sodium is attached, we are forming a salt, okay, of this particular acid. And since we have been H2 here, here becomes Na2. So, what are we having? Na2CO3. So, this is the salt of this weak acid. So, this is the salt of this weak acid forming the pH of a buffer. So, taking other examples using, uh, um, basic buffers, uh, we can have examples that we explain this, but most important things for us to know how the formula works. So, you can see how it is done.
So, let us start solving practice questions on. So, let us quickly go over to this question and solve it without stress. First of all, we have to locate the salt and the acid, or the salt and the base. You know, for the first formula, it has to do with salt and acid. The order for pH, salt and base. So, let us quickly know what we are giving. So, first of all, this is an acid. Yes, this is an acid, but it's a weak acid. It is a weak acid. You know, buffer has to do with weak acids and weak bases. And also, this is salt of this weak acid. You know, I, I told you how to get the the salt of a weak acid using the reaction up there. Now, this is called sodium ethanoate. Okay, now, moving further, how do we solve question under this aspect without stress? Now, they are asking us to get buffer, so the pH of a buffer. So, let's all remember the formula I gave. pH of the buffer is equal to pKa plus log concentration of salt over concentration of acid. Now, first of all, what was the parameter, okay, in the question? It was given to be 1.8 times 75 minus 5. Now, you can see that the Ka of this acid is constant, 1.8 times 10 raised to the power of minus 5. Now, moving further. Now, moving further to the next level, which is the weak acid, CH3COOH. Now, what's the concentration of this weak acid? This is the acid, the weak acid, to be precise. Uh, what's the concentration? They gave out to be 0.003 molar. And also, what was the salt in the question? It was given to be CH3COONa, which is called sodium ethanoate, and what was concentration? It is basically 0.009 molar. So, you can see on the parallel, all the parameters brought out in the question. So, pH becomes. Now, remember, pKa is simply minus log Ka, plus log concentration of salt over concentration of acid. So, pH becomes minus log 1.8 times 75 minus 5, plus log. Or what's the concentration of the salt? It was given to be 0.009. So, 0.009 over concentration of the acid to be 0.003. So, let's impute and solve. So, here, I think it's 4.74. Okay, here is 4.74, plus log. Now, when you do this, you're having three, 0.009 over 0.003, that's three. So, what are we having? Log 3 here is equal to 0.478. So, pH here becomes 4.74 plus 0.48. Right there. Okay, so what are we to do? 4.74 plus 0.48, we're having 5.22. So, with the pH of this buffer. So, you can see how questions under this aspect have been solved without stress. So, the next application here, you'll be solving it, and probably answering the comment shown below. So, here is the question you'll be solving, and I'm already answered in the comment section below. So, let's quickly go over to the next practice question.
So, let us quickly go over to this practice question. You know, we have to be careful here. This question is is very, very smart. So, how do we solve questions under this aspect? First of all, they are asking, determine pH. But let us leave that. Let us look at the compounds given in this question. Here, I can see that this particular compound is a base. Yes, ammonium hydroxide. And this is the salt of the base. Now, you can see the question now. I'm seeing base and salt. Which formula will I actually relate it? It is pOH. Yes, because we are dealing with a base and a salt. You know, if it was acid and the salt, it is pH. So, pOH becomes, uh, uh, or pKb plus log, um, um, salt concentration of salts, okay, over concentration of base. Now, we have to be careful here because I can consider in the question they gave me pKa, but not pKb. How do we get pKb? Remember the equation I gave initially in this video, which is pKa plus pKb is equal to 14. So, how do we get pKb? It's because of 14 minus pKa. Okay, it's coming here. So, what are we having? pKb. We're not because of 14 minus what's the pKa? It is 4.58. Okay, so I write that 4.58. So, what becomes pKb? It is equal to, uh, 14 minus 4.58, that is 9.42. So, 9.42 becomes the pKb value.
Now, also in this particular question, I was given volume. Now, this must be noted. Whenever you are giving volume in solving questions under pH and pOH of a buffer, please and please convert. Come. Now, they gave me volume alongside with the concentration. You know, recall we said whatever value that should be here should be in the form of concentration. But right now, we're giving volume, and this volume are not the same. So, we simply convert because I am seeing something here. They gave me volume, and they gave me concentration. I can convert it to number of moles, which is n. Same applies to this parameter, 0.005 molar, and the volume here was 600 milliliters. Now, what am I trying to say? We simply convert both parameters to number of moles. So, number of moles of the first term, a number of moles of the base, we have a formula that relates both of them, which is C times V over a thousand. What is C? Concentration. What is V? Volume. Over one thousand. Okay, so, and, uh, and because the volume was in milliliters, that is why I divided divided by 1000. It is almost in liters, and we know divide by 1000. If the volume was in the N cubed, I will not divide by 1000. But if the volume was in cm cubed, I would divide that 1000. So, what am I trying to say here? Number of moles of the base will not be equal. What's the concentration of the base? We have to be careful. This is the concentration of the base. I was the volume 400 milliliters. So, 0.0025 times volume, which is 400, over 8000. What are we having? And let's quickly do that for the salt, which is, uh, or C times V over 1000. So, number of moles of the salt do not because what's the condition of the salt? Uh, basically, it's the 0.005 times what's the volume? 600, okay, over 1000. So, what are we having to be the number of moles of both parameters? Because now we have to use number of moles in place of concentration. I'm coming there. So, let's quickly get our value. 0.0025 times 400 over 1000. So, we are having here to be 0.00, um, okay, we are having here to be, let's just take it in standard form, 1 times 10 without minus three moles. And here, um, 0.005 times 600 over 1000. So, we're having here to be 3 times 10 raised to the power of minus three moles. So, you can see we've got in both parameters in moles. So, my formula simply changed. So, pOH will not because of pKb plus log number of moles of salt over number of moles of base. So, what becomes pOH? What's our pKb? We've got in is not 5, it's not 4.58, but rather 9.42. You can see I've converted. Plus, uh, number of moles of salt, which is, um, 1 times 10 minus 3 over, sorry, that's for the base. Okay, this number of bones of the salt. So, here become 3 times 10 to the minus 3 over 1 times n minus 3. So, what are we having? Let's do this first of all. So, 3 times 10 point minus 3 over, uh, 1 time times 10 without minus 3. So, we are having 3. Okay, so pOH will not be equal to, uh, 4.92 plus log 3. Okay, because he is already 3. So, log 3, that's, uh, 0.48. So, pOH will not be equal to, um, 4.92 plus 0.48. It becomes 9.9. Now, but the question is asking for solving pH, not pOH. Remember our formula, or we said initial that pH plus pOH is equal to 14. So, pH will not be equal to, uh, 14 minus pOH. So, pH now becomes 14 minus what's that pOH value? Uh, pOH value here is 9.9. So, uh, 14 minus 9.9, we are having 4.1 to be the value for the pH. So, you can see how questions like this are being tackled without stress. Now, if you find this video helpful, do what to click the subscribe button and also share these videos with your friends. Thanks for watching.