A buffer is a solution composed of a weak acid with its conjugate base.Â
Buffers
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Buffers
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So a buffer is just basically a weak acid with its conjugate base. Great example we have H. C. L. O. Which is hipaa Cloris acid, this is a week oxy acid here when we talk about a conjugate base that is the same form as the weak acid except it has one less hydrogen and that hydrogen that it's missing is typically replaced with a metal. Usually from group one A. So Hipaa Clorox acid is H. C. L. O. And here we have sodium hypochlorite. The hydrogen has been replaced with a metal so we have a weak acid and its conjugate base together. They form a buffer. Now a buffer hot. What does it do? Well it causes my solution to resist big changes in ph it does this by keeping constant both my H plus ions and my hydroxide ions. It tries to keep them around the same level as much as possible and for as long as possible. Great example is the buffer system within our blood which is composed of carbonic acid and bicarbonate. Without this our blood would become very acidic or very basic depending on what we're drinking soda for example is very acidic. Without a buffer system in our blood our blood will become incredibly acidic pretty quickly just from a can of soda. Now how does it do this? Well remember our buffer is composed of a weak acid and a conjugate base so if we add strong base to our solution to our buffer solution then the buffer resists a ph change by having the weak acid neutralize it. Remember acids and bases are natural um I don't want to say enemies but their natural opposites of one another. If if they're both present together in the same space they try to neutralize one another. So by adding strong base the weak acid steps up to try to get rid of it. So here we have still our weak acid are hipaa Cloris acid. I add some N. A. O. H. Strong base. Here we undergo a neutralization reaction where we create sodium hypochlorite which is some of my conjugate base which is great because these two will be in existence with each other in solution if there's not a lot of N. A. O. H. And therefore stabilize my buffer plus water. If I had strong acid. What happens? Well if I had strong acid then the other part of my buffer, the conjugate base steps up to neutralize that. So here now the conjugate base steps up sodium hypochlorite steps up and tries to neutralize the hcl, thereby creating more weak acid. So more component of my buffer to help stabilize the ph Now here's the thing, the weak acid and the conjugate base can't keep this up forever. So the more strong acid and the more strong base you add the weaker the buffer is going to get because these two are neutralizing each other and these two are neutralizing each other add too much of this and then all of this will be gone and without a weak acid present, you would no longer have a buffer. All you'd have is conjugate base, add too much of the strong acid and you would destroy your conjugate base. Therefore you would no longer have a buffer because all you have left is weak acid being produced. So a buffer resists large changes in ph by stepping up and counteracting when a strong acid or a strong base is added up to a certain point. So these are the fundamentals of what a buffer is. Click on the next video where we can talk about what are the best ranges for which a buffer can operate.
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Buffers
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So we've understood the fundamentals when it comes to buffers but what classifies a good buffer versus a bad buffer. Doesn't really matter what the proportions of weak acid are to to conjugate base. Well in reality yes it does matter. We're gonna say the weak acid and conjugate base can be different from one another by up to a magnitude of 10. We're gonna say that this is called the buffer range. If they're different by more than a magnitude of 10 then it will not be a buffer. So remember with buffers we have a conjugate base to weak acid ratio. So we're gonna say conjugate base CB two weak acid ratio. They can't be more than a magnitude of 10 away from each other. So how do I check to see if they are you would just say conjugate base to weak acid is just conjugate base over weak acid at most. That can be different by up to a magnitude of 10. Which means that conjugate base could be one, weak acid could be up to 10. So this is 100.10 Or conjugate base weak acid conjugate bases, 10 weak acid is one. So this is 10. So when we say a buffer range in a magnitude of 10. That means a range when we do conjugate base over weak acid. When we plug in the numbers for each one they have to lie between .10 And up to 10. Okay, so they have to fall within that range. If the conjugate base to weak acid ratio is outside these numbers then it is not a good buffer. And in fact we may we can say that although it looks like we have a weak acid and conjugate base, it does not constitute a buffer because the range the ratio doesn't give me a value of 0.10 to 10. So remember just do conjugate base over weak acid and put the values that you're given and see do they lie between 100.10 and 10? So if we take a look at this example we have one molar of our week oxy acid and 0.10 of our conjugate base here are weak acid is 10 times the amount of the conjugate base. So it's at an exact limit can't be more than 10 fold difference between the both of them, they've reached their maximum, you would do conjugate base over weak acid if you wanted and you can see how that stacks up and yes we get 0.10 which is within that range of 0.10 to 10 Here for this one. Now I've changed up the concentration so now it's .05 molar of my weak acid and 10.50 Moeller of my conjugate base. So my conjugate base to a weak acid ratio So that equals 10. Yes, again we fall within this range. So anytime you're given a weak acid and its conjugate base with amounts you can check to see does it lie within the right buffer range? If it doesn't it doesn't constitute a buffer. Now that we've learned about buffer range, click on to the final video to learn about one more requirement when it comes to being a good effective buffer.
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Buffers
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So the last idea behind a good effective buffers. Pretty simple here we're going to see the more concentrated the weak acid and conjugate base. The components of a buffer then the better the buffer can counteract the strong acid, strong base added. Remember if I had strong acid, my conjugate base steps up to neutralize it. If I had strong base then my weak acid steps up to neutralize it. We call this buffer capacity. So basically the larger the amount of weak acid and conjugate base you start with, the better and longer they're able to combat against strong acid and strong base that you're adding. So here we have our first buffer amount where one molar of each versus we have .01 molar of these. Both represent buffers because both have equal amounts of both weak acid and conjugate base. And in fact when you're weak acid is equal to your conjugate base amount we say that this is called an ideal buffer. Remember they can be different by up to a magnitude of 10 but if they're exactly the same, that's great. But the first combination is better because the numbers are much larger which means they'll be better able to combat longer larger amount of strong acid and strong base being added. So remember an ideal buffer is when they're equal in amount But they can be up to a difference of 10 from one another. And remember a buffer itself is just a weak acid and its conjugate base. Now that we've talked about what a buffer is, click on to the next series of videos to learn about how exactly can we create a buffer?
Buffer Synthesis
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Buffer Synthesis
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so we know that a buffer is a weak acid and its conjugate base. Once you have a buffer you can easily calculate the ph of your solution by relying on the Henderson Hasselbach equation. So here are Henderson Hasselbach equation as P H equals P. K. A. Plus log of your conjugate base over the weak acid. Now the units that you can place in here are either polarity or moles, it's really just based on the information provided and those are the two most common units used for this particular equation, realize here that we're only talking about one P. K. A. So this relies on a mono product buffer. Remember we have di product and polly protic acid that exists. So they also have their own set of Henderson Hasselbach equations that we'll discuss later on for now we're just focused on mono product acids and their conjugate bases. Now, like we said, we know that a buffer is a weak acid and its conjugate base, realize that there are three different ways we can create any buffer. So if we take a look the first way is pretty explanatory, it's pretty obvious to make a buffer, we mix some weak acid and its conjugate base. Okay, so we have a weak acid and its conjugate base in this case a buffer is most ideal when both components are highly concentrated because remember better buffer capacity when it's more concentrated and equal to one another. So it'll fit within the buffer range. So highly concentrated buffer with equal amounts of both, weak acid and conjugate base represents an ideal buffer. Now here we have 0.10 moller of hipaa Cloris acid with 0.10 moller of sodium hypochlorite here it's an ideal buffer because both are equal in amount. Now. The other two ways may not be as obvious in the creation of a buffer. The second way to make a buffer is to mix strong acid and conjugate base. Not its conjugate base, but the conjugate base of a weak acid. In this case the strong species mixing with the weak species Has to be lower an amount. So here my strong thing has to be there has to be less of it. The week species must be higher in amount, whether that amount is in moles or in polarity. The weak conjugate base has to be higher in amount. So here we have one molar of hydrochloric acid which is a strong acid. And here we have 1.25 molar of Ammonia which is a weak base. The week basis hire an amount. Therefore a buffer has been created Later on. We'll talk about hydration of buffers and we'll see why this is why does this help to create a buffer. Now for the second one we have 1.50 moller of hydrochloric acid but we still only have 1.25 molar of ammonia. In this case the strong acid is greater in amount. Therefore no buffer has been created because there is an excess of strong acid. It does not allow the formation of a buffer because it completely destroys all of this base here. Next we're going to say the last way to make a buffer is mixing a strong base with a weak acid. So again in this case we have something weak and strong mixing and again whatever is weak in this case the weak acid has to be higher in amount. So the weak species again must be higher an amount. So the same basic logic as we saw before. But now instead of polarity we're looking at moles which again we can use polarity or moles to fit our understanding of what's going on. If we look at the first one we have 1.50 moles of nitrous acid versus 1.25 moles of sodium hydroxide. The weak acid is hire an amount. Therefore a buffer can be created and then below it are strong base now though is hire an amount so no buffer can be created because this weak acid component is totally destroyed by the excess strong base. And if you need if you want a buffer you need to have some weak acid and you need to have conjugate base. All of this has been destroyed. So remember a buffer itself is a weak acid and its conjugate base and when it comes to the formation of a buffer there are three routes we can take, one is the obvious weak acid and its conjugate base and then the other two not so obvious, mixing something weak with something strong. In those last two cases, whatever is weak must be greater in amount, whether that's in polarity or whether that's in moles.
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Buffer Synthesis Calculations 1
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So here it states which of the following compounds can result in the formation of a buffer. Now remember with buffer creation there are three methods we can use So method one we can have weak acid and its conjugate base in this situation. Remember they have to operate within a buffer range. The difference between them cannot exceed a magnitude of 10 next to make a buffer. We could use weak acid plus strong base in this case we have something week with something strong. So remember in this situation whatever is weak has to be higher in amount. And then finally the third way we can make a buffer is to react a strong acid plus a weak base. Again we have something weak with something strong. So again whatever is weak must be higher an amount. Now in this question we're dealing with volumes and polarity. Remember polarity itself equals moles over leaders. And if you multiply both sides by leaders we'll see that moles equals leaders times more clarity. Hear the word of when it's in between two numbers means multiply. So if I were to change the mls into leaders and multiply them by the polarity, I can get moles to do that. We would just divide all the mls by 1000 and we would have leaders. But this being analytical you could also keep them as middle leaders and multiplied by polarity. And in that case you'd be dealing with millie moles. Now remember we're dealing with small amounts of solutions and compounds. So it's not uncommon to see Milly moles pop up from time to time. So you can use either method and you'll get the same number proportionally here, I'll just change them to leaders and multiplied by malaria to get moles. So when I do that for the first one I get .0075 moles. So here we have chlorate acid which represents a strong acid. And here we have methyl amine, it's an amine, it has carbon, hydrogen and nitrogen. It's a weak base. And here we have .005 moles here. Unfortunately, we do not have a buffer because again, when it's weak and strong, the weak species must be higher an amount in order for a buffer to be created here, the strong acid is hire an amount. So no buffer is created next here we have When we divide by 1000 multiplied by modularity, we get .0025 moles of a weak acid. And then here we get .004 moles of a strong acid. I'm a strong base. Now, H two S. 03 is sulfurous acid. It's a dye protic acid. But remember it's a weak acid to begin with. And even if it were a strong acid like H two S 04, only the first acidic hydrogen would be strong. The next one would be very weak. I'm telling you this because although it has two acidic hydrogen, we would not double the polarity here because when it comes to die a protic acids at best. The first hydrogen could be strong like sulfuric acid or in this case both will be weak. So there's no point in multiplying the concentration times too. So it just stays that number. Again, a strong species is hire an amount. So this is not a buffer. Next we have .005 moles. NH four cl represents an ionic salt. It breaks up into n H four positive and cl minus cl minus comes from a strong acid. So it is neutral. NH four positive is a positive amine. So it's a weak acid. So this salt is a weak acid. And then here we have strong team hydroxide. This is a strong base because we have a group to a metal with o h minus remember strong bases when they ayan eyes. Each hydroxide is of equal strength. And because each hydroxide is of equal strength, that means that I have to take into account both of them, which means I'd have to double This concentration because there are two of them. And so here I would also have .005 moles of a strong base. Now remember when it's weak and strong, the weak must be higher an amount here, it's equal to the strong acid to the strong base. So does not make a buffer. The weak must be higher an amount. Then finally we have divide this by 1000 multiplied by 10000.20 Moeller gives me 0.10 moles of H F, which is a weak acid. And then here 40 mls I divide by 1000 multiplied by polarity gives me 10000.80 moles of a strong base. Here, the weak species is higher than the strong species, so this will make a buffer. So remember, a buffer is weak acid and conjugate base. And when it comes to the formation of a buffer, these are three methods to do so.
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Buffer Synthesis Calculations 1
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So here it says calculate the ph of a solution formed by mixing 130 mls of 1300.300 moller ethyl amine solution with 70 mls of a 700.500 moller ethel ammonium solution here were given the KB of ethyl amine as being 5.0 times 10 to the negative four. Alright realize here, what do we have? Well, we have here a neutral amine compound, A neutral mean represents a weak base and here we have an extra hydrogen present. That's why it has a positive charge and we gain an extra hydrogen. So this represents its conjugate acid. You could also say that a positive amine represents a weak acid and therefore this has one less H plus this represents its conjugate base. So whether you're looking at looking at it as weak base and conjugate acid or weak acid and conjugate base. Both pairs are saying that we have a buffer. So we know we have a buffer. So we should rely on the Henderson Hasselbach equation. So P H equals P K A plus log of conjugate base over weak acid. Now here P K is just the negative log of K A. But we don't have K. We have K B. So we have to convert K B to K A. So we're gonna say here K A times KB equals are constant KW we're looking for a K A K b is 5.0 times 10 to the negative four Kw 1.0 times 10 to the negative 14. Remember kw is temperature dependent. But if no temperature is given, we assume it's 25 degrees Celsius and therefore Kw is this number of 1.0 times 10 to the negative 14, Divide both sides by 5.0 times 10 to the -4. That'll give me my K. A. So negative log. And when we plug that in my K. Is 2.0 times 10 to the negative 11 plus log of. All right. So my conjugate base is this ethyl amine? Remember in the previous video I said that you could do moles equals leaders, times more clarity. Or we could do millie moles equals milliliters, times more clarity, proportionally. Nothing really changes. So you can use either method. So we're gonna keep it in Ml so we can see how we get millie moles. So multiply these two numbers together because of means multiply. So when I do that I'm gonna get 39 millie moles of ethel um of salome and then I multiply 70 mls times 700.500 moller. So that's gonna give me 35 million moles. So I'll get that amount 35 million moles of ethyl ammonium ion. So then when I punch that into my calculator you should get back 10.75 as the ph for this buffer solution. So just remember in questions like this you should be able to spot. Do you have a buffer in any capacity. If you do then you should use the Henderson Hasselbach equation using that is the quick way of getting the ph for your buffer solution. Now that you've seen this look to see, can you answer the practice question left here on the bottom? Don't worry. If you get stuck as always, click on the next video and see how I approach that same exact practice question.
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Problem
Problem
Which of the following molar ratios is the correct equilibrium ratio of BASE : ACID for a solution made of aniline (Kb = 3.8 x 10-10) and anilinium nitrate where the pH is 4.80?
A
1:2
B
3:5
C
7:2
D
2:1
E
5:3
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example
Buffer Synthesis Calculations 2
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So for this question it states you are asked to go into the lab and prepare a solution with a ph of 6.40 plus or minus 0.2, which weak acid would be the best choice. Alright, so for a question like this when they're giving you a ph or in this case a range for the ph and they're asking, what's the best buffer to use? They're really asking you what's the most ideal buffer you can use for this ph and remember the word ideal when it comes to a buffer means that our weak acid and our conjugate base concentrations are basically equal to one another. And if we're dealing with a buffer, think of the Henderson Hasselbach equation, P H equals P K A plus log of conjugate base over weak acid. Now if they are equal to each other weak acid and conjugate base then this here will equal one. and remember log of one Equals zero. So this drops out. So when we have an ideal buffer ph equals P K. A. Now remember that P K A. Just means negative log of K. A. And what we're gonna do here is we're gonna divide both sides by -1. So negative P H equals log of K. A. If I take the anti log of both sides, this becomes 10 to the negative ph and then this log cancels out when I do the anti log of that side and just is left with K. A. So K. A equals 10 to the negative ph this is the equation we rely upon with this equation here, we can take the ph plug it in to get a K A value. The weak acid with the closest K value to our answer would be the most ideal buffer to use at a ph of 6.40. So when I do that I get 3.9, 8 times 10 to the -7. The closest K value that I see to that is the first one carbonic acid. Now realize that R K a is not exactly 4.2 times 10 to the negative seven. That's because we have levels of uncertainty there. Remember it's plus or minus 70.2 but the K s are far enough apart that I know for sure this is the correct answer because the other ones are off by magnitudes. So just remember when they're asking you to find the best buffer to operate at a certain ph they're really asking you to think of this relationship because for an ideal buffer ph equals P K A. If you can't quite remember that formula. What you could have also done is you could have taken the negative log of every single K A to get a p K. And since p K equals P H. For an ideal buffer, you will just look for which P K A is closer to the given ph this would still give you option A is the best answer. But use that method, if you can't remember this particular equation and how it relates to an ideal buffer now that you've seen this, see if you can do example to left on the bottom of the page. When you're done, come back and see how I approach that same question to get my answer.
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example
Buffer Synthesis Calculations 2
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So here it says calculate the ph of a solution made by mixing 8.627 g of sodium baton await and enough 0.45 to moller butin OIC acid which is the formula H. C. 48702. To make 250 ml of solution here we're told the K. A. Button OIC acid is 1.5 times 10 to the negative five. All right. So we have butin OIC acid and this particular asset has a K. A. Less than one. So, we know it's a weak acid and then we have a second compound sodium utan await recognize that the names are very similar. That that's because sodium you tanouye represents the conjugate base form of the acid. Now we need to know what the conjugate base looks like in order to determine its moles because I need to use the Henderson Hasselbach equation P. H. Equals P. K. A. Plus log of conjugate base over weak acid. We already have the K. A. And we have volume and polarity of my weak acid from that I can find millie moles or moles here. I'll just use moles. So we have .250 L multiplied by .45 to moller Gives me .113 moles a beauty like acid. So we have the bottom portion here. Now realize what am I missing. I'm missing the moles of my conjugate base. I need to know what sodium baton away looks like. Remember the conjugate base looks just like the weak acid except it has one less hydrogen present it lost an H. Plus to become the conjugate base. Usually the H plus that is lost is replaced with a metal What metal? In this case, sodium? So butin OIC acid has this formula. So sodium you Tana weight would be n. a. c four H 702. We've lost that H plus and replaced it with an N. A. Here. When you add up the one sodium four carbons, seven hydrogen and two oxygen's from their atomic masses on the periodic table, you'll get the combined mass of sodium. Utan await as being 110.87 g per mole. So that's the weight of that compound from this will be able to change the g of it that we have into moles. So I have 8.6-7 g. And I'm gonna say for every one mole of it, We have 100 and 10.087 g of it. So that equals .078366 moles of my conjugate base. So take those molds and plug it into the Henderson, Hasselbach equation. And then when we plug all that in we'll get approximately 4.66 for the ph making e the closest and most um correct answer out of the options given. So just realizing this question, we have beauty in OIC acid and we have a cave less than one. So we know it's the weak acid. Then sodium you tanouye has a similar name, so they must share some type of relationship with each other. The fact that a medal is involved with it must mean it's the conjugate base. The H plus was lost and replaced by that group, won a medal to give us the conjugate base. Knowing all this allowed us to use the Henderson Hasselbach equation to find our final ph.