Buffers - Online Tutor, Practice Problems & Exam Prep

A buffer consists of a weak acid and its conjugate base, which work together to resist significant pH changes. The effectiveness of a buffer is determined by its buffer capacity, which increases with higher concentrations of the acid and base. The Henderson-Hasselbalch equation, $\mathrm{pH}=\mathrm{pK}+\log (\mathrm{conjugate}base/\mathrm{weak}acid)$, allows for pH calculation in monoprotic buffers. A good buffer maintains a conjugate base to weak acid ratio within a magnitude of 10.

Buffers

A buffer is just basically a weak acid with its conjugate base. A great example is hypochlorous acid, HClO. This is a weak oxyacid. Here, when we talk about a conjugate base, it is the same form as the weak acid except it has one less hydrogen. That hydrogen that it's missing is typically replaced with a metal usually from group 1a. Hypochlorous acid is HClO, and here we have sodium hypochlorite. The hydrogen has been replaced with a metal. We have a weak acid and its conjugate base. Together, they form a buffer. Now, a buffer, what does it do? Well, it causes my solution to resist big changes in pH. It does this by keeping constant both my H⁺ ions and my hydroxide ions. It tries to keep them around the same level as much as possible and for as long as possible. A 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 would 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. If we add a 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, I don't want to say enemies, but they're natural opposites of one another. If they're both present together in the same space, they try to neutralize one another. By adding a strong base, the weak acid steps up to try to get rid of it. Here, we still have our weak acid, our hypochlorous acid. I add some NaOH, a strong base. Here, we undergo a neutralization reaction with 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 NaOH and therefore stabilize my buffer plus water. If I add a strong acid, what happens? If I add strong acid, then the other part of my buffer, the conjugate base, steps up to neutralize that. 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. 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. Without a weak acid present, you would no longer have a buffer. All you'd have is the 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'd have left is weak acid being produced. A buffer resists large changes in pH by stepping up and counteracting when a strong acid or strong base is added up to a certain point. 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.

Buffers

So we've understood the fundamentals when it comes to buffers. But what classifies a good buffer versus a bad buffer? Does it really matter what the proportions of weak acid are to the conjugate base? Well, in reality, yes, it does matter. We're going to say the weak acid and conjugate base can be different from one another by up to a magnitude of 10. We're going to 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. Remember, with buffers, we have a conjugate base to weak acid ratio. We're going to say conjugate base (CB) to weak acid ratio. They can't be more than a magnitude of 10 away from each other. 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, they can be different by up to a magnitude of 10, which means that the conjugate base could be 1, weak acid could be up to 10. This is 0.10. Or conjugate base, weak acid. Conjugate base is 10, weak acid is 1. This is 10. When we say a buffer range in a magnitude of 10, that means a range when we do CBWA, when we plug in the numbers for each one, they have to lie between 0.10 and up to 10. K. 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 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. Remember, just do conjugate base over weak acid, input the values that you're given, and see, do they lie between 0.10 and 10. Here, if we take a look at this example, we have 1 molar of our weak oxyacid and 0.10 of our conjugate base. Here, our weak acid is 10 times the amount of the conjugate base. It's at an exact limit. It can't be more than tenfold difference between the both of them. You would do CBWA 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 0.05 molar of my weak acid and 0.50 molar of my conjugate base. So, my conjugate base to weak acid ratio 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.

Buffers

The last idea behind a good effective buffer is pretty simple. Here, we're going to see the more concentrated weak acid and conjugate base, the components of a buffer, then the better the buffer can counteract the strong acid or 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. 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.

Here we have our first buffer amount where 1 molar of each versus we have 0.01 molar of these. Both represent buffers because both have equal amounts of both weak acid and conjugate base. And in fact, when your 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 a longer and larger amount of strong acid and strong base being added. 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

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-Hasselbalch equation. Here, our Henderson-Hasselbalch equation is pH equals pKa plus log of your conjugate base over the weak acid. Now, the units that you can place in here are either molarity or moles. It's really just based on the information provided and those are the 2 most common units used for this particular equation. Realize here that we're only talking about 1 pKa, so this relies on a monoprotic buffer. Remember, we have diprotic and polyprotic acids that exist, so they also have their own set of Henderson-Hasselbalch equations that we'll discuss later on. For now, we're just focused on monoprotic acids and their conjugate bases. Like we said, we know that a buffer is a weak acid and its conjugate base. Realize that there are 3 different ways we can create any buffer.

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. 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. A highly concentrated buffer with equal amounts of both weak acid and conjugate base represents an ideal buffer. Now here, we have 0.10 molar of hypochlorous acid and 0.10 molar 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 in amount. Here my strong thing has to be there has to be less of it. The weak species must be higher in amount. Whether that amount is in moles or in molarity, the weak conjugate base has to be higher in amount. Here we have 1 molar of hydrochloric acid which is a strong acid and here we have 1.25 molar of ammonia which is a weak base. The weak base is higher in amount. Therefore, a buffer has been created.

Later on, we'll talk about titrations 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 molar 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. 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.

The weak species again must be higher in amount. The same basic logic as we saw before but now instead of molarity, we're looking at moles which again we can use molarity 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 higher in amount. Therefore, a buffer can be created. Then below it, our strong base now though is higher in amount. No buffer can be created because this weak acid component is totally destroyed by the excess strong base. If you want a buffer, you need to have some weak acid and you need to have conjugate base. All of these have been destroyed. Remember, a buffer itself is a weak acid and its conjugate base. When it comes to the formation of a buffer, there are 3 routes we can take. One is the obvious weak acid and its conjugate base, and then the other 2 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 molarity or whether that's in moles.

Buffer Synthesis Calculations 1

So here it states, which of the following compounds can result in the formation of a buffer? Now remember, with buffer creation, there are 3 methods we can use. Method 1, we can have a 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 a weak acid plus a strong base. In this case, we have something weak 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 with a weak base. Again, we have something weak with something strong. So again, whatever is weak must be higher in amount.

Now in this question, we're dealing with volumes and molarity. Remember molarity itself equals moles over liters. And if you multiply both sides by liters, we'll see that moles equals liters times molarity. Here, the word "when" in between two numbers means multiply. So if I were to change the mL's into liters and multiply them by the molarity, I can get moles. To do that, we would just divide all the mL's by 1000, and we would have liters. But this being analytical, you could also keep them as milliliters and multiply by molarity. And in that case, you'd be dealing with millimoles. Now remember, we're dealing with small amounts of solutions and compounds. So it's not uncommon to see millimoles pop up from time to time. You can use either method and you'll get the same number proportionally.

Here, I'll just change them to liters and multiply by molarity to get moles. So when I do that for the first one, I get 0.0075moles. Here we have chloric acid which represents a strong acid, and here we have methylamine. It's an amine. It has carbon, hydrogen, and nitrogen. It's a weak base. And here we'd have 0.005moles. Here, unfortunately, we do not have a buffer because again, when it's weak and strong, the weak species must be higher in amount in order for a buffer to be created. Here, the strong acid is higher in amount so no buffer is created.

Next, here we have when we divide by 1000 and multiply by molarity, we get 0.0025moles of a weak acid. And then here we get 0.004moles of a strong base. Now, H₂SO₃ is sulfuric acid. It's a diprotic acid. But remember, it's a weak acid to begin with. And even if it were strong acid like H₂SO₄, only the first acidic hydrogen would be strong. The next one would be very weak. I'm telling you this because although it has 2 acidic hydrogens, we would not double the molarity here. Because when it comes to diprotic 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 2. So it just stays that number. Again, the strong species is higher in amount so this is not a buffer.

Next, we have 0.005moles. NH₄Cl represents an ionic salt. It breaks up into NH₄⁺ and Cl^-. Cl^- comes from a strong acid, so it is neutral. NH₄⁺ is a positive amine, so it's a weak acid. This salt is a weak acid. And then here we have strontium hydroxide. This is a strong base because we have a group 2A metal with OH^-. Remember, strong bases when they ionize, 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 2 of them. Here, I would also have 0.005moles of a strong base. Now remember, when it's weak and strong, the weak must be higher in amount. Here, it's equal to the strong as to the strong base. It does not make a buffer. The weak must be higher in amount.

Then finally we have divide this by 1000 multiply by 0.20 molar gives me 0.010moles of HF which is a weak acid. And then here, 40 mL, I divide by 1000 multiplied by molarity gives me 0.080moles 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 our 3 methods to do so.

Buffer Synthesis Calculations 1

So here it says, calculate the pH of a solution formed by mixing 130 ml of 0.300 molar ethylamine solution with 70 ml of a 0.500 molar ethylammonium solution. Here we're given the Kb of ethylamine as being 5.0×10-4. Realize here what we have. We have here a neutral amine compound. A neutral amine represents a weak base. Here, we have an extra hydrogen present. That's why it has a positive charge and we gain an extra hydrogen. This represents its conjugate acid. You could also say that a positive amine represents a weak acid and therefore, this has one less H⁺. This represents its conjugate base. Whether you're looking at it as a weak base and conjugate acid or weak acid and conjugate base, both pairs are saying that we have a buffer. We know we have a buffer, so we should rely on the Henderson-Hasselbalch equation.

pH equals pKa plus log of conjugate base over weak acid. Now, here pKa is just the negative log of Ka. But we don't have Ka, we have Kb. So we have to convert Kb to Ka. So we're gonna say here, Ka∗Kb=Kw. We're looking for Ka. Kb is 5.0×10-4. K_w, 1.0×10-14. Remember, K_w is temperature dependent, but if no temperature is given, we assume it's 25 degrees Celsius and therefore K_w is this number of 1.0×10-14. Divide both sides by 5.0×10-4. That'll give me my Ka. So negative log and when we plug that in, my Ka is 2.0×10-11 plus log of alright. So my conjugate base is this ethylamine. Remember, in the previous video, I said that you could do moles equals liters times molarity, or we could do millimoles equals milliliters times molarity. Proportionally, nothing really changes so you can use either method. We're gonna keep it in ml so we can see how we get millimoles. Multiply these two numbers together because that means multiply. So when I do that, I'm gonna get 39 millimoles of ethylamine. And then I multiply 70 ml's times 0.500 molar. So that's gonna give me 35 millimoles. So then when I punch that into my calculator, you should get back 10.75 as the pH for this buffer solution. 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-Hasselbalch equation. Using that is the quick way of getting the pH for your buffer solution. Now that you've seen this, look to see if you can 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.

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

Buffer Synthesis Calculations 2

So for this question, it states that 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. 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.

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. If we're dealing with a buffer, think of the Henderson-Hasselbalch equation:

pH = pKa + log conjugate base weak acid

Now, if they are equal to each other, weak acid and conjugate base, then this here will equal 1. Remember, log of 1 equals 0. This drops out. When we have an ideal buffer, pH equals pKa.

Now, remember that pKa just means the negative logarithm of the acid dissociation constant k_a. We can transform this relationship:

Ka = 10 - PH

This is the equation we rely upon. With this equation, we can take the pH, plug it in to get a k_a value. The weak acid with the closest k_a value to our answer would be the most ideal buffer to use at a pH of 6.40. When I do that, I get 3.98 times 10 to the power of -7. The closest k_a value that I see to that is the first one, carbonic acid.

Now realize that our k_a is not exactly 4.2 times 10 to the -7. That's because we have levels of uncertainty there. Remember, it's plus or minus 0.2. But the k_a's are far enough apart that I know for sure this is the correct answer because the other ones are off by magnitudes.

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 pKa. 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 pKa. Since pKa equals pH for an ideal buffer, you will just look for which pKa is closer to the given pH. This would still give you option as 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 the example 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.

Buffer Synthesis Calculations 2

So here it says, calculate the pH of a solution made by mixing 8.627 grams of sodium butanoate and enough 0.452 molar butanoic acid, which has the formula of HC₄H₇O₂, to make 250 ml of solution. Here we're told the K_a of butanoic acid is 1.5 × 10^-5. All right. We have butanoic acid, and this particular acid has a K_a less than 1, so we know it's a weak acid. And then we have a second compound, sodium butanoate. Recognize that the names are very similar. That's because sodium butanoate 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-Hasselbalch equation. Equation. PH equals pKa plus log of conjugate base over weak acid. We already have the Ka and we have volume and molarity of my weak acid. From that, I can find millimoles or moles. Here, I'll just use moles. We have 0.250 liters, multiply it by 0.452 molar gives me 0.113 moles of butanoic acid. 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 butanoate looks like. Remember, the conjugate base looks just like the weak acid except it has one less hydrogen present. It lost an H⁺ to become the conjugate base. Usually, the H⁺ that is lost is replaced with a metal. What metal? In this case, sodium. Butanoic acid has this formula. Sodium butanoate would be NaC₄H₇O₂. We've lost that H⁺ and replaced it with an Na. Here when you add up the 1 sodium, 4 carbons, 7 hydrogens and 2 oxygens from their atomic masses on the periodic table, you'll get the combined mass of sodium butanoate as being 110.087 grams per mole. That's the weight of that compound. From this, we'll be able to change the grams of it that we have into moles. I have 8.627 grams. I'm going to say for every one mole of it, we have 110.087 grams of it. So that equals 0.07836 moles of my conjugate base. Take those moles and plug it into the Henderson-Hasselbalch equation. And then when we plug all that in, we'll get approximately 4.66 for the pH, making it the closest and most correct answer out of the options given. Just realize in this question, we have butanoic acid and we had a K_a of less than 1, so we know it's the weak acid. Then sodium butanoate has a similar name, so they must share some type of relationship with each other. The fact that a metal is involved with it must mean it's the conjugate base. The H⁺ was lost and replaced by that group 1a metal to give us the conjugate base. Knowing all these allowed us to use the Henderson-Hasselbalch equation to find our final pH.

Here’s what students ask on this topic:

What is a buffer and how does it work?

A buffer is a solution composed of a weak acid and its conjugate base. It works by resisting significant changes in pH when small amounts of strong acid or base are added. The weak acid neutralizes added bases, while the conjugate base neutralizes added acids. This dual action helps maintain a relatively stable pH. For example, in a buffer system with hypochlorous acid (HClO) and sodium hypochlorite (NaClO), the weak acid (HClO) neutralizes added bases, and the conjugate base (ClO^-) neutralizes added acids, thus stabilizing the pH.

How do you calculate the pH of a buffer solution?

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:

$\mathrm{pH}=\mathrm{pK}{a}_{}+log\left(\frac{\mathrm{conjugate} \mathrm{base}}{\mathrm{weak} \mathrm{acid}}\right)$

Here, pK_a is the acid dissociation constant of the weak acid. The concentrations of the conjugate base and the weak acid can be in molarity or moles. This equation is particularly useful for monoprotic acids and their conjugate bases.

What factors determine the effectiveness of a buffer?

The effectiveness of a buffer is determined by its buffer capacity and the ratio of the conjugate base to the weak acid. Buffer capacity increases with higher concentrations of the weak acid and its conjugate base. Additionally, a good buffer maintains a conjugate base to weak acid ratio within a magnitude of 10. This means the ratio should lie between 0.10 and 10. If the ratio falls outside this range, the buffer will not be effective in resisting pH changes.

What is the buffer range and why is it important?

The buffer range refers to the range of pH values over which a buffer effectively resists changes in pH. It is determined by the ratio of the conjugate base to the weak acid, which should lie between 0.10 and 10. This range is important because it ensures that the buffer can neutralize added acids and bases effectively. If the ratio falls outside this range, the buffer will not function properly, and significant pH changes can occur.

What is the Henderson-Hasselbalch equation and how is it used?

The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution. The equation is:

$\mathrm{pH}=\mathrm{pK}{a}_{}+log\left(\frac{\mathrm{conjugate} \mathrm{base}}{\mathrm{weak} \mathrm{acid}}\right)$

Here, pK_a is the acid dissociation constant of the weak acid. The equation allows you to determine the pH based on the concentrations of the conjugate base and the weak acid. It is particularly useful for monoprotic acids and their conjugate bases.