Diprotic Buffers - Online Tutor, Practice Problems & Exam Prep

Diprotic buffer systems involve acids with two acidic hydrogens, leading to two distinct acid dissociation constants (Ka1 and Ka2). The Henderson-Hasselbalch equation adapts for these systems, allowing calculations based on the forms present: pH = pKa1 + log([intermediate]/[acid]) or pH = pKa2 + log([base]/[intermediate]). Understanding which form to use is crucial for accurate pH determination. This knowledge is essential for mastering acid-base reactions and buffer capacity in various chemical contexts.

Diprotic Buffers

In our discussions of buffers up to this point, we've discussed monoprotic buffers where the acid only has one acidic hydrogen. Now, we're going to extend that further by taking a look at diprotic buffer systems. We're going to say here that a diprotic buffer can be approached in a way similar to monoprotic buffers. The key difference is that a diprotic acid has 2 acidic hydrogens. We're going to say because it has 2 acidic hydrogens, it's going to possess 2 pKa1 or 2 Ka values.

Just a quick refresher: if we're taking a look at a monoprotic buffer, here represents our acid form where it possesses the H+ ion. Ka can be thought of as measuring the strength of this acid. The higher the Ka, the stronger it is, the more likely it is to give away an H+. Giving away an H+ creates its base form or conjugate base form.

Now, a good example here we have hypochlorous acid which is HClO. Oxyacid. Here, the conjugate base, we've removed the H+ from it. Usually, by losing an H+ from our acid form, we replace it with a group 1 metal. In this way, we keep it as a neutral compound.

Here it's a conjugate base because again it has one less hydrogen. We know that the Henderson-Hasselbalch equation is just simply pH equals pKa plus log of conjugate base over weak acid. Here in a monoprotic system, we only have 1 Ka value associated with the weak acid. We know that this pKa is just one possible value. When we deal with diprotic systems, though, we're going to have 2 acidic hydrogens.

We'll have to choose wisely. Do we use pKa1 or pKa2 within our calculations? Click on to the next video and see how we approach diprotic buffer systems.

Diprotic Buffers

So now when we take a look at diprotic buffers, we have to pay attention to the fact that we'll have more than one k value associated with our acid forms. Here, we started with H2A, which is the generic form of any diprotic acid. Here it's fully protonated so it has its 2 acidic hydrogens. We're going to remove the 1st acidic hydrogen to give us A-. If we're talking about removing the 1st acidic hydrogen, then we're dealing with Ka1.

And the A- represents the middle or intermediate form of our initial diprotic acid. Now, when we're in the A- form, we can talk about removing that last and second H+ ion. Talking about removing the second H+ ion means that we're going to deal with Ka2. Removing it gives us A2- which represents the base form with no H+ whatsoever attached to it. Because we have the presence of 2 acidic hydrogens, we'll be dealing with 2 k values within our calculations.

We just have to make sure, which form are we going to use. So if we take a look here because of this, the presence of 2 kas, we have 2 different forms of the Henderson-Hasselbalch equation. If we're dealing with the acid form and the intermediate form, then Ka1 connects them together. So the Henderson-Hasselbalch equation becomes pH=pKa1+log(intermediateacid). If we're dealing with the intermediate form and the base form, then we're dealing with Ka2.

So the Henderson-Hasselbalch equation becomes pH=pKa2+log(baseintermediate). If we're to take a look here at these two examples that we have, so if we take a look at this first one, here I'm telling us that we're dealing with sulfuric acid. Sulfurous acid is H2SO3. This is the acid form of our acid. If we're going to remove its first acidic hydrogen, that means we're going to deal with Ka1.

And that will give us HSO3-, which is hydrogen sulfite or bisulfite. This is my intermediate form. Then finally, removing the second and which represents the last acidic hydrogen, we're dealing with Ka2. That would give me your sulfite ion. So this is the base form.

In this first question, we're dealing with the example here, we'd say pH=pKa1=pka-log(ka1)+log(intermediateacid), giving us 1.89 as our possible answer. Now in this one, we're given volume of molarity. Remember when we talked about simple buffer systems when we're dealing with monoprotic acids in earlier videos? We said that moles equals liters times molarity.

If we were to change these mL's into liters by dividing them by a 1,000 and then multiplying by the molarity, we'd get the moles of both of these compounds. It'd be that many liters times 0.10 molar. That'd be 0.03 moles. Then we have 0.020 liters times 0.20 molar. That'd give me 0.004 moles.

Now in this example, what do we have? We have sodium sulfite. Here, both acidic hydrogens are totally gone now. This must represent the base form. Here, we still have one acidic hydrogen remaining, so this is the intermediate form.

Here because we're dealing with the base form and the intermediate form, we know that the Henderson-Hasselbalch equation becomes now pH=pKa2+log(moles basemoles intermediate). And again, remember also from earlier videos, we said that when it comes to the Henderson-Hasselbalch equation, we could use either moles or molarity as the units within these formulas. So when we work this out, it gives us 7.07 as our final pH in terms of determining the pH of my total solution. We can see there is a difference when we do this calculation whether we're dealing with the acid form and its intermediate or the intermediate form with its base form. So we have to be wary when it comes to diprotic buffer systems.

Know what forms you're dealing with, which will therefore determine which pKa value to use, which at the end will determine what your final pH will be. As we continue our discussion of buffers, we'll continue to do calculations dealing with diprotic acids. And later on, we'll move on to polyprotic acids.

Monoprotic & Diprotic Buffer Calculations 1

Here it states, what is the pH of a solution consisting of 2.5 molar potassium dihydrogen phosphite and 2.75 molar of phosphorus acid? Now here we're given 2 Ka's, Ka1 and Ka2. And if we take a look at phosphorous acid, it's H₃PO₃. And you might think that it's a triprotic acid because it has 3 hydrogens in front of it. But in reality, because of its structure, phosphorous acid only has 2 acidic hydrogens.

It looks something like this. The only acidic hydrogens are these 2 here. It's actually a diprotic acid not a triprotic acid as it might seem. Now, here we're dealing with its fully acidic version when none of its hydrogens have come off yet. This is its acid form.

Then here we've lost one of the hydrogens so that only has 2. The hydrogen that we've lost has been replaced by a potassium ion, so this potassium here. This represents our intermediate form. Now remember, if we're dealing with the acid form and the intermediate form, that means we're talking about Ka1. We're going to have to use Ka1 in this question.

Here we're going to say pH equals pKa1 plus log of the intermediate form over the acid form. Here, that'd be negative log of 3.0 x 10^-2 plus log of 2.5 molar, the concentration of my intermediate form divided by 2.75 molar which is the molarity of the acid form. When we plug these in, that gives me a value of 1.48 as my pH. Remember, for questions like this, we have to ascertain for sure what forms we are dealing with. Are we dealing with the acid form with the intermediate form?

In that case, we're dealing with Ka1. Are we dealing with the intermediate form and the base form? Then we would be dealing with Ka2. Now, take the same approach as we move on to example 2. Try to work out the question yourself to see how to answer it.

But if you get stuck, don't worry. Just come back and take a look at the next video. Look to see what exactly we are dealing with here, what forms are involved and therefore what Ka to use.

Monoprotic & Diprotic Buffer Calculations 1

Here it states that sulfuric acid, which is H₂SO₃, is a major component in the creation of commercial fertilizers. Here it asks, what is the buffer component concentration ratio of a buffer that has a pH of 1.15? All right. We have a buffer here, so we know we're dealing with the Henderson-Hasselbalch equation: pH equals pKa plus log of our conjugate base over our weak acid.

When it asks for the component concentration ratio, it's really asking what is the ratio of conjugate base to weak acid. What is that value there? We run into the problem though. We're dealing with a diprotic acid here and that's illustrated by it having 2 Ka values as well. We don't know, are we dealing with pKa₁ or pKa₂ within this question?

How do we determine that? Well, we can take a look at all the forms that sulfuric acid takes when we begin to remove its H+ ions. Here we have sulfuric acid which is the acid form. We remove the 1st acidic hydrogen so that means we're dealing with Ka₁. That's HSO₃⁻.

Then we remove the last acidic hydrogen so we're dealing with Ka₂ to give us SO₃²⁻. Now, we're going to say here that the relationship between the acid form and the intermediate form, we're going to say we have this imaginary line here. We're going to say this imaginary line here, we're going to say this represents the line for pKa₁. Here if we took the negative log of Ka₁ which is this number right here, take the negative log of it, we'll get pKa₁ equals 1.86. Then we're going to say there's an imaginary line here which separates the intermediate form from the base form.

We take the negative log of this Ka₂ to give me pKa₂. That comes out to 7.17. These pKa values are important because we're going to say here, we can relate them in terms of pH. Basically, if our pH is equal to 1.86, so let's say pH equals pKa₁. What would that mean?

That would mean that my acid form would be equal to my intermediate form. How do we know that? Well, if we're dealing with a buffer, if my acid form which would be this form is equal to my intermediate form which is this form, all of this would just equal 1. Log of 1 would equal 0. That would just all drop out.

And so pH would equal pKa₁. And we're gonna say here, if my pH equaled pKa₂, that would mean that my intermediate form equals my basic form for the same exact reason. Now, if your pH is less than your pKa₁, what does that mean? That means you're dealing with a give and take between the acid form and the intermediate form. Our pH here is 1.15 which means we fall somewhere in here which means we're talking about removing that first acidic hydrogen in order to create this intermediate form.

If your pH happened to be a number that was greater than 1.86 less than 7.17, then you would fall somewhere within here. Now remember, this goes all the way back to our principal species when we're talking about the form that predominates depending on the pH and when we compare it to our pKa value. If we had a pH greater than this pKa₂, then this would be the dominant form. Now, going back, we said the pH is 1.15 so we know we're dealing with, basically the acid form as the predominant form. We're talking about removing its first acidic hydrogen to make the intermediate form.

That would mean our equation now is pH equals pKa₁ plus log of my intermediate form divided by my acid form. So pH 1 is 1.15. Then we're going to say here, so let me take myself out of the subject, guys. pH equals 1.15. pKa₁ equals 1.86 plus log of HSO₃⁻ divided by H₂SO₃.

Subtract both sides by 1.86. So when we do that, we're gonna get initially negative 0.71 equals log of the intermediate form divided by the acid form. Then all we do now is we take the inverse log, which just means here 10 to the negative 0.71 equals the ratio of my intermediate form divided by my acid form. So when we punch that number into our calculators, that's gonna give us 0.195 over H₂SO₃. So what is this number telling me?

Well, this number here is telling me since it's a ratio, whatever number I get is always going to be over 1. So what this ratio is telling me is that for every one weak acid form, we have 0.195 of the intermediate form. That's the ratio. This ratio is telling me that I have more of the acid, weak acid form than I do of the intermediate form. Our ratio here would be 0.195 to 1 to represent the ratio of my conjugate base or intermediate to my weak acid form.

Just remember, remember with polyprotic acids and diprotic acids, we have to be aware of which Ka we are dealing with. That will determine which pKa we're going to deal with within our Henderson-Hasselbalch equation to find either the pH, the ratio of the conjugate base to the weak acid, or possibly just the amount of the conjugate base by itself or the weak acid by itself. So now that we've seen these two examples, attempt to do the practice question that's left on the bottom of the page. Attempt it on your own. Come back and take a look and see how I approach that same question.

Calculate the pH of a solution made by mixing 8.627 g of sodium butanoate in enough 0.452 M butanoic acid, HC₄H₇O₂, to make 250.0 mL of solution. K_a = 1.5 x 10^-5. 

• A. 4.82
• B. 4.98
• C. 5.85
• D. 4.66
• E. 4.06

Here’s what students ask on this topic:

What is a diprotic buffer system?

A diprotic buffer system involves an acid that has two acidic hydrogens, meaning it can donate two protons (H⁺). This results in two distinct acid dissociation constants, Ka₁ and Ka₂. The system can be described using the Henderson-Hasselbalch equation, which adapts to the different forms present: pH = pKa₁ + log([intermediate]/[acid]) or pH = pKa₂ + log([base]/[intermediate]). Understanding which form to use is crucial for accurate pH determination in various chemical contexts.

How do you calculate the pH of a diprotic buffer system?

To calculate the pH of a diprotic buffer system, you need to identify the forms present: the fully protonated acid (H₂A), the intermediate form (HA^-), and the fully deprotonated base (A^2-). Use the Henderson-Hasselbalch equation: for the acid and intermediate forms, pH = pKa₁ + log([intermediate]/[acid]); for the intermediate and base forms, pH = pKa₂ + log([base]/[intermediate]). Ensure you use the correct pKa value and concentrations for accurate pH determination.

What are the key differences between monoprotic and diprotic buffer systems?

The key difference between monoprotic and diprotic buffer systems is the number of acidic hydrogens. Monoprotic acids have one acidic hydrogen and one dissociation constant (Ka), while diprotic acids have two acidic hydrogens and two dissociation constants (Ka₁ and Ka₂). This results in different forms and equations for pH calculations. Monoprotic systems use pH = pKa + log([base]/[acid]), whereas diprotic systems use either pH = pKa₁ + log([intermediate]/[acid]) or pH = pKa₂ + log([base]/[intermediate]).

Why is it important to choose the correct pKa value in diprotic buffer calculations?

Choosing the correct pKa value in diprotic buffer calculations is crucial because it ensures accurate pH determination. Diprotic acids have two dissociation constants, Ka₁ and Ka₂, corresponding to the removal of the first and second acidic hydrogens, respectively. Using the wrong pKa value can lead to incorrect pH calculations, affecting the buffer's effectiveness. For the acid and intermediate forms, use pKa₁; for the intermediate and base forms, use pKa₂.

How does the Henderson-Hasselbalch equation adapt for diprotic buffer systems?

The Henderson-Hasselbalch equation adapts for diprotic buffer systems by incorporating the two dissociation constants (Ka₁ and Ka₂). For the acid (H₂A) and intermediate (HA^-) forms, the equation is pH = pKa₁ + log([intermediate]/[acid]). For the intermediate (HA^-) and base (A^2-) forms, the equation is pH = pKa₂ + log([base]/[intermediate]). This adaptation allows for accurate pH calculations based on the specific forms present in the buffer system.