Amino Acids and Henderson-Hasselbalch - Online Tutor, Practice Problems & Exam Prep

The Henderson-Hasselbalch equation relates the pH of a solution to the pKa of ionizable groups, crucial for understanding amino acids. It helps determine the final pH of weak acid solutions and the ratio of conjugate base to conjugate acid. This equation is applied to each ionizable group of an amino acid, allowing for the calculation of average net charges. The equation is expressed as: $\mathrm{pH}=\mathrm{pKa}+\log \left(\frac{\mathrm{[conjugate\; base]}}{\mathrm{[conjugate\; acid]}}\right)$.

Amino Acids and the Henderson-Hasselbalch Equation

In this video, we're going to talk about amino acids and the Henderson-Hasselbalch equation. So recall that way back in our previous lesson videos, we reviewed the Henderson-Hasselbalch equation from your old chemistry courses. And really, this lesson here on amino acids and Henderson-Hasselbalch is just an extension of those older lesson videos, and a lot of the principles and concepts that we already covered are just going to be reviewed and applied directly to amino acids. But if you don't remember much about the Henderson-Hasselbalch equation, make sure you go back and rewatch those older lesson videos before you continue here.

Now that being said, recall that the Henderson-Hasselbalch equation expresses the relationship between the solution pH and the pK_a of the ionizable group. Typically, the Henderson-Hasselbalch equation is used to determine one of two different things. The first is the final pH of a weak acid solution after it reaches equilibrium, and the second is the ratio of the concentration of conjugate base to the concentration of conjugate acid when we're already provided with the pH. Moving forward with the practice problems in this topic, it's important to know that the Henderson-Hasselbalch equation is applied independently to each of the ionizable groups of an amino acid. The concentration of conjugate base to concentration of conjugate acid ratio can actually be used to calculate the average net charges of ionizable groups.

pH = pK a group + log [ conjugate base ] [ conjugate acid ]

Moving forward, we're going to cover an example of how to apply the Henderson-Hasselbalch equation specifically to amino acids, and then you guys will be able to get a lot of practice with that. So I'll see you guys in those videos.

Amino Acids and Henderson-Hasselbalch

Alright. So here we have an example problem that wants us to calculate the percentage of protonated amino group or NH₃⁺ in the R group of lysine specifically at a pH of 9.8. And then it tells us that the pKa of lysine's R group or lysine's pKa_R is equal to 10.8. And so notice over here, I've drawn the structure of lysine just for visualization purposes. And notice that we're only being asked about one of lysine's ionizable groups and that is the amino group in the R group of lysine. And so notice at the very end of lysine's R group, there is an amino group. And really, this question is just asking us what percentage of all of the lysines in solution would have a protonated amino group in its R group at a pH of 9.8. And so we can use the Henderson-Hasselbalch equation to help us determine that answer.

And so notice that we're given the pH of the solution as 9.8, and we're also given the relevant pKa for the ionizable group, which is 10.8. And so really the only thing that we're not given is this ratio of the final concentration of conjugate base over the final concentration of conjugate acid. And so we'll need to solve for this ratio here in order to calculate the percentage of the protonated amino group. And so again, notice that we're given the pH here of 9.8, so we can go ahead and plug that in down below, 9.8, And that is going to be equal to our pKa, which is again given to us as 10.8 over here, so we can plug that in down below, plus the log of the concentration of conjugate base, which I'll abbreviate with CB, over the concentration of conjugate acid, which I'll abbreviate with CA. And again, we're looking to solve for this yellow ratio right here. So, what we'll need to do is start to isolate for it.

So we can subtract 10.8 from both sides of the equation, and of course 9.8−10.8 is going to be negative 1. And so this is going to be equal to the log of the concentration of conjugate base over the concentration of conjugate acid. And, again, we're trying to isolate for this ratio here, so we want to get rid of this log, and the way that we get rid of the log is to take the antilog. And so if we take the antilog of one side of the equation, we have to take the antilog of the other side of the equation. And the antilog of a number is literally just going to be 10 raised to the power of the same exact number.

And so this is going to be equal to, remember the log is removed through the antilog, and so we're just going to have the concentration of conjugate base over the concentration of conjugate acid. And so, 10−1 is essentially the same thing as one over 10. So this is the ratio that we were trying to solve for. And, really, what this ratio means is that for every one conjugate base, base, so one conjugate base is at the top, so for every one conjugate base, there are going to be 10 conjugate acids. And, so, essentially, what this means is that there's going to be a total of 11 molecules, so a total of 11, and that's because 10 + 1 is 11. And so essentially, because we're being asked about the protonated amino group, we need to realize that this is really the conjugate acid form because it is protonated and it has an extra hydrogen.

And so this is the conjugate acid form, and because it's asking us about the percentage, all we need to do is take the ratio of the conjugate acid to the total. And so we have 10 conjugate acids and for every 10 conjugate acids, we have 11 total molecules. And so this is going to give us a ratio, but we want a percentage, so we just multiply this ratio by 100%, and of course this is going to come out to about 90.909%, if you type that in your calculator, which is about equal to 91%. And so essentially, this percentage here is the percentage of amino group in the R group of lysine that will be protonated at pH 9.8. And so, that matches with answer option b, so we can go ahead and indicate that b here is the correct choice. And so that concludes this example problem, and you guys will be able to get a lot more practice utilizing the Henderson-Hasselbalch equation with amino acids in our next couple of practice videos. So I'll see you guys there.

At pH 11.8, what is the % of protonated amino group in the R-group of Lysine. (Lysine's pK_R = 10.8)?

• A. 9%
• B. 45%
• C. 3%
• D. 86%

Draw Glu & calculate the % of -COOH in the R-group of Glu at pH 3.2. (Glu pK  _R = 4.1).

• A. 88.8%
• B. 58.1%
• C. 97.3%
• D. 21.6%

Draw Asp & calculate the pH at which two thirds (2/3) of Asp's R-group is dissociated. (Asp pK_R = 3.9).

• A. 4.2
• B. 3.5
• C. 7.4
• D. 8.9

Draw Arg & calculate the pH at which 23% of Arg's R-group is dissociated. (Arg pK  _R = 12.5).

• A. 11.98
• B. 9.93
• C. 8.41
• D. 12.67

What is the ratio of [conjugate base] to [conjugate acid] for each of Histidine's three ionizable groups at pH 7?

A) Amino group ratio: ________________

B) Carboxyl group ratio: ________________

C) R-group ratio: ________________

D) Use the ratios above to determine the average net charge of the ionizable groups & the entire His molecule.

1. Net charge of Amino groups: ____________                                         

2. Net charge of Carboxyl groups: ____________

3. Net charge of R-groups: ____________                                                

4. Net charge of His: ____________ (Hint: sum previous 3 charges).

Alright. So here's a 4 part practice problem broken up into parts a, b, c, and part d down below. And so what you'll notice is the first three parts to our problem, parts a, b, and c, are just asking us to perform pretty straightforward calculations using the Henderson Hasselbeck equation. And so I provided the Henderson Hasselbeck equation up above in this box. And that's because it's asking us what is the ratio of the concentration of conjugate base to the concentration of conjugate acid for each of histidine's 3 ionizable groups, specifically at a pH of 7. And so notice that we're given the structure of histidine and we're also given the pKas for each of histidine's 3 ionizable groups. And so, what you'll notice is part a is asking us for the ratio of the amino group, b is asking us for the carboxyl group ratio, and c is asking us for the R group ratio. And for all three of these calculations, notice that the pH of the solution is the same. It's a pH of 7. And so notice that the pH is going to be the same for all of the calculations. And really, the only thing that's going to change is the pKa value for the calculations, and that's because each group has its own for the first group here, the amino group, and the amino group is shown to have a pKa of 9. So we can go ahead and plug that, pKa value in. So the pKa value is 9 for the amino group. And so, we're going to be solving for this ratio of the concentration of conjugate base to the concentration of conjugate acid. And so we can subtract 9 from both sides of the equation, and what we end up getting is negative 2 on the left side of the equation is equal to the log of the concentration of conjugate base over the concentration of conjugate acid. And so again, we're trying to isolate for this ratio, so we need to get rid of this log, and the way that we get rid of the log is to take the antilog. And if we take the antilog of the left-right side of the equation, we have to take the antilog of the left side of the equation. And the antilog of a number is literally just 10 raised to the power of the same exact number, so 10 raised to the negative 2, and that is going to be equal to the concentration of conjugate base over the concentration of conjugate acid. And so, 10 raised to the negative two is the same thing as saying, 1 over 10 squared, which is the same thing as saying one over 100. And so this is essentially the ratio here that we were looking for. And so, we can go ahead and put up above that the amino group ratio is going to be 1 over 100. And so we know that the amino group is going to have a conjugate base, which is the top number, of an NH2. That's the conjugate base of the amino group. It's deprotonated. And the conjugate acid form of the amino group is going to be an NH3+ that is protonated. And so really what this one-to-one hundred ratio really means is that for every one molecule of NH2, there are 100 molecules of NH3+. And so now that we are finished with this ratio here for the amino group, we can move on to part b and do the same for the carboxyl group. And so for the carboxyl group, we're given a pKa of 2. So we can go ahead and plug that into our equation here. So literally just swapping out the pKa, so pKa of 2. And so we're just gonna perform the same calculation. So we're going to subtract 2 from both sides of the equation, so what we end up getting is 5 on the left, is equal to the log of the concentration of conjugate base over the concentration of conjugate acid. And, of course, we're going to be isolating for this ratio here, so we need to take, the antilog to get rid of this log here, and the antilog of both sides of the equation is what we need to do. So the antilog of 5 is just gonna be 10 raised to the 5th, raised to the same power here. And that's gonna be equal to the law, the concentration of conjugate base over the concentration of conjugate acid. Alright. So, 10 to the 5th, 5th is the same thing as just 100,000, and this is a ratio of 100,000 over 1. So that is the ratio of concentration of conjugate base over the concentration of conjugate acid that we're looking for, so we can go ahead and plug that in up above over here, 100,000 to 1. And so really what this ratio means is, we know that the carboxyl group has a conjugate base form that is going to be the carboxylate anion. It is deprotonated, so it's gonna be COO-, and the conjugate acid form of the carboxyl group is gonna be protonated. It's gonna be COOH with a neutral charge. And so this ratio of 100,000 to 1 just means that for every 100,000 molecules of conjugate base form of the carboxyl group, there's gonna be 1 molecule of conjugate acid form of the carboxyl group. And so now that we've calculated the ratio for the carboxyl group, we can do the same for the r group ratio down below. And again, all we're gonna be doing is swapping out the pKa. And so, the pKa for the r group is 6. So we can go ahead and plug that in here. And so we're going to do the same calculation, subtract 6 from both sides of the equation, so what we end up getting is 1 is equal to the log of the concentration of conjugate base over the concentration of conjugate acid. And again, we're gonna take the antilog of both sides of the equation, so what we end up getting is 10 raised to the 1 is equal to the concentration of conjugate base over the concentration of conjugate acid. And so, 10 raised to the 1 is the same thing as just 10, and this is a ratio of just 10 to 1. And so this is the R group ratio that we're looking for. It's literally just 10 to 1. And so if we take a look at the r group of histidine over here, notice that the group that's ionized is really this nitrogen atom that's here. And so the protonated, the conjugate acid form, which is on the bottom here, is really just going to be an NH with a positive charge on it. Then the deprotonated form of this nitrogen atom is going to just be a nitrogen, and it's going to be neutral. And so this ratio here means for every 10, nitrogen atoms in the r group of histidine, there's going to be one positively charged conjugate acid form of the R group. And so, essentially we have finished with parts a, b, and c, so now we can finally move on to part d. And part d down below wants us to use the same exact ratios, so it wants us to use the ratios that we just calculated above, to determine the average net charges of the ionizable groups and the entire histidine molecule at pH 7. And so, notice that the first one that we're going to calculate here is the net charge of the amino group. And so if we go back up to our ratio up above for the amino group, notice that it was 1 to 100 ratio. And so, because we're looking to calculate the net charges here, we're looking to calculate the average net charges, we're only interested in the groups that are gonna have a charge in them. And so notice for the amino group, it's the conjugate acid that has a charge on it. And so, what that means is we have a 100 charges. This ratio means that we have 100, conjugate acid forms of the amino group for every one of the, conjugate base form. And so there's a total of 101 molecules. We have 100 of this, and one of this. And so, we have a total of 101, so down below here we can say, we have 101 total molecules, and of those total, 100 of them, so 100 positive charges. And so this is really the ratio that we're looking for here. And so if you take your calculator and do 100 divided by 101, you'll get a ratio of about 0.9901. And so this rounds off to about, 0.99, and this is a positive number, which means that the net charge of our amino group is gonna be a positive 0.99. And so that is the answer for this first part over here. So now we can do the same thing for the net charge of the carboxyl group. So going up to the ratio for the carboxyl group, notice that it was 100,000 to 1, But this time, it's actually the conjugate base, which is the top number here. It's the conjugate base that has a charge on it. And remember, because we're determining the average net charge, we're only interested in the ones that have a charge on it. So out of the total amount of molecules, so a 100,000 here and one down below, that's 100,001, there are 100,000 charges. So we can go ahead and indicate that down below. So we can say, out of 100,001 total molecules, we have 100,000 of them that are going to have a negative charge. And so, essentially, what we need to do is take our calculators and do negative 100,000 divided by 100,001, and that comes out to an answer of about negative 0.9999, which, of course, rounds off to about negative one. And so, what this means is that our net charge on our carboxyl groups is gonna be negative 1. Alright. So now we need to do it for the r groups over here and essentially our r group ratio is 10 to 1, and this time it's the conjugate acid on the bottom that has a charge on it. And so, essentially, out of 11 total molecules, so ten of the conjugate acid I'm sorry. 10 of the conjugate base and 1 of the conjugate acid, 11 total. So out of 11 total, we can write that down here, so out of 11 total, we have only one that's going to have a charge on it, so we're only gonna have one positive charge. And so if you take your calculators and you do 1 divided by 11, you'll get an answer of 0.0909, which rounds off to about 0.09. And so this is a positive number here, so, this is gonna be a positive 0.09 charge on the r group. And so now that we've finished with the first three parts, we can move on to the net charge of the entire histidine molecule here, and it gives us this hint to just sum the previous three charges. And so essentially all we need to do is take the the charge here, the average net charge on the amino group, so 0.99, And, essentially, we can subtract 1 here for the net charge of the carboxyl group, so minus 1.00. And then, we can add on the charge for the r group here which is plus 0.09, 0.09. And so we can go ahead and add all of these numbers up together, type it in your calculator, and you'll get an answer of 0 0.08. And this is a positive number, so that means that we have a positive 0.08 charge on the histidine molecule. And so essentially what this means is that at a pH of 7, notice that our histidine molecule does have a positive charge on it, but it's a really small positive charge. It's only a positive 0.08 charge. And really the reason for this is, you'll notice that the, amino the amino group has a 0.99 charge. So that's about, a positive one charge. And the carboxyl group has a negative one charge. So that, is negative one. And so these two charges here, the 0.99 and the negative one, they pretty much cancel each other out. And the r group of histidine at a pH of 7 only has a really small positive charge, and so that's why, overall, histidine has such a small positive charge at neutral pH. So that's something interesting to keep in mind about histidine at neutral pH, 그것 will have a fraction of a charge. And so, 그 concludes this long problem, an I will see you guys in our next video.

Here’s what students ask on this topic:

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

The Henderson-Hasselbalch equation is a mathematical formula that relates the pH of a solution to the pKa of an ionizable group. It is expressed as:

$\mathrm{pH}=\mathrm{pKa}+log\left(\frac{\mathrm{[conjugate\; base]}}{\mathrm{[conjugate\; acid]}}\right)$

In biochemistry, this equation is crucial for understanding the behavior of amino acids and other weak acids in solution. It helps determine the final pH of a weak acid solution at equilibrium and the ratio of conjugate base to conjugate acid when the pH is known. This is particularly important for calculating the average net charges of ionizable groups in amino acids.

How do you apply the Henderson-Hasselbalch equation to amino acids?

To apply the Henderson-Hasselbalch equation to amino acids, you need to consider each ionizable group independently. For each group, use the equation:

$\mathrm{pH}=\mathrm{pKa}+log\left(\frac{\mathrm{[conjugate\; base]}}{\mathrm{[conjugate\; acid]}}\right)$

First, identify the pKa values of the ionizable groups in the amino acid. Then, determine the concentrations of the conjugate base and conjugate acid forms of each group. Plug these values into the equation to find the pH or the ratio of conjugate base to conjugate acid. This allows you to calculate the average net charge of the amino acid at a given pH.

What is the significance of pKa in the Henderson-Hasselbalch equation?

The pKa value in the Henderson-Hasselbalch equation represents the pH at which half of the ionizable group is in its protonated form and half is in its deprotonated form. It is a measure of the strength of an acid. In the context of amino acids, the pKa values of the ionizable groups (such as the carboxyl and amino groups) are crucial for predicting the ionization state of the amino acid at different pH levels. This information is essential for understanding the behavior of amino acids in biological systems, including protein structure and function.

How can the Henderson-Hasselbalch equation be used to calculate the average net charge of an amino acid?

To calculate the average net charge of an amino acid using the Henderson-Hasselbalch equation, follow these steps:

1. Identify the pKa values of all ionizable groups in the amino acid.
2. Determine the pH of the solution.
3. For each ionizable group, use the Henderson-Hasselbalch equation:

$\mathrm{pH}=\mathrm{pKa}+log\left(\frac{\mathrm{[conjugate\; base]}}{\mathrm{[conjugate\; acid]}}\right)$

Calculate the ratio of conjugate base to conjugate acid for each group.

Use these ratios to determine the fraction of each ionizable group that is protonated or deprotonated.

Sum the charges of all ionizable groups to find the average net charge of the amino acid at the given pH.

Why is it important to understand the Henderson-Hasselbalch equation in the study of amino acids?

Understanding the Henderson-Hasselbalch equation is crucial in the study of amino acids because it allows for the prediction of the ionization states of amino acids at different pH levels. This is important for several reasons:

• Protein Structure: The ionization state of amino acids affects protein folding and stability.
• Enzyme Activity: The activity of many enzymes depends on the ionization state of their amino acid residues.
• Buffer Systems: Amino acids can act as buffers, maintaining pH in biological systems.
• Drug Design: Understanding ionization helps in designing drugs that can interact effectively with biological molecules.

Overall, the Henderson-Hasselbalch equation provides a foundational understanding of biochemical processes involving amino acids.