Titrations of Amino Acids with Non-Ionizable R-Groups - Online Tutor, Practice Problems & Exam Prep

Amino acids with non-ionizable R groups are polyprotic acids, exhibiting multiple pKa values corresponding to their ionizable groups. Titration curves reveal inflection points, indicating pKa values, and equivalence points, marking complete neutralization. The isoelectric point (pI) is the average of two pKa values, where the net charge is zero. For alanine, the pI is calculated as (2.4+9.7)=6.05. Understanding these concepts is crucial for analyzing amino acid behavior in biochemical contexts.

Titrations Of Amino Acids With Non-Ionizable R-Groups

In this video, we're going to talk about the titrations of amino acids with nonionizable R groups. So recall that way back in our previous lesson videos, we reviewed titrations from your old chemistry courses. And really, this lesson here on the titrations of amino acids with nonionizable R groups is just an extension of those older lesson videos. A lot of the information that we're going to cover in this video is just going to be review. A lot of the principles and concepts that we already learned about titrations are just going to be applied specifically to the titrations of amino acids with nonionizable R groups. But if you don't remember much about titrations at all, be sure to go back and rewatch those older videos before you continue here. Now, that being said, it's important to note that all amino acids are polyprotic acids. We know that polyprotic acids are acids that have multiple acidic hydrogens, and we know that each acidic hydrogen has its own pKa value, so all amino acids have multiple pKa values. At least 2 pKa values, one for the alpha amino group on the backbone, and one for the alpha carboxyl group on the backbone.

Also, recall from our previous lesson videos that titrations can reveal the pKa values of weak acids, and that point on the titration curve is referred to as the inflection point, which is also called the midpoint. Because this is exactly where half of the molar equivalents of titrant have been added to neutralize half of an acid. And so the inflection point, or the midpoint, corresponds to each pKa value. Also, recall from our previous lesson videos that the equivalence point is also known as the endpoint because that is exactly where 100% of an acid has been neutralized, and that is the end of an acid. It represents the point of neutralization of an acid.

Down below in our example, we're going to do some titration curve review. You can see that we have a titration for an amino acid with a nonionizable R group. The surrounding titration curve, we have all of these colored boxes that correspond to the colors of specific regions found within our titration curve. We're just going to go around and fill out these boxes as we review the titration curve. We'll actually start off with this black box here in the bottom right, which is asking what does the black curve represent? The black curve is just referring to this black curve that we see going up through our graph. We know that this is going to be alanine's titration curve. We know that it's alanine because it says so at the top left here—it tells us that it's alanine's titration curve.

Now we'll move on to this green box here in the bottom left. This is asking what the green dot and lines represent. Notice that they're showing up at exactly 0.5 molar equivalents of titrant being added. So that means this has got to be a midpoint or an inflection point. Before this point, alanine's titration curve is curving downwards. After the titration curve, notice that the curve changes to an upward curve, so we know that this is an inflection point. Each inflection point and midpoint corresponds with a pKa value, so we can see we have a pKa value of 2.4 here. Carboxyl groups on the backbones of amino acids have pKa values in the ballpark range of about 2. This is really close to 2 and right in that ballpark range.

So, this must be the pKa value for the carboxyl group. Alright, moving on, we'll skip this yellow box really quickly and jump up to this dark blue box at the top left. You'll notice this is another midpoint because it's asking us what the dark blue dot and lines represent corresponding to this dark blue dot here and dark blue lines coming up and down. It's another midpoint because, again, what we see is a curve. Notice that it's curving downwards like this. After the point, it curves and changes its curvature to going upwards. This lets us know that this is an inflection point or a midpoint. Each midpoint corresponds with a pKa value, and this is a pKa value of 9.7. Amino groups on the backbones of amino acids have pKa values in the ballpark range of about 9 to 10.5.

this falls right in that ballpark range, so this must be the pKa value of the amino group. We can jump down to the yellow box here, which is asking us what the yellow dot and line represent. Here we see the yellow dot in the center and the yellow line going across to the left. This is referring to the isoelectric point. We know that the isoelectric point, also known as just the pI, is an average of the 2 pKa values for an amino acid with a nonionizable R group. Each pKa value corresponds with the y-axis of pH shown right here at this point. For amino acids with nonionizable R groups, we can take the average of these 2 pKa values, and that'll give us the pI, or the isoelectric point, which is what's showing up here, right with this yellow line here. We know from our previous lesson videos that the isoelectric point is the exact point where alanine's net charge is 0; it's going to have a neutral net charge.

We're moving on to the pink box at the top right, which is asking what the pink line represents. It's this pink vertical line going down like this. It appears at exactly 1 molar equivalent of titrant being added, which means that we must be neutralizing 100% of an acid. That means this has got to be an equivalence point or an endpoint because we're neutralizing 100% of an acid. Since we have the carboxyl group acid on the left, down below this pink line, this must be the neutralization of the carboxyl group.

This is the equivalence point of the carboxyl group. We're neutralized at this point here, this pink vertical line; we are essentially neutralizing 100% of the carboxylic acid, conjugate acid. This means that there's no more conjugate acid form of the carboxyl group, so no more COOH at any point after this pink line over here, moving to the right.

Now we're moving on to the next box over here, corresponding to the light blue boxes that we see in the background. We have 2 light blue boxes referring to the effective buffering ranges. We know that effective buffering ranges are within plus or minus one of the pKa. You can see that this blue box is referring to pKa plus or minus 1. This point up here and this point up here make up the effective buffering range, essentially, this whole area here between these pHs. The same applies to this one down here. The effective buffering range would be within plus or minus one of the pKa, so between 1.4 and 3.4. The effective buffering range is plus or minus 1 of the pKa.

Now for this red box, which is asking what the red line represents. It might be a little tough to see, but it's this red line over here on the far right. Note that it's showing up when we add an entire second molar equivalent of titrant. This means that we must be neutralizing another acid. This is going to be another equivalence point or another endpoint. Since we already covered the equivalence point of the carboxyl group, this must be the equivalence point of the amino group. We know that the equivalence points represent the end of an acid, and so that means that there's going to be no more of this conjugate acid form of the amino group. It has been completely neutralized at this red point over here on the far right.

Now you can see that we've filled out all of the boxes, and we've broken down the titration curve for alanine. In our next lesson video, we're going to talk about how to draw the structures of the predominant structures of amino acids just by looking at the titration curves of amino acids with nonionizable R groups. So I'll see you guys in that next lesson video.

Titrations Of Amino Acids With Non-Ionizable R-Groups

So now that we've reviewed titration curves again, we can talk about how to draw amino acids with non-ionizable R groups just from their titration curves. But before we get there, I first want to highlight 2 very important pieces of information that apply to all amino acids with non-ionizable R groups. The first piece of information is that the isoelectric point is always found at the carboxyl group equivalence point, just like we said in our previous video. The second important piece of information is that the titration curves for amino acids with non-ionizable R groups only have 2 inflection points and 2 equivalence points since amino acids with non-ionizable R groups only have 2 ionizable groups, the amino group and the carboxyl group, resulting in one set of inflection point and equivalence point for each of the ionizable groups. If there were a third ionizable group, such as with the amino acids with ionizable R groups, there would be an additional inflection point and equivalence point, but we'll talk about the titration curves of amino acids with ionizable R groups in another video later in our course.

Now, let's move on to our example where we will draw the predominant structure of alanine just from alanine's titration curve, noting that alanine's titration curves have these colored regions: a pink colored region over here, a yellow colored region up here, and a green colored region at the top. We're going to draw the predominant structure of alanine at each of these colored regions of the titration curve. Additionally, we will calculate the isoelectric point of alanine at the end.

Firstly, in the pink region, which corresponds with pHs below 2.4, the pH is below the pKa of the carboxyl group (2.4) and the amino group (9.7). In this region, the protonated form will predominate for both groups. The conjugate acid form of the amino group will be NH₃⁺, and the carboxyl group will be undissociated, not carrying a charge. Alanine's R group, a methyl group, is CH₃ and does not ionize. The net charge of the predominant structure in this region is +1.

Moving to the green dot in the middle, corresponding to the pKa of the carboxyl group (2.4), we find equal concentrations of its conjugate acid and base forms. However, the amino group, with a pKa much higher, remains fully protonated. The amino group contributes a +1 charge, and the half-ionized carboxyl group contributes a -0.5 charge. Thus, the net charge is +0.5.

In the yellow region between the two pKa's, we explore a scenario at pH 7, where the amino group remains protonated (NH₃⁺) and the carboxyl group is deprotonated (as carboxylate anion with a negative charge). The net charge of alanine's predominant structure in this region is 0, corresponding to the isoelectric point.

At the pKa of the amino group (9.7), the carboxyl group is fully dissociated, contributing a -1 charge, and the amino group, being half dissociated, contributes +0.5. The total charge at this pH is -0.5.

Finally, in the green region, where the pH exceeds both pKa's, both groups are deprotonated. The amino group (NH₂) carries no charge and the carboxyl group, being a carboxylate anion, has a -1 charge. The resulting net charge of the predominant structure in this region is -1.

To find the isoelectric point, we average the two pKa's (2.4 and 9.7), resulting in an isoelectric point (pI) of 6.05. This concludes our example, and we'll have more practice drawing structures of amino acids from titration curves in our upcoming practice video.

Draw the predominate structures of Leu at each of the indicated sections ( 1, 2, 3) on its titration curve.

So this practice problem wants us to draw the predominant structure of leucine at each of its indicated sections on the titration curve. And so the sections are 1, which is here, we have section 2, which is over here on the titration curve. Now, before we actually get to drawing these structures, we need to notice that there are these 2 pKa values. So, we've got one pKa value over here shown as pKa1, and we've got a second pKa over here shown as pKa2. And then, over here in the middle, what we have is the pI of alanine, or the isoelectric point of alanine. And so, up here in this first region that we're going to draw the structure in, we're referring to essentially this highlighted region here. And this highlighted region corresponds with the pH value on the y-axis that is higher than the pKa for both of the pKa values. And recall that when the pH is higher than the pKa, the conjugate base predominates. And so leucine only has 2 ionizable groups, the amino group and the carboxyl group, and both of these ionizable groups are going to be in their conjugate base form. So the conjugate base form of the amino group is just an NH2, which is not charged, so we'll leave it like that. And then we can draw in the carboxyl group. The conjugate base form of the carboxyl group is going to be a carboxylate anion which is negatively charged. And then recall that leucine's R group, so leucine's R group is a loose extension of valine, so it just has an extra CH2 in comparison to valine. So it's just going to look like this. So that is leucine structure at this particular region of its titration curve. So now, let's move on to the next region. And the next region is showing, pointing right here, which is the same location as the isoelectric point. So really, you can think of this whole second region as being the region in between all the pKa values, because the predominant structure is going to be the same in this entire region. And so it's going to be the region where the pI is, so we know that the overall charge has to be neutral in this green region. And so, basically what you'll see is that in this green region here, this is lower than the pKa for the amino group, which means that the amino group is going to be in its conjugate acid form. And the conjugate acid form of the amino group is NH3+ with a positive charge. And so, this region here is actually higher than the pKa of the carboxyl group. And when the pH is higher than the pKa of the carboxyl group, that means that the carboxyl group is going to be in its conjugate base form. And so we can go ahead and draw the carboxyl group in its conjugate base form. And the conjugate base form, of course, is just going to be the carboxylate anion. And then we can, of course, fill in our R group, which is again non-ionizable, so it'll be CH2CH3 with methyl groups coming off. And so this is the predominant structure here and its net charge is going to be 0. And so, keep that in mind here. We can keep track of the net charges below. So the net charge is 0 for this structure and for this structure over here, the net charge is negative 1. So that was up here in this region, the net charge was negative 1. And the net charge in this other region here is 0. So now let's look at the 3rd region, which is down here and this region down here. And notice that this region corresponds with pH values that are both below both pKa values, so pKa1 and pKa2. So this region is below both pKas. And when the pH is below the pKa, that means that the conjugate acid is going to predominate for both of those ionizable groups. And so that means that the amino group is going to be in its conjugate acid form, which is protonated like that with a charge. And then the carboxyl group is also going to be protonated. So the carboxyl group is going to be in its OH form which is neutral. And so again, we can draw in the R group and the R group is going to be just a CH2 with the CH3 and two lines coming off for the methyl groups. And then the overall net charge at this structure is just going to be positive 1. And so down here, we can put a positive one to help us remember the charge. So again, as the pH starts to increase, notice that as the pH increases, the charge of the predominant structure decreases. So there's this inverse relationship between the pH of the solution and the charge of the predominant structure. And so that concludes this practice problem, and I'll see you guys in our next one.

Calculate the pI of Ile using its titration curve. Mark the approximate position of the pI on the titration curve.

So this practice problem wants us to calculate the isoelectric point of isoleucine just by using its titration curve that's given here. And then it says to mark the approximate position of the isoelectric point on the titration curve. And so, recall that isoleucine's one-letter code is just I, and the mnemonic that helps us memorize isoleucine structure is GAVLYMP, and this is the group for the nonpolar amino acids. And so, isoleucine is represented by this I here, and its structure is really just an isomer of leucine, so it has the same number and types of atoms as leucine, but it has a different arrangement, and the arrangement of isoleucine actually comes from valine. So when we draw isoleucine's structure, so it'll have NH3+, it'll have the carboxyl group here, and the carboxyl group, we can put as a carboxyl group with a charge, and then its R group is going to essentially be, valine and a lopsided valine. So it's just gonna be, valine, which is just like this and it'll have, valine and then we just need to make it lopsided. So this is isoleucine's structure. And so notice that isoleucine does not have an ionizable R group. So, what this means is that it only has 2 pKa values and these are corresponding to the only 2 ionizable groups that it has. And so the amino group is going to have a pKa of about 9 to 10, so we know that this pKa2 here corresponds with the amino group. And the other, pKa must be for the carboxyl group, which is going to have a pKa around 2. And so, for non-ionizable amino acid R groups, the way that you calculate the isoelectric point is to just average the only 2 pKa values. So the isoelectric point is just going to be 9.6+2.42. So this is gonna be 12 divided by 2, which is equal to 6. And so the isoelectric point here is equal to 6. So that is the first part of the problem. Now the second part says to mark the approximate position of the isoelectric point on the titration curve. And so, recall that the isoelectric point is going to be the midpoint between the 2 pKa values. So over here, what we have is 9.6, and that's the pKa of the amino group. And over here, what we have is 2.4. And so between these two lines here, what we want to do is mark the approximate position, which is gonna be the exact midpoint. So right around here, which is about the midpoint between these 2 pKas, that's going to correspond with a horizontal point right on the titration curve. So this point right here should be the approximate position of the isoelectric point, and this, of course, should equal to a value of 6. And so this concludes this practice problem, and I'll see you guys in our next one.

Which of the following compounds would make for the best buffer at pH 8?

• A. Acetic acid, pKa = 4.8
• B. Tricine, pKa = 8.15
• C. Glycine, pKa = 9.9
• D. Tris, pKa = 8.3

Identify the region(s) on glycine's titration (I, II, III, IV, or V) that corresponds with each statement below:

a. Region where Gly predominant species has net charge of +1. ________

b. Region where the average net charge of Gly is +½. ________ 

c. Region where ½ of Gly's amino groups are ionized. ________

d. Region where the pH = pK_a of carboxyl group. ________

e. Region where the pH = pK_a of amino group. ________

f. Regions where Gly has its maximum buffering capacity. _____________

g. Region where the average net charge of Gly is 0. ________

h. Region where Gly's carboxyl group has been completely titrated. ________

i. Region where Gly has been completely titrated. ________

j. Region where Gly's predominant species is a zwitterion. ________

k. Region where the average net charge of Gly is -1. ________

l. Region where Gly is a 50:50 mixture of protonated & deprotonated carboxyl group. ________

m. Region indicating Gly's isoelectric point (pI). ________

n. Region indicating the end of Gly's titration. ________

o. Regions where Gly has poor buffering power. ____________________

So here's a pretty hefty practice problem with a bunch of different parts to it, but it's a lot more straightforward than you might think. And really, it's asking us to identify the region or the regions on glycine's titration curve, which is shown below, that corresponds with each of the statements that are shown below. Notice that on Glycine's titration curves, we have these different regions that are marked with Roman numerals I, II, III, IV, and V. The first statement here says, to indicate the region where glycine's predominant species has a net charge of positive 1. First, we need to do is identify where the pKas are for, glycine's 2 ionizable groups, because remember that glycine only has 2 ionizable groups. Recall that there's going to be a pKa at each of the midpoints. Here at region number II, what we have is a midpoint. This is actually going to be the pKa of the carboxyl group. We can put the carboxyl pKa right here. Up here, there's another midpoint; this is region IV. It's going to be the midpoint of the amino group. So it's going to be the amino group pKa, because the midpoints correspond with pKas. Basically, what we're looking for is when glycine has a plus one charge, what that means is that both of its ionizable groups have to be protonated, and they have to take on their conjugate acid form. If we want the ionizable groups to take on their conjugate acid form, that means that we're going to want the pH to be really, really low. When the pH is low, just like it is in this region here, region number I, that corresponds with a pH that is below both of these pKas, meaning that both of these groups are going to be protonated and in their conjugate acid form, and that means that they're going to have a positive charge at this region number I, a plus one charge. The region that we want to put in this blank is just region number I. Now, b says region where the average net charge of glycine is, plus 1 half, so positive 0.5. We know that at this pKa here, the carboxyl group pKa, the amino group is going to be completely charged. The amino group is going to contribute a complete plus one charge, because all of the amino groups are going to be positively charged at the pKa here. But we know that at the carboxyl group pKa, the carboxyl group is going to have an equal concentration of conjugate acid and conjugate base. That means that only half of the carboxyl groups are going to be ionized. If only half of the carboxyl groups are ionized, then there's only going to be a half charge, so a half of a negative charge. If all of the carboxyl groups were charged, there would be a negative one charge, but there's only half that are charged at the pKa, which is in region II. The total charge here at the pKa of the carboxyl group is just going to be a sum of these two charges, which is just going to be positive 0.5, which is what we're looking for, a positive 0.5 charge. That's going to occur at the carboxyl group pKa, which is region number II. We can put region II for our answer here, for b. Now, c says, the region where half of glycine's amino groups are ionized. And of course, that's going to be where the amino group pKa is because at the pKa, we know again that the concentration of conjugate acid is going to be equal to the concentration of conjugate base, specifically for the amino group. That's where half of glycine's amino groups are going to be ionized, so it'll be region IV. Now d says the region where the pH is equal to the pKa of the carboxyl group. And, of course, that's got to be, where the pH is equal to the pKa of the carboxyl group, that's got to be the carboxyl group pKa, and that's region II. Now e says the region where the pH is equal to the pKa of the amino group. And of course, that's going to have to be the exact pKa of the amino group, which is going to be region number IV. Now, f says the region where glycine has maximum buffering capacity. And recall that the maximum buffering capacity is found within plus or minus 1, so it's found within plus or minus 1 of the pKa. But the thing is that, glycine actually has 2 pKas, so that means that it actually has 2 different regions of maximum buffering capacity, and those are going to be at both of its pKas. So it's going to be both region number II, the carboxyl group pKa, and region number IV, which is the amino group pKa. Now g says the region where the average net charge of glycine is 0, and we know that when, glycine's net charge is equal to 0, that that's going to be where the isoelectric point is. And so, of course, that's going to be region number III here. So region number III corresponds with the isoelectric point or the pI. And so we can put in for g, that region number III here. So, h here says, the region where glycine's carboxyl group has been completely titrated. That's saying that, the equivalence point the equivalence point for the carboxyl group. And we know that the equivalence point for the carboxyl group is going to be where the isoelectric point is found. So it's essentially going to be at this region right here, which is equal to the equivalent, one molar equivalent of titrant being added. Because it corresponds with the pI, it's also going to be region number III here. Now, for I, the next one, it says the region where glycine has been completely titrated. When glycine has been completely titrated, that means that both its amino group and its carboxyl group have reached neutralization. The only point where that's true is going to be where you've added 2 molar equivalents. So it's going to be this region here, where this is going to be the equivalence point of the amino group, and that's where glycine has been completely titrated. So, I is going to be region V. Now j says where, glycine's predominant species is a zwitterion, and we know that when glycine is a zwitterion, it's going to have a neutral net charge, and that the neutral net charge is found where the isoelectric point is found. The isoelectric point or the pI is going to be region number III. So, the region where, glycine's predominant species is a zwitterion is going to be region III. Now, k says, where the average net charge of glycine is negative 1. For glycine to have a negative one charge, that means that both of its ionizable groups are going to have to be completely deprotonated. For glycine's both of glycine's groups to be completely deprotonated, it's going to have to have a really high pH because when the pH is high and above both of the pKas, for its ionizable groups, that means that the conjugate base is going to predominate and the conjugate base is deprotonated. That means that region V here is going to be, the region that's above both of the pKas, and that means that, both of these groups are going to be deprotonated and region V is going to be where glycine's going to have a net charge of negative one. So we can put in, region V for, k. Now, l says region where glycine is a 50:50 mixture of protonated and deprotonated carboxyl group. This is going to be where the pKa is of the carboxyl group because at the pKa of the carboxyl group, we know that there's an exact equal amount of conjugate acid and conjugate base, so there's going to be a 50:50 mixture of conjugate of protonated and deprotonated forms of the carboxyl group. So that is going to be region number II for, l. Now, m says, end region indicating glycine's isoelectric point, and we already indicated this previously when we talked about glycine having a net charge of 0. The glycine's region of isoelectric point is going to be region number III. So we can go ahead and mark that n. And, n says the region indicating the end of glycine's titration, and so the end of glycine's titration is another way to say that glycine has been titrated. It's really going to be the same region, region number V, which corresponds to the equivalence point of the amino group, which is region number V here. Alright. So we're getting close to the end, and our, the next one, which is our last one says the region where glycine has been, or has poor buffering power. That's just going to be the regions that are not associated with the pKas. There are three regions that are not associated with the pKas, and those are going to be the regions, number I, so it'll be region number I, region number III, and region number V are not associated with the 2 pKa. We know that the pKas are going to have good buffering effectiveness. So, it's these three regions, 1, 3, and 5, that have poor buffering capacity. So we can put those down here in Oh, so it'll be region I, region III, and region V. And so that concludes this practice problem, and I'll see you guys in our next video when we're going to talk about the titration of amino acids with ionizable R groups.

Here’s what students ask on this topic:

What is the significance of the isoelectric point (pI) in amino acids with non-ionizable R groups?

The isoelectric point (pI) of an amino acid is the pH at which the amino acid has a net charge of zero. For amino acids with non-ionizable R groups, the pI is calculated as the average of the pKa values of the carboxyl and amino groups. At this point, the amino acid is electrically neutral, which is crucial for understanding its behavior in different pH environments. For example, alanine has a pI of 6.05, calculated as (2.4 + 9.7) / 2. This concept is important in protein purification and electrophoresis, where the pI can influence the solubility and migration of amino acids and proteins.

How do you determine the pKa values from a titration curve of an amino acid with a non-ionizable R group?

To determine the pKa values from a titration curve of an amino acid with a non-ionizable R group, identify the inflection points on the curve. These points, also known as midpoints, occur where the curve changes direction, indicating the pH at which half of the amino acid's ionizable group is neutralized. For example, in alanine's titration curve, the first inflection point at pH 2.4 corresponds to the pKa of the carboxyl group, and the second inflection point at pH 9.7 corresponds to the pKa of the amino group. These pKa values are essential for calculating the isoelectric point and understanding the amino acid's ionization state at different pH levels.

What are the key features of a titration curve for an amino acid with a non-ionizable R group?

A titration curve for an amino acid with a non-ionizable R group has several key features: 1) Two inflection points, corresponding to the pKa values of the carboxyl and amino groups. 2) Two equivalence points, where 100% of each ionizable group is neutralized. 3) The isoelectric point (pI), where the net charge of the amino acid is zero, typically found between the two pKa values. 4) Effective buffering ranges, which are within ±1 pH unit of each pKa value. These features help in understanding the ionization states and buffering capacities of the amino acid at different pH levels.

How do you calculate the isoelectric point (pI) of an amino acid with a non-ionizable R group?

To calculate the isoelectric point (pI) of an amino acid with a non-ionizable R group, take the average of the pKa values of the carboxyl and amino groups. The formula is:

$\mathrm{pI}=\frac{({\mathrm{pK}}_{\mathrm{a1}}+{\mathrm{pK}}_{\mathrm{a2}})}{2}$

For example, for alanine, the pKa values are 2.4 (carboxyl group) and 9.7 (amino group). Thus, the pI is:

$\mathrm{pI}=\frac{(2.4+9.7)}{2}=6.05$

This pI value indicates the pH at which alanine has a net charge of zero.

What is the relationship between pH and the charge of an amino acid with a non-ionizable R group?

The charge of an amino acid with a non-ionizable R group depends on the pH relative to its pKa values. Below the pKa of the carboxyl group, the amino acid is fully protonated, carrying a positive charge. Between the pKa values of the carboxyl and amino groups, the amino acid is zwitterionic, with a net charge of zero. Above the pKa of the amino group, the amino acid is fully deprotonated, carrying a negative charge. For example, alanine has a positive charge below pH 2.4, a neutral charge around its pI of 6.05, and a negative charge above pH 9.7. This relationship is crucial for understanding amino acid behavior in different pH environments.