Identify the region(s) on glycine's titration (I, II, III, IV,...

Question

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

Graph showing glycine titration curve with pH vs OH- equivalents, indicating various regions.

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. ____________________

Accepted Answer

So here's a pretty hefty practice problem with a bunch of different parts to it, but it's a lot more straightforward than what 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. And so notice that on Glycine's titration curves that we have these different regions that are marked with roman numerals 1, 2, 3, 4, and 5. And so, the first statement here says, to indicate the region where glycine's predominant species has a net charge of positive 1. And so, first, what 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. And so, recall that there's going to be a pKa at each of the midpoints. And so here at region number 2, what we have is a midpoint. And so this is actually going to be the pKa of the carboxyl group. So what we can do is put the carboxyl pKa right here. And then up here, there's another midpoint, and this is going to be the midpoint, this is region 4. And so 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. And so, 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 and so they have to take on their conjugate acid form. And if we want the ionizable groups to take on their conjugate acid form, that means that we're gonna want the pH to be really, really low. And so when the pH is low, just like it is in this region here, region number 1, that corresponds with a pH that is below both of these pKas, a's, that means 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 1, a plus one charge. And so the region that we wanna put in this blank is just region number 1. Now, b says region where the average net charge of glycine is, plus 1 half, so positive 0.5. And so, of course, we know that, at this, p k a here, the carboxyl group p k a, the amino group is gonna be completely charged. So the amino group is gonna contribute a complete plus one charge, because all of the amino group is gonna be positively charged at the pKa here. But we know that at the carboxyl group p k a, that the carboxyl group is gonna have, a equal concentration of conjugate acid and conjugate base. And so that means that only half of the carboxyl groups are going to be ionized. And so if only half of the carboxyl groups are ionized, then there's only gonna be a half charge, so a half of a negative charge. Because 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 2. And so, what that means is 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 gonna be positive 0.5, which is what we're looking for, a positive 0.5 charge. So that's gonna occur at the carboxyl group pKA, which is region number 2. So we can put region 2 for our answer here, for b. Now, for c, it says, the region where half of glycine's amino groups are ionized. And of course, that's gonna be where the amino group p k a is because at the p k a, we know again that the concentration of conjugate acid is gonna be equal to the concentration of conjugate base, specifically for the amino group. And so that's where half of glycine's amino groups are gonna be ionized, so it'll be region 4. Now d says the region where the pH is equal to the p k a of the carboxyl group. And, of course, that's gotta be, where the pH is equal to the pKa of the carboxyl group, that's gotta be the carboxyl group pKa, and that's region 2. Now e says the region where the pH is equal to the p k a of the amino group. And of course, that's gonna have to be the exact p k a of the amino group, which is gonna be region number 4. 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 gonna be at both of its pKas. So it's gonna be both region number 2, the carboxyl group pKa, and region number 4, 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 gonna be where the isoelectric point is. And so, of course, that's gonna be region number 3 here. So region number 3 corresponds with the isoelectric point or the p I. And so we can put in for g, that region number 3 here. So, h here says, the region where glycine's carboxyl group has been completely titrated. And so, that's just saying that, the equivalence point the equivalence point for the carboxyl group. And so we know that the equivalence point for the carboxyl group is gonna be where the isoelectric point is found. So it's essentially gonna be at this region right here, which is equal to the equivalent, one molar equivalent of titrant being added. And so, because it corresponds with the p I, it's also going to be region number 3 here. Now, for I, the next one, it says the region where glycine has been completely titrated. And so when glycine has been completely titrated, that means that both its amino group and its carboxyl group have breached neutralization. And the only point where that's true is going to be where you've added 2 molar equivalents. So it's gonna be this region here, where this is gonna be the, equivalence point of the amino group, and that's where glycine has been completely titrated. So, I is gonna be region 5. Now j says where, glycine's predominant species is a zwitterion, and we know that when glycine is a zwitterion, that it's gonna have a neutral net charge, and that the neutral net charge is found where the isoelectric point is found. And so the isoelectric point or the p I is gonna be region number 3. So, the region where, glycine's predominant species is a zwitterion is gonna be region 3. Now, k says, where the average net charge of glycine is negative 1. And for glycine to have a negative one charge, that means that both of its ionizable groups are gonna have to be completely deprotonated. And so, in order for glycine's both of glycine's groups to be completely deprotonated, it's gonna 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 gonna predominate and the conjugate base is, deprotonated. And so that means that region 5 here is going to be, the region that's above both of the p k a's, and that means that, both of these groups are gonna be deprotonated and region 5 is gonna be where glycine's gonna have a neg net charge of negative one. So we can put in, region 5 for, k. Oops. We can put in region 5 here. Now, l says region where glycine is a 5050 mixture of protonated and deprotonated carboxyl group. And of course, this is gonna be where the p k a is of the, carboxyl group because at the p k a of the carboxyl group, we know that there's an exact, equal amount of conjugate acid and conjugate base, so there's gonna be a 5050 mixture of conjugate of protonated and deprotonated forms of the carbox group. So that is gonna be region number 2 for, l. Now, m says, end region indicating glycine's isoelectric point, and we already, actually indicated this previously when we talked about glycine having a net charge of 0. And so the glycine's, region of isoelectric point is gonna be region number 3. 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. And so it's really gonna be the same region, region number 5, which corresponds to the equivalence point of, the amino group, which is region number 5 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. And so that's just gonna be the regions that are not associated with the pKas. And so there are 3 regions that are not associated with the pKas and those are gonna be, the regions, number 1, so it'll be region number 1, region number 3, and region number 5 are not associated with the 2 pKa. So we know that the pKas are gonna have good buffering, effectiveness. And so, it's these 3 regions, 1, 3, and 5, that have poor buffering capacity. So we can put those down here in Oh, so it'll be region 1, region 3, and region 5. And so that concludes this practice problem, and I'll see you guys in our next video when we're gonna talk about the titration of amino acids with ionizable r groups.

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