What is the ratio of [conjugate base] to [conjugate acid] for ...

Question

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).

Accepted Answer

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 Hasselbalch equation. And so I provided the Henderson Hasselbalch 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 gonna change is the pKa value for the calculations, and that's because each group has its own p k a. And so we can perform the calculation for the first group here, the amino group, and the amino group is shown to have a p k a of 9. So we can go ahead and plug that, p k value in, so pKa value is 9 for the amino group. And so, we're gonna 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 is gonna have a conjugate base, which is the top number, of an NH 2. That's the conjugate base of the amino group. It's deprotonated. And the conjugate acid form of the amino group is gonna be an NH 3 plus that is protonated. And so really what this one to 100 ratio really means is that for every one molecule of NH 2, there are 100 molecules of NH 3 plus. 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 gonna 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 log, 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 to 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 minus, 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 or the pkr for the r group is 6, so we can go ahead and plug that in here. And so we're gonna 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 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. And 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 ten, 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 gonna 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 a 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 point 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 one 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 in it. So out of the total amount of molecules, so a 100,000 here and one down below, that's 100,000 and 1, there are 100,000 charges. So we can go ahead and indicate that down below. So we can say, out of 100,000 and one 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 a 100,001, and that comes out to an answer of about negative 0.999 9, which of course rounds off to about negative 1. 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 10 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 going to 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 0.08. 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 So that, is negative one, and so these two charges here, the 0.99 and 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, that it has a fraction of a charge. And so, that concludes this long problem, and I'll see you guys in our next video.

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