Fill in the appropriate groups for Asp at pH 4.3.

Question

Accepted Answer

So this practice problem wants us to fill in the appropriate groups for aspartic acid, specifically at a pH of 4.3. And so recall that aspartic acid's one letter code is just a d for phonetic sounding. And the mnemonic to help us memorize aspartic acid structure is dragons eat knights riding horses. And so the d here represents aspartic acid, which we know is an acidic negatively charged acidic amino acid, and it's gonna have a carboxyl group in its group, r group. And so that's what we're seeing here, a carboxyl group, and really, it's just alanine with a carboxyl group on it. And so, in this problem, all we need to do is to, compare the pH of the solution independently to each of the pKas. And in our previous problems, we only had 2 pKas to compare to, but in this problem, we have 3 pKas. So we have to independently compare the pH to each of the pKas. So let's start over here with pKA 1. So pKA 1 is 9.82, so that means that it's gotta be our alpha amino group because we know the alpha amino group, ranges p k a's range from 9 to 10.5. And so, let's go ahead and do that one first. So the p h is equal to 4 point 3, and that is less than a pka 1, which is equal to 9.82. And so recall that when the pH is less than the pKa, that means that the conjugate acid is going to predominate. And so the conjugate acid form of the amino group, so the amino group conjugate acid form, is just gonna be an NH 3+. And so that's what we can put into this group up here. We can put in an NH3, and we'll put the plus in red here to make it a little easier to see. So now we can compare this group over here which has a p k a, of 2.09 and this, of course, must be the alpha carboxyl group or the carboxyl group that's part of the backbone of the amino acid. And we know that those pKas are right around 2, so it's gotta be this 1. And so for this, all we need to do is say the pH is still equal to 4.3 and the pH of 4.3 is actually greater than pKa 2, which is equal to 2.09. And so, when the pH is greater than the pKa, that means that the conjugate acid's gonna be less concentration and the conjugate base is gonna predominate. So the conjugate base of the carboxyl group, I'll put alpha carboxyl group, is actually going to be the just the o minus, the deprotonated form. So it's gonna over here, what we can do is just put it in an o, and we'll put the minus in a different color. And so now, we've got these 2 groups, the next is to move on to the R group pKa, which is pKa 3 here. And so with pKa 3, we're gonna do the same thing. We're gonna compare the pH to the pka. So the pH hasn't changed, it's still 4.3. And a pH of 4.3 is greater than pka 3, which is equal to 3.86. And so, what that means is that when the pH is greater than the pKa, the conjugate base is gonna predominate. So this is another carboxyl group. It's not an alpha carboxyl group because remember, alpha means that it's part of it's extending off of the alpha carbon, which is this carbon right here. And so, this one is a carboxyl group that's in the r group. And so with this carboxyl group, the deprotonated form or the conjugate base form is gonna predominate, so it's also going to be just an o minus. So we can put in an o in here with a minus on it. And so, all we need to do this is, all of the appropriate charges in each of these groups. And to find the overall net charge, all we need to do is sum up all of these groups. So we've got 2 negative charge, so that's -two, and then we've got one positive charge, so that makes it -one. So the overall, net charge is just -one. And so that concludes this practice problem and I'll see you guys in our next video.

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Fill in the appropriate groups for Asp at pH 4.3.

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