[erv10s]

Question:

If I provided you with 100 mL of a 0.215 M solution of disodium glutamate, how much 0.1 M NaOH or 0.1 M HCl would you need to titrate the solution to its isoelectric point? What have you prepared by doing this titration?

Attempt:

0.215 mol disodium glutamate/L x 0.1 L = 0.0215 mol disodium glutamate

0.1 mol HCl/L x (x L of HCl) = 0.1x mol HCl (I'm using HCl to titrate since glutamate is a weak base)

pI = 2.2+4.3/2 = 3.25

pH = pKa + log A-/HA
3.25 = 2.2 + log 0.1x/.0215
1.05 = log 0.1x/.0215
10 ^ 1.05 = 0.1x/.0215
11.22 = 0.1x/.0215
.2412 = 0.1x
x = 2.4

2.4 L of HCl does not seem right, so what am I doing incorrectly? As for what is prepared by doing this titration, I'm not really sure. I was thinking maybe glutamic acid is prepared since the sodium ions would be replaced by H+. However, is there a particular type of solution that is being prepared?

Thanks in advance.

[Borek]

3.25 = 2.2 + log 0.1x/.0215

You have lost me here. Up to this moment I know what you are doing.

What have you prepared by doing this titration?

You use Henderson-Hasselbalch equation - why?

[erv10s]

3.25 = 2.2 + log 0.1x/.0215

Well, that was my guess at finding the concentration of the conjugate since I thought that would help in trying to find the amount of HCl used.

I did ask my professor about this question and have gotten to the point of understanding that disodium glutamate exists at a pH of 6.9. The isoelectric point is 3.25 so to get to that point HCl must be added. From drawing a titration curve, it seems that 1 mole of HCl is needed. And this where I get stuck again; I am not sure how to proceed with this information.

What have you prepared by doing this titration?

You use Henderson-Hasselbalch equation - why?
[/quote]

Is it just that a buffer solution is prepared?

[Borek]

You have to find out conjugates concentrations, you just did something funny. Henderson-Hasselbalch equation (yes, it is a buffer solution) gives you information about acid/base ratio, at the same time you know sum of their concentrations. That's two simultaneous equations and two unknowns. Just solve.

[erv10s]

I know I am making this problem much harder than it really is, but I am not following. Here is what I'm understanding so far:

The concentration of the base, disodium glutamate, is 0.215 M and 100 mL is given. The concentration of the acid, HCl, is 0.1 M but its volume is unknown.

You mentioned that the sum of the acid/base concentrations is known.
So, 0.215 M + 0.1 M = 0.315 M

However, I do not see how this can be put in the Henderson-Hasselbalch equation.

I was thinking something like 3.25 = 2.2 + log (x/0.315-x), but I am confusing myself because the acid/base concentrations are known and only the volume of acid is missing so the (x/0.315-x) ratio doesn't make sense to me.

[Borek]

You mentioned that the sum of the acid/base concentrations is known.
So, 0.215 M + 0.1 M = 0.315 M

No, it is sum of concentrations of acid and conjugated base - in your case hydrogen glutamate and glutamic acid.

I was thinking something like 3.25 = 2.2 + log (x/0.315-x), but I am confusing myself because the acid/base concentrations are known and only the volume of acid is missing so the (x/0.315-x) ratio doesn't make sense to me.

You are on the right track! Assuming there is x base and 0.215-x conjugated acid, sum of their concentrations is x+0.215-x=0.215, and that's what was given in the question. Ratio you have entered into Henderson-Hasselbalch equation is just calculated from the equation descriving sum of their concentrations, that's what I was referring to.

Now, once you will know x you have to calculate how much HCl should be used to convert your original gluamate to x base and 0.215-x conjugated acid. You can do it assuming that protonation reaction goes to completion and all added HCl is consumed.

[erv10s]

So,

3.25 = 2.2 + log (x/0.215-x)
10^(1.05) = x/0.215-x
11.22 = x/0.215-x
2.412 - 11.22x = x
2.412 = 12.22x
x= .1974  M
0.215-x = .0176 M

.1974 mol/L x 0.1 L disodium glutamate solution = 0.01974 mol

0.01974 mol = x (0.1 M HCl)
x = 0.1974 L --> 197.4 mL HCl is needed

[Borek]

x= .1974  M
0.215-x = .0176 M

So far, so good. But now think how much HCl is needed to get to this point. Remember, there are two steps of the protonation.

[erv10s]

Based off of a titration curve that I previously mentioned, is it correct to think that 1 mole of HCl is needed?

From the calculations, however, I was under the impression that 0.1974 M of the conjugate was equivalent to the amount of HCl added. Does the 0.0176 M also need to be considered?

[Borek]

You start with Glutamate^2-, you have to protonate it to HGlutamate^- first, then part of it to H₂Glutamate.

[erv10s]

Maybe I'm confused with what x and .215-x represent.

x = [Glu^2-] = .1974
0.215 -x = [Glu^-] = .0176

If so, then:

3.25 = 2.2 + log (Glu^-/Glu⁰) ?
3.25 = 2.2 + log (.0176/x + .0176)

Then use the value obtained from that calculation and add it to the .1974 from the first calculation to get the total HCl added to reach the isoelectric point.

I suppose I'm still not grasping the last part of this question.

[Borek]

Maybe I'm confused with what x and .215-x represent.

Seems like.

x = [Glu^2-] = .1974

No. You don't have Glutamate^2-, pH is too low for that. pH of the solution is given by pKa1. (That's not whole truth, difference between both pKa values is too small, but it seems like you have enough troubles without getting into that).