Author Topic: In Need of Urgent Help (Read 3465 times) Tweet Share

Bear · « **on:** January 06, 2008, 01:35:34 pm »

Doing some summer study ...
Using Henderson-Hassalbach Equation
(Remember pH = pKa + ([Base] / [Acid] )

Glycine is NH2 - Ch2 - COOH
Amino Group has pKa = 9.6
at pH = 9.6
Whats the charge on NH3+ ?

Their working was
9.6-9.6 = 0 = log10 ([NH2]/[Nh3+]
1 = [Nh2] / [Nh3+] = [0.5 NH2] / [0.5 NH3+ ]
Charge = +0.5
Which i totally don't get! Any ideas?

Collin Li · « **Reply #1 on:** January 06, 2008, 01:50:26 pm »

I was not aware that anything like this was on the course, but anywho, this is what I believe is going on:

You have an amino acid with two forms, the protonated HOOC-CH2-NH3+ acidic form, or the HOOC-CH2-NH2 conjugate base.

Using the formula given, subbing in pH = pKa = 9.6, you get:

$0 = \log_{10}\left(\frac{[\mbox{Base}]}{[\mbox{Acid}]}\right)$

$\implies 1 = \frac{[\mbox{Base}]}{[\mbox{Acid}]}$

$\implies [\mbox{Base}] = [\mbox{Acid}]$

This means there is a 1:1 ratio of the acid and the base. This means that there is 50% of the +1 charged species, and 50% of the neutral species. This means the overall 'average' charge on the amino acid is +0.5.

EDIT: By the way, this will be moved to the Chemistry board.

Bear · « **Reply #2 on:** January 06, 2008, 01:58:30 pm »

Thanks COBLIN

Collin Li · « **Reply #3 on:** January 06, 2008, 01:59:49 pm »

No problem

Sorry, can I just ask you, was that a VCE question? It doesn't look like it to me... are you a university student?

Bear · « **Reply #4 on:** January 06, 2008, 02:00:26 pm »

Can you show me what the charges on Nh3+ will be, say if the pH was 6.6 and 12.6 ?

Collin Li · « **Reply #5 on:** January 06, 2008, 02:07:39 pm »

Bear: I'm not sure if it's the charge on NH3+, it should be the charge of the overall solution (because both the neutral and acidic forms are present at the same time). The charge on NH3+ will always just be +1.

I will find the charge of the overall solution:

For pH = 6.6:

$6.6 - 9.6 = \log_{10}\left(\frac{[\mbox{Base}]}{[\mbox{Acid}]}\right)$

$\implies 10^{-3} = \frac{[\mbox{Base}]}{[\mbox{Acid}]}$

$\implies [\mbox{Base}] = \frac{[\mbox{Acid}]}{1000}$

This means there is a 1:1000 ratio of the base and the acid. This means that there is $\frac{1000}{1001}\times 100$ % of the +1 charged species, and $\frac{1}{1001}\times 100$ % of the neutral species. This means the overall 'average' charge on the amino acid is $\frac{1000}{1001}\times (+1) + \frac{1}{1001}\times 0 = + 0.999$ to 3 significant figures.

This makes sense. In a more acidic solution (lower pH), the protonated species is more abundant, so we have a higher positive charge.

For pH = 12.6:

$12.6 - 9.6 = \log_{10}\left(\frac{[\mbox{Base}]}{[\mbox{Acid}]}\right)$

$\implies 10^{3} = \frac{[\mbox{Base}]}{[\mbox{Acid}]}$

$\implies \frac{[\mbox{Base}]}{1000} = [\mbox{Acid}]$

This means there is a 1000:1 ratio of the base and the acid. This means that there is $\frac{1}{1001}\times 100$ % of the +1 charged species, and $\frac{1000}{1001}\times 100$ % of the neutral species. This means the overall 'average' charge on the amino acid is $\frac{1}{1001}\times (+1) + \frac{1000}{1001}\times 0 = + 0.001$ to 3 significant figures.

This makes sense. In a more basic solution (high pH), the protonated species is less abundant, so we have a smaller positive charge.

Bear · « **Reply #6 on:** January 06, 2008, 02:15:08 pm »

Thanks for that, but in the next bit of working they started calculating and they started getting charges in NEGATIVE
Like Carboxyl group has pKa = 2.3
at pH = 2.3, charge = -0.5 etc...
why is it suddenly negative? i thought because the carboxyl group is a negative charged ion, thats why..and Nh3+ is a postitively charged ion.

Collin Li · « **Reply #7 on:** January 06, 2008, 02:23:02 pm »

Ah, because the amino acid can also take this combination:

H2N-CH2-COOH (neutral or "acidic")
H2N-CH2-COO- (conjugate base - after it has donated its proton)

The neutral amino acid can act as an acid or a base. In this case, the amino acid is acting as an acid (it was acting as a base before).

So now the acid is the neutral form, and the base is the negatively charged species, so when you find out the ratio for them, you will end up with an overall negative charge in the solution.

jeremiahk · « **Reply #8 on:** January 06, 2008, 04:26:24 pm »

Nothing on the new study design indicates or suggests this sort of increase in difficulty of acid/base questions. However I have had a few students present these sorts of "equilibrium" changes with using K values in more advanced ways. They ALL were from high performing schools.

Bear · « **Reply #9 on:** January 06, 2008, 04:56:57 pm »

Can you please show me how do you know whether the amino acid is acting as a acid or base? Because all the information we got was the formula for Glycine. Can you show me the working when we are finding the overall charge when the amino acid is acting as a base? Say, when pH = 4.3? Thanks Coblin.

Collin Li · « **Reply #10 on:** January 06, 2008, 05:02:44 pm »

Quote from: Bear on January 06, 2008, 04:56:57 pm

Can you please show me how do you know whether the amino acid is acting as a acid or base? Because all the information we got was the formula for Glycine. Can you show me the working when we are finding the overall charge when the amino acid is acting as a base? Say, when pH = 4.3? Thanks Coblin.

Sorry, I wouldn't have a clue. You have two pKa values, so it might be competing. I'd roughly guess that you take the midpoint of the two pKa's ( $\frac{9.6 + 2.3}{2} = 5.95$ ) and if the pH is below 5.95, you'd work with pKa = 2.3, and if the pH is above 5.95, you'd work with pKa = 9.6.

This is a total guess, by the way. I never really did this in great detail.

Collin Li · « **Reply #11 on:** January 06, 2008, 05:12:16 pm »

What we could try is analysing both situations, and getting a three-way ratio.

Say the pH is 4.3:

Looking at the NH2 (amine) group

$4.3 - 9.6 = \log_{10}\left(\frac{[\mbox{HOOC-CH2-NH2}]}{[\mbox{HOOC-CH2-NH3+}]}\right)$

$\implies 10^{-5.3} = \frac{[\mbox{HOOC-CH2-NH2}]}{[\mbox{HOOC-CH2-NH3+}]}$

$\implies [\mbox{HOOC-CH2-NH2}] = \frac{[\mbox{HOOC-CH2-NH3+}]}{10^{5.3}}$

Now, we could look at the COOH (carboxy) group:

$4.3 - 2.3 = \log_{10}\left(\frac{[\mbox{H2N-CH2-COO-}]}{[\mbox{H2N-CH2-COOH}]}\right)$

$\implies 10^{2} = \frac{[\mbox{H2N-CH2-COO-}]}{[\mbox{H2N-CH2-COOH}]}$

$\implies [\mbox{H2N-CH2-COOH}] = \frac{[\mbox{H2N-CH2-COO-}]}{10^{2}}$

Now, with both of these ratios, we can combine them to get:

$\implies \frac{[\mbox{HOOC-CH2-NH3+}]}{10^{5.3}} = [\mbox{HOOC-CH2-NH2}] = [\mbox{H2N-CH2-COOH}] = \frac{[\mbox{H2N-CH2-COO-}]}{10^{2}}$
(The two species in the middle are the same thing, just written back to front.)

$\implies \frac{[\mbox{HOOC-CH2-NH3+}]}{10^{5.3}} = [\mbox{glycine}] = \frac{[\mbox{H2N-CH2-COO-}]}{10^{2}}$

So, we have a $10^{5.3}\mbox{:}10^2$ ratio between the charged species. This means that there will be an overall charge of $\left[\frac{10^{5.3}}{10^2 + 10^{5.3}} \times (+1)\right] + \left[\frac{10^{2}}{10^2 + 10^{5.3}} \times (-1)\right] = + 0.999$ to 3 significant figures. The acidic form strongly overpowers the basic form.

Bear · « **Reply #12 on:** January 06, 2008, 05:44:41 pm »

Actually...i think we find the charge of the carboxyl group, and the amino group, and add it together to find the net charge on glycine..?

Collin Li · « **Reply #13 on:** January 06, 2008, 05:48:57 pm »

Possibly. I'm not really sure how this works, because I've never done these calculations for a substance with two different pKa values.

Bear · « **Reply #14 on:** January 06, 2008, 05:55:20 pm »

Bloody hell. Who could have thought unit 1/2 chemistry was so hard

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