Biochem problem

I really should know how to do this....

Composition and transformation of substance.

Moderators: Calilasseia, ADParker

Biochem problem

#1  Postby Rachel Bronwyn » Jan 29, 2016 7:00 pm

What the hell am in forgetting from first year that's preventing me from helping a student with this question?

I feel like such an asshole.

Determine the relative ratio of all four ionic species of histidine at pH 1.8. 1.8 just so happens to be pKa1....

The [A-/HA] for forms A/B, B/C and C/D at pH 1.8 are: 1, 6.31e-5 and 3.16e-8.

Given that 1.8 is pKa1, how the hell does one go from three A-/HA concentrations to a relative ratio of the concentrations of all four ionic species at this pH?
User avatar
Rachel Bronwyn
THREAD STARTER
 
Name: a certain type of girl
Posts: 12014
Age: 30
Female

Canada (ca)
Print view this post

Ads by Google


Re: Biochem problem

#2  Postby campermon » Jan 29, 2016 7:11 pm

Yes.

You should really know how to do this.

:shifty:
Scarlett and Ironclad wrote:Campermon,...a middle aged, middle class, Guardian reading, dad of four, knackered hippy, woolly jumper wearing wino and science teacher.
User avatar
campermon
RS Donator
 
Posts: 17032
Age: 48
Male

United Kingdom (uk)
Print view this post

Re: Biochem problem

#3  Postby TopCat » Jan 29, 2016 10:43 pm

So you've got three equilibria, with three pKa values, but instead of giving you the pKa values, they've given you the relative concentration of base over acid in each case. The diprotonated form is the one with the lowest pKa of 1.8 - let's call this H3A2+. The next one is H2A+. The neutral one is HA, and the final one is A-

Ok so far?

So the first [ b ]/[ a ] is: [H2A+] / [H3A2+] = 1 (since the pH = pKa, right?). Let's suppose the relative [H3A2+] is 1.

So [H2A+] = [H3A2+] x 1 = 1

The next [ b ]/[ a ] is given by: [HA] / [H2A+] = 6.31x10-5. Multiplying by [H2A+] gives:

[HA] = [H2A+] 6.31x10-5 = [H3A2+] x 1 x 6.31x10-5

The final [ b ]/[ a ] is given by: [A-] / [HA] = 3.16x10-8. Multiplying by [HA]:

[A-] = [HA] x 3.16x10-8

But as before, we have [HA] from the previous step too, so:

[A-] = [H3A2+] x 1 x 6.31x10-5 x 3.16x10-8

So the relative concentrations of all four species are:

1 : 1 : 6.31x10-5 : 2x10-12

Does that help?
TopCat
 
Posts: 595
Age: 55
Male

Country: England
United Kingdom (uk)
Print view this post

Re: Biochem problem

#4  Postby Rachel Bronwyn » Jan 30, 2016 2:56 am

The solution literally pinged into my mind while I was at work and I nearly leapt up from my desk and screamed I FUCKING GOT IT.

Thank you though TopCat!
User avatar
Rachel Bronwyn
THREAD STARTER
 
Name: a certain type of girl
Posts: 12014
Age: 30
Female

Canada (ca)
Print view this post

Re: Biochem problem

#5  Postby TopCat » Jan 30, 2016 2:51 pm

No probs.

Couple of things come to mind - I did wonder if the way you stated the problem made it harder than it needed to be.

Rather than A/B, B/C and C/D, it's really B/A, C/B, D/C that are the respective [base]/[acid] values, where A, B, C, D refer to the sequentially deprotonated forms.

It's a small point, but it's more obvious what you have to multiply by each time if you get the fractions the right way up and referring to the right species.

If you're helping a younger student with this stuff, might be an idea to make sure that (s)he knows how to calculate the pKa values from the information given too.
TopCat
 
Posts: 595
Age: 55
Male

Country: England
United Kingdom (uk)
Print view this post

Re: Biochem problem

#6  Postby Rachel Bronwyn » Jan 30, 2016 5:40 pm

We calculated the pKas and then I realised I hadn't a goddamn clue how I was going to resolve the rest of the problem and pulled an "I'm going to get back to you on this one."

You won't be surprised to hear we answered the same question for pHs that correspond with pKa2 and pKa3 as well. The results demonstrate an increase in basicity favours deprotonated forms over protonated forms of a compound. Surprise!
User avatar
Rachel Bronwyn
THREAD STARTER
 
Name: a certain type of girl
Posts: 12014
Age: 30
Female

Canada (ca)
Print view this post


Return to Chemistry

Who is online

Users viewing this topic: No registered users and 1 guest