Jump to content

What is meant by the Pka value of a side chain in an amino acid?


scilearner

Recommended Posts

I know how to get the answer for this question using hesselback equation but I don't understand the question. I have some questions about the questiuon.

 

1. The PKa value of the sulphydryl (-SH) group of cysteine is 8.33. Calculate the fraction of anion to free sulfhydryl group at PH 7.O.

 

So is this Pka value referring to

 

SH + H20 <---> H+ + S- reaction

 

Why is this reaction significant. How does getting the PH of the equlibrium constant of this reaction provide any meaningful value.

 

Also in this amino acid there is only one SH group(meaning two atoms) so how can an equilibrium reaction occur.

 

My question is related to PKa values. I don't understand how getting the PH of the equilibrium constant can be meaningful and that two atom problem.

 

Thank you so much

Link to comment
Share on other sites

A Pka value is not a pH value. Both, however are equal to [ce] - log_{10}[X] [/ce] where X is the Ka for pKa and X is the hydrogen ion concentration for pH.

 

Here is the Henderson-Hasselbalch Equation:

 

[math] pH=pKa+ log_{10}([A^-]/[HA])[/math]

 

In this problem, we need to treat the -SH group as a weak acid that can lose it's proton. The anion form is represented by [ce] [A^-] [/ce] and the sulfhydyl is represented by [ce] [HA] [/ce]. Technically, the brackets indicate concentration, but since you're only interested in the ratio, it's irrelevant whether the solution is 1M, 10M or 0.005M.

 

They've given you the pH and pKa values to plug in. Then you just need to remove the [ce] log_{10} [/ce] part of the term to get the ratio of anion to sulfhydryl.

 

I don't understand what you mean by "how can an equilibrium reaction occur?"

 

[ce] R-SH <-> R-S^- + H^+ [/ce] is an equilibrium reaction. The equilibrium constant represents the tendency toward the right or left of that equilibrium arrow.

Edited by UC
Link to comment
Share on other sites

A Pka value is not a pH value. Both, however are equal to [ce] - log_10[X] [/ce] where X is the Ka for pKa and X is the hydrogen ion concentration for pH.

 

Here is the Henderson-Hasselbalch Equation:

 

[math] pH=pKa+log_10([A^-]/[HA])[/math]

 

In this problem, we need to treat the -SH group as a weak acid that can lose it's proton. The anion form is represented by [ce] [A^-] [/ce] and the sulfhydyl is represented by [ce] [HA] [/ce]. Technically, the brackets indicate concentration, but since you're only interested in the ratio, it's irrelevant whether the solution is 1M, 10M or 0.005M.

 

They've given you the pH and pKa values to plug in. Then you just need to remove the [ce] log_10 [/ce] part of the term to get the ratio of anion to sulfhydryl.

 

I don't understand what you mean by "how can an equilibrium reaction occur?"

 

[ce] R-SH <-> R-S^- + H^+ [/ce] is an equilibrium reaction. The equilibrium constant represents the tendency toward the right or left of that equilibrium arrow.

 

Thank you so much for your answer. I understand PKa now but I have a problem.

 

In an amino acid there is only one molecule of SH. How can there be a large group SH molecules in one amino acid. Is the question talking about like one amino acid or mole of amino acid. I know I'm not understanding something obvious but please help

Thank you!!

Link to comment
Share on other sites

Thank you so much for your answer. I understand PKa now but I have a problem.

 

In an amino acid there is only one molecule of SH. How can there be a large group SH molecules in one amino acid. Is the question talking about like one amino acid or mole of amino acid. I know I'm not understanding something obvious but please help

Thank you!!

 

Oh, they just mean an arbitrary solution containing a bunch of it. It's just the convention to speak about it as if it's one thing when discussing functional groups and pKa, etc.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.