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Offline CAFOzevl

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Biochemistry Question
« on: September 06, 2015, 10:35:42 PM »
Hello all,

I have no idea if this is an appropriate post (and if it is whether it's in the right area), but I was hoping to get some help with a biochemistry question! Thank you in advance for your time.

A compound is known to have a free amino group with a pKa of 8.8, and one other ionizable group with a pKa between 5 and 7. To 100 mL of a 0.2 M solution of this compound at pH 8.2 was added 40 mL of a solution of 0.2 M hydrochloric acid. The pH changed to 6.2. The pKa of the second ionizable group is...?

I know the answer involves use of the Henderson-Hasselbalch equation and probably involves calculating the change in the moles of acid and base, but the second ionizable group thing is throwing me off. I appreciate the assistance !

Offline Babcock_Hall

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Re: Biochemistry Question
« Reply #1 on: September 07, 2015, 11:07:30 AM »
Welcome to the forum.  It is a forum rule that you must show an attempt before anyone can help you.  With what group or groups do you suppose the HCl will react?

Offline CAFOzevl

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Re: Biochemistry Question
« Reply #2 on: September 07, 2015, 11:54:07 AM »
Hi Babcock_Hall ,

Thanks for getting back to me!

HCl will fully dissociate into H+ and Cl- because it is a strong acid. I imagine after addition of HCl the free amino group with pKa 8.8 will interact with the free H+ ions from HCl such that this amino group will be fully protonated (with a net 1+ charge). I deduce this because the resultant pH is 6.2, which is lower than the pKa 8.8.

Please let me know if you need any more of my thoughts on this problem before you can assist me with the calculations. Thank you!

Offline Babcock_Hall

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Re: Biochemistry Question
« Reply #3 on: September 07, 2015, 12:28:28 PM »
I agree that the group with the pKa of 8.8 will be fully protonated at pH 6.2.  My advice is to calculate how much acid will be consumed just in this process.  You will need to use the Henderson Hasselbalch equation.

Offline CAFOzevl

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Re: Biochemistry Question
« Reply #4 on: September 07, 2015, 12:53:55 PM »
Hi again Babcock_Hall,

OK, here's my stab at it:

1) First things first, converting Molar to moles for each compound involved.
 1a) 0.1L*0.2mol/L => 0.02mol unknown compound
 1b) 0.04L*0.2mol/L => 0.0008mol HCl

2) Find original mol of base::acid
pH=pKa+log(base/acid)
8.2=8.8+log(base/acid)
10^-0.6=base/acid
0.25=base/acid
Therefore, there are 0.004mol of the base and 0.016mol of the acid ORIGINALLY.

3) Find change in mol of base::acid
Change is from 1b) above, i.e. 0.0008 mol acid added, 0.0008 mol base subtracted. The final amounts of base and acid are 0.0032 mol and 0.0168 mol, respectively.

4) I just thought of this now... maybe you can confirm if I'm correct? I think I need to plug in the new base::acid amounts and the new pH, then calculate the new pKa...right? (maybe?)
6.2=pKa+log(0.0032/0.0168)
6.2=pKa+.72
5.5=pKa

If 4) above is correct, could you explain why? I just guessed. I don't understand the reasoning behind it, it just seemed to work with the numbers I had up until that point. Thanks!

Online mjc123

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Re: Biochemistry Question
« Reply #5 on: September 07, 2015, 01:11:28 PM »
Quote
1b) 0.04L*0.2mol/L => 0.0008mol HCl
Is this correct?

Offline CAFOzevl

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Re: Biochemistry Question
« Reply #6 on: September 07, 2015, 01:17:34 PM »
Hi mjc123,

Damn! No, it's not  :-[

So redoing everything...

1)
 1a) 0.1L*0.2mol/L => 0.02mol unknown compound
 1b) 0.04L*0.2mol/L => 0.008mol HCl

2)
pH=pKa+log(base/acid)
8.2=8.8+log(base/acid)
10^-0.6=base/acid
0.25=base/acid
Therefore, there are 0.004mol of the base and 0.016mol of the acid ORIGINALLY.

3)
Change is from 1b) above, i.e. 0.008 mol acid added, 0.008 mol base subtracted. The final amounts of base and acid are -0.004 mol and 0.024 mol, respectively.
I'm confused by the negative moles...so... I assume just count it as 0?

4)
6.2=pKa+log(0/0.024)
6.2=pKa+0
6.2=pKa

Is that right...? Something seems wrong.  :(

Online mjc123

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Re: Biochemistry Question
« Reply #7 on: September 07, 2015, 03:50:08 PM »
Quote
I'm confused by the negative moles...so... I assume just count it as 0?
No. You can't have negative amounts of anything. Have you heard of the concept of the limiting reagent? If you have 0.004 mol base and you add 0.008 mol acid, what do you get?

Offline CAFOzevl

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Re: Biochemistry Question
« Reply #8 on: September 07, 2015, 05:01:17 PM »
Hello again mjc123,

Thanks for your continued assistance.

I have heard of limiting reagents but obviously forgot to apply that knowledge in this case! So if you had 0.004mol base and added 0.008mol acid you'd have 0.000 mol base and 0.02 mol acid with 0.004mol of H+ floating around...? But then my answer would be the same. Save me!
« Last Edit: September 07, 2015, 05:24:07 PM by CAFOzevl »

Online mjc123

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Re: Biochemistry Question
« Reply #9 on: September 08, 2015, 06:48:19 AM »
Remember you have "one other ionizable group". What does that do?

Offline CAFOzevl

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Re: Biochemistry Question
« Reply #10 on: September 08, 2015, 10:12:54 AM »
Hi mjc123,

The other ionizable group I guess could act as a base? It would depend on its pKa I suppose. If the pKa were below 6.2 it would act as an acid, or if above 6.2 act as a base. As I said in my original post, the second ionizable group part of the question is what is the biggest stumbling block for me, unfortunately.

Offline Babcock_Hall

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Re: Biochemistry Question
« Reply #11 on: September 08, 2015, 10:38:31 AM »
Your starting pH was 8.2, and you know that this group has a pKa between 5 and 7.  What does that tell you about its state?

Offline CAFOzevl

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Re: Biochemistry Question
« Reply #12 on: September 08, 2015, 10:49:00 AM »
Hello Babcock_Hall,

That would mean it is protonated, i.e. acting as an acid, at pH 8.2. Correct?

Offline Babcock_Hall

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Re: Biochemistry Question
« Reply #13 on: September 08, 2015, 10:57:42 AM »
Suppose for argument's sake that the pKaa of this group is 6.  Then by the Henderson Hasselbalch equation, you can show for yourself that it would be more than 99% in the conjugate base form.

Offline CAFOzevl

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Re: Biochemistry Question
« Reply #14 on: September 08, 2015, 11:22:10 AM »
Ugh, right! Sorry. In your example where the pKa of the group is 6, you would find that 99% is in conjugate base form by...
8.2=6+log(base/acid)
2.2=log(base/acid)
158.5=base/acid
Therefore % protonated is (1/(158.5+1)) which is 0.006 or 0.6% in protonated form, 99.4% in deprotonated form.

But how do I apply this to the original question? I don't know the pKa... that's what I'm trying to find!

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