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### Topic: Biochemistry Question  (Read 14537 times)

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#### mjc123

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« Reply #15 on: September 08, 2015, 12:40:23 PM »
If it's initially in the deprotonated form, and you have 0.004 mol H+ floating around, what happens? (Clue: how much H+ is floating around in the final state at pH 6.2?) So work out [base] and [acid] and apply HH at pH 6.2 to find pKa.

#### CAFOzevl

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« Reply #16 on: September 08, 2015, 02:06:04 PM »
Well if it's in the deprotonated form then it will attack the "loose" H+ and 0.004 mol of the base will be converted to acid (and there will be no more H+ floating around). But how can I work out the [base] and [acid]? Here's the equation I came up with...

6.2=pKa+log((x-0.004)/(y+0.004))

How do I solve this?

#### Babcock_Hall

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« Reply #17 on: September 08, 2015, 02:16:50 PM »
What you actually have is one molecule, with two basic sites.

#### CAFOzevl

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« Reply #18 on: September 08, 2015, 02:21:31 PM »
You guys love to make me sweat haha.

OK so it would be
6.2=pKa+log((1-0.004)/(1+0.004))
6.2=pKa+log(.992)
6.2=pKa-0.0035
6.2035=pKa

That doesn't seem right...

#### Babcock_Hall

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« Reply #19 on: September 08, 2015, 03:06:16 PM »
The concentration of this compound is 0.2 M and the amount is 0.02 moles.  I am not why you used 1 in your calculation.

#### CAFOzevl

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« Reply #20 on: September 08, 2015, 03:22:23 PM »
Oh. So are you saying it would be:
6.2=pKa+log((0.02-0.004)/(0.02+0.004))
6.2=pKa+log(.667)
6.2=pKa-0.176
6.376=pKa

?

#### Babcock_Hall

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« Reply #21 on: September 08, 2015, 03:36:54 PM »
Offhand, I would say that it does not look correct.  You have 0.004 mol strong acid and 0.020 mole weak base at the beginning.  Once the acid reacts with the base, what do you have?

#### CAFOzevl

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« Reply #22 on: September 08, 2015, 03:53:50 PM »
I have no idea. I thought there were 0.020 moles of the compound, not of the base form of the compound necessarily? Since I have given it a legitimate go and still cannot deduce the answer, could you please tell me how to solve this problem? If not I can just go ask my professor during his office hours later this week, but I would have to go during my lunch break from work so it would be more convenient to have the solution from you. I understand if that violates the forum policy, though. Either way, thank you for your help so far.

#### Babcock_Hall

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« Reply #23 on: September 08, 2015, 04:44:46 PM »
Your compound existed as a mixture of HB1- and B2- before the acid was added.  The addition of the acid consumed the remaining B2-, converting it to HB1-.  This did not consume all of the HCl however.  The remaining HCl reacted with HB1- to form some H2B.  Unless I am missing something you now have 0.004 mole H2B and 0.016 mole HB1-.  It just requires one final application of the Henderson-Hasselbalch equation to find the pKa.  BTW the charge states I have chosen to write above assume that the diprotonated form of the molecule (H2B) is neutral.  That is just an arbitrary choice on my part that should not affect the math.

EDT
The method we are using here separates the two acid-base reactions.  That is a simplification, but (given the difference in the two pKa values) I suspect we are on reasonably safe grounds.  Others here might be able to comment further on this question.
« Last Edit: September 08, 2015, 05:32:49 PM by Babcock_Hall »

#### Babcock_Hall

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« Reply #24 on: September 09, 2015, 09:12:53 AM »
You know the final pH (6.2) and you know the ratio of H2B to HB1-.  The pKa is the only unknown.

#### markyo8

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« Reply #25 on: September 29, 2015, 11:28:56 PM »
Hello Babcock_Hall,
I'm sorry for digging up an old thread, but I had the exact same problem and was having difficulties figuring out how to solve this problem.
First of all thank you very much for your guidance, I found it to be extremely beneficial reading through this thread.
When you state that the remaining HCl reacts with HB- to form some H2B, would some of the HB- also be used up in the reaction?
So then the left over 0.004mol of H+ would react with the 0.016 mole of HB- to form 0.004mol of H2B and 0.012 mole of HB-.
And from there using the pH and the Henderson-Hasselbalch equation we can find the pKa.
Is that how it works? I'm not sure and when rounded, the answer is the same. I just wanted to ask to be sure for future references if HB- also gets used up.

Thank you very much.

#### mjc123

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« Reply #26 on: September 30, 2015, 07:10:10 AM »
Quote
When you state that the remaining HCl reacts with HB- to form some H2B, would some of the HB- also be used up in the reaction?
Of course. "A reacts with B to give C. Is any B used up?" How could it not be?
Quote
So then the left over 0.004mol of H+ would react with the 0.016 mole of HB- to form 0.004mol of H2B and 0.012 mole of HB-.
No, you're performing the subtraction twice. 0.016 mol HB- is what is left after reaction with the 0.004 mol H+. That is what Babcock_Hall told you: "The remaining HCl reacted with HB1- to form some H2B.  Unless I am missing something you now have 0.004 mole H2B and 0.016 mole HB1-."

#### Babcock_Hall

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« Reply #27 on: September 30, 2015, 10:59:26 AM »
markyo8,

I agree with mjc123; you are double-counting the strong acid.

#### mayleekin

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« Reply #28 on: September 09, 2018, 05:35:35 PM »
THIS IS THE SOLUTION:

1. HH Equation:
1a.) 8.2=8.8+log(B/A)
1b.) solving for ratio of B:A= (B/A)=.251
1c.) %B= (.251/.351)*100%= 20%, thus
%A= 80%

2.Amount Reacted:
KNOWN: .1L, .2M of compound
2a.) .1L*.2M= .02 moles

By using percentages, we have
2b.) .8*.02= .004 moles of base
.2*.02= .016 moles of acid

KNOWN: .04L, .2M of HCl
2c.) .04L* .2M= .008 moles HCl

.004 moles of base will get protonated by .004 moles of H+ from HCl
This forms .004 moles of acid
Adding the original acid concentration gives .02 moles of acid

There are still .004 moles of H+ left
This can protonate the acid
This can occur because this is an ionizable group with pKa from 5-7
See top of thread to see original problem

Thus, .004 moles of the H+ protonates .004 of the .02 moles of the original acid

This is your new Base to Acid ratio for the HH eq.
(.016/.004)
The .016 comes from .02mol original acid - .004 mol H+
The .004 on the denominator comes from .004 mol formation of the new acid

3. PLUG INTO HH Eq.
3a.) 6.2 (given)=pKa + log(.016/.004)
3b.) Solve for pKa

4. Thus the pKa of the second ionizable group is 5.6

Hopefully this is easy to understand.