Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: CAFOzevl on September 06, 2015, 10:35:42 PM
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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 !
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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?
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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!
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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.
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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!
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1b) 0.04L*0.2mol/L => 0.0008mol HCl
Is this correct?
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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. :(
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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?
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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!
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Remember you have "one other ionizable group". What does that do?
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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.
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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?
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Hello Babcock_Hall,
That would mean it is protonated, i.e. acting as an acid, at pH 8.2. Correct?
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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.
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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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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.
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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?
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What you actually have is one molecule, with two basic sites.
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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...
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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.
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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
?
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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?
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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.
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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.
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You know the final pH (6.2) and you know the ratio of H2B to HB1-. The pKa is the only unknown.
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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.
I just had one question about the last step.
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.
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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?
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-."
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markyo8,
I agree with mjc123; you are double-counting the strong acid.
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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.