Help with pI/pH problem? Any input Appreciated!!!

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confused2021
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Help with pI/pH problem? Any input Appreciated!!!

Post by confused2021 »

Hello. My question is as follows :Calculate the pI of Leucine pKa (carboxyl) = 2.4; pKa (amino) = 9.6; What is the ratio of COO- : COOH at pH 3.4?
For pI I get 6 (9.6 + 2.4/2 = 6). I am having great difficulty with the 2nd part of the question. I get 63.8% COO-: 36.2% COOH. I deduced this since 50% is dissociated at pH2.4 and 100% is dissociated at pH6. There are 3.6 units from 2.4 to 6. Since it is one above 2.4, I divided 1/3.4 and multiplied it by 50 to get 13.8%. This added to the 50% dissociated at 2.4pH equals 63.8% dissociated and 36.2% not dissociated.

Any help on how to reason this out to get the right answer would be greatly appreciated!!!!
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ChenBeier
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Re: Help with pI/pH problem? Any input Appreciated!!!

Post by ChenBeier »

I dont know how this can be: (9.6 + 2.4/2 = 6). I got 10.8 you missed some brackets. (9.6+2.4)/2=6
But it was asked what is the ratio of COOH and COO-.
According Henderson Hasselbalch pH = pks- log( c(COOH)/c(COO-))
3.4 = 2.4- log( c(COOH)/c(COO-)) => 1 = - log( c(COOH)/c(COO-)) => c(COOH)/c(COO-) = 0.1 what means 1 to 10.
confused2021
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Re: Help with pI/pH problem? Any input Appreciated!!!

Post by confused2021 »

Thanks for your reply. 9.6 plus 2.4 equals 12 and 12 divided by 2 equals 6. Looking up the pI of Leucine, I also saw it was 6 so I am pretty sure that is the right answer.

The 2nd part of the question asks about when the pH= 3.4. I do not believe that the pka is 2.4 when the pH is 3.4. Therefore I do not have the pka when the pH=3.4. This is why I am having difficulty solving the problem. It is unlikely that there is a 1 to 10 ratio at pH 3.4 because there is a 50 to 50 ration at pH = 2.4. Let me know if that makes sense.

Anybody who can provide any input on how to solve the 2nd part of this question will be greatly appreciated!!!!
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ChenBeier
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Re: Help with pI/pH problem? Any input Appreciated!!!

Post by ChenBeier »

The pka is a fixed number. It will not change. Change is pH and concentration of acid and salt. Otherwise the HH-equation is not valid.
Regarding pI i wrote you missed the brackets. 9.6 + 2.4/2 = 9.6 +1.2 = 10.8. But (9.6 + 2.4 )/2 =12/2 = 6.

Yes at pH 2.4 the concentration of acid and salt are equal, what means that the logarithm of 1 = 0 => pH = pKa
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Re: Help with pI/pH problem? Any input Appreciated!!!

Post by confused2021 »

Thank you for explaining. Are you 100% confident that the way to do the problem is using Henderson Hasselback equation and utilizing a pka=2.4?
3.4=2.4 + logA-/HA gets me a 10 to 1 ratio of dissociated acid (A-) to undissociated acid (HA). Is this the correct answer?
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Re: Help with pI/pH problem? Any input Appreciated!!!

Post by ChenBeier »

Yes that is what I am saying.
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Re: Help with pI/pH problem? Any input Appreciated!!!

Post by confused2021 »

Thanks!!! I am having difficulty understanding why 50% dissociated at pH2.4 and 90% has dissociated at pH3.4. That seems like a substantial increase over a short pH difference.
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Re: Help with pI/pH problem? Any input Appreciated!!!

Post by ChenBeier »

It is the same substance the aminoacid. It is a change of equilibrium nothing more. The same happens at high pH if you have more free amin.
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