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