(I wanna a biological view of this question) Hi. Can you help me with getting th
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(I wanna a biological view of this question)
Hi. Can you help me with getting the answer to this question (an a good easy guide) because the one that is uploaded in the solutions manual for”Biochemstry 4th. Edition –by voet and voet”. Chapter 4, problem 5P" seems to be wrong (and kind of confuesing). I have big trouble with the henderson hasselbalch equation, (still cant understand it), so if you perhaps could explain it in a "basic" and "structural" way. I already got help once in here by, but I couldnt understand it. (Notice you have to find it for PH=4 ,PH=7, and PH=10. NOT PH=4,7)
5. What are the concentrations of the various ionic species in a 0.1M solution of lysine at pH 4,7, and 10?Explanation / Answer
The pKas for lysine are 2.16,9.06 and 10.54. At pH 4.0, we can ignore the two higher
pKas. At tLys pH, lysine's side chain and the a-amino group are protonated and thus both are
positively charged. So, the ionic species of lysine at pH =4.0 depend critically on the
ionization state of the carboxyl group. When the carboxyl group is in the carboxylate form,
Lystidine will be in the 1+ ionic form. When the carboxyl group is protonated, and hence
uncharged, lysine will carry a 2+ charge. For the a-carboxyl group:
pH = pKa + log [COO-] / [COOH]
or, [COO-] / [COOH]= 10pH- pKa = 104-2.16 = 101.84 or [COO- ] =69.18 x´[COOH] and,
[COO- ] +[COOH] = 0.1M(given)
Thus, 69.18 ´[COOH] +[COOH] = 70.18 [COOH] = 0.1M and,
[COOH] = 0.1M/70.18 = 2.03M
[COO- ] = 69.18 ´[COOH] =0.0014M
So, at pH = 4.0, [Lys 2+ ] = 2.03 M and [lys1+] = 0.0041 M
The concentration of uncharged Lysine and negatively charged lysine are both small but
can be estimated as follows. The concentration of uncharged lysine will be approximately
equal to the concentration of the unprotonated side-chain species. Using the Henderson-
Hasselbalch equation with pKa = 9.06.0, pH = 4.0, and [lys1+] = 0.0041 M, we find that
pH = 9.06 + log[lys0] /[lys 1+]
[lys0] /[lys 1+] = 10^ pH - 9.06 = 10^4-9. 06 and,
[Lys 0] = [Lys 1+] * 8.7 = 3.57M
The concentration of negatively charged Lysine is calculated using the Henderson-Hasselbalch
equation with pKa = 10.54, pH = 4.0, and [Lys0] = 3.57M. In Lys case
pH = 10.54 + log[Lys1-]/[Lys0]
or,
[Lys1-]/[Lys0] = 10^pH-10.54 = 2.88 and,
[Lys1- ] = [Lys 0] * 2.88 = 3.57 * 2.88 = 1.02M
At pH = 4.0 we have: [Lys2+] = 2.03 M , [Lys0] = 3.57 M,
[Lys1+] = 0.0041 M, [Lys1-] = 1.02 M.
similarly it can be calculated for the other 2 pH by using the above method
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