A) The acidic form of the buffer MES has a formula weight of 195.24 and a pKa of

Question

A) The acidic form of the buffer MES has a formula weight of 195.24 and a pKa of 6.1. The basic form, available as the potassium salt, has a formula weight if 233.33. Outline how you would use these two to make 1.00 L of a 0.120 M solution of MES with a pH of 6.4
B) Draw the following peptide as it would appear in approximately neutral solution. Using the pKa values for the side chains from the table in your book, and using estimated values for the pKa if the N-terminal amino group of 8.9 and of the C-terminal carboxyl group of 3.5, calculate the isoelectric pH of the peptide.
KPERIHYDCW A) The acidic form of the buffer MES has a formula weight of 195.24 and a pKa of 6.1. The basic form, available as the potassium salt, has a formula weight if 233.33. Outline how you would use these two to make 1.00 L of a 0.120 M solution of MES with a pH of 6.4
B) Draw the following peptide as it would appear in approximately neutral solution. Using the pKa values for the side chains from the table in your book, and using estimated values for the pKa if the N-terminal amino group of 8.9 and of the C-terminal carboxyl group of 3.5, calculate the isoelectric pH of the peptide.
KPERIHYDCW
B) Draw the following peptide as it would appear in approximately neutral solution. Using the pKa values for the side chains from the table in your book, and using estimated values for the pKa if the N-terminal amino group of 8.9 and of the C-terminal carboxyl group of 3.5, calculate the isoelectric pH of the peptide.
KPERIHYDCW

Explanation / Answer

There are multiple questions here. Both the questions are completely unrelated . i am allowed to answer only 1 at a time. I will answer 1st question for you.Please ask other as different question.

pka of Mes = 6.1
use:
pH= pKa + log( [salt]/[acid] )
6.4 = 6.1 + log ([salt]/ 0.120)
[salt]= 0.24 M
so concetration of salt should be 0.24 M
Since volume of solution is 1L,number of moles of salt = 0.24 mol

number of moles= mass /molar mass
0.24 = mass/233.33
mass = 55.9 gm
So tomake this buffer you must add 55.9 gm of salt in the solution

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