(Solved): The structure of anserine at pH 7 is shown below. It has three ionizable groups ...

The structure of anserine at pH 7 is shown below. It has three ionizable groups: the carboxyl group pka1 = 2.64, imidazole nitrogen pka2 = 7.04, amino group pka3 = 9.49. H?N H?C NH a Complete the paragraphs to describe the titration curve for anserine: Starting at acidic pH, all three groups will be protonated, thus the molecule will have a +2 charge As base is added, the carboxylic acid group will deprotonate with a midpoint at pH 2.64 After the first deprotonation the molecule will have a +1 -? charge. As the pH approaches 7, the imidazole nitrogen will be 50% deprotonated. After this deprotonation the molecule will be uncharged Previous a Complete the paragraphs to describe the titration curve for anserine: Starting at acidic pH, all three groups will be protonated, thus the molecule will have a +2 charge As base is added, the carboxylic acid group will deprotonate with a midpoint at pH 2.64 After the first deprotonation the molecule will have a +1 charge. As the pH approaches 7, the imidazole nitrogen will be 50 % deprotonated. After this deprotonation the molecule will be uncharged Finally, as the pH passes 9.49, the amino group will be 50 % deprotonated 11.5, the molecule will have a -1 charge 2 Regarding the shape of the graph; in the region of each pk?, pH remains relatively unaffected as increments of OH- are added the weak acid and its conjugate base are acting as a buffer Correct J and by about pH ?. ??. This is because Starting at acidic pH, all three groups will be protonated and thus the molecule will have a +2 charge. As base is added, the carboxyl group deprotonated with a midpoint at 2.64. This will result in the molecule having a +1 charge after the addition of one equivalent of base. As th approaches 7.04, the imidazole nitrogen will be deprotonated leaving the molecule uncharged after the addition of two equivalents of base. Finally, as the pH passes 9.49, the amino group will be deprotonated and by about pH 11.5, the molecule will have a -1 charge. Regarding the shape of the curve; in the region of each pK a pH remains relatively unaffected as increments of OH are added. This is beca weak acid and its conjugate base are acting as a buffer. The titration curve for anserine: Previou Starting at acidic pH, all three groups will be protonated and thus the molecule will have a +2 charge. As base is added, the carboxyl group will be deprotonated with a midpoint at 2.64. This will result in the molecule having a +1 charge after the addition of one equivalent of base. As the pH approaches 7.04, the imidazole nitrogen will be deprotonated leaving the molecule uncharged after the addition of two equivalents of base. Finally, as the pH passes 9.49, the amino group will be deprotonated and by about pH 11.5, the molecule will have a -1 charge. Regarding the shape of the curve; in the region of each pK a, pH remains relatively unaffected as increments of OH are added. This is because th weak acid and its conjugate base are acting as a buffer. The titration curve for anserine: 12 10 4 2 O 8 0 PK1 = 2.64 0.5 T 1 pK2 = 7.04 LO N pK3 = 9.49 3 b The isoelectric point of anserine is the pH where the net charge on the molecule is zero. Calculate the isoelectric point for anserine. pH = 8.265 The isoelectric point, pl, is the pH at which the molecule is uncharged. Clearly, this will happen at a pH at which the carboxyl group's -1 charge is balanced by positive charges from either the imidazole group or the amino group. Therefore, the isoelectric point must be between the pK a's of t imidazole and amino groups. Thus, the sum of the protonated imidazole group and the protonated amino group must be equivalent to a charge of +1. Let I and IH+ be the unprotonated and protonated imidazole groups respectively. Let A and AH be the unprotonated and protonated amino groups. [IH*] + [AH+] = one equivalent That is, the sum of the positively charged species for the imidazole and the amino groups must sum to the one equivalent of negative charge from the carboxylate group. Correct The concentrations of the imidazole species, protonated and unprotonated, must sum to one equivalent. The same is true for the amino species. Thus, and so, [I] + [IH+] = [A] + [AH+] [I] + [IH+] = [IH*] + [AH*] [A] + [AH+] = [IH*] + [AH*] (1) (2) concentrations of the imidazole species, protonated and unprotonated, must sum to one equivalent. The same is true for the amino species. 0, [I] + [IH+] = [A] + [AH+] [I] + [IH+] = [IH+] + [AH+] [A] + [AH+] = [IH+] + [AH+] equations (1) and (2) we can see that: [AH+] = [I] or [IH+] = [A] (3) enderson-Hasselbalch equations for each are: M IH¹] pH = pK?2 + log10 uting (3) into these we have: [AH+] [IH*] pH =pK?2 + log10 pH=pKa3 + log10 JAH (1) (2) =pK?3 + log10 [A] AH] Substituting (3) into these we have: [AB+] and S: pH=pK?2 + log10 pH =pK?3 + log10 olving these two equations for the log terms, which are inversely related, and setting them equal we have: [AH*] IH pH - PK32 = log10 pH -pH + PK?3 = -log10 [IH+] [AH+1 J pH-PK?2 = -pH + PK?3 pK a2 + PK?3 2 pl = 8.27 IH+] AH] = pl = = log10 7.04 +9.49 2 IH+] pH-pK?2 pH = pl = 8.27 pK?2 + PK83 2 Submit = -pH + pK?s = = pl = ven a 0.100 M solution of anserine at its isoelectric point and ready access to 0.300 M HCl, 0.300 M NaOH and distilled water, describe the p 1 of 0.0500 M anserine buffered solution, pH 7.20. lume of stock solution needed ume of HCI needed to adjust the imidazole group ume of HCI needed to titrate the amino group = al HCI needed 7.04 + 9.49 2 F mL = mL mL 1- pka + pho 2 16.0500 and 7.04 +9.49 2 isoelectric point and ready access to 0.300 HC 0.300 ac and led describe buffered solution, pH 7.20 -p+pKs = -log10 pH-pK=-pH+pKas pK+pKss 2 pH = pl= 8.27 TH] [AH] Total HCI needed = = pl = mL log10 7.04+ 9.49 Given a 0.100 M solution of anserine at its isoelectric point and ready access to 0.300 M HCl, 0.300 M NaOH and distilled water, describe the preparation of I L of 0.0500 Manserine buffered solution, pH 7.20. Volume of stock solution needed = mL Volume of HCI needed to adjust the imidazole group = Volume of HCI needed to titrate the amino group = [AH IH+ mL mL 8.27 0.100 M solution of anserine at its isoelectric point and ready access to 0.300 M HCl, 0.300 M NaOH and distilled water, describe the preparation of 0500 M anserine buffered solution, pH 7.20. of stock solution needed = of HCI needed to adjust the imidazole group = of HCI needed to titrate the amino group = Cl needed = mL mL mL mL