Question

Both please! Quick!

24) The following is a titration curve of histidine. (8 pts) 10- Histidine pK2 9.17 Part A What is the predominant charge of histidine at the following pH values: p1 = 7.59 아-…- pH pH= 1.00 pH= 4.50 pH= 7.59 1-82 2 pH= 9.50 Part B If you have a 500 mM solution of histidine, what is the molarity of the protonated R-grou 1.0 2.0 3.0 OH (equivalents) pH is 7.00?

Answer 1

24. Part A

i) pH=1:

At pH 1, His has 1 posively charged side chain, 1 posively charged amine group and a uncharged carboxyl group (COOH). So the net charge = 1+1+0 = +2

ii) ph=4.50:

Since pK₁ is 1.82, His loses a proton from the carboxyl group at pH 1.82.

At pH 4.50, His has 1 posively charged side chain, 1 positiely charged amine group and a negatively charged carboxyl group (COOH). So the net charge = 1+1-1 = +1

iii)pH=7.59:

Since pK_R is 6.0, His loses a proton at pH 6.0.

At pH 7.59, His has uncharged side chain, 1 positively charged amine group and a negatively charged carboxyl group (COOH). So the net charge = 0+1-1 = 0. It exists as a zwitter ion.

iv)pH=9.50:

Since pK₂ is 9.17, His loses a proton from the amine group at pH 9.17.

At pH 9.50, His has uncharged side chain, 1 uncharged amine group and a negatively charged carboxyl group (COOH). So the net charge = 0+0-1 = -1

Part B:

pH= 7; pK_a= 6.0;

[His]+[His^-]=500nM

[His]=500-[His^-]-------------1

According to Henderson–Hasselbalch equation pH=pK_a + log([His^-]/[His])

7=6+log([His^-]/[His])

log([His^-]/[His])=7-6

log([His^-]/[His])=1

([His^-]/[His])=10¹

([His^-]/[His])=10

Using 1,

[His^-]/(500-[His^-])=10

[His^-]=10*(500-[His^-])

[His^-]=5000-10[His^-]

[His^-]+10[His^-]=5000

11[His^-]=5000

[His^-]=5000/11

[His^-]=454.5mM

Moarity of protonated R-group at pH=454.5mM