12. Properties of a Buffer The amino acid glycine is often used as the main ingr
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Question
12. Properties of a Buffer The amino acid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine, which has a pKa of 9.6, can exist either in the protonated form (-NH3+) or as the free base (-NH2), because of the reversible equilibrium
(a) In what pH range can glycine be used as an effective buffer due to its amino group?
(b) In a 0.1 M solution of glycine at pH 9.0, what fraction of glycine has its amino group in the -NH3+ form?
(c) How much 5 M KOH must be added to 1.0 L of 0.1 M glycine at pH 9.0 to bring its pH to exactly 10.0?
(d) When 99% of the glycine is in its -NH3+ form, what is the numerical relation between the pH of the solution and the pKa of the amino group?
Explanation / Answer
The solutions are as follows,
(a) Buffers are effective for pH = pKa +/- 1. Therefore, this is effective for pH of 8.6 to 10.6.
(b) We will use the Hendersen-Hasselbalch equation,
pH = pKa + log(A-/HA)
We have given value,
pH = 9.0
pKa = 9.6
Thus,
9 = 9.6 + log[base]/[acid]
-0.6 = log [NH2]/[NH3+]
0.25 = [NH2]/[NH3+]
or, 0.25/[1 + 0.25] x 100 = 20% of total amino group is in the form of NH2
Thus, 100-20 = 80% is in the protonated form (NH3+).
(c) Given is the concentrations of glycine before the additon of NaOH. Now we have to find how much more NH3+ form we lose to reach pH = 10.0.
Set y as total concentration converted to NH2, then,
R R-NH3+ <--> R-NH2 + H+
I | .1 - y | | y | 0
C | -x | | +x | | +x |
E | .1 -x -y | | y+x | | x |
Then you know that x is equal to [H+] from abive which is 10^-10, or 1E-10.
Given is Ka = 10^-9.6, which is 2.51E-10.
Algebraically, we solve for y to find that if (y+1E-10)(1E-10)/(.1-y-1E-10) = 2.51E-10
then y = 0.07151 M.
So the change in concentration of the free base is 0.07151 M - 0.02006 M = 0.05145 M
1.0 L * 0.05145 M = 0.05145 mol glycine converted
Since the glycine loses one proton to every one NaOH molecule, and we know that 0.05145 mol NaOH were added.
thus, 0.05145 mol / (5 mol/L) = 0.0102 L = 10.2 mL of NaOH.
(d) We will use the Hendersen-Hasselbalch equation,
pH = pKA + log(A-/HA)
when 99% (0.99) glycine is in NH3+ form, then,
[NH2] = 1-0.99 = 0.01
[HA] = 0.99
Feeding the value in the above equation we get,
pH = 9.6 + log [0.01]/[0.99]
= 7.6
Thus the pH = 7.6
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