Question

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:
R-NH3+ left right double arrow R−NH2 + H+
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?

Answer 1

Answer:

Explanation:

From the information given:

a)

The function of a buffer is effective between the region of  about one pH unit below and one unit above its pKa value.

For glycine; The pH range that is best effective for glycine to be used as a buffer is: (9.6 - 1.0) to (9.6 + 1.0)

= 8.6 to 10.6 pH region

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?

According to Henderson hasselbalch equation:

pH = pKa + log [A⁻]/[HA]

9.0 = 9.6 + log [A⁻]/[HA]

[A⁻]/[HA] = $10 ^{9.6-9.0}$

[A⁻]/[HA] = $10^{0.6}$

[A⁻]/[HA] = 0.25 = 1/4

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?

Using Henderson Hasselbalch equation

pH = pKa + log[A-]/[HA]

1 mole of KOH would result in 1 mole of [A-] salt formation

The number of moles of glycine = Molarity  x Litre = 0.1 x 1 = 0.1 moles

From (B) :

[A⁻]/[HA] = 0.25

[A⁻] = 0.25 [HA]

we know that, equal volume of [base]+[acid] = 0.1

For [A-] from above:

0.25[acid] + [acid] = 0.1

1.25[acid] = 0.1

[acid] = 0.1/1.25

=  0.08 mol,

NOW; [base] = 1 - 0.08 mol

= 0.02 mol

For pH at  10.0

pH = pKa + log [A⁻]/[HA]

10.0 = 9.6 + log[A-]/[HA]

10.0 -9.6 =  log[A-]/[HA]

0.4 = log[A-]/[HA]

[A-]/[HA]  = $10^{0.4}$

[A-] = 2.511[HA]

Now;

2.5[acid] + [acid] = 0.1

[acid] = 0.03 mol,

[base] = 1-  0.03 mol  = 0.07 mol

Thus,  the difference between the two moles of [base] is  (0.07 - 0.02) = 0.05 mol will be the amount of base required to be added

Finally, the volume of KOH = number of moles/Molarity of KOH

= 0.05/5

= 0.010 L

= 10 mL

Therefore; the volume of KOH needed to add would be 10 mL in order to bring the pH  from 9.0 to exactly pH 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?

Using Henderson Hasselbalch equation again:

$pH = pKa + log [\dfrac{NH_2}{NH_3^+}]$

$pH = pKa + log [\dfrac{0.01}{0.99}]$

pH = pKa + (-1.99)

pH = pKa - 1.99

Therefore ,  the numerical relation between the pH of the solution and the pKa of the amino group is approximately 2.00 pH units