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?

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] =

[A⁻]/[HA] =

[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]  =

[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 + (-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

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.