Describe the preparation of 2.00 L of 0.100 M glycine buffer, pH 9.0, from glycine and 1.00 M NaOH.

a) What mass of glycine is required? The appropriate pKa of glycine is 9.6.
b) What volume of 1.00 M NaOH is required? The appropriate pKa of glycine is 9.6.

To prepare the glycine buffer with a pH of 9.0, we can follow the steps below:

a) Calculate the mass of glycine required:

Step 1: Determine the desired concentration of the buffer:
The concentration of the buffer is given as 0.100 M.

Step 2: Calculate the number of moles of glycine needed:
Using the formula for molarity (M = mol/L), we can rearrange it to determine the number of moles:
mol = M × L

Here, the desired concentration of the buffer is 0.100 M, and the desired volume is 2.00 L. Plugging these values into the formula:
mol = 0.100 mol/L × 2.00 L = 0.200 mol

Step 3: Use the molecular weight of glycine to calculate the mass:
The molecular weight of glycine is 75.07 g/mol.

mass = mol × molecular weight
mass = 0.200 mol × 75.07 g/mol = 15.014 g

Therefore, a mass of approximately 15.014 grams of glycine is required to prepare 2.00 L of 0.100 M glycine buffer.

b) Calculate the volume of 1.00 M NaOH required:

Step 1: Determine the desired volume of NaOH:
Since the appropriate pKa of glycine is 9.6 and we want to achieve a pH of 9.0, we need to add a base (in this case, NaOH) to the solution.

Step 2: Calculate the number of moles of NaOH needed:
We can use the balanced equation between NaOH and glycine to calculate the moles of NaOH needed. The equation is:
NaOH + C2H5NO2 ⇌ C2H5NO2- + Na+

From the balanced equation, it is apparent that every mole of NaOH reacts with one mole of glycine. Since we need to adjust the pH, we need to convert half of the glycine (which will be in the form of glycine-), so we divide the number of moles of glycine (from part a) by 2:
moles of NaOH = moles of glycine / 2
moles of NaOH = 0.200 mol / 2 = 0.100 mol

Step 3: Calculate the volume of 1.00 M NaOH:
Using the formula M = mol/L, we can rearrange it to solve for the volume:
L = mol / M

Here, the number of moles of NaOH required is 0.100 mol and the concentration of NaOH is 1.00 M. Plugging these values into the formula:
L = 0.100 mol / 1.00 mol/L = 0.100 L = 100 mL

Therefore, 100 mL (or 0.100 L) of 1.00 M NaOH is required to prepare 2.00 L of 0.100 M glycine buffer.

To prepare a 2.00 L solution of 0.100 M glycine buffer with a pH of 9.0, you will need to follow the following steps:

a) Calculate the mass of glycine required:
1. Determine the number of moles of glycine needed using the formula:
Moles of glycine = (Volume of solution in liters) x (Molarity of glycine buffer)

Moles of glycine = (2.00 L) x (0.100 mol/L)

2. Calculate the required concentration of glycine based on the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])

Rearranging the equation:
[A-]/[HA] = 10^(pH - pKa)

Considering that the pKa of glycine is 9.6 and the desired pH is 9.0:
[A-]/[HA] = 10^(9.0 - 9.6)

[A-]/[HA] = 10^(-0.6)

[A-]/[HA] = 0.25

3. Since glycine is a zwitterion (an amino acid with both a positive and negative charge), the concentration of A- and HA will be equal in a glycine buffer. Therefore, the overall concentration of glycine needed is twice the concentration of the buffer solution:
Concentration of glycine = 2 x (0.100 M)

4. Calculate the moles of glycine required using the formula:
Moles of glycine = (Concentration of glycine) x (Volume of solution in liters)

Moles of glycine = (2 x 0.100 mol/L) x (2.00 L)

5. Finally, calculate the mass of glycine using its molar mass, which is approximately 75.07 g/mol:
Mass of glycine = (Moles of glycine) x (Molar mass of glycine)

Mass of glycine = (0.400 mol) x (75.07 g/mol)

Mass of glycine ≈ 30.028 g

Therefore, approximately 30.028 grams of glycine will be required.

b) Calculate the volume of 1.00 M NaOH required:
1. Determine the number of moles of NaOH needed to neutralize the glycine:
Moles of NaOH = Moles of glycine

Moles of NaOH ≈ 0.400 mol

2. Calculate the volume of 1.00 M NaOH required using the formula:
Volume of NaOH = (Moles of NaOH) / (Molarity of NaOH)

Volume of NaOH ≈ 0.400 mol / 1.00 mol/L

Volume of NaOH ≈ 0.400 L

Therefore, approximately 0.400 liters (or 400 mL) of 1.00 M NaOH will be required.