Desribe the preparation of 2.00 L of 0.100M pH 9.00 glycine starting with the sodium chloride salt of zwitterionic glycinhe and 1.00M NaOh. If a reaction that generated 2.00 mmoles of base were carried out in 100 mL of the above buffer, what would be the final solution pH? ( the pka values of the carboxyl and protonated alpha amino groups of glycine are 2.35 and 9.78, respectively.)
To prepare 2.00 L of 0.100M pH 9.00 glycine buffer starting with the sodium chloride salt of zwitterionic glycine, as well as 1.00M NaOH, follow these steps:
Step 1: Calculate the moles of glycine needed
Using the formula for molarity, we can determine the number of moles required for the desired concentration and volume:
Moles of glycine = Molarity × Volume
Moles of glycine = 0.100 mol/L × 2.00 L
Moles of glycine = 0.200 mol
Step 2: Calculate the volume of sodium chloride salt needed
Since the glycine is in the form of the sodium chloride salt, we need to determine the volume required to obtain the desired number of moles. To do this, we can use the formula:
Volume of sodium chloride salt = Moles of glycine / Concentration of sodium chloride salt
The concentration of the sodium chloride salt is not given, so we can assume it remains constant at 1.00M for this calculation. Hence:
Volume of sodium chloride salt = 0.200 mol / 1.00 mol/L
Volume of sodium chloride salt = 0.200 L (or 200 mL)
Step 3: Prepare the sodium chloride salt solution
Measure 200 mL of the sodium chloride salt solution and transfer it to a container. This will provide the required moles of glycine.
Step 4: Prepare the NaOH solution
Since the NaOH is given at a concentration of 1.00M, we can directly use it to prepare the buffer solution.
Step 5: Calculate the amount of NaOH required to neutralize glycine
The given reaction generates 2.00 mmoles of base (NaOH) in 100 mL of the buffer solution. As the amount of NaOH used is proportional to the amount of glycine present, we can calculate the amount of NaOH required for the desired buffer volume as follows:
Amount of NaOH required = (2.00 mmoles / 100 mL) × 2000 mL
Amount of NaOH required = 40.0 mmoles
Step 6: Calculate the amount of glycine neutralized by NaOH
Since glycine is a zwitterionic compound, the equation for neutralization with NaOH is as follows:
NaOH + H2NCH2COOH → Na+ + H2O + H2NCH2COO-
From this equation, we can see that one mole of NaOH neutralizes one mole of glycine. Therefore, the amount of glycine neutralized by NaOH is equal to the amount of NaOH used:
Amount of glycine neutralized = 40.0 mmoles
Step 7: Calculate the remaining glycine concentration
To find the concentration of glycine remaining in the buffer, we subtract the amount neutralized by NaOH from the initial amount of glycine:
Remaining amount of glycine = Initial amount - Amount neutralized
Remaining amount of glycine = 0.200 moles - 0.040 moles
Remaining amount of glycine = 0.160 moles
Step 8: Calculate the final solution pH
To determine the final solution pH, we compare the remaining glycine concentration with the pKa values of the carboxyl and protonated alpha amino groups of glycine:
pH = pKa + log ([A-] / [HA])
For the carboxyl group:
pKa (carboxyl) = 2.35
[H+] (carboxyl) = [A-] = 10^(-pH) = 10^(-9)
[HA] (carboxyl) = Remaining amount of glycine = 0.160 moles
pH = 2.35 + log (0.160 / 0.000000001) = 2.35 + 8.8 = 11.15
For the protonated alpha amino group:
pKa (protonated alpha amino) = 9.78
[H+] (protonated alpha amino) = [A-] = 10^(-pH) = 10^(-9)
[HA] (protonated alpha amino) = Remaining amount of glycine = 0.160 moles
pH = 9.78 + log (0.160 / 0.000000001) = 9.78 + 8.8 = 18.58
Therefore, the final solution pH is 11.15 for the carboxyl group and 18.58 for the protonated alpha amino group.
To prepare 2.00 L of 0.100 M pH 9.00 glycine buffer starting with the sodium chloride salt of zwitterionic glycine and 1.00 M NaOH, you would follow these steps:
1. Calculate the molar mass of glycine: Glycine has a molar mass of 75.07 g/mol.
2. Calculate the moles of glycine needed: Since we want to prepare a 0.100 M solution, we need (2.00 L) x (0.100 mol/L) = 0.200 mol of glycine.
3. Determine the amount of sodium chloride needed: Glycine will be prepared from its sodium chloride salt. Each molecule of sodium chloride contributes one molecule of glycine. Therefore, the amount of sodium chloride needed will also be 0.200 mol.
4. Dissolve the sodium chloride: Weigh out 0.200 mol of sodium chloride and dissolve it in enough water to make a final volume of 2.00 L.
5. Convert sodium chloride salt to zwitterionic glycine: The sodium chloride will react with NaOH to form zwitterionic glycine. The reaction equation is:
NaCl + NaOH → H2O + Na2CO3
Na2CO3 + CO2 + H2O → 2NaHCO3
NaHCO3 + H2CO3 ⇌ NaH2CO3 + H2O
NaH2CO3 + H2CO3 ⇌ 2H2CO3
6. Prepare 1.00 M NaOH solution: Dissolve an appropriate amount of NaOH in water to prepare a 1.00 M solution.
7. Add NaOH solution to sodium chloride solution: Slowly add the prepared 1.00 M NaOH solution to the sodium chloride solution while monitoring the pH. Stop adding NaOH when the pH reaches 9.00.
Now, for the second part of the question:
If a reaction that generated 2.00 mmol of base were carried out in 100 mL of the above buffer, we need to determine the final pH of the solution.
1. Calculate the initial concentrations of the buffer components:
- Sodium chloride (NaCl): 0.200 mol / 2.00 L = 0.100 M
- Glycine: 0.200 mol / 2.00 L = 0.100 M
- Sodium hydroxide (NaOH): 2.00 mmol / 0.100 L = 20.0 mM
2. Determine the initial ratio of protonated to deprotonated glycine: Since the pKa of glycine's carboxyl group is 2.35, and the pH is 9.00, the ratio of protonated (NH3+) to deprotonated (NH2) glycine can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
9.00 = 2.35 + log([NH2]/[NH3+])
Rearranging the equation:
[NH2]/[NH3+] = antilog(9.00 - 2.35) = antilog(6.65)
[NH2]/[NH3+] = 4.17 * 10^6
3. Determine the change in glycinhe concentration: Since the reaction generated 2.00 mmol of base, it also consumed 2.00 mmol of deprotonated glycine (NH2). The concentration of deprotonated glycine decreases by 2.00 mmol / 0.100 L = 20.0 mM.
4. Calculate the final concentration of deprotonated glycine: Subtract the change in concentration from the initial deprotonated glycine concentration:
[NH2]final = [NH2]initial - change in concentration
[NH2]final = 0.100 M - 20.0 mM = 80.0 mM
5. Calculate the final pH using the Henderson-Hasselbalch equation again:
pH = pKa + log([A-]/[HA])
pH = 9.78 + log(80.0 mM / 20.0 mM)
pH = 9.78 + log(4.0)
pH ≈ 10.30
Therefore, the final solution pH would be approximately 10.30.