Refer to Appendix 6 which shows the calculations involved in determining the volumes of 0.20 M acetic acid and 0.20 M sodium acetate needed to make 20.00 mL of a buffer with pH = 5.00. Study the method used: set up two equations with two unknowns; the first is the Henderson-Hasselbalch equation, the second is the fact that the volumes of HA and A− solutions should sum to 20 mL. In the Henderson-Hasselbalch equation, use the same volumes for the ratio, since the concentrations of the HA and A− solutions are the same, and the change in the ratio is due to the different volumes. Once you understand the method, use it to calculate the volumes of 0.20 M acetic acid and 0.20 M sodium acetate needed to make 20.00 mL of a buffer having pH = 4.00 and again to make 20.00 mL of a buffer with pH = 6.00. Your TA will check the accuracy of your calculations before you mix your buffer solutions. 2. Between the three buffers (pH 4, pH 5, pH 6), which will stay in the buffer region for the greatest addition of base. You may use a titration curve to explain this.
Refer to Appendix 6 which shows the calculations involved in determining the volumes of 0.20 M acetic acid and 0.20 M sodium acetate needed to make 20.00 mL of a buffer with pH = 5.00. Study the method used: set up two equations with two unknowns; the first is the Henderson-Hasselbalch equation, the second is the fact that the volumes of HA and A− solutions should sum to 20 mL. In the Henderson-Hasselbalch equation, use the same volumes for the ratio, since the concentrations of the HA and A− solutions are the same, and the change in the ratio is due to the different volumes. Once you understand the method, use it to calculate the volumes of 0.20 M acetic acid and 0.20 M sodium acetate needed to make 20.00 mL of a buffer having pH = 4.00 and again to make 20.00 mL of a buffer with pH = 6.00. Your TA will check the accuracy of your calculations before you mix your buffer solutions. 2. Between the three buffers (pH 4, pH 5, pH 6), which will stay in the buffer region for the greatest addition of base. You may use a titration curve to explain this.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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1. Refer to Appendix 6 which shows the calculations involved in determining the
volumes of 0.20 M acetic acid and 0.20 M sodium acetate needed to make 20.00
mL of a buffer with pH = 5.00. Study the method used: set up two equations
with two unknowns; the first is the Henderson-Hasselbalch equation, the second
is the fact that the volumes of HA and A− solutions should sum to 20 mL. In the
Henderson-Hasselbalch equation, use the same volumes for the ratio, since the
concentrations of the HA and A− solutions are the same, and the change in the
ratio is due to the different volumes. Once you understand the method, use it to
calculate the volumes of 0.20 M acetic acid and 0.20 M sodium acetate needed to
make 20.00 mL of a buffer having pH = 4.00 and again to make 20.00 mL of a
buffer with pH = 6.00. Your TA will check the accuracy of your calculations
before you mix your buffer solutions.
2. Between the three buffers (pH 4, pH 5, pH 6), which will stay in the buffer region
for the greatest addition of base. You may use a titration curve to explain this.
APPENDIX 6 attached below:
Please help me answer this.
![Experiment#10 - Buffers
Appendix 6: Calculations for 20 mL of a 5.00 pH buffer, specific to this experiment
Henderson-Hasselbalch equation where [] denotes concentration in moles/liter.
pH = pka + log
([A-]
[HA]
Substituting the desired value 5.000 for pH and the pKa of acetic acid 4.757 gives
5.000 4.757 + log
log
[A-]
[HA]
This results in 1 equation with two unknowns; we will deal with that later by using a
second equation with the same variables. Now apply algebraic manipulation.
log
= 5.000 -4.757 = 0.243
Then use the definition of logarithm/antilogarithm.
[A-]
= 100.243 1.750
[HA]
Rearranging, we find that the concentration of A- is 1.750 that of HA.
[A-] = 1.750 [HA]
This is all in one beaker, so the volumes are exactly the same, which allows us to state
Moles of A- = 1.750 moles of HA
Molarity of the acetic acid solution and molarity of the sodium acetate solution we are
using are EQUAL, so to obtain the moles of acetate ion and moles of acetic acid involved,
we must use different volumes to get the required number of moles of each reactant.
VA 1.750 VHA
(a)
Now we are finally ready to create the second equation. Since we want a total of 20.00
mL of buffer solution, the second equation (in millilters) is
VA- + VHA = 20.00
(b)
Substitution of equation (a) into equation (b) gives
1.750 VHA + VHA = 20.00
and further algebraic manipulation yields
VHA = 20.00/2.750 = 7.27 mL of acetic acid solution, back substitute into equation (b) to
obtain VA- = 20.00 -7.27 = 12.73 mL of sodium acetate solution.
10-10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf76ebb0-a1fb-4e27-be44-e5f7107968b4%2F8889a014-fa3b-49a1-ba9b-0ab16af5334e%2Fwj44jia_processed.png&w=3840&q=75)
Transcribed Image Text:Experiment#10 - Buffers
Appendix 6: Calculations for 20 mL of a 5.00 pH buffer, specific to this experiment
Henderson-Hasselbalch equation where [] denotes concentration in moles/liter.
pH = pka + log
([A-]
[HA]
Substituting the desired value 5.000 for pH and the pKa of acetic acid 4.757 gives
5.000 4.757 + log
log
[A-]
[HA]
This results in 1 equation with two unknowns; we will deal with that later by using a
second equation with the same variables. Now apply algebraic manipulation.
log
= 5.000 -4.757 = 0.243
Then use the definition of logarithm/antilogarithm.
[A-]
= 100.243 1.750
[HA]
Rearranging, we find that the concentration of A- is 1.750 that of HA.
[A-] = 1.750 [HA]
This is all in one beaker, so the volumes are exactly the same, which allows us to state
Moles of A- = 1.750 moles of HA
Molarity of the acetic acid solution and molarity of the sodium acetate solution we are
using are EQUAL, so to obtain the moles of acetate ion and moles of acetic acid involved,
we must use different volumes to get the required number of moles of each reactant.
VA 1.750 VHA
(a)
Now we are finally ready to create the second equation. Since we want a total of 20.00
mL of buffer solution, the second equation (in millilters) is
VA- + VHA = 20.00
(b)
Substitution of equation (a) into equation (b) gives
1.750 VHA + VHA = 20.00
and further algebraic manipulation yields
VHA = 20.00/2.750 = 7.27 mL of acetic acid solution, back substitute into equation (b) to
obtain VA- = 20.00 -7.27 = 12.73 mL of sodium acetate solution.
10-10
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