Calculate the millimoles of HCl added to your buffer [line 7.] by using the molarity of HCl (0.10 M) and the volume (in mL) of 0.10 M HCl added to each tube.
Calculate the millimoles of HCl added to your buffer [line 7.] by using the molarity of HCl (0.10 M) and the volume (in mL) of 0.10 M HCl added to each tube.
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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Calculate the millimoles of HCl added to your buffer [line 7.] by using the molarity of HCl (0.10 M) and the volume (in mL) of 0.10 M HCl added to each tube.
![After Addition of 0.10 M HCI
Tube 1A
Tube 2A
Tube 3A
Tube 4A
0.50 mL
1.50 mL
3.00 mL
4.50 mL
1. Measured pH of Buffer
6.55
5.79
4.25
3.20
After Addition of 0.10 M NAOH
Tube 1B
Tube 2B
Tube 3B
Tube 4B
0.50 mL
1.50 mL
3.00 mL
4.50 mL
2. Measured pH of Buffer
7.05
7.39
8.29
9.99
3. mmoles of phosphate (HPO, + H2PO4) in 5.00 mL of 0.10 M buffer solution
0.50 mmoles
2-.
[HPO,]
4. Ratio
(From Part A)
0.5011
[H,PO, ]
Calculations for Line 5 & Line 6 Below Using Ratio (Show All Work Here OR on the Back Side!)
mmoles of H2PO4+mmoles of HPO4–2=0.50 mmoles from eqn (1): mmoles of H2PO4+0.5011×(mmoles of H2PO4-)
=0.50 mmoles 1.5011 mmoles of H2PO4–0.50 mmolesmmoles of H2PO4–=0.50 mmoles1.5011mmoles of H2PO4-0.3
33 mmoles
Calculation for mmoles of each HPO42:
-2.
(mmoles of HP04-2)-0.5011×(mmoles of H2P04-)mmoles of HPO4–2=0.5011×(0.333 mmoles)mmoles of HPO4-2-0.
167 mmoles
5. mmoles of H,PO, in 5.00 mL of 0.10 M bưffer solution
0.333mmoles
6. mmoles of HPO,²- in 5.00 mL of 0.10 M buffer solution
0.167 mmoles
Tube 1A
Tube 2A
Tube 3A
Tube 4A
After Addition of 0.10 M HCI
0.50 mL
1.50 mL
3.00 mL
4.50 mL
7. mmoles of HCI Added
8. mmoles of HPO.2- [line 6.- line 7.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbd285f6f-67ec-4ffa-8e9b-700f12fa780d%2F9ca00a5f-f6c6-4cd2-9db1-9c5458390f4c%2Fyq16xq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:After Addition of 0.10 M HCI
Tube 1A
Tube 2A
Tube 3A
Tube 4A
0.50 mL
1.50 mL
3.00 mL
4.50 mL
1. Measured pH of Buffer
6.55
5.79
4.25
3.20
After Addition of 0.10 M NAOH
Tube 1B
Tube 2B
Tube 3B
Tube 4B
0.50 mL
1.50 mL
3.00 mL
4.50 mL
2. Measured pH of Buffer
7.05
7.39
8.29
9.99
3. mmoles of phosphate (HPO, + H2PO4) in 5.00 mL of 0.10 M buffer solution
0.50 mmoles
2-.
[HPO,]
4. Ratio
(From Part A)
0.5011
[H,PO, ]
Calculations for Line 5 & Line 6 Below Using Ratio (Show All Work Here OR on the Back Side!)
mmoles of H2PO4+mmoles of HPO4–2=0.50 mmoles from eqn (1): mmoles of H2PO4+0.5011×(mmoles of H2PO4-)
=0.50 mmoles 1.5011 mmoles of H2PO4–0.50 mmolesmmoles of H2PO4–=0.50 mmoles1.5011mmoles of H2PO4-0.3
33 mmoles
Calculation for mmoles of each HPO42:
-2.
(mmoles of HP04-2)-0.5011×(mmoles of H2P04-)mmoles of HPO4–2=0.5011×(0.333 mmoles)mmoles of HPO4-2-0.
167 mmoles
5. mmoles of H,PO, in 5.00 mL of 0.10 M bưffer solution
0.333mmoles
6. mmoles of HPO,²- in 5.00 mL of 0.10 M buffer solution
0.167 mmoles
Tube 1A
Tube 2A
Tube 3A
Tube 4A
After Addition of 0.10 M HCI
0.50 mL
1.50 mL
3.00 mL
4.50 mL
7. mmoles of HCI Added
8. mmoles of HPO.2- [line 6.- line 7.]
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