[HA] H2PO4 [A-] -2 HPO4° 10. To prepare 50 mL of a 50 mM potassium phosphate buffer at the pH of your assigned buffer, calculate the ratios of the monobasic and dibasic species you would use with the Henderson-Hasselbalch equation. If you were preparing the buffer from 1 M stock solutions of the weak acid and the conjugate base, what volume would you use of each? Total moles = 0.050 M (0.050 L) = 0.0025 moles (moles HA + moles A-) PH = pKa + log [A-] 8.6-7.21 10° = [A] [HA] [HA] 8.6 = 7.21 + log (A / HA) = 1.39 A¯ HA Adibasic)=0.0025 mol 1.39 2.39 = 1.45 x 10° moles À A 1000 mL = 1.45 mL 1 mot 1K HA (monobasic) = 0.0025 mol = 12.39 1.05 x 10³ moly ĦA / X 1000 mL = 1.05 mL 1 mot 14 Initial pH 8.60 1. Record which buffer was assigned to you: A 2. Initial pH is (circle) above / below the pKa. I will be adding: (Acid/Base) *The pKa of the phosphate buffer is 7.21 Solution 50 mM Buffer 150 ML pH after 2nd Addition of Acid/Base 6.92 pH after 1st Addition of Acid/Base 7.95 Distilled Water 6.18 1.49 1.09 5 mM Buffer 8.06 6.30 3.04
[HA] H2PO4 [A-] -2 HPO4° 10. To prepare 50 mL of a 50 mM potassium phosphate buffer at the pH of your assigned buffer, calculate the ratios of the monobasic and dibasic species you would use with the Henderson-Hasselbalch equation. If you were preparing the buffer from 1 M stock solutions of the weak acid and the conjugate base, what volume would you use of each? Total moles = 0.050 M (0.050 L) = 0.0025 moles (moles HA + moles A-) PH = pKa + log [A-] 8.6-7.21 10° = [A] [HA] [HA] 8.6 = 7.21 + log (A / HA) = 1.39 A¯ HA Adibasic)=0.0025 mol 1.39 2.39 = 1.45 x 10° moles À A 1000 mL = 1.45 mL 1 mot 1K HA (monobasic) = 0.0025 mol = 12.39 1.05 x 10³ moly ĦA / X 1000 mL = 1.05 mL 1 mot 14 Initial pH 8.60 1. Record which buffer was assigned to you: A 2. Initial pH is (circle) above / below the pKa. I will be adding: (Acid/Base) *The pKa of the phosphate buffer is 7.21 Solution 50 mM Buffer 150 ML pH after 2nd Addition of Acid/Base 6.92 pH after 1st Addition of Acid/Base 7.95 Distilled Water 6.18 1.49 1.09 5 mM Buffer 8.06 6.30 3.04
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
Problem 1P
Related questions
Question
- Using your data, explain your observations for the pH changes upon the addition of acid/base to your 50 mM buffer, the 5 mM buffer, and the distilled water. Complete the calculations below. Use your notebook for scratch paper if needed.
For each solution, you will need to calculate the total moles of buffer (if applicable), the moles of monobasic, the moles of dibasic, and the moles for each addition of acid/base. Determine the expected pH for the first addition of acid/base, the second addition of acid/base and compare to your measured values.
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50 mM Buffer Calculations |
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A- moles |
HA moles |
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Moles total = |
2.5 x 10^-3 |
1.45 x10 ^-3 |
1.5 x10 ^-3 |
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Addition of Acid/Base |
1st addition |
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moles added = |
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2nd addition |
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moles added = |
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5mM Buffer Calculations |
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A- moles |
HA moles |
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Moles total = |
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Addition of Acid/Base |
1st addition |
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moles added = |
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2nd addition |
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moles added = |
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Water Calculations |
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A- moles |
HA moles |
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Moles total = |
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Addition of Acid/Base |
1st addition |
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2nd addition |
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![[HA]
H2PO4
[A-]
-2
HPO4°
10. To prepare 50 mL of a 50 mM potassium phosphate buffer at the pH of your assigned
buffer, calculate the ratios of the monobasic and dibasic species you would use with the
Henderson-Hasselbalch equation. If you were preparing the buffer from 1 M stock
solutions of the weak acid and the conjugate base, what volume would you use of each?
Total moles =
0.050 M (0.050 L)
= 0.0025 moles
(moles HA + moles A-)
PH = pKa + log [A-]
8.6-7.21
10°
=
[A]
[HA]
[HA]
8.6 = 7.21 + log (A / HA)
= 1.39 A¯
HA
Adibasic)=0.0025 mol
1.39
2.39
= 1.45 x 10° moles À
A
1000 mL
= 1.45 mL
1 mot
1K
HA (monobasic) = 0.0025 mol
=
12.39
1.05 x 10³ moly ĦA / X
1000 mL
=
1.05 mL
1 mot
14](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa5644016-779f-4740-bb26-bd379069d798%2F516072dd-c27b-48d6-b3e9-6b2a97a83479%2F7e4x6au_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[HA]
H2PO4
[A-]
-2
HPO4°
10. To prepare 50 mL of a 50 mM potassium phosphate buffer at the pH of your assigned
buffer, calculate the ratios of the monobasic and dibasic species you would use with the
Henderson-Hasselbalch equation. If you were preparing the buffer from 1 M stock
solutions of the weak acid and the conjugate base, what volume would you use of each?
Total moles =
0.050 M (0.050 L)
= 0.0025 moles
(moles HA + moles A-)
PH = pKa + log [A-]
8.6-7.21
10°
=
[A]
[HA]
[HA]
8.6 = 7.21 + log (A / HA)
= 1.39 A¯
HA
Adibasic)=0.0025 mol
1.39
2.39
= 1.45 x 10° moles À
A
1000 mL
= 1.45 mL
1 mot
1K
HA (monobasic) = 0.0025 mol
=
12.39
1.05 x 10³ moly ĦA / X
1000 mL
=
1.05 mL
1 mot
14
Transcribed Image Text:Initial pH
8.60
1. Record which buffer was assigned to you:
A
2. Initial pH is (circle) above / below the pKa. I will be adding:
(Acid/Base) *The pKa of the phosphate buffer is 7.21
Solution
50 mM Buffer
150 ML
pH after 2nd
Addition of
Acid/Base
6.92
pH after 1st
Addition of
Acid/Base
7.95
Distilled Water
6.18
1.49
1.09
5 mM Buffer
8.06
6.30
3.04
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