Experiment 7 Irresistible Buffers
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Feb 20, 2024
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Experiment 7
Irresistible Buffers?
Total Points: 22
Student’s Name:
Adam Miller
Lab Section: 804
The Effect of Acid on Solutions (6 pts)
Beakers 1-6
Beaker
1
2
3
4
5
6
Contents
DI H
2
O
50 mL NaCl
0.35 g NaAc
+
50 mL HAc
0.60 g NaAc
+
50 mL HAc
1.00 g NaAc
+
50 mL HAc
2.00 g NaAc
+
50 mL HAc
mL of HCl
added
pH
pH
pH
pH
pH
pH
Black initial
solvent, 0
7.54
4.80
4.28
4.50
4.87
5.06
1
1.87
1.62
3.94
4.24
4.60
4.96
2
3.23
4.00
4.36
4.72
3
1.97
3.67
4.22
4.68
4
3.00
3.98
4.55
5
1.89
3.87
4.42
6
3.43
4.31
7
2.60
4.23
8
1.77
4.19
9
4.00
10
3.89
1.
(2 pts) Which beaker had the least amount of pH change? Explain why this occurred.
Beaker 6, because it had the greatest amount of NaAc
2.
(2 pts) Write the reaction occurring in beaker 1.
HCl
(aq)
+ H
2
0
(l)
<-> Cl
-
(aq)
+ H
3
0
+
(aq)
Experiment 7
Irresistible Buffers?
Total Points: 22
The Effect of Base on Solutions (6 pts)
Beakers 7-12
Beaker
7
8
9
10
11
12
Contents
DI H
2
O
NaCl
0.35 g NaAc
+
50 mL HAc
0.60 g NaAc
+
50 mL HAc
1.00 g NaAc
+
50 mL HAc
2.00 g NaAc
+
50 mL HAc
mL of NaOH
pH
pH
pH
pH
pH
pH
Blank, 0
6.00
9.41
4.40
4.55
4.68
5.04
1
10.22
11.53
4.45
4.62
4.88
5.17
2
4.66
4.85
5.10
5.31
3
4.92
5.05
5.42
5.58
4
5.31
5.42
5.55
5.71
5
5.95
6.27
6.23
6.59
6
10.87
10.75
11.45
11.50
7
8
9
10
Post-lab Questions
NOTE: You must show your work for all calculations; no work, no credit.
1.
(2 pts) Another buffer found in blood is based on the equilibrium between dihydrogen phosphate and monohydrogen phosphate. The reaction is shown below:
H
2
PO
4
-
(aq)
+ H
2
O
(l)
→ H
3
O
+
(aq)
+ HPO
4
2-
(aq)
If the pH of a blood sample was 7.11, what would you calculate as the ratio of [H
2
PO
4
-
] to [HPO
4
2-
]?
(K
a1
= 7.5 × 10
-3
, K
a2
= 6.2 × 10
-8
, K
a3
= 3.6 × 10
-13
)
pH = pKa + log [A
-
]/[HA]
-log(6.2E-8) = 7.2076
7.11 = 7.2076 + log [HPO
4
2-
]/[H
2
PO
4
-
]
[HPO
4
2-
]/[H
2
PO
4
-
] = 10
-.0976
= .799
[H
2
PO
4
-
]/[HPO
4
2-
] = 1/.799 = 1.25
Experiment 7
Irresistible Buffers?
Total Points: 22
2.
(4 pts) If you needed to perform a reaction in a controlled pH environment that was fairly basic, you might choose to use the ammonium and ammonia buffer system. How many grams of solid ammonium chloride would you have to add to 2.750 liters of 0.150M NH
3
to obtain a buffered solution of pH = 8.00?
(K
b
of NH
3
= 1.8 × 10
-5
)
pOH = pKb + log [NH
4
]/[NH
3
]
6.00 = 4.74 + log [NH
4
]/[NH
3
]
1.26 = log [NH
4
]/[NH
3
]
10
1.26
= 18.197
18.197 = [NH
4
]/[.150M]
2.729 = x / 2.750
x = 7.50mol
7.50 * 53.492 = 401 grams
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