Experiment 8 Identification of Unknown Diprotic Acid
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Experiment 8
Identification of Unknown Diprotic Acid
Total Points: 49
Student’s Name:
Adam Miller
Lab Section: 804
NOTE: You must show your work for all calculations; no work, no credit.
Standardization of NaOH (3 pts)
Molar Mass of KHP:
Trial
Mass of KHP
Initial Buret
Reading
Final Buret
Reading
Total Volume
Used
Molarity of
NaOH
1
.570
0
30.4 mL
30.4 mL
.0918 M
2
.539
0
27.9 mL
27.9 mL
.0946 M
3
.567
0
29.4 mL
29.4 mL
.0944 M
Average Molarity of NaOH: .0936 M
(3 pts) Calculate the [NaOH] and the average [NaOH}.
Mass KHP / MM KHP * 1(ratio) / volume(L) NaOH = M NaOH
Trial 1 = .570 / 204.22 g * 1 / .0304 L = .0918M NaOH
Trial 2 = .539 / 204.22 g * 1 / .0279 L = .0946M NaOH
Trial 3 = .567 / 204.22 g * 1 / .0294 L = .0944M NaOH
Average = (.0918M + .0946M + .0944M) / 3 = .0936 M NaOH
Diprotic Acid Table (10 pts)
Buret Reading
Initial Reading
0
at First Endpoint
10.9
at Second Endpoint
22.3
pK
a1
= 2.54
K
a1
= 10
-2.34
= .00288
pK
a2
= 5.99
K
a2
= 10
-5.99
= 1.023E-6
Determination of Unknown Acid
Molecular Weight (show work)
117.62g
Mass of Unknown Acid
.120g
Identity of Your Acid
Maleic
(8 pts) Attach your graph with the pK
a
labeled to your report. Using pK
a
values and the molecular weight,
determine the identity of the unknown acid (justify your answer).
pKa
1
= 5.45mL = 2.54pH -> 10
-2.34
= .00288
pKa
2
= 16.35mL = 5.99pH -> 10
-5.99
= 1.023E-6
Experiment 8
Identification of Unknown Diprotic Acid
Total Points: 49
.0936 M NaOH = x/.0109L
.0010 mol NaOH * 1 mol H
2
A / 1 mol NaOH = .0010 mol H
2
A
.0010 mol H
2
A / .120 g = .008502
1/.008502 = 117.62g
Maleic Acid = 116g for Molecular Weight, 117.62g is closer to 116g than the 134g for Malic Acid
Post-lab Questions
NOTE: You must show your work for all calculations; no work, no credit.
1.
(5 pts)
a.
Calculate the pH of a 0.0653M solution of hydrogen sulfide (H
2
S) where:
H
2
S
(aq) → H
+
(aq)
+ HS
-
(aq)
K
a1
= 1.0 × 10
-7
HS
-
(aq)
→ H
+
(aq) + S
2-
(aq)
K
a2
= 1.0 × 10
-19
H
2
S
(aq) → H
+
(aq)
+ HS
-
(aq)
I
.0653
0
0
x
2
/ .0653-x = 1.0E-7
C
-x
+x
+x
x = 8.0758E-5
E
.0653 – x
x
x
pH = 4.09
b. Calculate [S
2-
] in the solution.
HS
-
(aq)
→ H
+
(aq) + S
2-
(aq)
I
8.0758E-5 8.0758E-5
0
(x(8.0758E-5+x) / (8.0758E-5 – x) = 1.0E-19
C
-x
+x
+x
x = 1.0E-19
E
8.0758E-5 – x. 8.0758E-5 + x x
2.
(2 pts) Why does a diprotic acid dissociate in a stepwise manner?
Diprotic means 2 hydrogens and each step is a hydrogen being removed. A diprotic acid loses an H
+
after its first dissociation. What remains has a negative charge so the H
+
is held tighter and makes it harder to remove the second hydrogen and dissociating in a stepwise matter.
Experiment 8
Identification of Unknown Diprotic Acid
Total Points: 49
3.
(18 pts) Calculate the pH at each of the following points in the titration of 50.00 mL of a 0.150M acetic acid using 0.275M NaOH. (Show work below.) (K
a
= 1.8 × 10
-5
) (You can also refer to pages 792-794 from your textbook for help.)
a.
Initial pH: 0 mL of base added
b.
Halfway to equivalence point
c.
5.00 mL before equivalence point
d.
At equivalence point
e.
5.00 mL beyond equivalence point
f.
a.
HA
(aq) + H
2
O
(l)
→ A
-
(aq)
+ H
3
O
(aq)
I
.150
-
0 0
C
-x
-
+x
+x
E .150 – x
-
x x
1.8E-5 = x
2
/(.150 – x)
x = .001634
-log(.001634) = pH = 2.79
b.
pKa = pH
-log(1.8E-5) = 4.74
c.
.05 L * .150 = .0075 mol Acetic acid
Equivalence point volume = 5 mL before equivalence point
27.3 – 5.00 = 22.3 mL
Moles NaOH = .275 * .0223 = .00613mol
HA
(aq) + OH
-
(aq)
→ A
-
(aq)
+ H
2
O
(l)
I
.0075
0
0 -
C -.00613
+.00613
+.00613
-
E .00138 0 .00613 -
pH = pKa + log(.00613/.00138) = 5.39
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Experiment 8
Identification of Unknown Diprotic Acid
Total Points: 49
d.
HA
+
OH
→
A
-
+
H
2
O I
.0075
0
0
-
moles of A
-
= .0075
/ .07727 L = .097 M C
-.0075
+.0075
+.0075
-
E
0
0
.0075
-
A
-
+
H
2
O
→
HA
+
OH
1.0e-14 / 1.8e-5 = 5.5556e-10 I
.097
-
0
0
x
2
/ (.097 – x ) = 5.5556e-10 C
-x
-
+x
+x
x = 7.31e-
6 E
.097-x
-
.0075
+x
-log (7.34e-6) = 5.134 = pOH
14 – 5.134 = 8.87
= pH
e.
OH
-
+
HA
→
A
-
+
H
2
O I
0
.0075
0
-
.00137 / .08227 = .016652
= [OH
-
]
C
+.00887
-.00887
-
-
pOH = -log (.016652) = 1.7785 E
.00137
0
.0075
-
14 – 1.7785 = 12.22
=
pH
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C Solved: Item 14 Acid Refers To X +
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Part B
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Use the naming conventions and table of polyatomic ions to complete the following formula-name pairs of oxyacids and organic acids.
Match the acid names and chemical formulas in the left column to the appropriate blanks in the chemical formulas and acid names on the right. Make certain each acid
name and chemical formula is complete before submitting
Document Sharing
answer.
User Settings
• View Available Hint(s)
Course Tools
>
Reset
Help
manganic acid
Acid name:
; Chemical formula: H2 MnO4
formic acid
Acid name: iodous acid; Chemical formula:
acetic acid
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Chemistry 1223 Molar Weight of a Solid Acid Laboratory Exercise
An unknown acid is chosen from the list in the 1223 Lab Procedure. Determine the Identify the unknown diprotic acid.
The procedure for identifying an unknown acid is exactly as described in the lab manual.
Part 1: Standardization of NaOH. 1.4378 g of oxalic acid was dissolved in 250.0 mL of water and a 25.00 mL aliquote was titrated with
NaOH. 24.73 mL of titrant was required.
Determine the concentration of sodium hydroxide.
Answer:
Part 2: Titration of the diprotic unknown acid. 1.9875 g of the unknown acid was dissolved in 250.0 ml of water and 25.00 ml of this
solution was titrated with the NaOH. 29.50 mL of titrant was required.
Determine the Molar Mass of the diprotic unknown.
units g/mol
Answer:
MyASUS
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Next Page
Page 4 of 15
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Concentration Concentration
of H+ ions
Solution
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of OH ions
100%
High
Low
E
8%
8.5
High
High
Low
100%
11
Low
3%
3.5
High
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Which two solutions would be the BEST choice for their experiment?
A. Solutions 1 and 3 because they are a strong acid and a strong base.
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C. Solutions 2 and 3 because they are a weak base and a strong base.
D. Solutions 3 and 4 because they are a strong base and a weak acid.
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Name of Sample Vinegar
Vinegar Sample 1
Vinegar Sample 2
Vinegar Sample 3
Initial pH Reading
pH = 3.211
pH = 3.102
pH = 3.105
Initial Buret Reading
2.89 mL
3.00 mL
1.61 mL
Midpoint pH Reading
NOT NEEDED
pH = 4.895
pH = 4.687
Midpoint Buret Reading
NOT NEEDED
12.92 mL
9.50 mL
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pH = 8.238
pH = 8.531
pH = 8.934
Final Buret Reading
20.12 mL
20.31 mL
19.00 mL
Volume of NaOH Dispensed
Calculated Molarity of Acetic Acid in the Vinegar Sample
Volume NaOH Dispensed at Midpoint Reading
NOT NEEDED
Moles of NaOH Consumed at Midpoint Reading
[A–]
NOT NEEDED
Moles of Acetic…
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c. 500
d. 100
e. 200
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1.
1
2
3
O = HA O= A
= H,0*
%3D
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HBRO
HCIO
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Which one of the following solutions will have the greatest concentration of hydroxide ions?
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12
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shown.
Concentration
Concentration
Solution
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pH
of H+ ions
of OH- ions
1
100 %
high
low
2
8 %
8.5
low
high
3
100%
11
low
high
4
3%
3.5
high
low
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Solutions 2 and 3 because they are a weak base and a strong base.
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1.What reagent is used as the indicator for the neutralization?
a)citric acid
b)sodium hydroxide
c)sodium citrate
d)phenolphthalein
2.
What is the mole ratio of NaOH to citric acid for this neutralization reaction?
a)2 mole NaOH/1 mole citric acid
b)1 mole NaOH/1 mole citric acid
c)1 mole NaOH/2 mole citric acid
d)1 mole NaOH/3 mole citric acid
e)3 mole NaOH/1 mole citric acid
3.If 35.00 mL of 0.1104 M NaOH were used to reach the neutralization point, how many moles of NaOH were dispensed?
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. Calculate the moles of H+ delivered to eachsample based on your experimental dataand record these values in Data Table 2.
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Initial HCl volume in Syringe
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1.0ml
1.0ml
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.54ml
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2.3 x 10³ M
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