Lab 2
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CHEM 282 Experiment #2: Separation of an Acid, a Base, and a Neutral Compound by Extraction Sophie Wolkoff (20107258) & Kaelen Partridge (20127197) TA: Bily Deng January 30, 2020
Experimental: Before the experiment, a flow chart was provided in the lab manual in step 1 which displayed each stage in the separation scheme in order to summarize the actions to separate a mixture into its constituents. Step 2 of the experiment began with the extraction of base from the unknown three-component mixture provided. This mixture was dissolved in ether and placed into a separatory funnel. 5% hydrochloric acid was subsequently added and mixed in order to separate the solution into layers, releasing the pressure in the stoperred and stopcocked funnel after each inversion, swirl and shake of the solution. The lower layer was drained into a flask and the process was repeated twice, adding 5% hydrochloric acid first and then using water. The lower layer was drained into the same flask each time, labelled as hydrochloric acid extract. This liquid was set aside, containing the hydrochloride salt of the organic amine. In step 3, the acid and neutral compounds were separated from within the mixture. Using the ether layer remaining in the separatory funnel, 5% sodium hydroxide was added and mixed into the solution to separate the layers, venting after each inversion of the stoperred and stopcocked funnel. The lower layer was drained into a separate flask and this process was repeated twice, first with 5% sodium hydroxide solution and then with water, draining into the same flask each time. The flask was labelled as sodium hydroxide extract and set aside, containing the sodium salt of the organic acid. The neutral compound was then isolated by pouring the ether layer through the neck of the funnel, as to not come into contact with other substances, into a flask and adding anhydrous sodium sulfate. The solution was swirled and left to stand
for about 15 minutes, subsequently being filtered by gravity filtration into a previously weighed beaker. The sodium sulfate was rinsed with ether and placed in the fume hood to allow the ether to evaporate, eventually leaving the neutral compound residue which was weighed for percent recovery. Step 4 involved the use of the collected hydrochloric acid extract, cooling the solution in an ice-water bath and stirring in 10% aqueous sodium hydroxide to make the solution basic. The precipitate that formed within the flask was collected using vacuum filtration and transferred to a preweighed watch glass to be air-dried and weighed to report the percent recovery. A similar process was performed in step 5, cooling the sodium hydroxide extract in an ice-water bath and gradually stirring in 10% hydrochloric acid to make the solution acidic. The precipitate formed within the mixture was collected using vacuum filtration and transferred to a pre-weighed watch glass to be air-dried and weighed for the percent recovery. One of the compounds produced in the lab, acid, base, and or neutral, were chosen to be identified by melting point measurement using reference samples in step 6. The three substances, one experimental and two reference, were placed in a melting point apparatus to confirm their identity by performing mixed melting point determinations.
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Results: Masses of Products
Starting mass of the mixture: 1.500 g Mass of Watch Glass/Beaker (g) Mass of Watch Glass/Beaker +
Isolated Compound (g) Mass of Isolated Compound (g) Percent Recovery Basic Compound 31.394 31.852 0.458 30.53% Acidic Compound 33.201 33.230 0.029 1.933% Neutral Compound 50.198 50.280 0.082 5.467% Percent recovery calculations
Basic compound:
Pure
I
m
m
p
a
u
ss
r
e
o
m
f
i
a
s
s
o
s
l
a
o
t
f
e
d
m
c
ix
o
t
m
ur
p
e
o
(
u
g
n
)
d
(
g
) × 100% =
× 100% = 30.53%
Acic compound:
Pure
I
m
m
p
a
u
ss
r
e
o
m
f
i
a
s
s
o
s
l
a
o
t
f
e
d
m
c
ix
o
t
m
ur
p
e
o
(
u
g
n
)
d
(
g
) × 100% =
× 100% = 1.933%
Neutral compound:
Pure
I
m
m
p
a
u
ss
r
e
o
m
f
i
a
s
s
o
s
l
a
o
t
f
e
d
m
c
ix
o
t
m
ur
p
e
o
(
u
g
n
)
d
(
g
) × 100%
= × 100% = 5.467%
Melting Point Comparison
Our basic product (°C)
Ethyl p-Aminobenzoate (°C)
P-Chloroaniline
(
°C)
Experimental melting point: 78.6-81.5 90.4-93.4 68.5-72.9 Literature melting point: N/A 88-90 70-72 Based on the melting point of our basic compound, compared with the melting points of the known basic compounds, it is most likely that our unknown compound is p-
Chloroaniline. Although the melting point for our unknown compound is slightly higher than that of p-chloroaniline, it is still closer than the melting point of ethyl p-
aminobenzoate, indicating that p-chloroaniline is the best option of the two. Discussion: Part A
Liquid-liquid extraction, otherwise known as solvent extraction, is a method used to separate the components within a liquid mixture by pulling the compounds, or solutes,
within a solution from one solvent to another.² This involves the countercurrent movement of solutes from within solutions using other immiscible solvents that the solutes are soluble in.¹ Extraction methods differ depending upon the density of the solvents being used, since solvents that are more or less dense than water will require different or additional glassware.² The most common method is performed using repetitive extraction with a separatory funnel. Effective separation strategies by extraction can be planned by using the pKa values of the compounds that are used. Using these acid and base properties of
substances, and recognizing how organic functional groups will react to these conditions, can assist in planning a highly functional extraction process. Firstly, it is understood that stronger acids have higher values of Ka, describing their hydrogen ion concentration, and therefore lower values of pKa.³ This is opposite for weak acids, with smaller Ka and larger pKa values. Strong bases have small pKb values, describing their hydroxide ion concentration, which relates to higher values of pKa.³ This is opposite according to weak bases. When analyzing a reaction of an acid being extracted from a solution using a base, it is important for the conjugate base made from the reaction to have a higher pKa
than the initial acid that was extracted. This is significant as this higher value of pKa of the conjugate base compared to the acid will lead to a product of higher stability.
⁴
If a base chosen to extract an acid is strong, the pKa of the conjugate base produced will be
higher than the acid it was made from, regardless if it was strong or weak. Using this information, it is understood that a strong base cannot be chosen to extract two different
acids with differing pKa values within one solution. Therefore, a weak base must also be
used in order to block one reaction from proceeding with one of the acids. Only the reaction with the strong acid will continue because it will produce a conjugate base product of less reactivity and higher stability.
⁴
This allows the two acids of differing strength to be separated and extracted. Part B
Our results of this experiment were not entirely consistent with the theory. Our percent yield for each product was quite low, and the total percent yield only added up
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to 37.93%. Considering that the compound was made up of one third of acid, base and neutral compound, it was expected that the percent yield for each would be around 33.33%. Our yield for the basic compound was 30.53%, which is quite close to the expected, however the yields for the acid and neutral compounds were significantly lower than the expected, showing inconsistencies with the theory. It is possible that our yields were so low due to errors during the experiment, leading to the loss of some of the product. In relation to the melting points, our results showed that the unknown basic compound was most likely to be p-chloroaniline. Although our unknown’s experimental melting point did not exactly match the melting point of p-chloroaniline, they came quite close to each other, indicating the likelihood of them being the same and proving that our experimental findings for the melting point data was quite consistent with the theory. Overall, we were successful at isolating the basic compound from the mixture, however we were not as successful at isolating the neutral and acidic compounds, as shown by their very small final yields. We were also quite successful at using the melting points to identify what the unknown base was. Questions: 1.
Throughout this experiment, there were multiple instances in which experimental errors could have occurred. Firstly, when performing the liquid-liquid extractions by draining the lower layer out of the separatory funnel, it is possible that too much solution was drained out. If this occurred, then there would be incorrect compounds mixed into the layers, resulting in impure compounds after filtration.
Another potential source of error could have occurred during the vacuum filtration
process. When scraping the crystals off of the filter paper, it is possible that amounts of crystal stuck to the paper and thus were not included when we weighed the yield. Finally, another potential source of error is that we did not dry the product for enough time after filtration and the crystals were still slightly damp, which added excess weight to the final yield. 2.
In part 1 of the lab, the step by step separation scheme of a liquid-liquid extraction is shown, starting with a mixture dissolved in ethyl ether solution. The first stage involves separating the base using an acid solution of 5% hydrochloric acid. This process gives a proton from the acid to the base, effectively forming a salt. The salt product is then able to dissolve from the organic layer into the aqueous layer where it can be collected as HCl extract.(1) This aqueous solution is then mixed with a base solution of 10% sodium hydroxide which now makes the mixture neutral. This is performed so that the base can be isolated from the solution as a solid or oil.(2) The previous diethyl ether solution now only contains the acid and neutral components which must be separated in the second stage. This next process involves adding a base of 5% sodium hydroxide. The base accepts a proton from the acid to become a salt, which can then be separated into the aqueous layer and collected as NaOH extract.(3) An acid solution of 10%
hydrochloric acid is then added to this aqueous solution to neutralize the base present and allow the acid to be isolated as a solid.(4) (1) R’-NH
2 (ether)
+ HCl
(
aq)
⇌
R’-NH
3
+ (aq)
+ Cl
-
(aq)
(2) R’-NH
3
+
+ Cl
-
(aq)
+ NaOH
(
aq)
⇌
R’-NH
2 (s)
+ NaCl
(
aq)
+ H
2
O
(
l)
(3) R-COOH
(ether)
+ NaOH
(
aq)
⇌
R-COO
-
Na
+
(aq)
+ H
2
O
(
l) (4) R-COO
-
Na
+
(aq)
+ HCl
(
aq)
⇌
R-COOH
(
s)
+ NaCl
(
aq) 3. 4.
A better percent recovery of the base product could have been obtained if a number of adjustments were made in the performance of the experiment. One of these adjustments could be allowing the solution in the separatory funnel to settle
for a longer period of time so as to give the compounds within the solution
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enough time to separate completely. This would ensure that the maximum amount of compound was collected into each layer. Another possible method that could have helped produce more base is extracting the solution repeatedly with 5% hydrochloric acid. If we were to perform additional extractions, along with
the extractions written into the procedure, we would be able to ensure with more accuracy that most of the basic compound in the ether solution was separated. This occurs by the base interacting with the acid to form a salt in order to separate into the aqueous layer and be dispensed. Another modification that would have helped in this process is further shaking the separatory funnel, as this would have increased the interaction between the hydrochloric acid molecules and the basic compound, and would allow more of the basic compound to flow into the aqueous layer. Additionally, in part 4 of the lab, if the flask containing the aqueous solution from the funnel was left in the ice-water bath for a longer length of time, this could have also helped to produce a higher yield of base. This is because it would allow for more base crystals to be formed, filtered and eventually weighed. The number of vacuum filtrations could have also increased the yield, as additional crystals could have accidentally gone through the filtration. If filtration were performed repeatedly, the lost crystals would have been collected and would increase the final weight recovered. 5.
Similar to the base product, a better percent recovery of the acid product could have been obtained by modifying certain steps of our experimental procedure. Firstly, the compounds present in the solution would have been able to
completely separate into their respective layers if the contents within the separatory funnel could sit for a longer period of time. Ensuring that all compounds have divided completely into the layers would allow for the collection of a more accurate extract. Another adjustment to the procedure could include performing multiple extraction of the solution using 5% sodium hydroxide. This repeated process could guarantee that most to all of the acid compound in the solution would react with the base provided to form a salt that would separate into the aqueous layer. Additionally, if the separatory funnel containing the solution and solvent were shaken more, this would also ensure that all of the acid
and base compound contained in the funnel came into contact with each other to form salt and dissolve into the lower layer. Both of these methods could help to verify that the maximum amount of acid and base components interacted and were present within the aqueous layer. More acid yield could have been produced if the flask containing the aqueous solution collected remained in the ice-water bath for a prolonged amount of time in step 5, much like the previous base product. This is because more crystals would have likely been formed and subsequently been collected by vacuum filtration to increase the final weight obtained. The amount of vacuum filtrations performed could have also affected this value as the potential for crystals to be lost in the water removed from filtration would be reduced, and the amount of additional crystals collected would increase.
References: 1.
Thornton, J. D. EXTRACTION, LIQUID-LIQUID. http://www.thermopedia.com/content/752/
(accessed Jan 29, 2020). 2.
Finchsigmate, K. Liquid-Liquid Extraction. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Demos,_Tech niques,_and_Experiments/General_Lab_Techniques/Liquid-Liquid_Extract ion
(accessed Jan 29, 2020).
3.
Helmenstine, A. M. pH, pKa, Ka, pKb, and Kb Explained. https://www.thoughtco.com/ph-pka-ka-pkb-and-kb-explained-4027791
(accessed Jan 29, 2020). 4.
Lancashire, R. J. Experiment 8 - Separation of an Unknown Mixture by Acid/Base Extraction. http://wwwchem.uwimona.edu.jm/lab_manuals
/
c1901exp8.html
(accessed Jan 29, 2020).
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8. What is the name of the following compound?
Headings
Pages
CH2CH3
Results
<-CH,CH, ci°
ČH2CH3
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c. N,N,N-triethylanilinium chloride
d. N,N,N-triethylaniline chlorine
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a. N,N,N-trimethylanilinium chloride
b. N,N,N-diethylanilinium chloride
move your content around.
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both questions go together... thank you!
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Should be 3 answers in total. Will thumbs up!
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In RUN 2, 20.0 mL of 5.00 M NAOH is combined with 10.0 mL of 0.010 M CO and 70.0 mL of wate
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[crystal orange] vs time (RUN 1)
In[crystal orange] vs time (RUN 1)
0.0012
0.001
-2
y--0.4463x- 7.3334
y =-0.0002x + 0.0007
R- 0.6565
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0.0008
4
0.0006
-6
0.0004
-8
0.0002
0.0002
-12
time (hr)
time (hr)
1/[crystal orange] vs time (RUN 1)
1/[crystal orange] vs time (RUN 2)
12000
25000
10000
20000
8000
15000
6000
y = 2000x + 1000
R =1
10000
y 4000x + 1000
R =1
4000
2000
5000
time (hr)
time (hr)
Figure 4. Kinetic traces for the reaction between crystal orange and sodium hydroxide.
What is the order with respect to CO?
| Select ]
What is the order with respect to NaOH?
| Select ]
What is the value of the averaged rate constant for the reaction above? Give…
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O Dr. Waybell retain reject Ho and computes n=0.005. She has made a Type Il error.
O Dr. Waybell rejects Ho and computes 7=0.005. She has made a Type Il error.
Dr. Waybell retain reject Ho and computes n=0.05. She has made a Type Il error.
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For the following questions (Q1-Q5), consider the compound tartrazine (Yellow 5 food dye) that has a chemical formula of C16H9N4Na3O9S2. Ana weighs out 0.1328 g of tartrazine powder on the balance for her experiment and makes a stock solution by dissolving her powder in 500.0 mL of solution.
Concentration (µM or µmol/L)
Absorbance
0.00
0.001
9.94
0.252
19.9
0.506
29.8
0.738
39.8
1.018
slope (m): 0.02534 1/(µM)
y-intercept (b): -0.0009147
Q4. Calculate the molar extinction coefficient (molar absorptivity) of tartrazine in units of 1/(M·cm). Assume a cuvette path length of 1.00 cm.
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Number 2,3,and 4
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pH
pH = 4.389
pH = 4.874
=
pH = 5.339
pKa
Answer Bank
Using the equation to calculate the quotient [AT]/[HA] at three different pH values.
[A]
[HA]
+ log
[A-]
[HA]
[A-]
[HA]
=
=
=
LA I
[HA]
1
0.3273
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