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Embed code for: Solubility Lab Report
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(Notebook Pages 12-21)
Julia Troncoso, Antonio McNealy, Wessley Putney, Matthew Kazee
Laboratory Instructor: Dr. Mark Italia
CHM 2210 L
September 15, 2016
The purpose of this laboratory experiment was to explore the concept of solubility as it applies to organic molecules. A solid that completely dissolves in a solvent is said to be soluble; in turn, if a significant amount of solid solute remains intact in a solvent, the substance is said to be insoluble. For liquids, the terms miscible and immiscible are used in place of soluble and insoluble, respectively. If two substances are immiscible when mixed together, distinct layers form within the mixture. It is important to know the solubility properties of organic molecules. The composition and structure of an organic molecule greatly influences the solubility of the substance in a particular solvent. Being familiar with the structure of a molecule allows a prediction about solubility to be made. When making predictions about solubility, the rule that “like dissolves like” can be applied. This rule is based off of the idea that polar solutes more readily dissolve in polar solvents than they do in nonpolar solvents. Likewise, nonpolar solvents will dissolve nonpolar molecules. The polarity of a molecule can be established by the presence of polar functional groups that contain more electronegative atoms. The idea that substances that are alike in polarity will be soluble (or miscible) was tested throughout this experiment and the solubility properties of various organic molecules were noted.
This lab was performed in four units (A-D). Each exercise was done in order to establish the solubility of different solutes, and their solubility in various solvents.
For part A of this lab, four test tubes were used. Approximately 40.0 mg of benzophenone was added to each test tube. After the benzophenone was added to each test tube, different solvents were added in order to check the solubility of the organic compound. 1 mL of deionized water was added to the first test tube. 1 mL of methyl alcohol was added to the second test tube. 1 mL was of hexane was added to the third test tube, and the last test tube acted as a control. Once all of the test tubes contained both the solute and solvent, each test tube was swirled rapidly for about one minute. If the compound dissolved completely, the compound is soluble. If the compound partially dissolves then the compound is considered partially soluble. If none of the compound dissolves, the compound is insoluble. After the results were calculated, the same procedure was done using malonic acid and biphenyl instead of the benzophenone.
For part B of this lab, six test tubes were required. 1 mL of water was added to three of the test tubes, and 1 mL of hexane was to the other three test tubes. Once all the solvents were in the test tubes, alcohols were added dropwise to the solvents until approximately 20 drops were added. 20 drops were added of 1-octanol to one of the test tubes containing water and one of the test tubes containing hexane. 20 drops of 1-butanol were added to two test tubes (one water, and one hexane). Finally, 20 drops of methyl alcohol were added to the last two test tubes. If the alcohol was partially soluble, the drops would dissolve at first but eventually a second layer would form separating the two liquids. If it were insoluble then the drops would not dissolve, and if it were soluble it would have dissolved completely.
For part C of this lab, four test tubes were used. 1 mL of two different liquids was added to each test tube, and the miscibility needed to be determined for each. Once both liquids were added, they were shaken for about 10-20 seconds. If one layer was formed it was miscible, and if two layers were formed then it was immiscible. Test tube one contained water and ethyl alcohol. Test tube two contained water and methylene chloride. Test tube three contained water and hexane. Lastly, test tube four contained hexane and methylene chloride.
For part D of this lab, six test tubes were required. Approximately 30 mg of benzoic acid was added to three of the test tubes. The three test tubes were labeled 1, 2, and 3. 1 mL of water was added to test tube 1. 1 mL of 1.0M NaOH to test tube 2, and 1 mL of 1.0M HCl was added to test tube 3. Each mixture was stirred for 10-20 seconds. After the mixture was stirred, the solubility was recorded. 6.0M HCl was added dropwise to test tube 2 until the mixture became acidic. Once the mixture became acidic, the solubility was recorded. The experiment was then repeated using ethyl 4-aminobenzoate instead of benzoic acid. Then 6.0M of NaOH were added dropwise to test tube 3 until the mixture became basic. The results were then recorded again for the solubility of test tube 3.
Since the data accumulated for this series of experiments was more qualitative as opposed to quantitative, there are no resulting calculations to be included.
Results (and Report)
Part A. Solubility of Solid Compounds
Solubility of Solids
Methanol (moderately polar)
Malonic Acid (polar)
In Part A, Benzophenone was found to be insoluble in water, soluble in methanol (methyl alcohol), and partially soluble in hexane. This reflects the fact that benzophenone is mostly non-polar, except for the oxygen double-bonded to its central carbon. Being mostly non-polar makes it insoluble in water (which is polar), soluble in methanol (which is moderately polar). The difference in polarity between benzophenone (mostly non-polar) and hexane (non-polar) result in benzophenone being partially soluble in hexane.
Malonic acid was found to be soluble in water, soluble in methanol, and insoluble in hexane. This reflects the fact that malonic acid is polar and exhibits hydrogen bonding. Being polar with hydrogen bonding makes malonic acid soluble in water (which is polar with hydrogen bonding), and methanol (which is moderately polar with hydrogen bonding). The polar nature of malonic acid makes it insoluble in hexane (which is non-polar and does not exhibit hydrogen bonding).
Biphenyl was found to be insoluble in water, partially soluble in methanol, and soluble in hexane. As with the other solutes discussed, this reflects the fact that biphenyl is non-polar and does not exhibit hydrogen bonding.
Part B. Solubility of Different Alcohols
Solubility of Alcohols
1-Octanol (8:1 carbon to -OH ratio)
1-Butanol (4:1 carbon to -OH ratio)
Methanol (1:1 carbon to -OH ratio)
In Part B, 1-Octanol was found to be insoluble in water, but soluble in hexane. 1-Butanol was found to be insoluble in water and soluble in hexane. Methanol was found to soluble in water and insoluble in hexane. These results reflect the fact that the ratio of carbons to -OH group is significant in determining an alcohol’s relative polarity, and therefore its solubility in polar and non-polar solvents. Since the carbon to -OH ratio in octanol is 8:1, the relatively long, non-polar carbon chain blunts the polarizing effect of the -OH group, and octanol is a mostly non-polar molecule. Methanol, in contrast, has a 1:1 carbon to -OH group ratio, and the single-carbon methyl group does little to blunt the polarity of the -OH group. Therefore, methanol is a relatively more polar molecule than octanol. Butanol, with a 4:1 carbon to -OH group ratio, is more polar than octanol, but less polar than methanol. The solubility results support this concept, as the relatively more polar methyl alcohol is soluble in water, and the relatively less polar butanol and octanol are soluble in hexane.
Part C. Miscible or Immiscible Pairs
Miscible or Immiscible Pairs
Ethanol (moderately polar)
Methylene Chloride (non-polar)
** Not tested
In Part C, the polarity and hydrogen bonding capacity of the tested liquids determined whether they were miscible or immiscible. Water (which is polar with hydrogen bonding) was found to be miscible with ethanol (which is moderately polar with hydrogen bonding), and immiscible with hexane and methylene chloride (which are both non-polar with no hydrogen bonding). Methylene chloride was found to be miscible with hexane, as both substances are non-polar.
Part D. Solubility of Organic Acids and Bases
Solubility of Organic Acids and Bases
1.0 M NaOH
1.0 M HCl
(Add 6.0 M HCl)
(Add 6.0 M NaOH)
In Part D, benzoic acid was found to be insoluble in water and 1.0 M HCl solution. Benzoic acid was soluble in 1.0 M NaOH solution, but the benzoic acid precipitated out of the solution after 6.0 M HCl was added. Benzoic acid being mostly insoluble in water reflects the fact that benzoic acid is a weak acid (not easily dissociated) and is only slightly polar (a non-polar, six-carbon ring bonded to a smaller polar carboxyl group). Benzoic acid is soluble in NaOH solution because as a weak acid, the presence of excess -OH ions (from the NaOH) results in dissociation of the benzoic acid and the formation of its conjugate base, sodium benzoate. As sodium benzoate is ionic (and therefore polar), it easily dissolves in the aqueous NaOH solution. This dynamic is further illustrated by the addition of 6.0 M HCl to the solution, which results in excess hydronium ion, shifting the equilibrium back to favoring the formation of the non-ionic, mostly non-polar (and mostly insoluble) benzoic acid.
The solubility results of ethyl 4-aminobenzoate reflect the opposite of the mechanism by which benzoic acid and sodium benzoate are respectively insoluble and soluble in aqueous solution. Ethyl-4 aminobenzoate is insoluble in water and NaOH solution, but soluble in HCl solution (and consequently insoluble after 6.0 M NaOH is added). Ethyl 4-aminobenzote is a weak base which does not dissociate easily in water, nor NaOH solution with its excess hydroxide ion. However, in HCl solution ethyl 4-aminobenzoate picks up an additional proton to form its conjugate acid, ethyl 4-aminobenzoate hydrochloride (with an ionic bond to a chloride ion in solution), and is consequentially ionic and easily soluble in aqueous solution. When 6.0 M NaOH is then added in sufficient quantity the equilibrium shifts back, and the ethyl 4- aminobenzoate hydrochloride gives up its proton (and chloride ion) and becomes ethyl 4-aminobenzoate again, which is mostly non-polar and insoluble in the aqueous solution.
Generally speaking, the results were consistent with expectations. Since molecule polarity is a relative spectrum more than binary (polar/non-polar) values, some variation can exist in solubility of substances that are not completely polar or non-polar. Additionally, variations in technique (vigorousness of stirring, duration of stirring) create the potential for variation in solubility results. Also, the solutes were not weighed on a scale, and the solvents were not precisely pipetted to 0.1 ml, so the starting quantities were not as absolutely precise as they could have been. All these possible sources of variation could account for differences between measured and expected solubility.
Again, since polarity (and therefore, solubility) is a spectrum of relative values, the starting quantities and particular technique are important to the final results. 1-Butanol, for instance, is somewhere between absolute polar and absolute non-polar, and is therefore especially susceptible to variations in technique. If the mixing was not particularly vigorous, or the dropwise addition of butanol to water not particularly meticulous, the result could have been skewed toward insoluble, rather than partially soluble. Generally speaking, the results were within expectation.
In this section, the results did not vary greatly from expectation. Keeping with the idea that solubility is sensitive to mixing technique and starting quantities, the quantities of liquids were transferred with unmarked pipettes, which could have realistically been inaccurate by as much as 0.5 ml. A variation of this amount could have skewed the qualitative results.
The results of this section were consistent with expectations. Additionally, the experiment was set up in such a way that over-titrating the solutions would not have varied the results; for instance, adding too much 6.0 M NaOH solution would have the same result as adding exactly enough of that solution. That being said, the previously discussed variations in solubility related to substances that are not completely polar or non-polar apply to benzoic acid and ethyl 4-aminobenzoate. Therefore, the expected solubility of those substances, especially in water, could be difficult to precisely qualify.
1. For each of the following pairs of solutes and solvents, predict whether the solute would be soluble or insoluble.
a. Malic acid in water
b. Naphthalene in water
c. Amphetamine in ethyl alcohol
d. Aspirin in water
e. Succinic acid in hexane
f. Ibuprofen in diethyl ether
g. 1-Decanol (n-decyl alcohol) in water
2. Predict whether the following pairs of liquids would be miscible or immiscible.
a. Water and methyl alcohol
Miscible (both are polar)
b. Hexane and benzene
Miscible (both are nonpolar)
c. Methylene chloride and benzene
Immiscible (Methylene chloride is polar and benzene is nonpolar)
d. Water and toluene
Immiscible (water is polar and toluene is nonpolar)
e. Cyclohexane and water
Immiscible (cyclohexane is nonpolar and water is polar)
f. Ethyl alcohol and isopropyl alcohol
Miscible (both are alcohols)
3. Would you expect ibuprofen to be soluble or insoluble in 1.0M NaOH? Explain.
Ibuprofen is expected to be soluble in 1.0M NaOH because ibuprofen is an acid and NaOH is a base. When the two substances are mixed, and acid base reaction forms a soluble carboxylic salt.
4. Thymol is very slightly soluble in water and very soluble in 1.0M NaOH. Explain.
Thymol is a polar molecule, but it is not nearly as polar as water (which is highly polar). They are technically both “polar” but the level of polarity differs so greatly that it makes thymol only slightly soluble in water. Thymol is soluble in NaOH because it reacts to form a soluble salt. Therefore, thymol is soluble in NaOH.
5. Although cannabinol and methyl alcohol are both alcohols, cannabinol is very slightly soluble in methyl alcohol at room temperature. Explain.
Since cannabinol and methyl alcohol are both in the same group (alcohols), it would be natural to expect them to be soluble upon mixture. However, cannabinol is not completely soluble in methyl alcohol. This is because cannabinol is a lot bigger in structure than methyl alcohol (CH3OH) and it has more nonpolar substituents making it only slightly soluble in methyl alcohol, which is more polar. e not precisely pipetted to 0.1 ml, so the starting quantities were not as absolutely precise as they could have been. All these possible sources of variation could account for differences between measured and expected solubility.
Again, since polarity (and therefore, solubility) is a spectrum of relative values, the starting quantities and particular technique are important to the final results. 1-Butanol, for instance, is somewhere between absolute polar and absolute non-polar, and is therefore especially susceptible to variations in technique. If the mixing was not particularly vigorous, or the dropwise addition of bu