Solubility - AP Chemistry

Iodine ( $\mathrm{I_2}$ ) is a nonpolar molecular solid whose particles interact mainly via London dispersion forces. A student attempts to dissolve $\mathrm{I_2(s)}$ in liquid carbon tetrachloride ( $\mathrm{CCl_4}$ ), which is also nonpolar. Is $\mathrm{I_2}$ likely to be soluble in $\mathrm{CCl_4}$ , and why?
Answer: Yes; both are nonpolar, so dispersion forces between solute and solvent can be favorable. The skill being tested is evaluating solubility of nonpolar substances in nonpolar solvents using London dispersion forces. Iodine (I₂) is a nonpolar molecule where particles interact via dispersion forces, and carbon tetrachloride (CCl₄) is similarly nonpolar. The 'like dissolves like' rule applies here, as both can form comparable dispersion interactions, making dissolution favorable. No chemical reaction is needed; the similarity in forces allows I₂ to mix into CCl₄. A tempting distractor is choice A, which mistakenly states nonpolar substances dissolve only in polar solvents, based on the misconception that opposites attract in solubility rather than similarities. Remember, to assess nonpolar solubility, check if both solute and solvent rely primarily on dispersion forces for compatibility.
Benzene ( $\mathrm{C_6H_6}$ ) is a nonpolar solvent with London dispersion forces as its primary intermolecular force. A student attempts to dissolve methanol ( $\mathrm{CH_3OH}$ ), a polar molecule capable of hydrogen bonding, in benzene. Is methanol likely to be very soluble in benzene, and why?
Answer: No; methanol is polar and hydrogen-bonding, while benzene is nonpolar and cannot provide comparable attractions. This question tests the incompatibility between polar and nonpolar substances. Methanol has very limited solubility in benzene because methanol is polar and forms strong hydrogen bonds with other methanol molecules, while benzene is nonpolar and can only provide weak dispersion forces. The energy required to break the hydrogen bonds between methanol molecules is not compensated by the weak dispersion forces that would form between methanol and benzene. This energy imbalance strongly favors methanol molecules remaining associated with each other rather than dispersing in benzene. Choice C incorrectly states that opposites attract in solubility; the correct principle is "like dissolves like" - substances with similar intermolecular forces are mutually soluble. The key strategy is to recognize that hydrogen-bonding substances are generally incompatible with nonpolar solvents because the energy cost of breaking hydrogen bonds is too high.
A student tries to dissolve solid sucrose ( $\mathrm{C_{12}H_{22}O_{11}}$ ), which has many O–H groups capable of hydrogen bonding, in hexane ( $\mathrm{C_6H_{14}}$ ), a nonpolar solvent with only London dispersion forces. Is sucrose likely to be soluble in hexane, and why?
Answer: No; sucrose is highly polar and hydrogen-bonding, while hexane is nonpolar, so solute–solvent IMFs are too weak. This question tests the skill of predicting solubility based on the compatibility of intermolecular forces between solute and solvent using the 'like dissolves like' principle. Sucrose is a highly polar molecule with multiple O-H groups that enable strong hydrogen bonding, while hexane is nonpolar and relies solely on London dispersion forces. The mismatch in polarity means that the solute-solvent interactions are weak dispersion forces, which are insufficient to overcome the strong hydrogen bonds in sucrose and the dispersion forces in hexane. Therefore, sucrose is not likely to dissolve well in hexane, as the nonpolar solvent cannot effectively stabilize the polar solute molecules. A tempting distractor is choice A, which incorrectly assumes sucrose can form hydrogen bonds with hexane, misunderstanding that hydrogen bonding requires compatible polar groups like O-H or N-H in both solute and solvent. To predict solubility effectively, always compare the dominant intermolecular forces of the solute and solvent to ensure they can form strong attractive interactions.
A student adds naphthalene, C10H8(s), to ethanol, CH3CH2OH(l). Naphthalene is largely nonpolar (dispersion forces), while ethanol is moderately polar but has a nonpolar ethyl group and can provide dispersion interactions. Is naphthalene likely to be more soluble in ethanol than in water, and why?
Answer: Yes; ethanol can provide significant dispersion interactions and has less polarity than water, making it a better match for naphthalene. This question tests understanding of how solvent polarity affects solubility of nonpolar solutes. Naphthalene (C10H8) is a nonpolar aromatic hydrocarbon that interacts primarily through London dispersion forces. While naphthalene has very low solubility in highly polar water, it shows better solubility in ethanol because ethanol has both polar (OH group) and nonpolar (ethyl group) regions. The ethyl portion of ethanol can interact favorably with naphthalene through dispersion forces, while ethanol's reduced polarity compared to water means less energy is required to create a cavity for the nonpolar solute. Choice C incorrectly suggests polar solvents repel nonpolar solutes, misunderstanding that the issue is insufficient favorable interactions rather than repulsion. The strategy is to recognize that moderately polar solvents with significant nonpolar regions often dissolve nonpolar solutes better than highly polar solvents do.
A student mixes acetone, (CH3)2CO(l), with water, H2O(l). Acetone is polar due to its C=O group and can accept hydrogen bonds from water; water is polar and hydrogen-bonding. Is acetone likely to be soluble (miscible) in water, and why?
Answer: Yes; acetone is polar and can engage in dipole interactions and hydrogen bonding (as an acceptor) with water. This question tests understanding of polar molecular solubility and hydrogen bonding interactions. Acetone ((CH3)2CO) has a polar carbonyl group (C=O) with a significant dipole moment, making the molecule polar overall despite having two methyl groups. The oxygen atom in acetone has lone pairs that can accept hydrogen bonds from water molecules, while water can also interact with acetone through dipole-dipole forces. These favorable interactions allow acetone and water to be completely miscible, as the acetone-water interactions effectively replace water-water hydrogen bonds. Choice A incorrectly claims acetone cannot interact with water due to lacking an O-H bond, confusing hydrogen bond donation with hydrogen bond acceptance. The strategy is to identify polar functional groups and remember that molecules can participate in hydrogen bonding as acceptors even without O-H, N-H, or F-H bonds.
A student tries to dissolve solid naphthalene ( $\mathrm{C_{10}H_8}$ ), a nonpolar aromatic hydrocarbon, in toluene ( $\mathrm{C_7H_8}$ ), a nonpolar aromatic solvent dominated by London dispersion forces. Is naphthalene likely to be soluble in toluene, and why?
Answer: Yes; both are largely nonpolar and interact mainly through dispersion forces, so mixing is favorable. The skill being assessed is evaluating solubility of nonpolar substances in nonpolar solvents using dispersion forces. Naphthalene is a nonpolar aromatic hydrocarbon relying on London dispersion forces, similar to toluene, which is also nonpolar and aromatic. This compatibility allows strong dispersion force interactions between solute and solvent, facilitating dissolution. Therefore, naphthalene is likely soluble in toluene, as both share the same type of intermolecular forces. Choice B is a tempting distractor, incorrectly claiming nonpolar solutes need polar solvents to induce dipoles, which contradicts the 'like dissolves like' rule. To solve solubility questions, identify shared intermolecular force types between solute and solvent for favorable outcomes.
Ammonia ( $\mathrm{NH_3}$ ) is a polar molecule that can hydrogen-bond with water. A student bubbles $\mathrm{NH_3(g)}$ into liquid water. Is $\mathrm{NH_3}$ likely to be soluble in water, and why?
Answer: Yes; ammonia is polar and can form hydrogen bonds with water, increasing solute–solvent attractions. The skill being tested is evaluating gas solubility in polar solvents based on hydrogen-bonding potential. Ammonia is polar and can form hydrogen bonds with water using its N-H groups and lone pair. These strong interactions allow NH3 to integrate into water's hydrogen-bond network, enhancing solubility. Therefore, NH3 is likely soluble in water. Choice B is a tempting distractor, wrongly generalizing that no gases dissolve in liquids due to particle spacing, ignoring specific intermolecular attractions. When predicting gas solubility, check for compatible forces like polarity or hydrogen bonding between gas and solvent molecules.
Propane ( $\mathrm{C_3H_8}$ ) is a nonpolar gas with only London dispersion forces. It is brought into contact with liquid water (polar, hydrogen bonding). Is propane likely to be very soluble in water at room temperature, and why?
Answer: No; propane is nonpolar and cannot form strong attractions with polar water, so solubility is low. The skill tested is predicting gas solubility in liquids based on polarity and intermolecular forces. Propane is a nonpolar gas with London dispersion forces, while water is polar with hydrogen bonding, leading to weak solute-solvent interactions. These weak forces cannot compete with water's strong hydrogen bonds, resulting in low solubility for propane. Therefore, propane is not very soluble in water at room temperature. Choice C is a tempting distractor, mistakenly suggesting C-H bonds in propane can hydrogen-bond with water, confusing weak C-H with true hydrogen bonding requiring O-H, N-H, or F-H. When evaluating gas solubility, consider if the gas molecules can form strong enough attractions with the solvent to disrupt its structure.
A student tries to dissolve solid sucrose ( $\mathrm{C_{12}H_{22}O_{11}}$ ), which has many polar $\mathrm{O\!-\!H}$ groups capable of hydrogen bonding, in liquid hexane ( $\mathrm{C_6H_{14}}$ ), which is nonpolar and interacts primarily through London dispersion forces. Is sucrose likely to be soluble in hexane, and why?
Answer: No; sucrose is polar while hexane is nonpolar, so intermolecular attractions between them are too weak. The skill being tested is predicting the solubility of a solute in a solvent based on their polarities and intermolecular forces. Sucrose is a polar molecule with many O-H groups that enable hydrogen bonding, while hexane is nonpolar and relies on London dispersion forces. According to the 'like dissolves like' principle, polar solutes dissolve best in polar solvents, but here the mismatch in polarity means weak solute-solvent interactions compared to the strong hydrogen bonds in sucrose. Therefore, sucrose is not likely to dissolve in hexane, as the attractions between sucrose molecules are stronger than those with hexane. A tempting distractor is choice A, which incorrectly assumes sucrose can hydrogen-bond to hexane, overlooking the misconception that nonpolar molecules like hexane cannot participate in hydrogen bonding. To predict solubility effectively, always evaluate whether the solute and solvent have compatible intermolecular forces, such as both being polar or both nonpolar.
Water is a polar solvent capable of strong hydrogen bonding. A student tries to dissolve vegetable oil (a mixture of large, mostly nonpolar triglycerides) in water at room temperature. Is the oil likely to be soluble in water, and why?
Answer: No; water forms strong hydrogen bonds with itself and has weak attractions to mostly nonpolar oil molecules. This question tests understanding of why nonpolar substances don't dissolve in polar solvents. Vegetable oil is essentially insoluble in water because oil molecules are large, nonpolar hydrocarbons that cannot form favorable interactions with polar water molecules. Water molecules form strong hydrogen bonds with each other, creating a cohesive network that effectively excludes nonpolar molecules. The energy required to disrupt water's hydrogen bonding network is not compensated by the weak dispersion forces that would form between water and oil molecules. This results in phase separation rather than dissolution. Choice D incorrectly suggests that covalent bonds within oil molecules must break for dissolution; dissolution involves separating molecules from each other, not breaking bonds within molecules. The key principle is that highly polar, hydrogen-bonding solvents like water cannot accommodate large nonpolar solutes because the energy cost of disrupting solvent-solvent interactions is too high.
A student tries to dissolve solid sodium chloride, NaCl(s), in hexane, C6H14(l), at room temperature. Hexane is nonpolar and cannot form strong ion–dipole interactions. Is NaCl likely to be soluble in hexane, and why?
Answer: No; NaCl is ionic while hexane is nonpolar, so the solvent cannot stabilize separated ions effectively. This question tests understanding of solubility based on matching intermolecular forces between solute and solvent. Sodium chloride (NaCl) is an ionic compound that requires strong ion-dipole interactions with a polar solvent to overcome its lattice energy and dissolve. Hexane (C6H14) is a nonpolar hydrocarbon that can only provide weak London dispersion forces, which are insufficient to stabilize separated Na+ and Cl- ions in solution. The energy required to break apart the ionic lattice cannot be compensated by the weak interactions between ions and nonpolar hexane molecules. Choice A incorrectly claims hexane can form ion-dipole attractions, but nonpolar molecules lack the dipole moment necessary for such interactions. The key strategy is to remember that ionic compounds require polar solvents with strong dipole moments to dissolve, following the principle that "like dissolves like."
Methanol ( $\mathrm{CH_3OH}$ ) is a polar, hydrogen-bonding solvent. A student adds a small amount of glycerol ( $\mathrm{HOCH_2CHOHCH_2OH}$ ), which has three O–H groups and is very polar. Is glycerol likely to be soluble in methanol, and why?
Answer: Yes; both are polar and can hydrogen-bond, so strong solute–solvent interactions favor solubility. The skill tested is predicting solubility of polar, hydrogen-bonding molecules in similar solvents. Glycerol is highly polar with three O-H groups for hydrogen bonding, matching methanol's polar and hydrogen-bonding nature. This allows strong solute-solvent hydrogen bonds and dipole interactions, favoring miscibility. Therefore, glycerol is likely soluble in methanol due to these compatible forces. Choice A is incorrect and tempting, as it misinterprets multiple O-H bonds as causing self-association that prevents dissolution, overlooking similar self-association in the solvent. To predict such solubilities, confirm that both solute and solvent can engage in the same strong intermolecular forces like hydrogen bonding.
Methanol ( $\mathrm{CH_3OH}$ ) is a polar solvent capable of hydrogen bonding. A student attempts to dissolve sodium chloride ( $\mathrm{NaCl}$ ), an ionic solid, in methanol. Is $\mathrm{NaCl}$ likely to be soluble in methanol, and why?
Answer: Yes; methanol is polar, so ion–dipole attractions can form and some dissolution is expected. The skill being tested is predicting solubility of ionic solids in polar organic solvents capable of hydrogen bonding. Sodium chloride (NaCl) is ionic, and methanol is polar with hydrogen-bonding ability, allowing ion-dipole and hydrogen-bond interactions to stabilize Na⁺ and Cl⁻. Though less effective than water, methanol's polarity enables some dissolution. No chemical reaction is involved; it's due to solvation forces. A tempting distractor is choice C, which limits NaCl solubility to water only, based on the misconception that hydrogen bonding must be identical to water's for ionic dissolution. To generalize, test solubility in polar solvents by considering their ability to form ion-dipole attractions similar to water.
Sodium nitrate ( $\mathrm{NaNO_3}$ ) is an ionic compound. A student compares its solubility in ethanol ( $\mathrm{C_2H_5OH}$ ), which is polar but less polar than water, versus in hexane ( $\mathrm{C_6H_{14}}$ ), which is nonpolar. Which statement best predicts whether $\mathrm{NaNO_3}$ is likely to be soluble in ethanol, and why?
Answer: Yes; ethanol is polar and can provide ion–dipole attractions, so some dissolution is expected. The skill being tested is comparing solubility of ionic compounds in solvents of varying polarity. Sodium nitrate (NaNO₃) is ionic, and ethanol is polar with the ability to form ion-dipole attractions, though less strongly than water. Compared to nonpolar hexane, ethanol provides better stabilization for the ions, allowing some solubility. The polarity of ethanol enables it to partially disrupt the ionic lattice, making dissolution possible. A tempting distractor is choice B, which falsely claims molecular solvents cannot dissolve ionic compounds, stemming from the misconception that only water can solvate ions effectively. When evaluating solubility in organic solvents, always compare the solvent's polarity to water and consider if ion-dipole interactions are feasible.
Acetone ( $\mathrm{(CH_3)_2CO}$ ) is a polar molecule with a carbonyl group that can act as a hydrogen-bond acceptor. A student mixes acetone with liquid water, which is polar and can hydrogen-bond. Is acetone likely to be soluble (miscible) in water, and why?
Answer: Yes; acetone can accept hydrogen bonds from water and also has dipole–dipole interactions with water. The skill being tested is assessing miscibility of polar molecular liquids based on hydrogen bonding and dipole-dipole interactions. Acetone has a polar carbonyl group that can accept hydrogen bonds from water, and both are polar, enabling dipole-dipole attractions. Water, being a strong hydrogen-bond donor and acceptor, forms favorable interactions with acetone, leading to miscibility. This compatibility overcomes any solute-solute or solvent-solvent forces, allowing them to mix completely. A tempting distractor is choice C, which incorrectly asserts that acetone cannot interact because it doesn't donate hydrogen bonds, ignoring the misconception that only donors are needed when acceptors can still participate effectively. A useful strategy is to identify specific functional groups in molecules that enable hydrogen bonding or dipole interactions for predicting solubility.
A student compares dissolving paraffin wax (a mixture of long nonpolar hydrocarbons with primarily London dispersion forces) in liquid toluene ( $\mathrm{C_7H_8}$ , largely nonpolar) versus in liquid water (polar). Is paraffin wax more likely to be soluble in toluene, and why?
Answer: Yes; both wax and toluene are largely nonpolar, so dispersion interactions make mixing more favorable. The skill being tested is comparing solubility of nonpolar substances in nonpolar versus polar solvents. Paraffin wax consists of nonpolar hydrocarbons relying on London dispersion forces, and toluene is largely nonpolar, allowing compatible dispersion interactions. In contrast, polar water offers weak attractions to wax, making toluene the better solvent for dissolution. The 'like dissolves like' principle favors mixing in toluene due to similar nonpolar natures. A tempting distractor is choice E, which claims toluene is polar and repels wax, stemming from the misconception that aromatic rings confer significant polarity. A key strategy is to classify solvents as polar or nonpolar based on their molecular structure before predicting solute compatibility.
Carbon tetrachloride ( $\mathrm{CCl_4}$ ) is a nonpolar molecular liquid (London dispersion forces). A student attempts to dissolve acetone ( $\mathrm{(CH_3)_2CO}$ ), a polar molecule with a strong dipole (but no O–H bond), in $\mathrm{CCl_4}$ . Is acetone likely to be highly soluble in $\mathrm{CCl_4}$ , and why?
Answer: No; acetone is polar while $\mathrm{CCl_4}$ is nonpolar, so solute–solvent attractions are relatively weak. The skill tested is predicting solubility by matching intermolecular forces between polar and nonpolar substances. Acetone is a polar molecule with a significant dipole moment but no O-H for hydrogen bonding, while CCl4 is nonpolar with only London dispersion forces. The polarity mismatch leads to weak dipole-induced dipole interactions, which are not strong enough to favor mixing over the separate pure substances. Therefore, acetone is not highly soluble in CCl4, as the nonpolar solvent cannot adequately interact with the polar solute. Choice A is a tempting distractor, incorrectly stating that polar solutes are most soluble in nonpolar solvents due to dipole alignment, which confuses the 'like dissolves like' principle. A transferable strategy is to classify both solute and solvent as polar or nonpolar and expect high solubility only when they match.
A student tries to dissolve potassium permanganate ( $\mathrm{KMnO_4}$ ), an ionic compound, in water versus in toluene ( $\mathrm{C_7H_8}$ ), a nonpolar solvent. Which outcome is most likely, and why?
Answer: It dissolves better in water; water's polarity provides ion–dipole attractions that stabilize the ions. The skill being assessed is comparing ionic solubility in polar versus nonpolar solvents. Potassium permanganate is ionic, and water provides ion-dipole stabilization for K+ and MnO4- ions, unlike nonpolar toluene's weak dispersion forces. This makes dissolution favorable in water but not in toluene. Therefore, it dissolves better in water. Choice D is a tempting distractor, misapplying 'like dissolves like' to suggest ionic solutes prefer nonpolar solvents, which is inverted. To compare solvents, identify which one can form the strongest stabilizing interactions with the solute's components.
A student attempts to dissolve calcium carbonate ( $\mathrm{CaCO_3}$ ), an ionic solid with a high lattice energy, in liquid water (polar). Is $\mathrm{CaCO_3}$ likely to be very soluble in water, and why?
Answer: No; strong ionic attractions in the solid (high lattice energy) are not fully overcome by hydration, so solubility is low. This question tests understanding of how lattice energy affects ionic solubility. Calcium carbonate has exceptionally high lattice energy due to the small, highly charged Ca²⁺ and CO₃²⁻ ions, which create very strong electrostatic attractions in the solid. While water can provide ion-dipole interactions for hydration, the hydration energy is insufficient to overcome CaCO₃'s large lattice energy, resulting in very low solubility (Ksp ≈ 10⁻⁹). This explains why limestone and chalk (forms of CaCO₃) persist in nature despite water exposure. Choice A incorrectly assumes all ionic compounds are highly soluble in water, but solubility depends on the balance between lattice energy and hydration energy—compounds with high lattice energies relative to hydration energies have low solubility. When predicting ionic solubility, consider both ion charges and sizes: small, highly charged ions typically form less soluble compounds.
A student attempts to dissolve potassium iodide ( $\mathrm{KI}$ ), an ionic compound, in acetone ( $\mathrm{(CH_3)_2CO}$ ), a moderately polar solvent that cannot donate hydrogen bonds but can accept them. Is $\mathrm{KI}$ expected to be very soluble in acetone, and why?
Answer: No; acetone's polarity may not provide sufficient ion–dipole stabilization to overcome the ionic lattice. This question evaluates ionic solubility in polar aprotic solvents. Potassium iodide is ionic, and acetone is moderately polar but cannot donate hydrogen bonds, providing weaker ion-dipole stabilization. This may not suffice to overcome KI's lattice energy, leading to limited solubility. Thus, KI is not expected to be very soluble in acetone. Choice A is a tempting distractor, overgeneralizing that any polar solvent fully dissolves any ionic solid, neglecting solvent-specific properties like hydrogen-bonding ability. For ionic compounds, assess the solvent's polarity and hydrogen-bonding capacity to determine if it can effectively hydrate ions.

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