Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane ( \(\mathrm{C}_{7} \mathrm{H}_{16}\), nonpolar solvent): (a) vegetable oil (nonpolar) (b) isopropyl alcohol (polar) (c) potassium bromide (ionic)

Short Answer

Expert verified
(a) Vegetable oil would be more soluble in heptane (nonpolar). (b) Isopropyl alcohol would be more soluble in water (polar). (c) Potassium bromide would be more soluble in water (polar).

Step by step solution

01

- Understanding Solubility

The solubility of a substance depends on the 'like dissolves like' principle. Polar and ionic substances are more soluble in polar solvents, while nonpolar substances are more soluble in nonpolar solvents.
02

- Analyzing Vegetable Oil

Vegetable oil is a nonpolar substance. Based on the 'like dissolves like' principle, it would be more soluble in a nonpolar solvent, such as heptane, rather than polar solvents like water.
03

- Analyzing Isopropyl Alcohol

Isopropyl alcohol is a polar molecule because it has a hydroxyl (-OH) group that can form hydrogen bonds. Thus, it would be more soluble in a polar solvent like water.
04

- Analyzing Potassium Bromide

Potassium bromide is an ionic compound which is typically soluble in polar solvents due to the strong interactions between ions and polar molecules. Therefore, it would be more soluble in water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Like Dissolves Like Principle
When it comes to understanding why certain substances dissolve easily in one solvent but not in another, the 'like dissolves like' principle is fundamental. Simply put, this rule means that substances with similar chemical properties tend to be more soluble in each other. For instance, polar substances, which have uneven distribution of electron density and thus a slight charge difference across their molecules, dissolve well in polar solvents because the solvent's polarity can interact with the substance's charges. This is why substances like sugar, which have polar characteristics, dissolve well in water, a polar solvent.

On the flip side, nonpolar substances lack a significant charge difference across their molecules, which means they don't mix with the charges in polar solvents. Instead, they dissolve in nonpolar solvents, like oils and hydrocarbons, because there are no charges to prevent the substances from mixing freely. This principle explains why oil (a nonpolar substance) won't mix with water (a polar solvent) but will mix with other oils. In the exercise, vegetable oil, a nonpolar substance, is correctly predicted to be more soluble in heptane, a nonpolar solvent, because they share similar nonpolar characteristics.
Polar and Nonpolar Substances
The difference between polar and nonpolar substances hinges on the distribution of electrical charges within their molecules. Molecules are made up of atoms bonded together, and these bonds can be either polar or nonpolar. In polar molecules, there is an uneven distribution of electrons between the atoms, which leads to partial positive and negative charges on different parts of the molecule. An example is water (H_{2}O), which has a bent shape causing a difference in the distribution of electrons. This polar nature of water enables it to dissolve other polar substances and ionic compounds, which also exhibit charge separations.

Nonpolar molecules, in contrast, have an equal distribution of electrons, causing the molecule to have no significant charge difference across its structure. Hydrocarbons, like the heptane mentioned in the exercise, consist of carbon and hydrogen atoms linked with nonpolar bonds, making them nonpolar substances. As a result, nonpolar substances do not dissolve well in polar solvents like water but will dissolve in nonpolar solvents such as other hydrocarbons or oils. In the exercise provided, isopropyl alcohol, a polar molecule due to its hydroxyl group, is correctly recognized as being more soluble in water than in hydrocarbon solvents like heptane.
Ionic Compounds Solubility
Ionic compounds, like potassium bromide (KBr), are composed of positively charged ions (cations) and negatively charged ions (anions) held together by strong ionic bonds. The solubility of these ionic compounds in different solvents is highly influenced by the solvent's polarity. Because polar solvents have a partial positive and partial negative charge, they can effectively separate and stabilize the individual ions from the ionic lattice, allowing the ionic compound to dissolve.

In a polar solvent like water, the positive end of the water molecules is attracted to the anions, while the negative end is attracted to the cations, effectively surrounding each ion and pulling them into the solution. This process is called solvation and is the major reason why ionic compounds, such as potassium bromide, dissolve well in water, as correctly predicted in step four of the exercise. However, this solvation process cannot occur in nonpolar solvents like hydrocarbons, because they lack the charge properties necessary to separate the ions in the compound.

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Most popular questions from this chapter

The vapor pressure of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is 94 torr at \(20^{\circ} \mathrm{C}\). The vapor pressure of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), is 44 torr at the same temperature. (a) Calculate the mole fraction of methanol and of ethanol in a solution of \(50.0 \mathrm{g}\) of methanol and \(50.0 \mathrm{g}\) of ethanol. (b) Ethanol and methanol form a solution that behaves like an ideal solution. Calculate the vapor pressure of methanol and of ethanol above the solution at \(20^{\circ} \mathrm{C}\). (c) Calculate the mole fraction of methanol and of ethanol in the vapor above the solution.

When \(\mathrm{KNO}_{3}\) is dissolved in water, the resulting solution is significantly colder than the water was originally. (a) Is the dissolution of \(\mathrm{KNO}_{3}\) an endothermic or an exothermic process? (b) What conclusions can you draw about the intermolecular attractions involved in the process? (c) Is the resulting solution an ideal solution?

The Henry's law constant for \(\mathrm{O}_{2}\) is \(1.3 \times 10^{-3} \mathrm{M} /\) atm at \(25^{\circ} \mathrm{C}\). Assuming ideal solution behavior, what mass of oxygen would be dissolved in a 40-L aquarium at \(25^{\circ} \mathrm{C}\), assuming an atmospheric pressure of \(1.00 \mathrm{atm}\), and that the partial pressure of \(\mathrm{O}_{2}\) is 0.21 atm?

The Henry's law constant for \(\mathrm{CO}_{2}\) is \(3.4 \times 10^{-2} \mathrm{M} / \mathrm{atm}\) at \(25^{\circ} \mathrm{C}\). Assuming ideal solution behavior, what pressure of carbon dioxide is needed to maintain a \(\mathrm{CO}_{2}\) concentration of \(0.10 \mathrm{M}\) in a can of lemon-lime soda?

Arrange the following solutions in order by their decreasing freezing points: \(0.1 \mathrm{m} \mathrm{Na}_{3} \mathrm{PO}_{4}, 0.1 \mathrm{m} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) \(0.01 \mathrm{m} \mathrm{CO}_{2}, 0.15 \mathrm{m} \mathrm{NaCl},\) and \(0.2 \mathrm{m} \mathrm{CaCl}_{2}\).

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