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Chapter 13: Problem 34

Which of the following in each pair is likely to be more soluble in water: (a) cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) or glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\), (b) propionic acid \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right)\) or sodium propionate \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COONa}\right),(\mathbf{c}) \mathrm{HCl}\) or ethyl chloride \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\right) ?\) Explain in each case.

Short Answer

Expert verified
(a) Glucose, (b) Sodium propionate, (c) HCl

Step by step solution

01

Determine the polarity of each compound

First, let's assess the polarity of each compound in the pairs. Polar molecules have regions of partial positive and negative charge due to the uneven distribution of electron density in the molecule, while nonpolar molecules have an even distribution of electron density. (a) cyclohexane (C6H12) - nonpolar; glucose (C6H12O6) - polar (has hydroxyl groups) (b) propionic acid (CH3CH2COOH) - polar (has carboxylic acid group); sodium propionate (CH3CH2COONa) - ionic (has an ionic bond between sodium and carboxylate group) (c) HCl - polar (due to the difference in electronegativity between hydrogen and chlorine); ethyl chloride (CH3CH2Cl) - polar but less polar than HCl (large alkyl group with chlorine)
02

Analyze the interactions between each compound and water molecules

Now, let's analyze the possible interactions between each compound and water, a polar solvent: (a) cyclohexane - weak van der Waals forces with water; glucose - hydrogen bonding with water (b) propionic acid - hydrogen bonding with water; sodium propionate - ion-dipole interactions with water (c) HCl - strong ion-dipole interactions with water due to dissociation into ions; ethyl chloride - weaker dipole-dipole interactions with water
03

Determine which compound is more soluble in water for each pair

Based on polarity and interactions with water, we can determine which compound is more soluble in water: (a) Glucose is more soluble in water than cyclohexane, because it is polar and can form hydrogen bonds with water molecules. (b) Sodium propionate is more soluble in water than propionic acid, because it is ionic and can form strong ion-dipole interactions with water molecules. (c) HCl is more soluble in water than ethyl chloride, because it can form strong ion-dipole interactions with water molecules when completely dissociated, while ethyl chloride can only form weaker dipole-dipole interactions.

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Key Concepts

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

Polarity and Solubility in Water
Polarity is a key factor in determining a molecule's solubility in water. Polar molecules have regions with slight positive and negative charges due to differences in electron distribution. This allows them to interact better with water, which is also a polar solvent. For instance, glucose is more soluble in water than cyclohexane. This is because glucose is a polar molecule with hydroxyl (-OH) groups that can form attractions with water. Conversely, cyclohexane is nonpolar, lacking the partial charges needed to interact with water effectively.
Hydrogen Bonding
Hydrogen bonding is a special kind of attraction that occurs when hydrogen is bonded to a highly electronegative element like oxygen, nitrogen, or fluorine. This interaction is crucial for solubility in water.
  • In glucose, the abundance of hydroxyl groups allows it to form hydrogen bonds with water, increasing its solubility.
  • Propionic acid can also form hydrogen bonds due to its carboxyl group, making it more soluble than compounds without such groups.

Hydrogen bonds are surprisingly strong for dipole interactions and significantly impact a substance's ability to dissolve in water.
Ion-Dipole Interactions
Ion-dipole interactions occur between ions and polar molecules. These are even stronger than hydrogen bonds and are crucial for the solubility of ionic compounds.
  • Sodium propionate, an ionic compound, is more soluble in water due to these interactions. The sodium ions interact strongly with water's polar molecules.
  • HCl, though covalent in structure, dissociates into ions in water. These ions engage in ion-dipole interactions, improving its solubility.

The strength of ion-dipole interactions often leads to better solubility compared to compounds that can only engage in dipole or hydrogen bonding.

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Most popular questions from this chapter

List the following aqueous solutions in order of decreasing freezing point: \(0.040 \mathrm{~m}\) glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right), 0.020 \mathrm{~m} \mathrm{KBr}\),

Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of \(5.00 \mathrm{~g}\) of lauryl alcohol in \(0.100 \mathrm{~kg}\) of benzene freezes at \(4.1^{\circ} \mathrm{C}\). What is the molar mass of lauryl alcohol from this data?

Describe how you would prepare each of the following aqueous solutions, starting with solid \(\mathrm{KBr}\) : (a) \(0.75 \mathrm{~L}\) of \(1.5 \times 10^{-2} M \mathrm{KBr},(\mathbf{b}) 125 \mathrm{~g}\) of \(0.180 \mathrm{~m} \mathrm{KBr},(\mathbf{c}) 1.85 \mathrm{~L}\) of a solution that is \(12.0 \% \mathrm{KBr}\) by mass (the density of the solution is \(1.10 \mathrm{~g} / \mathrm{mL}),\) (d) a \(0.150 \mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough \(\mathrm{KBr}\) to precipitate \(16.0 \mathrm{~g}\) of AgBr from a solution containing \(0.480 \mathrm{~mol}\) of \(\mathrm{AgNO}_{3}\).

You take a sample of water that is at room temperature and in contact with air and put it under a vacuum. Right away, you see bubbles leave the water, but after a little while, the bubbles stop. As you keep applying the vacuum, more bubbles appear. A friend tells you that the first bubbles were water vapor, and the low pressure had reduced the boiling point of water, causing the water to boil. Another friend tells you that the first bubbles were gas molecules from the air (oxygen, nitrogen, and so forth) that were dissolved in the water. Which friend is mostly likely to be correct? What, then, is responsible for the second batch of bubbles? [Section 13.4]

Indicate whether each statement is true or false: \((\) a) \(\mathrm{NaCl}\) dissolves in water but not in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) because benzene is denser than water. (b) NaCl dissolves in water but not in benzene because water has a large dipole moment and benzene has zero dipole moment. (c) NaCl dissolves in water but not in benzene because the water-ion interactions are stronger than benzene-ion interactions.
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