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Choose the member of each set that you would expect to be more soluble in water. Explain your answer. (a) naphthalene, \(\mathrm{C}_{10} \mathrm{H}_{8}\), or hydrogen peroxide, \(\mathrm{H}-\mathrm{O}-\mathrm{O}-\mathrm{H}\) (b) silicon dioxide or sodium hydroxide (c) chloroform, \(\mathrm{CHCl}_{3}\), or hydrogen chloride (d) methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), or methyl ether, \(\mathrm{H}_{3} \mathrm{C}-\mathrm{O}-\mathrm{CH}_{3}\)

Short Answer

Expert verified
(a) Naphthalene or Hydrogen peroxide? (b) Silicon dioxide or Sodium hydroxide? (c) Chloroform or Hydrogen chloride? (d) Methyl alcohol or Methyl ether? Answer: (a) Hydrogen peroxide (b) Sodium hydroxide (c) Hydrogen chloride (d) Methyl alcohol

Step by step solution

01

(a) Naphthalene vs Hydrogen peroxide

Naphthalene (\(\mathrm{C}_{10} \mathrm{H}_{8}\)) is a nonpolar molecule with a hydrocarbon structure, meaning it will not readily form hydrogen bonds with water. Hydrogen peroxide (\(\mathrm{H}-\mathrm{O}-\mathrm{O}-\mathrm{H}\)) is a polar molecule with oxygen atoms capable of forming hydrogen bonds with water. Therefore, hydrogen peroxide is more soluble in water.
02

(b) Silicon dioxide vs Sodium hydroxide

Silicon dioxide is a covalent network solid and is insoluble in water. Sodium hydroxide (\(\mathrm{NaOH}\)) is an ionic compound that readily dissociates into ions when dissolved in water, making it highly soluble. As a result, sodium hydroxide is more soluble in water.
03

(c) Chloroform vs Hydrogen chloride

Chloroform (\(\mathrm{CHCl}_{3}\)) is a polar molecule, but its ability to form hydrogen bonds is limited as the highly electronegative chlorine atoms have a stronger effect on the molecule. Hydrogen chloride (\(\mathrm{HCl}\)) is also a polar molecule, but when dissolved in water, it ionizes into \(\mathrm{H}^{+}\) and \(\mathrm{Cl}^{-}\) ions and forms strong hydrogen bonds with water molecules. Hence, hydrogen chloride is more soluble in water.
04

(d) Methyl alcohol vs Methyl ether

Methyl alcohol, or methanol (\(\mathrm{CH}_{3} \mathrm{OH}\)), is a polar molecule with an oxygen atom that forms strong hydrogen bonds with water molecules, making it highly soluble in water. Methyl ether (\(\mathrm{H}_{3} \mathrm{C}-\mathrm{O}-\mathrm{CH}_{3}\)) is also a polar molecule, but its ability to form hydrogen bonds with water is weaker as its central oxygen atom is unable to interact efficiently with the hydrogen atoms of water molecules. Thus, methyl alcohol is more soluble in water.

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Key Concepts

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

Hydrogen Bonding
Hydrogen bonding is a type of interaction that significantly influences the solubility of substances in water. It occurs when a hydrogen atom, which is covalently bonded to a highly electronegative atom like oxygen, nitrogen, or fluorine, forms an additional attraction to another electronegative atom. This results in relatively strong intermolecular forces.
Consider substances like hydrogen peroxide ( H−O−O−H ) that can establish hydrogen bonds with water. The presence of oxygen atoms allows it to bond strongly with the water molecules, facilitating its dissolution. Meanwhile, naphthalene ( C_{10}H_{8} ), being nonpolar, lacks this capability, making it less soluble in water.
  • Hydrogen bonds are typically stronger than ordinary dipole-dipole attractions, but weaker than covalent or ionic bonds.
  • This bonding type is crucial for the solubility of molecules like alcohols in water due to the presence of -OH groups.
For instance, methyl alcohol ( CH_{3}OH ) is more soluble in water due to its ability to form hydrogen bonds, while methyl ether's ( (CH_{3})_{2}O ) inability to form such strong interactions with water makes it less soluble.
Polar and Nonpolar Molecules
The polarity of a molecule dictates how it interacts with water molecules. Polar molecules have an uneven distribution of charge, meaning they have a partial positive and negative side. This polarity enables them to dissolve easily in water, a polar solvent, due to "like dissolves like" principle.
For example, hydrogen peroxide is polar, owing to its -OH groups, which makes it highly soluble in water. Conversely, naphthalene, with its symmetrical and hydrocarbon-only structure, is nonpolar, and thus does not dissolve well in water.
  • Water is a polar molecule, and so it attracts other polar molecules but repels nonpolar ones.
  • Polar molecules tend to form hydrogen bonds, further enhancing their solubility in water.
Sodium hydroxide ( NaOH ) is ionic but behaves as a polar substance due to its dissociation into ions, aiding its solubility in water, unlike nonpolar silicon dioxide.
Ionic Compounds
Ionic compounds are composed of positive and negative ions held together by strong ionic bonds. These materials typically dissolve well in water, as the ionic bonds are weakened in the presence of a polar solvent like water, allowing the ions to separate and interact with water molecules.
When sodium hydroxide ( NaOH ) is dissolved in water, it separates into sodium ( Na^{+} ) and hydroxide ions ( OH^{-} ), which interact with water, making it highly soluble.
  • Ionic compounds dissociate in water to form their constituent ions.
  • This allows the water molecules to surround and stabilize the individual ions, increasing solubility.
Hydrogen chloride ( HCl ) behaves similarly; it dissolves in water by ionizing into H^{+} and Cl^{-} ions, significantly increasing its solubility compared to a covalent network solid like silicon dioxide, which does not easily interact with water.

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Most popular questions from this chapter

Show that the following relation is generally valid for all solutions: $$ \text { molality }=\frac{\text { molarity }}{d-\frac{\mathrm{MM}(\mathrm{molarity})}{1000}} $$ where \(d\) is solution density \(\left(\mathrm{g} / \mathrm{cm}^{3}\right)\) and \(\mathrm{MM}\) is the molar mass of the solute. Using this equation, explain why molality approaches molarity in dilute solution when water is the solvent, but not with other solvents.

The freezing point of a \(0.20 \mathrm{~m}\) solution of aqueous \(\mathrm{HF}\) is \(-0.38^{\circ} \mathrm{C}\). (a) What is \(i\) for the solution? (b) Is the solution made of (i) HF molecules only? (ii) \(\mathrm{H}^{+}\) and \(\mathrm{F}^{-}\) ions only? (iii) Primarily HF molecules with some \(\mathrm{H}^{+}\) and \(\mathrm{F}^{-}\) ions? (iv) primarily \(\mathrm{H}^{+}\) and \(\mathrm{F}^{-}\) ions with some HF molecules?

Pure benzene boils at \(80.10^{\circ} \mathrm{C}\) and has a boiling point constant, \(k_{\mathrm{b}}\), of \(2.53^{\circ} \mathrm{C} / \mathrm{m} .\) A sample of benzene is contaminated by naphthalene, \(\mathrm{C}_{10} \mathrm{H}_{8}\). The boiling point of the contaminated sample is \(81.20^{\circ} \mathrm{C}\). How pure is the sample? (Express your answer as mass percent of benzene.)

Consider the process by which lead chloride dissolves in water: $$ \mathrm{PbCl}_{2}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ (a) Using data from tables in Chapter 8, calculate \(\Delta H\) for this reaction. (b) Based only on thermodynamic data, would you expect the solubility of \(\mathrm{PbCl}_{2}\) to increase if the temperature is increased?

In a sugar mill, sugar cane is shredded, mixed with water and crushed between rollers. The resulting juice contains \(7-11 \%\) sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\). Assume that \(25 \mathrm{~L}\) of collected juice (of density \(1.0 \mathrm{~g} / \mathrm{mL}\) ) yields \(11 \%\) (by mass) sucrose. What is the molality of the sucrose solution after \(33 \%\) (by mass) of the water content of the juice has been removed?

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