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Chapter 12: Problem 5

Using the concept of hydration, describe the process of dissolving a sodium chloride crystal in water.

Short Answer

Expert verified
Water hydrates NaCl by surrounding ions with water molecules, dissolving the crystal.

Step by step solution

01

Understanding Hydration

Hydration involves the interaction of water molecules with ions or polar molecules. In the case of sodium chloride (NaCl), water molecules will surround its constituent ions due to their polar nature.
02

Recognizing Polar Water Molecules

Water molecules are polar because they have a partial negative charge near the oxygen atom and a partial positive charge near the hydrogen atoms. This polarity allows water molecules to interact with ions.
03

Formation of Ion-Dipole Interactions

When a sodium chloride crystal is placed in water, the water molecules will interact with the Na鈦� and Cl鈦� ions. The partially negative oxygen atoms of water molecules are attracted to the Na鈦� ions, while the partially positive hydrogen atoms are attracted to the Cl鈦� ions. These interactions are known as ion-dipole interactions.
04

Breaking the Ionic Lattice

These ion-dipole interactions overcome the electrostatic forces holding the Na鈦� and Cl鈦� ions together in the crystal lattice, causing the crystal to break apart into individual ions.
05

Solvation of Ions

The individual sodium (Na鈦�) and chloride (Cl鈦�) ions are surrounded by a 'hydration shell' of water molecules, effectively solvating the ions and allowing them to disperse throughout the water.

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Key Concepts

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

Hydration
Hydration is a fundamental part of dissolving ionic compounds, playing a key role when substances like sodium chloride (NaCl) enter water. In this process, each ion in the compound is surrounded by several water molecules.
Water, being polar, has a distinct arrangement where the oxygen end is slightly negative and the hydrogen end is slightly positive. This distinct structure of water is essential for hydration, as it enables the effective interaction with ions from ionic compounds.
  • Hydration begins the moment an ionic substance is introduced to water.
  • The water molecules immediately start to interact with the ions.
  • These interactions lead to the formation of a surrounding layer of water, called the hydration shell.
This shell effectively isolates each ion from others, allowing it to move more freely in the solution.
Ion-Dipole Interactions
Ion-dipole interactions are a specific type of attraction that occurs between charged ions and the partial charges in polar molecules, such as water. In the context of sodium chloride dissolution, these interactions are critical.
Once a sodium chloride crystal is placed in water, the polar water molecules quickly align themselves around the ions.
  • Positive sodium ions (Na鈦�) attract the partially negative oxygen ends of water molecules.
  • Negative chloride ions (Cl鈦�) attract the partially positive hydrogen ends of the water molecules.
This process of attraction is called an ion-dipole interaction and is strong enough to weaken the ionic bonds within the NaCl crystal. This effectively breaks apart the crystal, allowing the ions to enter the solution.
Sodium Chloride Dissolution
Dissolution of sodium chloride in water is a process driven by the forces between ions and water molecules. When NaCl dissolves, the crystal lattice composed of sodium and chloride ions breaks down. This is primarily due to ion-dipole interactions, where water molecules effectively "pull" at the ions.
The dissolution process can be broken down as follows:
  • The sodium ions (Na鈦�) and chloride ions (Cl鈦�) separate as water surrounds them.
  • As the water molecules continue to interact with each individual ion, they form a hydration shell.
  • This shell facilitates the equal dispersion of ions throughout the solution.
Thus, sodium chloride goes from being a solid crystal to a dispersed form in a liquid, resulting in a homogeneous mixture.
Polar Water Molecules
Water molecules exhibit polarity due to their V-shaped structure, which causes an uneven distribution of charges. This arises because the molecule has a partial negative charge near the oxygen atom and partial positive charges at the hydrogen atoms.
This polarity is incredibly useful in dissolving ionic compounds like sodium chloride.
  • The partial charges allow water to interact closely with various ions.
  • These interactions enable the breakdown of bonds within ionic crystals.
The polar nature of water molecules is why they are considered a "universal solvent." It allows them to dissolve many substances by surrounding and isolating their ions, thus facilitating the dissolution process.

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Most popular questions from this chapter

A solution is made up by dissolving \(15.0 \mathrm{~g} \mathrm{Na}_{2} \mathrm{CO}_{3}\) " \(10 \mathrm{H}_{2} \mathrm{O}\) in \(100.0 \mathrm{~g}\) of water. What is the molality of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in this solution?

A \(0.0182-\mathrm{g}\) sample of an unknown substance was dissolved in \(2.135 \mathrm{~g}\) of benzene. The molality of this solution, determined by freezing-point depression, was \(0.0698 \mathrm{~m} .\) What is the molecular mass of the unknown substance?

Equal numbers of moles of two soluble, substances, substance \(\mathrm{A}\) and substance \(\mathrm{B}\), are placed into separate \(1.0-\mathrm{L}\) samples of water. a. The water samples are cooled. Sample A freezes at \(-0.50^{\circ} \mathrm{C}\), and Sample \(\mathrm{B}\) freezes at \(-1.00^{\circ} \mathrm{C}\). Explain how the solutions can have different freezing points. b. You pour \(500 \mathrm{~mL}\) of the solution containing substance \(\mathrm{B}\) into a different beaker. How would the freezing point of this \(500-\mathrm{mL}\) portion of solution \(\mathrm{B}\) compare to the freezing point of the \(1.0-\mathrm{L}\) sample of solution \(\mathrm{A}\) ? c. Calculate the molality of the solutions of \(\mathrm{A}\) and \(\mathrm{B}\). Assume that \(i=1\) for substance \(\mathrm{A}\). d. If you were to add an additional \(1.0 \mathrm{~kg}\) of water to solution B, what would be the new freezing point of the solution? Try to write an answer to this question without using a mathematical formula. e. What concentration (molality) of substances \(\mathrm{A}\) and \(\mathrm{B}\) would result in both solutions having a freezing point of \(-0.25^{\circ} \mathrm{C}\) ? f. Compare the boiling points, vapor pressure, and osmotic pressure of the original solutions of \(\mathrm{A}\) and \(\mathrm{B}\). Don't perform the calculations; just state which is the greater in each case.

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