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We have learned in this chapter that many ionic solids dissolve in water as strong electrolytes, that is, as separated ions in solution. What properties of water facilitate this process?

Short Answer

Expert verified
The properties of water that facilitate the dissolution of ionic solids as strong electrolytes are due to its molecular structure and polarity. Water molecules have a bent shape with an oxygen atom bonded to two hydrogen atoms, giving them a dipole moment and making them polar. This polarity allows water molecules to surround and interact with positive and negative ions of the ionic solid, stabilizing them in the solution and leading to the dissolution of the solid as separated ions.

Step by step solution

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Steps to explain the properties of water in dissolving ionic solids as strong electrolytes

To discuss the properties of water that facilitate the dissolution of ionic solids as strong electrolytes, we will follow these steps: 1. Describe the structure and properties of water molecules 2. Explain the concept of polarity in water molecules 3. Describe how water interacts with positive and negative ions

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Key Concepts

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

Polarity
Water is a unique and crucial molecule largely because of its polarity. Polarity arises from the way the atoms are arranged within a molecule. Water consists of two hydrogen atoms and one oxygen atom, arranged in a bent shape. In this arrangement, the oxygen atom has a higher electronegativity than the hydrogen atoms, meaning it attracts the shared electrons more strongly.
This creates a partial negative charge on the oxygen and a partial positive charge on the hydrogens. Consequently, water molecules become polar, having positive and negative poles. Polarity enables water to effectively interact with various substances, including ionic compounds. This interaction is foundational to water's ability to dissolve many substances, acting as a solvent.
Ionic Solids Dissolution
When ionic solids, like salt ( ext{NaCl}), are introduced to water, the dissolution process begins due to water's polar nature. The solid comprises ions held together by ionic bonds. In water, these bonds are disrupted by the interaction with polar water molecules.
The positive ends of water molecules surround the negatively charged ions, while the negative ends surround the positively charged ions. This interaction helps to break apart the ionic solid into its individual ions, dispersing them throughout the solution. This process continues until the entire ionic solid is dissolved, forming a homogeneous mixture known as a solution.
Therefore, water's polarity is pivotal in breaking the ionic bonds and facilitating the dissolution of ionic solids.
Electrolytes
Ionic compounds dissolved in water typically form solutions containing separated ions, which are known as electrolytes. These electrolytes are significant because they conduct electricity when in solution. The ability of a solution to conduct electricity is reliant on these mobile ions, as they transport electric charge throughout the solution.
Strong electrolytes dissociate completely into ions when in water, such as sodium chloride ( ext{NaCl}). This complete dissociation ensures that there are sufficient ions present to carry electrical current efficiently. The behavior of electrolytes in water is crucial for various biological and physical processes, ranging from nerve impulses in the body to applications in electrochemistry.
Understanding electrolytes can elucidate why certain solutions conduct electricity and others do not, providing insight into the fundamental principles of chemistry.
Hydration Process
The hydration process is vital in dissolving ionic solids and maintaining ions in solution. As ionic compounds dissolve in water, the individual ions are surrounded by water molecules. This is known as hydration.
When an ion becomes hydrated, water molecules cling to the ion because water's oxygen atom, having a partial negative charge, is attracted to positive ions, while the hydrogen part, with a partial positive charge, surrounds negative ions. This attraction stabilizes the ions and keeps them from rejoining to form the solid.
  • Hydration shells form around the ions, effectively reducing the attraction between the opposite charged ions.
  • This prevention of reformation into the solid allows the ions to stay evenly distributed in the solution.
In summary, the hydration process is crucial for maintaining the stability and uniformity of ionic solutions, directly tied to the solvent capabilities of water.

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Most popular questions from this chapter

(a) How would you prepare \(175.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{AgNO}_{3}\) solution starting with pure solute? (b) An experiment calls for you to use \(100 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HNO}_{3}\) solution. All you have available is a bottle of \(3.6 \mathrm{M} \mathrm{HNO}_{3}\). How would you prepare the desired solution?

A sample of solid \(\mathrm{Ca}(\mathrm{OH})_{2}\) is stirred in water at \(30^{\circ} \mathrm{C}\) until the solution contains as much dissolved \(\mathrm{Ca}(\mathrm{OH})_{2}\) as it can hold. A \(100-\mathrm{mL}\) sample of this solution is withdrawn and titrated with \(5.00 \times 10^{-2} \mathrm{M} \mathrm{HBr}\). It requires \(48.8 \mathrm{~mL}\) of the acid solution for neutralization. What is the molarity of the \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution? What is the solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in water, at \(30^{\circ} \mathrm{C}\), in grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) per \(100 \mathrm{~mL}\) of solution?

A sample of \(5.53 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is added to \(25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3}\). (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of \(\mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{HNO}_{3}\), and \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) are present after the reaction is complete?

(a) How many milliliters of \(0.120 \mathrm{M} \mathrm{HCl}\) are needed to completely neutralize \(50.0 \mathrm{~mL}\) of \(0.101 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize \(0.200 \mathrm{~g}\) of \(\mathrm{NaOH}\) ? (c) If \(55.8 \mathrm{~mL}\) of \(\mathrm{BaCl}_{2}\) solution is needed to precipitate all the sulfate ion in a 752 -mg sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), what is the molarity of the solution? (d) If \(42.7 \mathrm{~mL}\) of \(0.208 \mathrm{M} \mathrm{HCl}\) solution is needed to neutralize a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\), how many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) must be in the solution?

You are presented with three white solids, \(A, B\), and \(C\), which are glucose (a sugar substance), \(\mathrm{NaOH}\), and AgBr. Solid A dissolves in water to form a conducting solution. B is not soluble in water. C dissolves in water to form a nonconducting solution. Identify \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). [Section 4.2]

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