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The following compounds are water-soluble. What ions are produced by each compound in aqueous solution? (a) \(\mathrm{KOH}\) (c) \(\mathrm{LiNO}_{3}\) (b) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
(a) KOH: K鈦, OH鈦; (b) K鈧係O鈧: 2K鈦, SO鈧劼测伝; (c) LiNO鈧: Li鈦, NO鈧冣伝; (d) (NH鈧)鈧係O鈧: 2NH鈧勨伜, SO鈧劼测伝.

Step by step solution

01

Determine the Ions for KOH

When potassium hydroxide (\(\text{KOH}\)) dissolves in water, it splits into ions. Potassium (\(\text{K}^+\)) as a cation and hydroxide (\(\text{OH}^-\)) as an anion are formed. So, the ions produced are \(\text{K}^+\) and \(\text{OH}^-\).
02

Determine the Ions for LiNO3

Lithium nitrate (\(\text{LiNO}_3\)) dissolves in water to form ions. Lithium ion (\(\text{Li}^+\)) is the cation, and the nitrate ion (\(\text{NO}_3^-\)) is the anion. Therefore, the ions in the solution are \(\text{Li}^+\) and \(\text{NO}_3^-\).
03

Determine the Ions for K2SO4

When potassium sulfate (\(\text{K}_2\text{SO}_4\)) dissolves, it forms two potassium ions (\(2\text{K}^+\)) and one sulfate ion (\(\text{SO}_4^{2-}\)). Thus, the ions are \(2\text{K}^+\) and \(\text{SO}_4^{2-}\).
04

Determine the Ions for (NH4)2SO4

Ammonium sulfate (\((\text{NH}_4)_2\text{SO}_4\)) dissociates in water into two ammonium ions (\(2\text{NH}_4^+\)) and one sulfate ion (\(\text{SO}_4^{2-}\)). The resulting ions are \(2\text{NH}_4^+\) and \(\text{SO}_4^{2-}\).

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Key Concepts

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

Aqueous Solution
An aqueous solution is a mixture where a solute is dissolved in water. Water acts as the solvent, meaning it is the substance that does the dissolving. Why use water? Because it is a universal solvent, known for its ability to dissolve many substances. This happens due to water's polar nature, where one end is slightly positive and the other is slightly negative.

When ionic compounds like potassium hydroxide (KOH) or lithium nitrate (LiNO鈧) are added to water, they dissolve, meaning they mix with water so perfectly that they are no longer visible. This forms an aqueous solution.
  • The solute (in this case, the ionic compound) separates into individual ions when in water.
  • These ions are then surrounded by water molecules, which stabilize them in solution.
Water's role is crucial in this process, as its polarity is key to pulling apart the ions of the solute and keeping them dissolved.
Water-Soluble Compounds
Water-soluble compounds inherently dissolve in water to produce a solution. Understanding what makes a compound water-soluble involves looking at its molecular structure and the bonds within it. Typically, compounds made of ionic bonds, such as salts, are water-soluble.

For example, potassium sulfate ( K鈧係O鈧) and ammonium sulfate ( (NH鈧)鈧係O鈧) are considered salts with ionic bonds that easily dissolve in water:
  • Ionic compounds dissociate into ions because water, as a solvent, separates these ions and stabilizes them in solution.
  • This solubility is crucial for various biological and industrial processes.
Not all compounds are water-soluble, but many ionic compounds are due to their interaction with water molecules. This is central to understanding how substances like KOH and LiNO鈧 behave in aqueous environments.
Dissociation of Ionic Compounds
Dissociation refers to the process by which an ionic compound breaks apart into its individual ions when dissolved in water. It's like the ions are freed from their solid crystal lattice and can move around independently. This is an essential concept in chemistry because it explains how ionic solutions conduct electricity.

Let's break down some examples:
  • KOH dissociates into K鈦 and OH鈦 ions.
  • LiNO鈧 becomes Li鈦 and NO鈧冣伝 ions.
  • K鈧係O鈧 splits into two K鈦 ions and one SO鈧劼测伝 ion.
  • (NH鈧)鈧係O鈧 releases two NH鈧勨伜 ions and one SO鈧劼测伝 ion.
During dissociation, the charge of the ions is important. It allows attraction between the ions and polar water molecules, pulling the crystal lattice apart.

Understanding dissociation helps explain why solutions of these ionic compounds can carry an electrical current, crucial for many practical applications.

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Most popular questions from this chapter

Write a balanced equation for the reaction of sulfur trioxide with water.

Identify each of the following substances as a likely oxidizing or reducing agent: \(\mathrm{HNO}_{3}, \mathrm{Na}, \mathrm{Cl}_{2}, \mathrm{O}_{2} \mathrm{KMnO}_{4}\).

Describe how to prepare zinc chloride by (a) an acid-base reaction, (b) a gas- forming reaction, and (c) an oxidation-reduction reaction. The available starting materials are \(\mathrm{ZnCO}_{3}, \mathrm{HCl}, \mathrm{Cl}_{2}, \mathrm{HNO}_{3}\) \(\mathrm{Zn}(\mathrm{OH})_{2}, \mathrm{NaCl}, \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2},\) and \(\mathrm{Zn} .\) Write complete balanced equations for the reactions chosen.

Balance the following equations, and name each reactant and product: (a) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{MgO}(\mathrm{s})+\mathrm{Fe}(\mathrm{s})\) (b) \(\mathrm{AlCl}_{3}(\mathrm{s})+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{NaCl}(\mathrm{aq})\) (c) \(\mathrm{NaNO}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{s})+\mathrm{HNO}_{3}(\mathrm{aq})\) (d) \(\mathrm{NiCO}_{3}(\mathrm{s})+\mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)\)

Balance the following equations: (a) \(\mathrm{Cr}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})\) (b) \(\mathrm{Cu}_{2} \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g})\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}(\ell)+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\)

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