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Specify what ions are present in solution upon dissolving each of the following substances in water: (a) \(\mathrm{FeCl}_{2}\), (b) \(\mathrm{HNO}_{3}\), (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\), (d) \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Short Answer

Expert verified
When each substance dissolves in water, the following ions are present in the solution: (a) \(\mathrm{FeCl}_{2}\): 1 \(\mathrm{Fe}^{2+}\) and 2 \(\mathrm{Cl}^{-}\) ions. (b) \(\mathrm{HNO}_{3}\): 1 \(\mathrm{H}^{+}\) and 1 \(\mathrm{NO}_{3}^{-}\) ions. (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\): 2 \(\mathrm{NH}_{4}^{+}\) and 1 \(\mathrm{SO}_{4}^{2-}\) ions. (d) \(\mathrm{Ca}(\mathrm{OH})_{2}\): 1 \(\mathrm{Ca}^{2+}\) and 2 \(\mathrm{OH}^{-}\) ions.

Step by step solution

01

(a) Identify ions for \(\mathrm{FeCl}_{2}\)

When \(\mathrm{FeCl}_{2}\) dissolves in water, it dissociates into its ions. The ions are: 1. Iron(II) ion: \(\mathrm{Fe}^{2+}\) 2. Chloride ion: \(\mathrm{Cl}^{-}\) Since there are two chloride ions in the formula, we have: 1 \(\mathrm{Fe}^{2+}\) and 2 \(\mathrm{Cl}^{-}\) ions in the solution.
02

(b) Identify ions for \(\mathrm{HNO}_{3}\)

When \(\mathrm{HNO}_{3}\) dissolves in water, it dissociates into its ions. The ions are: 1. Hydrogen ion: \(\mathrm{H}^{+}\) 2. Nitrate ion: \(\mathrm{NO}_{3}^{-}\) So we have 1 \(\mathrm{H}^{+}\) and 1 \(\mathrm{NO}_{3}^{-}\) ions in the solution.
03

(c) Identify ions for \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

When \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) dissolves in water, it dissociates into its ions. The ions are: 1. Ammonium ion: \(\mathrm{NH}_{4}^{+}\) 2. Sulfate ion: \(\mathrm{SO}_{4}^{2-}\) Since there are two ammonium ions in the formula, we have: 2 \(\mathrm{NH}_{4}^{+}\) and 1 \(\mathrm{SO}_{4}^{2-}\) ions in the solution.
04

(d) Identify ions for \(\mathrm{Ca}(\mathrm{OH})_{2}\)

When \(\mathrm{Ca}(\mathrm{OH})_{2}\) dissolves in water, it dissociates into its ions. The ions are: 1. Calcium ion: \(\mathrm{Ca}^{2+}\) 2. Hydroxide ion: \(\mathrm{OH}^{-}\) Since there are two hydroxide ions in the formula, we have: 1 \(\mathrm{Ca}^{2+}\) and 2 \(\mathrm{OH}^{-}\) ions in the solution.

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Key Concepts

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

Chemical Compound Dissolution
Chemical compound dissolution refers to the process where a solid, liquid, or gas solute becomes incorporated into a liquid solvent, typically water, forming a solution. When a chemical compound dissolves in water, it breaks down into its constituent ions in a process called dissociation. Dissolution is driven by the interaction between the solvent molecules and the solute particles, which can overcome the forces holding the solute together.

In the context of ionic compounds, dissolution involves separating the ions that form the compound's crystal lattice structure. For example, when \(\mathrm{FeCl}_{2}\) is dissolved in water, the compound dissociates into one iron(II) ion \(\mathrm{Fe}^{2+}\) and two chloride ions \(\mathrm{Cl}^{-}\).

Factors affecting dissolution include the nature of the solute and solvent, temperature, and pressure. A compound's solubility in water determines how well it can dissolve. Soluble compounds like \(\mathrm{HNO}_{3}\) dissociate completely in water, facilitating various chemical reactions.
Ion Identification
Ion identification is crucial in understanding how substances behave in aqueous solutions. Ions are charged atoms or molecules and can be either positive (cations) or negative (anions). When identifying ions formed during the dissolution of a compound, it's essential to know the chemical formula of the substance because it indicates the type and ratio of ions present in the solution.

For instance, in calcium hydroxide \(\mathrm{Ca(OH)}_{2}\), which will dissociate into one calcium ion \(\mathrm{Ca}^{2+}\) and two hydroxide ions \(\mathrm{OH}^{-}\). The identification process involves examining the compound's formula and determining the number of each type of ion.

Let's take another example, ammonium sulfate \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\). This compound dissociates into two ammonium ions \(\mathrm{NH}_{4}^{+}\) and one sulfate ion \(\mathrm{SO}_{4}^{2-}\). Understanding these individual ions helps predict the behavior of solutions and their possible interactions in chemical reactions.
Aqueous Solution Chemistry
Aqueous solution chemistry focuses on reactions and behaviors of substances when dissolved in water. It's fundamental in understanding how substances interact with one another when ions form in solutions. In aqueous solutions, ions are dispersed throughout, allowing them to react with one another frequently. This predictability forms the basis for many chemical reactions, including neutralization, precipitation, and redox reactions.

For example, when an acid like \(\mathrm{HNO}_{3}\) dissolves in water, it donates hydrogen ions \(\mathrm{H}^{+}\) to the solution, making it acidic. In contrast, bases like \(\mathrm{Ca(OH)}_{2}\) donate hydroxide ions \(\mathrm{OH}^{-}\), resulting in a basic solution.

The balance of ions in an aqueous solution determines its properties, such as pH value, electrical conductivity, and reactivity. Studying aqueous solution chemistry helps in fields ranging from biochemistry to environmental science, where the properties of water solutions are critically important.

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Most popular questions from this chapter

(a) Calculate the molarity of a solution made by dissolving \(12.5\) grams of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) in enough water to form exactly \(750 \mathrm{~mL}\) of solution. (b) How many moles of \(\mathrm{KBr}\) are present in \(150 \mathrm{~mL}\) of a \(0.112 \mathrm{M}\) solution? (c) How many milliliters of \(6.1 \mathrm{M} \mathrm{HCl}\) solution are needed to obtain \(0.150 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

The concept of chemical equilibrium is very important. Which one of the following statements is the most correct way to think about equilibrium? (a) If a system is at equilibrium, nothing is happening. (b) If a system is at equilibrium, the rate of the forward reaction is equal to the rate of the back reaction. (c) If a system is at equilibrium, the product concentration is changing over time. [Section 4.1]

Determine the oxidation number of sulfur in each of the following substances: (a) barium sulfate, \(\mathrm{BaSO}_{4}\), (b) sulfurous acid, \(\mathrm{H}_{2} \mathrm{SO}_{3}\), (c) strontium sulfide, \(\mathrm{SrS}\), (d) hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\). (e) Locate sulfur in the periodic table in Exercise 4.47; what region is it in? (f) Which region(s) of the period table contains elements that can adopt both positive and negative oxidation numbers?

(a) How many milliliters of a stock solution of \(6.0 \mathrm{MHNO}_{3}\) would you have to use to prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HNO}_{3}\) ? (b) If you dilute \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L}\), what will be the concentration of the diluted solution?

Which of the following solutions is the most basic? (a) \(0.6 \mathrm{M} \mathrm{NH}_{3}\), (b) \(0.150 \mathrm{M} \mathrm{KOH}\), (c) \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\).

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