Question

Calculate the concentrations of each of the ions in (a) \(0.060 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2},\) (b) \(0.15 \mathrm{M} \mathrm{FeCl}_{3}\), (c) \(0.22 M\) \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (d) \(0.60 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)

Short answer

The concentrations of the ions are as follows: (a) Ca2+: 0.060 M, OH-: 0.120 M; (b) Fe3+: 0.15 M, Cl-: 0.45 M; (c) Cr3+: 0.44 M, SO4^2-: 0.66 M; (d) NH4+: 1.20 M, SO4^2-: 0.60 M.

Step-by-step solution

Identifying the Ions in Each Compound

Determine the ions and their respective charges that are produced when each of the given compounds is dissolved in water. Ca(OH)2 produces Ca2+ and OH- ions, FeCl3 produces Fe3+ and Cl- ions, Cr2(SO4)3 produces Cr3+ and SO4^2- ions, and (NH4)2SO4 produces NH4+ and SO4^2- ions.

Determining the Ion Concentrations for Ca(OH)2

For every 1 mole of Ca(OH)2 that dissolves, 1 mole of Ca2+ ions and 2 moles of OH- ions are produced. Multiply the molarity of Ca(OH)2 by the number of each ion the compound produces to find their concentrations. Ca2+: 0.060 M, OH-: 0.060 M * 2 = 0.120 M.

Determining the Ion Concentrations for FeCl3

For every 1 mole of FeCl3 that dissolves, 1 mole of Fe3+ ions and 3 moles of Cl- ions are produced. Multiply the molarity of FeCl3 by the number of each ion the compound produces to find their concentrations. Fe3+: 0.15 M, Cl-: 0.15 M * 3 = 0.45 M.

Determining the Ion Concentrations for Cr2(SO4)3

For every 1 mole of Cr2(SO4)3 that dissolves, 2 moles of Cr3+ ions and 3 moles of SO4^2- ions are produced. Multiply the molarity of Cr2(SO4)3 by the number of each ion the compound produces to find their concentrations. Cr3+: 0.22 M * 2 = 0.44 M, SO4^2-: 0.22 M * 3 = 0.66 M.

Determining the Ion Concentrations for (NH4)2SO4

For every 1 mole of (NH4)2SO4 that dissolves, 2 moles of NH4+ ions and 1 mole of SO4^2- ions are produced. Multiply the molarity of (NH4)2SO4 by the number of each ion the compound produces to find their concentrations. NH4+: 0.60 M * 2 = 1.20 M, SO4^2-: 0.60 M

Key concepts

Molarity

Molarity is a term that describes the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. Calculating molarity is fundamental in chemistry because it allows scientists and students to know how much of a compound is present in a given volume of solution. This is especially important in reactions, where the amount of reactants determines the amount of product formed.

For example, in a solution with a molarity of 0.060 M for calcium hydroxide, \(Ca(OH)_2\), there are 0.060 moles of \(Ca(OH)_2\) per liter of solution. By understanding molarity, students can then begin to calculate the concentrations of individual ions that result when ionic compounds like \(Ca(OH)_2\) dissolve in water.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In ion concentration calculations, stoichiometry helps us determine the ratio in which compounds dissociate into their constituent ions.

For instance, one mole of \(Ca(OH)_2\) dissociates into one mole of \(Ca^{2+}\) ions and two moles of \(OH^-\) ions. The stoichiometric coefficients in the formula tell us the ratio of ions produced. By applying stoichiometric principles, we can derive the concentration of each ion in solution from the molarity of the initial compound. This step is crucial in solving exercises like the ones provided, where different ionic compounds produce ions in different ratios upon dissolution.

Solubility

Solubility refers to the ability of a substance to dissolve in a solvent, such as water. It is significant in the context of ionic compounds, as their solubility determines the degree to which they will dissociate and produce ions. Not all ionic compounds have the same solubility. Some are highly soluble and dissociate completely, while others have limited solubility and therefore do not fully separate into ions.

The solubility principles do not directly affect the stoichiometric calculations for the given problem since it is assumed that the compounds are sufficiently soluble to produce the stated molarity. However, in the real world, solubility constraints are often critical, especially when dealing with saturated solutions or precipitate formation.

Ionic Compounds

Ionic compounds are chemical compounds composed of cations (positively charged ions) and anions (negatively charged ions). The ionic bond is the electrostatic force that holds the ions together in an ionic compound. When ionic compounds dissolve in water, they break apart into their constituent ions. This process is known as dissociation.

The concept of ionic compounds is central to understanding the exercises at hand. Each compound \(Ca(OH)_2, FeCl_3, Cr_2(SO_4)_3, (NH_4)_2SO_4\) is an ionic compound and will dissociate according to its chemical formula. This provides the foundation for calculating ion concentrations, as it is necessary to identify the ions present and in what ratio they occur to apply molarity and stoichiometry principles effectively.