Question

Calculate the concentration of ions in these saturated solutions: (a) \(\left[\mathrm{I}^{-}\right]\) in AgI solution with \(\left[\mathrm{Ag}^{+}\right]=9.1 \times 10^{-9} \mathrm{M}\) (b) \(\left[\mathrm{Al}^{3+}\right]\) in \(\mathrm{Al}(\mathrm{OH})_{3}\) with \(\left[\mathrm{OH}^{-}\right]=2.9 \times 10^{-9} \mathrm{M}\)

Short answer

Therefore, the ion concentration for \(\left[\mathrm{I}^{-}\right]\) in AgI solution is \(9.1 \times 10^{-9} \mathrm{M}\) and \(\left[\mathrm{Al}^{3+}\right]\) in \(\mathrm{Al}(\mathrm{OH})_{3}\) solution is \(9.67 \times 10^{-10} \mathrm{M}\).

Step-by-step solution

Formulating Ionic Equilibrium for AgI Solution

To calculate the concentration of \(\mathrm{I}^{-}\) in AgI solution, we use the equilibrium expression for the dissolution of AgI in water: \(AgI_{(s)} \leftrightarrow Ag^{+}_{(aq)} + I^{-}_{(aq)}\). From this equilibrium expression, we can deduce that the concentration of \(\mathrm{I}^{-}\) and \(\mathrm{Ag}^{+}\) are equal in a saturated solution of AgI.

Calculating [I-] in AgI Solution

Since \(\left[\mathrm{Ag}^{+}\right]=9.1 \times 10^{-9} \mathrm{M}\), and the concentration of \(\mathrm{I}^{-}\) ions is equal to that of \(\mathrm{Ag}^{+}\) ions in the AgI solution, we can determine that \(\left[\mathrm{I}^{-}\right]=9.1 \times 10^{-9} \mathrm{M}\) in the AgI solution.

Formulating Ionic Equilibrium for Al(OH)3 Solution

To calculate the concentration of \(\mathrm{Al}^{3+}\) in \(\mathrm{Al}(\mathrm{OH})_{3}\) solution, we use the equilibrium expression for the dissolution of \(\mathrm{Al}(\mathrm{OH})_{3}\) in water: \(\mathrm{Al}(\mathrm{OH})_{3_{(s)}} \leftrightarrow Al^{3+}_{(aq)} + 3OH^{-}_{(aq)}\). From this equilibrium expression, we can deduce that the concentration of \(\mathrm{Al}^{3+}\) is equal to one third of the \(\mathrm{OH}^{-}\) concentration in a saturated solution of \(\mathrm{Al}(\mathrm{OH})_{3}\).

Calculating [Al3+] in Al(OH)3 Solution

Since \(\left[\mathrm{OH}^{-}\right]=2.9 \times 10^{-9} \mathrm{M}\), and the concentration of \(\mathrm{Al}^{3+}\) ions is equal to one third of the \(\mathrm{OH}^{-}\) concentration in the \(\mathrm{Al}(\mathrm{OH})_{3}\) solution, we can determine that \(\left[\mathrm{Al}^{3+}\right] = \frac{\left[\mathrm{OH}^{-}\right]}{3} = \frac{2.9 \times 10^{-9} \mathrm{M}}{3} = 9.67 \times 10^{-10} \mathrm{M}\) in the \(\mathrm{Al}(\mathrm{OH})_{3}\) solution.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial for grasping how substances dissolve and interact in solution. In a chemical reaction that reaches equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. For dissolution processes, this concept is applied when ionic compounds like silver iodide (AgI) or aluminum hydroxide (Al(OH)₃) dissolve in water to form ions.

At equilibrium, certain concentrations of these ions will be present in the solution, and these concentrations will remain constant as long as the system's conditions don't change. By understanding that for every mole of ionic compound that dissolves, a stoichiometric amount of ions is produced, students can calculate the ion concentrations in saturated solutions using the given equilibrium constants or solubility products.

Solubility Product

The solubility product (K_sp) is an important concept when discussing the solubility of ionic compounds in water. This is a type of equilibrium constant specific to the dissolution of sparingly soluble salts. It represents the maximum product of the molar concentrations of the ions involved in a saturated solution at a given temperature.

For example, for a generic salt AB that dissociates into A⁺ and B^- ions, the K_sp would be expressed as K_sp = [A⁺][B^-]. While the saturation point tells us the maximum concentration of the ions before the salt starts to precipitate, understanding K_sp allows us to calculate how much of a substance will dissolve, and it also helps in predicting the formation of a precipitate.

Molarity

Molarity is a measure of the concentration of a solute in a solution. It is denoted by the unit M (mol/L), which stands for moles per liter. When working out ionic concentration calculations, molarity provides a clear quantification of how much of a substance is dissolved in a given volume of solvent.

To calculate molarity, one simply divides the number of moles of solute by the volume of the solution in liters. This straightforward formula, M = moles of solute / liters of solution, is fundamental in chemistry for preparing solutions of desired concentrations and for calculating the concentrations of each ion in a saturated solution.

Dissociation of Ionic Compounds

Ionic compounds dissociate in water to form ions, a process that is essential to understanding their behavior in solution. This dissociation is influenced by the nature of the ionic compound and the solvent, as well as temperature and other factors.

When an ionic compound like AgI or Al(OH)₃ dissolves in water, it separates into its respective cations and anions. The extent to which it dissociates is described by the stoichiometry of the dissolution reaction. For instance, if Al(OH)₃ dissociates into one Al³⁺ ion and three OH^- ions, this stoichiometry helps determine the proportions of the ions in the solution and is pivotal for solving problems related to ionic concentration calculations. Understanding this process is fundamental for predicting the behavior of ions in aqueous solutions.