Problem 83 - Chemistry Matter and Change

Question

At \(298 \mathrm{K}, K_{\text { sp }}\) for cadmium iodate \(\left(\mathrm{Cd}\left(\mathrm{IO}_{3}\right)_{2}\right)\) equals \(2.3 \times 10^{-8} .\) What are the molar concentrations of cadmium ions and iodate ions in a saturated solution at 298 \(\mathrm{K} ?\)

Step-by-Step Solution

Answer

The molar concentrations of cadmium ions (Cd2+) and iodate ions (IO3-) in a saturated solution of cadmium iodate (Cd(IO3)2) at 298 K are approximately \(1.34 \times 10^{-3}\) M and \(2.68 \times 10^{-3}\) M, respectively.

Step 1: Write the chemical equation for the dissociation of cadmium iodate and Ksp expression

Cadmium iodate dissociates into its constituent ions in water as follows: \[ \mathrm{Cd}(\mathrm{IO}_3)_2 \rightleftharpoons \mathrm{Cd}^{2+} + 2 \mathrm{IO}_3^- \] Now, we can write the Ksp expression: \[ K_{sp} = [\mathrm{Cd}^{2+}] [\mathrm{IO}_3^-]^2 \]

Step 2: Define the concentrations of ions

Let x be the molar concentration of cadmium ions (Cd2+). As there are two iodate ions (IO3-) for each cadmium ion in the dissociation process, the molar concentration of iodate ions will be 2x. So, we have: \[[\mathrm{Cd}^{2+}] = x\] \[[\mathrm{IO}_3^-] = 2x\]

Step 3: Substitute the concentrations into the Ksp expression and solve for x

Now, we can plug the concentrations (x and 2x) into the Ksp expression: \[ K_{sp} = (x) (2x)^2 = 4x^3\] We are given the value of Ksp at 298 K as \(2.3 \times 10^{-8}\). We'll substitute the given Ksp value in the expression and solve for x: \[2.3 \times 10^{-8} = 4x^3\] To find x, we first divide both sides by 4: \[x^3 = \frac{2.3 \times 10^{-8}}{4}\] Then, we take the cube root of both sides: \[x = \sqrt[3]{\frac{2.3 \times 10^{-8}}{4}}\] Now, use a calculator to find the value of x: \[x \approx 1.34 \times 10^{-3}\]

Step 4: Determine the molar concentrations of cadmium ions and iodate ions

We have found the value of x, which represents the concentration of cadmium ions: \[[\mathrm{Cd}^{2+}] \approx 1.34 \times 10^{-3} \ \mathrm{M}\] The concentration of iodate ions (IO3-) is 2x: \[[\mathrm{IO}_3^-] \approx 2(1.34 \times 10^{-3})\ \mathrm{M} \approx 2.68 \times 10^{-3} \ \mathrm{M}\] So, the molar concentrations of cadmium ions (Cd2+) and iodate ions (IO3-) in a saturated solution at 298 K are approximately \(1.34 \times 10^{-3}\) M and \(2.68 \times 10^{-3}\) M, respectively.

Key Concepts

Chemical Equilibrium

In chemistry, chemical equilibrium refers to the state in which the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentrations of reactants and products remain constant over time, even though they are not equal.
This concept is crucial in understanding how reactions reach a state where they appear to have "stopped" progressing.
However, the reaction is still occurring at the molecular level, with molecules constantly reacting back and forth.

• Equilibrium state is dynamic, not static.
• The equilibrium constant (K) quantifies the balance.

The solubility product constant, often denoted as \( K_{sp} \), is a special type of equilibrium constant for dissociation reactions involving ionic compounds. It helps predict the extent to which a compound will dissolve in water. In our exercise, the \( K_{sp} \) value for cadmium iodate determines the point at which cadmium and iodate ions reach equilibrium in a saturated solution.

Dissociation Reactions

Dissociation reactions involve the separation of a compound into its various parts, usually ions, when dissolved in a solvent. These reactions are typical for ionic compounds placed in water. In the given exercise, cadmium iodate \( \mathrm{Cd} \left( \mathrm{IO}_{3} \right)_{2} \) dissociates into cadmium ions \( \mathrm{Cd}^{2+} \) and iodate ions \( \mathrm{IO}_{3}^{-} \).
This can be represented by the equation:\[\mathrm{Cd(\mathrm{IO}_3)_2 \rightleftharpoons \mathrm{Cd}^{2+} + 2 \mathrm{IO}_3^-}\]

• Ionic compounds separate into ions in water.
• Dissociation creates equal and opposite reactions.

Each type of ion formed from the dissociation influences how much of the compound can ultimately dissolve. With two iodate ions produced for every cadmium ion, this determines the concentration relationships we'll use to solve for equilibrium solutions using \( K_{sp} \). Understanding this ratio is key to solving the exercise.

Molar Concentration

Molar concentration is a measure of the amount of a solute present in a solution, generally represented in moles per liter (M). It tells us how densely molecules or ions are packed into a solution.
In the context of our exercise with cadmium iodate, determining the molar concentration of ions in a saturated solution involves understanding the stoichiometry of its dissociation reaction.
For cadmium iodate, if the molar concentration of cadmium ions \( \mathrm{Cd}^{2+} \) is \( x \), then the concentration of iodate ions \( \mathrm{IO}_{3}^{-} \) will be \( 2x \) due to the 1:2 ratio.

• Molarity is vital for quantifying reactions.
• Stoichiometry influences ion concentration.

By substituting these relationships into the expression for \( K_{sp} \), we solve the equilibrium problem, resulting in the specific molar concentrations of the ions at equilibrium.

Ionic Compounds

Ionic compounds consist of positively and negatively charged ions held together by ionic bonds.
These bonds are strong due to the electrostatic forces between oppositely charged particles.
When ionic compounds dissolve in water, their bonds break, and the compound dissociates into individual cations and anions.

• Positive ions are called cations.
• Negative ions are known as anions.

For example, cadmium iodate breaks into \( \mathrm{Cd}^{2+} \) and \( \mathrm{IO}_{3}^{-} \) ions. Understanding how ionic compounds dissociate helps predict their behavior in solutions, which is essential when determining solubility and the resulting concentrations of ions. The ability of these compounds to dissolve and affect solution properties is fundamental in chemistry, often expressed through equilibrium and \( K_{sp} \) calculations.