Question

Cadmium metal is added to \(0.350 \mathrm{L}\) of an aqueous solution in which \(\left[\mathrm{Cr}^{3+}\right]=1.00 \mathrm{M} .\) What are the concentrations of the different ionic species at equilibrium? What is the minimum mass of cadmium metal required to establish this equilibrium? $$\begin{array}{r} 2 \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{Cd}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Cr}^{2+}(\mathrm{aq})+\mathrm{Cd}^{2+}(\mathrm{aq}) \\ K_{\mathrm{c}}=0.288 \end{array}$$

Short answer

The equilibrium concentrations are for Cd^2+ is 0.12 M, for Cr^2+ is 0.24 M and for Cr^3+ is 0.76 M. The minimum mass of cadmium metal required to establish this equilibrium in 0.350 L of solution is 4.43 g.

Step-by-step solution

Write the Equilibrium Expression

The law of mass action states that any equilibrium expression for a reaction is the product of the concentrations of the products divided by the product of the concentrations of the reactants. In this case, this is: \[ K_c = \frac{[\mathrm{Cd^{2+}}][\mathrm{Cr^{2+}}]^2}{[\mathrm{Cr^{3+}}]^2} \]

Set up the ICE Table

The Initial, Change, Equilibrium (ICE) table to keep track of the concentrations of each species. Initially, 'Cd' is a solid and '[Cd^2+]' is nil in the solution. Assuming that 'x' moles of 'Cd' dissolves, the ICE table will be :Initial: [Cr^3+] = 1 M, [Cr^2+] = 0, [Cd^2+] = 0 Change: [Cr^3+] = -2x, [Cr^2+] = +2x, [Cd^2+] = +x Equilibrium: [Cr^3+] = 1-2x M, [Cr^2+] = 2x M, [Cd^2+] = x M

Substitute the Equilibrium Concentrations into the \(K_c\) Expression

Substitute the equilibrium concentrations into the \(K_c\) expression: \[0.288 = \frac{x(2x)^2}{(1-2x)^2}\], The equation will have to be solved for x.

Solve for x

On solving, we get x = 0.12 M. Now we substitute x into the equilibrium concentrations to find these concentrations: [Cd^2+] = x = 0.12 M, [Cr^2+] = 2x = 2 * 0.12 = 0.24 M, [Cr^3+] = 1 - 2x = 1 - 2*0.12 = 0.76 M.

Find the Mass of Cadmium

Finally, to calculate the minimum mass of cadmium required, use the molar mass of cadmium (112.4 g/mol) and the fact that x is in moles per liter to obtain the mass in grams: Mass of Cd = (0.350 L)(0.12 mol/L)(112.4 g/mol) = 4.43 g.

Key concepts

Equilibrium Expression

When learning about chemical reactions, one key concept is understanding how to express the state of equilibrium. In a state of equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentration of reactants and products remains constant over time. To quantify this state, we use the equilibrium expression based on the law of mass action.

This law states that for a balanced chemical equation, the equilibrium constant (K) is the ratio of the product concentrations to the reactant concentrations, each raised to the power of their coefficients in the balanced equation. For a reaction like \(2\mathrm{Cr}^{3+}(aq) + \mathrm{Cd}(s) \rightleftharpoons 2\mathrm{Cr}^{2+}(aq) + \mathrm{Cd}^{2+}(aq)\), the equilibrium expression becomes \(K_c = \frac{[\mathrm{Cd^{2+}}][\mathrm{Cr^{2+}}]^2}{[\mathrm{Cr^{3+}}]^2}\).

In the given example, the equilibrium constant (Kc) is 0.288, which tells us the ratio of product to reactant concentrations at equilibrium. A higher Kc value implies a greater concentration of products at equilibrium, whereas a lower Kc value suggests a reaction that favors the reactants.

ICE Table

The Initial, Change, Equilibrium (ICE) table is an invaluable tool for organizing and calculating the changes in concentrations of reactants and products as a reaction approaches equilibrium. The table is typically set up in three rows corresponding to the initial concentrations (I), the changes in concentrations (C), and the final equilibrium concentrations (E).

In our example, the ICE table helps us track how the concentration of \(\mathrm{Cr}^{3+}\) decreases and how the \(\mathrm{Cr}^{2+}\) and \(\mathrm{Cd}^{2+}\) concentrations increase as cadmium metal dissolves. By defining 'x' as the change in concentration for cadmium dissolved (since it starts at zero), we can express the changes and eventual equilibrium concentrations for all species involved in the reaction. This structured approach simplifies the complex process of calculating concentration changes during the approach to equilibrium.

Law of Mass Action

The law of mass action is a principle that provides the relationship between the concentrations of reactants and products at equilibrium for a reversible chemical reaction at a given temperature. It's the theoretical foundation on which we write equilibrium expressions.

The general form of the law can be stated as: for a reaction where \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant (K) is given by \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\), where the concentrations (denoted by square brackets) are measured in molarity (moles per liter).

The equilibrium constant is specific to a particular reaction at a specific temperature and provides insight into the direction and extent of the reaction. When solving for equilibrium concentrations, as we do with the cadmium and chromium ions example, the law of mass action is the guiding principle for setting up both the equilibrium expression and the ICE table calculations.

Molar Mass Calculation

The concept of molar mass is fundamental in stoichiometry and chemical calculations, and it proves critical when determining the mass of a substance required to reach equilibrium, as shown in our example.

Molar mass, which is the mass of one mole of a substance, is expressed in grams per mole (g/mol), and it's numerically equal to the substance's average atomic or molecular weight. For cadmium, with an average atomic mass of about 112.4, its molar mass is 112.4 g/mol.

To find the mass of cadmium we need to establish equilibrium in a reaction, we multiplied the required moles of cadmium (determined from the ICE table) by its molar mass. This step is pivotal in translating the abstract quantities from our calculations into tangible amounts that one could measure in a laboratory setting.