Question

Calculate the concentrations of \(\mathrm{Cd}^{2+}, \mathrm{Cd}(\mathrm{CN})_{4}^{2-},\) and \(\mathrm{CN}^{-}\) at equilibrium when \(0.50 \mathrm{~g}\) of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) dissolves in \(5.0 \times 10^{2} \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{NaCN}\)

Short answer

At equilibrium: \([\text{Cd(CN)}_{4}^{2-}] = 0.00422\text{ M}\), \([\text{Cd}^{2+}] = 0\), \([\text{CN}^-] = 0.50\text{ M}\).

Step-by-step solution

Calculate Moles of Cd(NO3)2

Calculate the moles of Cd(NO3)2 from the mass. The molar mass of Cd(NO3)2 is approximately \(236.42\text{ g/mol}\). Use the formula: \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \). Hence, \( \text{moles of Cd(NO3)2} = \frac{0.50 \text{ g}}{236.42 \text{ g/mol}} \approx 0.00211 \text{ mol} \).

Initial Concentration of Cd(NO3)2

The total volume of the solution is \(5.0 \times 10^{2}\text{ mL} = 0.500\text{ L} \). The initial concentration is calculated by dividing the moles by the volume in liters: \([\text{Cd}^{2+}]_0 = \frac{0.00211 \text{ mol}}{0.500 \text{ L}} = 0.00422 \text{ M} \).

Initial Concentration of CN^-

The initial concentration of CN^- from NaCN is already given as \(0.50 \text{ M}\).

Write the Equilibrium Reaction

Write the balanced equation for the formation of the complex: \(\text{Cd}^{2+} + 4\text{CN}^- \rightleftharpoons \text{Cd(CN)}_{4}^{2-}\). This equation shows how the species interact at equilibrium.

Expression of Initial and Equilibrium Concentrations

Let \(x\) be the change in concentration of \(\text{Cd}^{2+}\). Initially, \([\text{Cd}^{2+}] = 0.00422\), \([\text{CN}^-] = 0.50\), \([\text{Cd(CN)}_{4}^{2-}] = 0\). At equilibrium, \([\text{Cd}^{2+}] = 0.00422 - x\), \([\text{CN}^-] = 0.50 - 4x\), and \([\text{Cd(CN)}_{4}^{2-}] = x\).

Apply the Equilibrium Constant Expression

The equilibrium expression for the formation constant \(K_f\) is: \(K_f = \frac{[\text{Cd(CN)}_{4}^{2-}]}{[\text{Cd}^{2+}][\text{CN}^-]^4}\). Given \(K_f = 1.0 \times 10^{23}\), substitute in the equilibrium expressions: \(1.0 \times 10^{23} = \frac{x}{(0.00422 - x)(0.50 - 4x)^4}\).

Solving the Equilibrium Constant Equation

Because \(K_f\) is very large, we can assume \(x \approx 0.00422 \text{ M}\) and \(0.50 - 4x \approx 0.50 \text{ M}\) for simplicity. Solve for \(x\) using this approximation: \(x \approx 0.00422 \). So, \( [\text{Cd(CN)}_{4}^{2-}] = 0.00422 \text{ M} \), \( [\text{Cd}^{2+}] = 0\), and \( [\text{CN}^-] = 0.50 \text{ M} \).

Verify the Approximations

Check the validity of approximations: substituting back to ensure \(0.00422 \ll 0.50\) and \(x\) remains small compared to initial concentrations confirms our approach.

Key concepts

Equilibrium Concentration

Chemical equilibrium is a state where the concentrations of reactants and products remain constant over time. In chemical equilibrium problems, like the one in our exercise, the key is to determine these concentrations precisely. This involves knowing the initial concentrations and how they change as the reaction proceeds.

In our case, we have an equilibrium reaction where

• initial concentration for \[\mathrm{Cd(NO}_3)_2\] is determined by converting its mass to moles, as calculated: \[\frac{0.00211\ \text{mol}}{0.500\ \text{L}} = 0.00422\ \text{M}.\]

The initial concentration of \([\mathrm{CN}^-]\) was given as \(0.50\ \text{M}\).

As equilibrium is reached, these concentrations change from their initial values. This is where we introduce a variable, often \(x\), to represent the shift in concentration. As indicated:

• At equilibrium: \[\mathrm{[Cd}^{2+}] = 0.00422 - x\]
• \[\mathrm{[CN}^-] = 0.50 - 4x\]
• \[\mathrm{[Cd(CN)}_4^{2-}] = x\]

Reaction Stoichiometry

Understanding reaction stoichiometry is crucial for solving equilibrium problems, particularly when it comes to determining how concentrations change. Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction.

In the given balanced reaction:

• \(\mathrm{Cd}^{2+} + 4\mathrm{CN}^- \rightleftharpoons \mathrm{Cd(CN)}_4^{2-}\)

This shows a 1:4 ratio of \(\mathrm{Cd}^{2+}\) to \(\mathrm{CN}^-\). This stoichiometric relationship is essential for determining how much each reactant is lost as the complex ion \(\mathrm{Cd(CN)}_4^{2-}\) is formed.

When calculating the equilibrium concentrations, we account for these ratios to set up expressions. Each decrease in \(\mathrm{Cd}^{2+}\) by \(x\) moles results in four times that decrease in \(\mathrm{CN}^-\). Thus:

• If the concentration of \(\mathrm{Cd}^{2+}\) decreases by \(x\), the concentration of \(\mathrm{CN}^-\) decreases by \(4x\).

This stoichiometric linkage helps us to form the equilibrium expressions needed to solve for the unknowns.

Equilibrium Constant (Kf)

The equilibrium constant \(K_f\) is a fundamental concept that helps quantify the ratio of the concentrations of products to reactants at equilibrium. It is a measure of the extent to which a reaction will proceed. A large \(K_f\) value, like the \(1.0 \times 10^{23}\) given in our exercise, indicates a reaction that heavily favors the formation of products.

For our equilibrium reaction:

• \(K_f = \frac{[\mathrm{Cd(CN)}_4^{2-}]}{[\mathrm{Cd}^{2+}][\mathrm{CN}^-]^4}\)

This expression relates the concentrations of products and reactants at equilibrium, allowing us to calculate unknown concentrations.

The assumption made in the exercise, that changes in concentration due to the formation of \(\mathrm{Cd(CN)}_4^{2-}\) (denoted by \(x\)) are small compared to initial values, simplifies the math while remaining realistic for such a large \(K_f\).

Thus:

• The equilibrium concentrations of \(\mathrm{Cd}^{2+}\) approaches zero, as nearly all \(\mathrm{Cd}^{2+}\) forms the complex ion.
• \([\mathrm{Cd(CN)}_4^{2-}] \approx 0.00422\ \text{M}\)

This approximation highlights the reactive tendency of systems with such powerful \(K_f\) values.