Question

For the reaction \(\mathrm{CdC}_{2} \mathrm{O}_{4}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)\) (a) calculate \(K .\left(K_{\mathrm{sp}}\right.\) for \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) is \(\left.1.5 \times 10^{-8} .\right)\) (b) calculate \(\left[\mathrm{NH}_{3}\right]\) at equilibrium when \(2.00 \mathrm{~g}\) of \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) are dissolved in \(1.00 \mathrm{~L}\) of solution.

Short answer

#tag_title#Step 2: Relate the equilibrium constant to the solubility product constant# #tag_content#We are given the solubility product constant, \(K_{sp}\), for the reaction involving the precipitation of \(\mathrm{CdC}_{2} \mathrm{O}_{4}\): \(K_{sp} = [\mathrm{Cd}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]\) Since \(\mathrm{Cd}^{2+}\) is completely complexed with \(\mathrm{NH}_{3}\), we can relate the concentration of \(\mathrm{Cd}^{2+}\) to that of \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\), which gives us: \([\mathrm{Cd}^{2+}] = [\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]\) Now we can substitute the expression for \([\mathrm{Cd}^{2+}]\) into the \(K_{sp}\) expression: \(K_{sp} = [\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]\) Now, divide the initial expression for \(K\) by the modified \(K_{sp}\) to find a new expression for \(K\): \(K = \frac{[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]}{[\mathrm{NH}_{3}]^4} \times \frac{1}{[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]} = \frac{1}{[\mathrm{NH}_{3}]^4}\) #tag_title#Step 3: Calculate the equilibrium constant, K# #tag_content#Given the solubility product constant (\(K_{sp}\)) of \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) as \(2.5 \times 10^{-14}\), we can now calculate the equilibrium constant, \(K\): \(K = \frac{K_{sp}}{[\mathrm{NH}_{3}]^4} = \frac{2.5 \times 10^{-14}}{[\mathrm{NH}_{3}]^4}\) #tag_title#Step 4: Determine the concentration of NH3 at equilibrium# #tag_content#We are given that 2.00 grams of \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) are dissolved in 1.00 L of solution to form the complex ion \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\). First, we need to find the number of moles of \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) in the solution: moles of \(\mathrm{CdC}_{2} \mathrm{O}_{4} = \frac{2.00\,\mathrm{g}}{\mathrm{Molar\,mass\,of\,CdC}_{2}\mathrm{O}_{4}}\) The molar mass of \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) is approximately 238 g/mol, therefore: moles of \(\mathrm{CdC}_{2} \mathrm{O}_{4} = \frac{2.00}{238} \approx 8.4 \times 10^{-3}\,\mathrm{mol}\) Since 1 mole of \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) reacts with 4 moles of \(\mathrm{NH}_3\), we can find the moles of \(\mathrm{NH}_3\) at equilibrium using the stoichiometry of the reaction: moles of \(\mathrm{NH}_{3} = 4 \times 8.4 \times 10^{-3} = 33.6 \times 10^{-3}\,\mathrm{mol}\) Now, we can find the concentration of \(\mathrm{NH}_3\) in the 1.00 L solution: \([NH_3] = \frac{33.6 \times 10^{-3}\,\mathrm{mol}}{1.00\,\mathrm{L}} = 33.6 \times 10^{-3}\,\mathrm{M}\) #tag_title#Answer# #tag_content#The concentration of \(\mathrm{NH}_{3}\) at equilibrium is \(33.6 \times {\space} 10^{-3}\,\,\mathrm{M}\).

Step-by-step solution

Write the expression for the equilibrium constant#

Since we are given the reaction: \(\mathrm{CdC}_{2} \mathrm{O}_{4}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)\) we can write the equilibrium constant expression as: \(K = \frac{[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]}{[\mathrm{NH}_{3}]^4}\)

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is fundamental for students diving into the intricacies of chemical reactions. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, meaning the concentrations of reactants and products remain constant over time. However, this does not imply that the amounts of reactants and products are equal, but rather that their concentrations do not change anymore.

The position of equilibrium is described by the equilibrium constant (\( K \)), which provides a ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients. In the given exercise, the equilibrium expression is derived from a balanced chemical equation involving the dissociation of a solid into its aqueous ions. The importance of understanding this concept cannot be overstated, as it lays the groundwork for predicting the extent of a reaction and calculating equilibrium concentrations.

Solubility Product Constant

The solubility product constant (\( K_{sp} \)) is a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. It is the product of the molar concentrations of the ions produced when the solid dissolves, each raised to the power of its stoichiometric coefficient in the balanced equation for the dissolution. The magnitude of the solubility product gives an indication of a compound's solubility: the smaller the constant, the less soluble the compound.

In our exercise, the solubility product constant for cadmium oxalate (\( \text{CdC}_2\text{O}_4 \)) is provided, which allows us to predict the concentration of ions at equilibrium. This value is crucial in determining the direction in which the reaction will proceed to reach equilibrium, especially in a scenario involving a common ion effect or a change in conditions.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of a reactant or product in a reaction mixture when the chemical reaction has reached equilibrium. These concentrations can be found by setting up an equilibrium expression and using the value of the equilibrium constant. It's essential for students to understand how to calculate these concentrations to predict the outcome of chemical reactions in real-world scenarios.

For the given problem, once we know the solubility product constant, we can calculate the concentration of ions in solution at equilibrium. This concept highlights the dynamic nature of chemical equilibrium where concentrations stop changing, providing a snapshot of the reaction's disposition at a specific moment.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation. It allows chemists to predict the amounts of substances consumed and produced in a given reaction, which is especially important when considering reaction yield and efficiency.

In the context of our exercise, the stoichiometric ratios are used to relate the amount of cadmium oxalate to the amount of ammonia (\( \text{NH}_3 \)) required to reach equilibrium. The stoichiometric coefficients from the balanced chemical equation are necessary when setting up the equilibrium constant expression and are utilized in the calculations involved in finding the equilibrium concentrations. Mastery of stoichiometry is indispensable for students because it serves as the backbone for solving a wide array of problems in chemistry.