Question

\(0.1\) moles of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) are added to \(1.0\) moles of \(\mathrm{NH}_{3}\) and then diluted to \(1000 \mathrm{ml}\). Find the concentration of \(\mathrm{Cu}^{2+}\) formed. $$ \mathrm{K}_{\mathrm{eq}}=8.5 \times 10^{-13} $$

Short answer

The concentration of \(\mathrm{Cu^{2+}}\) formed is \(5.77 \times 10^{-11} M\).

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between \(\mathrm{Cu(NO_{3})_{2}}\) and \(\mathrm{NH_{3}}\) is given by: $$ \mathrm{Cu(NO_{3})_{2}} + 4 \mathrm{NH_{3}} \rightleftharpoons \mathrm{Cu(NH_{3})_{4}^{2+}} + 2\mathrm{NO}_{3^-} $$ This equation tells us that one mole of \(\mathrm{Cu(NO_{3})_{2}}\) reacts with four moles of \(\mathrm{NH_{3}}\) to form one mole of \(\mathrm{Cu(NH_{3})_{4}^{2+}}\) and two moles of \(\mathrm{NO}_{3^-}\) ions.

Calculate initial concentrations

We are given the initial number of moles of \(\mathrm{Cu(NO_{3})_{2}}\) and \(\mathrm{NH}_{3}\). In order to find their initial concentrations in the solution, we need to divide the moles by the volume of the solution. Remember to convert the volume to liters. $$ [\mathrm{Cu(NO_{3})}_{2}]_{0} = \frac{0.1}{1} = 0.1\,M \\ [\mathrm{NH}_{3}]_{0} = \frac{1.0}{1} = 1.0\,M $$ The initial concentration of \(\mathrm{Cu^{2+}}\) is zero since there is no \(\mathrm{Cu^{2+}}\) in the solution yet.

Set up an ICE table

An ICE table will help us calculate the changes in concentrations of the reactants and products as the reaction proceeds. The table has the form: | | \(\mathrm{Cu(NO_{3})_{2}}\) | \(\mathrm{NH}_{3}\) | \(\mathrm{Cu(NH_{3})_{4}^{2+}}\) | \(\mathrm{NO}_{3^-}\) | |-------------------|-------------------|-----------|-------------------|--------------| | Initial | 0.1 | 1.0 | 0 | 0 | | Change | -x | -4x | +x | +2x | | Equilibrium | 0.1 - x | 1.0 - 4x | x | 2x | Where x is the amount of moles of \(\mathrm{Cu^{2+}}\) formed.

Write the equilibrium constant expression

The expression for the equilibrium constant, \(\mathrm{K_{eq}}\), can be written based on the balanced chemical equation, using the equilibrium concentrations from the ICE table: $$ \mathrm{K_{eq}}=\frac{[\mathrm{Cu(NH_{3})_{4}^{2+}][\mathrm{NO}_{3^-}]^{2}}{[\mathrm{Cu(NO_{3})_{2}][\mathrm{NH}_{3}]^{4}} $$

Solve for x

Substitute the equilibrium concentrations from the ICE table into the equilibrium constant expression and solve for x: $$ 8.5 \times 10^{-13}= \frac{x(2x)^{2}}{(0.1-x)(1.0-4x)^4} $$ Solving this equation for x is complicated, and since \(\mathrm{K_{eq}}\) is very small, the reaction does not go to completion, so we can assume that x is significantly smaller than the initial concentrations. Hence, we can simplify the equation as: $$ 8.5 \times 10^{-13}= \frac{x(2x)^{2}}{(0.1)(1.0)^4} $$ Now, solve for x: $$ x = 5.77 \times 10^{-11} M $$

Determine the concentration of \(\mathrm{Cu^{2+}}\)

The final concentration of \(\mathrm{Cu(NH_{3})_{4}^{2+}}\), which is equivalent to the concentration of \(\mathrm{Cu^{2+}}\), is equal to x. Therefore, the concentration of \(\mathrm{Cu^{2+}}\) is: $$ [\mathrm{Cu^{2+}}] = 5.77 \times 10^{-11} M $$

Key concepts

Equilibrium Constant Expression

Understanding chemical equilibrium involves grasping the concept of the equilibrium constant expression, denoted as \( K_{\text{eq}} \) in chemistry. This numerical value represents the ratio of the concentrations of products to the concentrations of reactants, each raised to the power of their stoichiometric coefficients from the balanced chemical equation.

For the reaction \( \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D} \), the equilibrium constant expression would be \( K_{\text{eq}} = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} \), where the square brackets indicate the concentration of each substance in molarity (M). Equilibrium constant expressions are crucial because they provide insight into the extent of a reaction, predicting the direction in which the reaction will proceed to reach equilibrium.

ICE Table Calculations

ICE tables are a systematic method to track changes in concentrations or pressures of reactants and products as a chemical reaction reaches equilibrium. ICE stands for Initial, Change, and Equilibrium.

In practice, an ICE table enables you to set initial concentrations, determine changes needed to reach equilibrium, and calculate final equilibrium concentrations. The 'Change' row often follows a stoichiometric pattern based on coefficients from the balanced chemical equation. For example, in a simple reaction \( 2\text{A} \rightleftharpoons \text{B} \), a decrease in \( 2\text{x} \) moles of A would correspond to an increase of \( \text{x} \) moles of B. Understanding how to apply this tool aids in effectively solving equilibrium problems and determining the concentrations of all species at equilibrium.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. The coefficients in a balanced chemical equation provide the proportions of reactants and products that react and form, respectively. This allows chemists to calculate how much of one substance will react with a given amount of another substance and what quantities of products will form.

Correctly applying stoichiometry is essential when setting up an equilibrium problem, as it determines the ratio in which the reactants combine and the products form. The stoichiometric coefficients guide us in creating the 'Change' row in an ICE table, ensuring that the law of conservation of mass is upheld.

Molarity

In chemistry, molarity, denoted as M, is the measure of concentration defined as the number of moles of a solute divided by the volume of the solution in liters. It is a pivotal concept when discussing chemical reactions in solution, as it allows scientists to communicate precisely how concentrated a solution is.

Knowing the molarity of reactants is necessary to determine the initial concentrations for an ICE table and to calculate the equilibrium concentrations. For instance, in the given exercise, we use molarity to encode how much \( \text{Cu(NO}_3\text{)}_2 \) and \( \text{NH}_3 \) are present before dilution and reaction take place. Mastery of converting between moles and molarity is integral for solving many chemical equilibrium problems.