Question

A solution is prepared by adding 0.10 mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to 0.50 \(\mathrm{L}\) of \(3.0 M\) \(\mathrm{NH}_{3}\) . Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text { overall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{8}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

Short answer

The concentrations of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) and \(\mathrm{Ni}^{2+}\) in the solution are both \(9.975 \times 10^{-4}\) M.

Step-by-step solution

Identify moles and initial concentrations

Given, 0.10 mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\mathrm{Cl}_{2}\) is added to 0.50 L of 3.0 M \(\mathrm{NH}_{3}\). The initial concentration of ammonia can be calculated as follows: Initial moles of $\mathrm{NH}_{3} = \text{Initial concentration} \times \text{Volume} \\ \text{Initial moles of } \mathrm{NH}_{3} = 3.0 \ \text{M} \times 0.50 \ \text{L} = 1.5 \ \text{moles}$

Select a change table to represent the reaction

We can set up a change table (also called an ICE table) to represent the initial concentrations of the species, the change, and the equilibrium concentrations. Reaction: $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$ | Species | Initial Concentration| Change | Equilibrium Concentration | |--------------|-----------------|-------|--------------------| | \(\mathrm{Ni}^{2+}\) | 0 | +x | x | | \(\mathrm{NH}_{3}\) | 1.5 M | -6x | 1.5 - 6x | | \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) | 0 | -x | x |

Substitute the equilibrium concentrations in the overall equilibrium constant expression

Now, substitute the equilibrium concentrations in the overall equilibrium constant expression is \(5.5 \times 10^{8} =\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}\): $$5.5 \times 10^{8} =\frac{x}{x(1.5 - 6x)^{6}}$$

Solve for the concentration of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\)

Rearrange the equation: $$x = - \frac{1}{6} + \sqrt{\frac{1}{36} + 5.5 \times 10^{8}}$$ Solve for the concentration of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\): \(x = 9.975 \times 10^{-4} \ \text{M}\) Thus, the concentration of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(9.975 \times 10^{-4}\) M.

Solve for the equilibrium concentration of \(\mathrm{Ni}^{2+}\)

Using the change table, we find that the equilibrium concentration of \(\mathrm{Ni}^{2+}\) is also equal to the concentration of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\): Equilibrium concentration of \(\mathrm{Ni}^{2+} = x = 9.975 \times 10^{-4} \ \text{M}\) Thus, the concentration of \(\mathrm{Ni}^{2+}\) in the solution is \(9.975 \times 10^{-4}\) M.

Key concepts

Coordination Chemistry

In coordination chemistry, we explore the fascinating world of complex ions and coordination compounds. These are formed when metal atoms or ions bind with one or more ligands. Ligands are molecules or ions that can donate pairs of electrons to metals, creating a coordinate covalent bond.

Take, for example, the complex \( \mathrm{Ni} \left(\mathrm{NH}_{3}\right)_{6}^{2+} \). Here, nickel (Ni) acts as the central metal ion, while ammonia (\( \mathrm{NH}_{3} \)) acts as the ligand. Each ligand donates an electron pair, forming a stable compound.

The metal-ligand bond is unique because it involves the donation of electrons from the ligand, creating an efficient way for the metal ion to achieve greater stability. Understanding coordination chemistry is essential for grasping how these bonding interactions occur. This includes studying the nature and geometry of coordination compounds, the type of ligands involved, and the metal’s oxidation state. These factors all influence the properties and reactivity of the complexes.

ICE Table

An ICE table is a valuable tool in chemistry, especially for solving equilibrium problems. ICE stands for Initial, Change, and Equilibrium, which are the stages we use to outline the concentrations of reacting species in a solution.

The process starts with the *Initial* concentrations. For the given reaction, elements such as \( \mathrm{Ni^{2+}}, \mathrm{NH_{3}}, \) and the complex ion \( \mathrm{Ni} \left(\mathrm{NH}_{3}\right)_{6}^{2+} \) have specified starting concentrations.

In the *Change* phase, we identify how these concentrations change as the reaction proceeds towards equilibrium. Typically, a reaction will shift right or left, consuming reactants and forming products. By using assigned variables such as \( x \), which denote these changes, we can quantify the amount reacted or produced.

Lastly, the *Equilibrium* state represents the stable concentrations of all species. Here, computations are made using the initial concentrations and the changes identified, applying these in equilibrium constant expressions to solve for unknowns.

Equilibrium Constant

The equilibrium constant is a crucial concept that describes the ratio of product concentrations to reactant concentrations once a chemical reaction has reached equilibrium. This constant, represented as \( K_{eq} \), provides insight into the position of equilibrium and the extent to which a reaction proceeds.

For our equation involving \( \mathrm{Ni} \left(\mathrm{NH}_{3}\right)_{6}^{2+} \), the equilibrium constant expression is written as: \[K_{\text{overall}} = \frac{\left[ \mathrm{Ni} \left(\mathrm{NH}_{3}\right)_{6}^{2+} \right]}{\left[ \mathrm{Ni}^{2+} \right] \left[ \mathrm{NH}_{3} \right]^{6}}\]

Given \( K_{\text{overall}} = 5.5 \times 10^{8} \), this high value signifies that the formation of the complex ion is favored significantly over the dissociation back into reactants.

Understanding equilibrium constants allows us to predict the direction and extent of reactions. A larger \( K_{eq} \) means there are more products at equilibrium, while a smaller \( K_{eq} \) indicates reactants are favored.

Solution Concentration Calculations

Effective calculation of solution concentrations involves understanding and applying the principles of molarity, stoichiometry, and altering concentrations through reactions or dilutions.

In our problem, we initially calculate the concentration of \( \mathrm{NH_{3}} \) using the molarity formula: \( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \). Given that \( 1.5 \) moles of \( \mathrm{NH_{3}} \) are in a \( 0.50 \) L solution, its initial concentration is \( 3.0 \ \text{M} \).

As the reaction proceeds, the concentrations of species change and are affected, as shown in the ICE table, requiring us to solve for equilibrium concentrations using the equation: \[ x = 9.975 \times 10^{-4} \ \text{M} \]
These calculations help us find the concentration of both \( \mathrm{Ni}^{2+} \) and \( \mathrm{Ni} \left(\mathrm{NH}_{3}\right)_{6}^{2+} \), key components to solving the overall chemical equilibrium in the solution. By understanding each stage of concentration calculation, from initial values to equilibrium, students can accurately deduce these complex parameters.