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 \(3.0 \mathrm{M} \mathrm{NH}_{3} .\) Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text {owendl }}\) 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 equilibrium concentrations of the species in the solution are approximately [Ni²⁺] ≈ \(1.5 \times 10^{-11}\) M and [Ni(NH₃)₆²⁺] ≈ 0.20 M.

Step-by-step solution

Write down the given information

We are given the following information about the solution: - Initial moles of Ni(NH3)6Cl2 = 0.10 mole - Volume of the solution = 0.50 L - Concentration of NH3 = 3.0 M - K_owendl = 5.5 x 10^8

Set up the ICE table

The ICE table helps us keep track of the initial, change, and equilibrium concentrations of the species involved in the reaction. Write the table with the given information for the concentrations of Ni2+, NH3, and Ni(NH3)6^2+. ``` [Ni2+] [NH3] [Ni(NH3)6^2+] Initial: 0 3.0 0.20 Change: +x -6x +x Equilibrium: x 3.0-6x 0.20+x ``` Note that the initial concentration of Ni(NH3)6^2+ is calculated as 0.10 moles / 0.50 L = 0.20 M.

Use the K_owendl value to write an equation

Now write the equation for K_owendl using the equilibrium concentrations from the ICE table. \[ K_owendl = \frac{[Ni(NH3)6^{2+}]}{[Ni^{2+}][NH3]^6} \] Substitute the equilibrium concentrations from the ICE table. \[ 5.5 \times 10^8 = \frac{(0.20 + x)}{x(3.0 - 6x)^6} \]

Solve for x

Now we will solve this equation for the variable x. Since the K_owendl value is quite large, it means that the reaction proceeds almost to completion. This allows us to make certain reasonable approximations. First, let's assume x is small relative to 0.20 and 3.0. So, we can simplify the equation as follows: \[ 5.5 \times 10^8 = \frac{0.20}{x(3.0)^6} \] Now, solve for x. \[ x = \frac{0.20}{(5.5 \times 10^8)(3.0)^6} \] \[ x \approx 1.5 \times 10^{-11} \]

Calculate the equilibrium concentrations

Now we can calculate the equilibrium concentrations of Ni2+ and Ni(NH3)6^2+ using the calculated value of x. - [Ni2+] = x = 1.5 x 10^(-11) M - [Ni(NH3)6^2+] = 0.20 + x ≈ 0.20 M These are the concentrations of Ni2+ and Ni(NH3)6^2+ in the solution at equilibrium.

Key concepts

Complex Ion Formation

Complex ion formation involves the combination of simple ions into more complex structures, often leading to highly stable species. In the given exercise, the nickel ion \(\mathrm{Ni}^{2+}\) forms a complex with ammonia, \(\mathrm{NH}_3\). This process is represented by the equation: \[\mathrm{Ni}^{2+}(a q) + 6 \mathrm{NH}_3(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_3\right)_6^{2+}(a q)\]Here, one nickel ion (\mathrm{Ni}^{2+}) coordinates with six ammonium ions to form a stable complex ion \(\mathrm{Ni}\left(\mathrm{NH}_3\right)_6^{2+}\).
This step-wise complex formation is typical in coordination chemistry where ligands like ammonia donate electron pairs to the central metal ion. Under conditions where nearly all \(\mathrm{Ni}^{2+}\) precipitates out as \(\mathrm{Ni}\left(\mathrm{NH}_3\right)_6^{2+}\), understanding this equilibrium is vital since it determines the availability of free nickel ions in solution.

Equilibrium Constant (K)

The equilibrium constant \(K\) offers a quantitative measure of the position of equilibrium for a chemical reaction at a given temperature. In the problem, we have the equilibrium constant \(K_{owendl}\) for the formation of \(\mathrm{Ni}\left(\mathrm{NH}_3\right)_6^{2+}\) ion, with a value \(5.5 \times 10^8\).
This large value of \(K\) indicates that the equilibrium lies significantly toward the product side, meaning the complex ion formation is favored.
The mathematical relationship is expressed as: \[K = \frac{\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_3\right]^6}\]
It shows the concentrations of products over reactants. In this particular case, since \(K\) is very large, the concentration of \(\mathrm{Ni}^{2+}\) without ligands is practically negligible. Understanding \(K\) is crucial for predicting which form nickel will predominantly exist in at equilibrium.

ICE Table Method

The ICE table method is a systematic approach used to track the Initial concentrations, the Change in concentrations, and the Equilibrium concentrations of species in a reaction. It simplifies solving equilibrium problems by organizing information clearly.
In the exercise:- **I**nitial concentrations: we begin with known concentrations of \(\mathrm{Ni}(\mathrm{NH}_3)_6^{2+}\) and \(\mathrm{NH}_3\). - \(\mathrm{Ni}(\mathrm{NH}_3)_6^{2+}\): 0.20 M - \(\mathrm{NH}_3\): 3.0 M - **C**hange: \(x\) amount of \(\mathrm{Ni}^{2+}\) is produced, impacting both reactants and products. - Ammonia decreases by \(-6x\), reflecting its stoichiometric consumption. - **E**quilibrium: sums the initial with the change. - \(\mathrm{Ni}^{2+}\) is \(x \), approximate as small, determining \(0.20 + x\) By using the ICE table, assumptions like \(x\) is small (relative to initial values) help simplify calculations, demonstrating equilibrium calculations even in complex ion scenarios.