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 concentration of Ni(NH3)6²⁺ at equilibrium is \(9.9 \times 10^{-3} \text{ M}\) and the concentration of Ni²⁺ at equilibrium is \(0.1901 \text{ M}\).

Step-by-step solution

Write down the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction is already provided in the exercise: $$ \text { Ni}^{2+}(aq) + 6 \text { NH}_{3}(aq) \rightleftharpoons \text { Ni(NH}_{3})_{6}^{2+}(aq) $$

Set up the equilibrium expression using the balanced equation.

We have been given the equilibrium constant expression: $$ K_{\text{overall}} = \frac{[\text{Ni(NH}_3)_6^{2+}]}{[\text{Ni}^{2+}][\text{NH}_3]^6} $$ with \(K_{\text{overall}} = 5.5 \times 10^8\).

Set up an ICE (initial, change, equilibrium) table for the reaction.

To set up an ICE table, we first determine the initial concentrations of each species: - Initial concentration of NH3 = \(\frac{0.50 \text{ L} \times 3.0 \text{ M}}{0.50 \text{ L}} = 3.0 \text{ M}\) - Initial concentration of Ni²⁺ = \(\frac{0.10 \text{ mole}}{0.50 \text{ L}} = 0.20 \text{ M}\) - Initial concentration of Ni(NH3)6²⁺ = 0 | | NH₃ | Ni²⁺ | Ni(NH₃)₆²⁺ | |------|----------|----------|------------| | I | 3.0 M | 0.20 M | 0 | | C | -6x | -x | +x | | E | 3.0-6x M | 0.20-x M | x M |

Solve the equilibrium expression for the concentrations of Ni(NH3)6²⁺ and Ni²⁺ at equilibrium.

Now we can substitute the concentrations from the equilibrium row of the ICE table into the equilibrium expression and solve for x: $$ 5.5 \times 10^8 = \frac{x}{(0.20-x)(3.0-6x)^6} $$ Assuming that x is small, we can simplify and solve for x (concentration of Ni(NH3)6²⁺): $$ 5.5 \times 10^8 = \frac{x}{(0.20)(3.0)^6} $$ $$ x = \frac{(5.5 \times 10^8)(0.20)(3.0)^6}{1} = 9.9 \times 10^{-3} \text{ M} $$ Now we can substitute the value of x back into our equilibrium expression to find the concentration of Ni²⁺: $$ [\text{Ni}^{2+}] = 0.20 - x = 0.20 - 9.9 \times 10^{-3} \text{ M} = 0.1901 \text{ M} $$ So the concentration of Ni(NH3)6²⁺ at equilibrium is \(9.9 \times 10^{-3} \text{ M}\) and the concentration of Ni²⁺ at equilibrium is \(0.1901 \text{ M}\).

Key concepts

ICE Table

The ICE table is a strategic method used to determine the concentrations of reactants and products under equilibrium in a chemical reaction. It stands for Initial, Change, and Equilibrium, representing the stages of the reaction as it moves towards equilibrium.

**Setting up the ICE Table**
Before setting up the table, we need to calculate the initial concentrations of the species involved. Once we have these values, we represent them in the I (Initial) row. Then, the C (Change) row indicates the change in concentration as the reaction progresses. Finally, the E (Equilibrium) row shows the concentrations when equilibrium is reached.

Here's how it works for a reaction like \(\text{Ni}^{2+} + 6 \text{NH}_3 \rightleftharpoons \text{Ni(NH}_3)_6^{2+}\):

• Calculate initial concentrations: These values come from the given moles and volume of the solution.
• Define the change in terms of \(x\): For every molecule of the complex formed, \(x\) amount of \(\text{Ni}^{2+}\) and \(6x\) of \(\text{NH}_3\) are used up. The complex \(\text{Ni(NH}_3)_6^{2+}\) has a change of \(+x\).
• Write equilibrium concentrations: Subtract or add the changes to the initial concentrations to find the equilibrium row.

Using the ICE table method simplifies the process of establishing the equilibrium concentrations, crucial for reactions like complex ion formation, helping in further calculations like using equilibrium constants.

Equilibrium Constant

An equilibrium constant \(K\) is a numerical value that indicates the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. This ratio remains constant at a given temperature. For the complex ion formation reaction in the exercise, the overall equilibrium constant \(K_{\text{overall}}\) is given as \(5.5 \times 10^8\).

**Understanding Equilibrium Constant Expression**
The equilibrium constant expression is derived from the balanced chemical equation, where for the reaction:
\[\text{Ni}^{2+} + 6 \text{NH}_3 \rightleftharpoons \text{Ni(NH}_3)_6^{2+}\]
The expression is:
\[K_{\text{overall}} = \frac{[\text{Ni(NH}_3)_6^{2+}]}{[\text{Ni}^{2+}][\text{NH}_3]^6}\]

**Applying the Equilibrium Constant**
We use the equilibrium constant to find unknown concentrations of chemical species at equilibrium. In our example, knowing the value of \(K_{\text{overall}}\) allows us to set up an equation based on the concentrations from the ICE table and solve for the equilibrium concentration of \(\text{Ni(NH}_3)_6^{2+}\) and \(\text{Ni}^{2+}\).

**Importance**

• Determines which way a reaction will favor (product vs. reactant favored).
• Enables prediction of the extent to which a reaction will proceed.
• Essential for calculating concentrations and bench-marking reaction dynamics under various conditions.

The equilibrium constant is a pivotal concept in understanding and computing equilibrium states in chemical reactions.

Complex Ion Formation

Complex ion formation is a process where a complex ion, an ion composed of a central metal atom or ion bonded to surrounding molecules or ions, known as ligands, is formed in solution. This concept is crucial for understanding how metals interact in aqueous environments.

**Characteristics of Complex Ion Formation**

• Involves coordination bonds: Ligands donate electron pairs to the metal ion.
• Results in a stable, solvated ion: Example, \(\text{Ni(NH}_3)_6^{2+}\) which forms from \(\text{Ni}^{2+}\) and \(NH_3\).
• The formation can be expressed with an equilibrium equation: As in the discussed exercise, where nickel ions react with ammonia.

**How Complex Ions Affect Chemical Equilibrium**
When complex ions form, they can greatly affect the concentrations of species in solution and shift the equilibrium position of the involved reaction.

• The stabilizing effect of complex formation means lower concentrations of free metal ions at equilibrium, impacting solubility.
• The large equilibrium constants typically associated with complex ion formation indicate that the reaction tends to favor the product side, forming complexes.

**Applications and Relevance**
Complex ions are a centerpiece in many chemistry fields, including biochemistry (such as enzyme functions), industrial processes, and analytical chemistry (like titrations or colorimetric analysis). Understanding their formation and equilibrium behavior enhances grasp of reaction mechanisms and solution chemistry.