Problem 64 - Chemistry An Atoms First Approach

Question

In the presence of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\) forms the complex ion \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .\) If the equilibrium concentrations of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) are \(1.8 \times 10^{-17} M\) and \(1.0 \times 10^{-3} M,\) respectively, in a \(1.5-M \mathrm{NH}_{3}\) solution, calculate the value for the overall formation constant of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\). $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K_{\mathrm{overall}}=?$$

Step-by-Step Solution

Answer

The overall formation constant of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) is approximately \(1.097 \times 10^{14}\).

Step 1: List the given equilibrium concentrations

We are given that $$[\mathrm{Cu}^{2+}] = 1.8 \times 10^{-17} M$$ $$[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}] = 1.0 \times 10^{-3} M$$ And the concentration of the solution $$[\mathrm{NH}_{3}] = 1.5 M$$

Step 2: Insert equilibrium concentrations into the K expression

Now that we have the equilibrium concentrations, we can plug them into our K expression: $$K_{\mathrm{overall}}=\frac{[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]}{[\mathrm{Cu}^{2+}][\mathrm{NH}_{3}]^4}=\frac{1.0 \times 10^{-3}}{(1.8 \times 10^{-17})[(1.5)^4]}$$

Step 3: Calculate K

Now, we can calculate the K value for the complex ion: $$K_{\mathrm{overall}}=\frac{1.0 \times 10^{-3}}{(1.8 \times 10^{-17})[(1.5)^4]}=\frac{1.0 \times 10^{-3}}{(1.8 \times 10^{-17})(5.0625)}$$ $$K_{\mathrm{overall}}=\frac{1.0 \times 10^{-3}}{9.1125 \times 10^{-17}}\approx 1.097 \times 10^{14}$$ Thus, the overall formation constant of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) is approximately \(1.097 \times 10^{14}\).

Key Concepts

Complex Ion

A complex ion is formed when a metal ion binds with surrounding molecules or ions, known as ligands. These ligands donate their electron pairs to the metal ion, forming a coordination bond. In the case of our exercise, the complex ion is \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\), which consists of a central copper ion \(\mathrm{Cu}^{2+}\) surrounded by four ammonia molecules \(\mathrm{NH}_3\).
This type of structure allows the metal ion to stabilize, as the electrons from the ligands fill the vacant orbitals of the metal ion.
Ammonia acts as a neutral ligand, contributing to the formation of the complex ion without changing its charge. As a result, the copper ion holds a 2+ charge within this complex.

Equilibrium Concentrations

In any chemical reaction, equilibrium concentrations mean the amounts of reactants and products present when the reaction is in a state where the forward and backward reactions occur at the same rate. In our problem, it's crucial to understand the concentrations of \([\mathrm{Cu}^{2+}]\), \([\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]\), and \([\mathrm{NH}_{3}]\).

• The copper ion \([\mathrm{Cu}^{2+}]\) is given as \(1.8 \times 10^{-17} M\).
• The complex ion \([\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]\) has a concentration of \(1.0 \times 10^{-3} M\).
• Ammonia \([\mathrm{NH}_{3}]\) is in a much larger concentration of \(1.5 M\).

These values are plugged into the equilibrium expression to help calculate the formation constant, capturing the balance point for the chemical reaction.

K Overall

The formation constant, or \(K_{\mathrm{overall}}\), is a special equilibrium constant that quantifies the stability of a complex ion in solution.
In our scenario, the formula used to compute \(K_{\mathrm{overall}}\) is derived from the equilibrium reaction:
\(\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q)\rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\).
The expression is:
\[K_{\mathrm{overall}}=\frac{[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]}{[\mathrm{Cu}^{2+}][\mathrm{NH}_{3}]^4}\]
By substituting the equilibrium concentrations into this formula, we've calculated the \(K_{\mathrm{overall}}\) to be approximately \(1.097 \times 10^{14}\).
A high \(K_{\mathrm{overall}}\) value suggests a highly stable complex ion, showing that at equilibrium, the species exist largely as the complex rather than dissociated into \(\mathrm{Cu}^{2+}\) and \(\mathrm{NH}_3\) reactants. This indicates the strong affinity between copper ions and ammonia.