Question

A solution is prepared by adding \(0.10 \mathrm{~mol} \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) 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. \(K_{\text {owerall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{\mathrm{x}}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ 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 Ni(NH3)6^2+ and Ni^2+ are approximately 0.1976 M and 2.42 x 10^(-3) M, respectively.

Step-by-step solution

Find the initial concentrations of Ni(NH3)6^2+ and NH3

From the given data, 0.10 mol of Ni(NH3)6Cl2 is added to 0.50 L of 3.0 M NH3 solution. So, the initial concentrations of both Ni(NH3)6^2+ and NH3 can be calculated as follows: -- For Ni(NH3)6^2+: Initial concentration of Ni(NH3)6^2+ = (0.10 mol) / (0.50 L) = 0.20 M -- For NH3: Initial concentration of NH3 = 3.0 M (since no reaction happened yet)

Write the equilibrium expression for the reaction

The given value of the overall equilibrium constant is 5.5 x 10^8. Therefore, the equilibrium expression for the given reaction is: \(5.5 \times 10^{8} = \frac{[Ni(NH_{3})_{6}^{2+}]}{[Ni^{2+}][NH_{3}]^{6}}\)

Solve for Ni(NH3)6^2+ and Ni^2+ using the equilibrium expression

Let's denote the change in the Ni^2+ concentration and the Ni(NH3)6^2+ concentration at equilibrium as "x". Then: [Ni(NH3)6^2+] = 0.20 - x [Ni^2+] = x [NH3] = 3 - 6x (because 6 moles of NH3 reacts with each mole of Ni^2+) Substitute these expressions into the equilibrium constant equation: \(5.5 \times 10^{8} = \frac{(0.20 - x)}{x(3 - 6x)^{6}}\)

Make an assumption and solve for x

Because the equilibrium constant is very large, we can make an assumption that x is small compared to 0.20 and 3. So, we can approximate: 0.20 - x ≈ 0.20 3 - 6x ≈ 3 Now, we can simplify the equilibrium equation: \(5.5 \times 10^{8} = \frac{0.20}{x(3)^{6}}\) Solve for x: x ≈ 2.42 x 10^(-3)

Calculate the equilibrium concentrations

Use the value of x obtained in Step 4 to find the equilibrium concentrations of Ni(NH3)6^2+ and Ni^2+: [Ni(NH3)6^2+] ≈ 0.20 - x = 0.20 - 2.42 x 10^(-3) ≈ 0.1976 M [Ni^2+] ≈ x = 2.42 x 10^(-3) M Hence, the equilibrium concentrations of Ni(NH3)6^2+ and Ni^2+ are approximately 0.1976 M and 2.42 x 10^(-3) M, respectively.

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a vital concept in chemistry that helps us understand how chemical reactions behave over time. It gives us a quantitative measure of the concentrations of products and reactants at equilibrium.

For the reaction in the exercise, the equilibrium constant \( K_{\text{overall}} \) is specifically calculated using the formula:

• \[ K_{\text{overall}} = \frac{[\text{Ni}(\text{NH}_3)_6^{2+}]}{[\text{Ni}^{2+}][\text{NH}_3]^6} \]

This formula represents the ratio of the concentration of the complex ion \( \text{Ni}(\text{NH}_3)_6^{2+} \) to the product of the concentrations of free nickel ions \( \text{Ni}^{2+} \) and ammonia raised to the sixth power.

A larger equilibrium constant, such as the one given \( 5.5 \times 10^8 \), indicates that the reaction heavily favors the formation of the products at equilibrium. This constant is crucial in manipulating and understanding the behavior of the reactions under different initial conditions.

Chemical Solution

A chemical solution consists of a solute dissolved in a solvent. In the given exercise, we look at a solution involving nickel ammine complex ions. Here's what's happening in our scenario:

• Firstly, you have the salt \( \text{Ni}(\text{NH}_3)_6\text{Cl}_2 \), which can dissociate in an aqueous solution to give \( \text{Ni}(\text{NH}_3)_6^{2+} \).
• A solvent \( 3.0 \text{ M } \text{NH}_3 \) is present, which acts as the environment in which this initial dissociation occurs.

The significance of these solutions is anchored in the interactions that occur between the nickel ions and ammonia molecules, stabilizing the complex and altering its reactivity. This idea is central in coordination chemistry, which studies how metal ions bond with other atoms or molecules.

Initial Concentration

Initial concentration refers to the concentration of substances before any shift towards equilibrium occurs. This is vital to determine because these values form the basis of calculating equilibrium concentrations later on.

In the given chemical setup:

• The initial concentration of \( \text{Ni}(\text{NH}_3)_6^{2+} \) is calculated by dividing the moles (0.10 mol) by the volume (0.50 L), yielding \( 0.20 \text{ M} \).
• Meanwhile, ammonia \( \text{NH}_3 \) has an initial concentration set by the prepared solution, as no reaction has occurred—the value here is \( 3.0 \text{ M} \).

Understanding initial concentrations allows us to set up an equilibrium expression and make assumptions about the changes that occur as equilibrium is established. These values help put bounds on possible changes and simplify the solutions to equilibrium conditions when constants are exceedingly large.