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 $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

Step by step solution

01

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)

02

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: \(5.5 \times 10^{8} = \frac{[Ni(NH_{3})_{6}^{2+}]}{[Ni^{2+}][NH_{3}]^{6}}\)

03

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: \(5.5 \times 10^{8} = \frac{(0.20 - x)}{x(3 - 6x)^{6}}\)

04

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)

05

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.

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