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

Step by step solution

01

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

02

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.

03

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. \[ 5.5 \times 10^8 = \frac{(0.20 + x)}{x(3.0 - 6x)^6} \]

04

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. \[ x = \frac{0.20}{(5.5 \times 10^8)(3.0)^6} \] \[ x \approx 1.5 \times 10^{-11} \]

05

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.

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