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

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

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 \(K_{\text{overall}} = 5.5 \times 10^8\).

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 |

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 \(0.1901 \text{ M}\).