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

Step by step solution

01

Identify moles and initial concentrations

Given, 0.10 mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\mathrm{Cl}_{2}\) is added to 0.50 L of 3.0 M \(\mathrm{NH}_{3}\). The initial concentration of ammonia can be calculated as follows: Initial moles of $\mathrm{NH}_{3} = \text{Initial concentration} \times \text{Volume} \\ \text{Initial moles of } \mathrm{NH}_{3} = 3.0 \ \text{M} \times 0.50 \ \text{L} = 1.5 \ \text{moles}$

02

Select a change table to represent the reaction

We can set up a change table (also called an ICE table) to represent the initial concentrations of the species, the change, and the equilibrium concentrations. Reaction: $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$ | Species | Initial Concentration| Change | Equilibrium Concentration | |--------------|-----------------|-------|--------------------| | \(\mathrm{Ni}^{2+}\) | 0 | +x | x | | \(\mathrm{NH}_{3}\) | 1.5 M | -6x | 1.5 - 6x | | \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) | 0 | -x | x |

03

Substitute the equilibrium concentrations in the overall equilibrium constant expression

Now, substitute the equilibrium concentrations in the overall equilibrium constant expression is \(5.5 \times 10^{8} =\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}\): $$5.5 \times 10^{8} =\frac{x}{x(1.5 - 6x)^{6}}$$

04

Solve for the concentration of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\)

Rearrange the equation: $$x = - \frac{1}{6} + \sqrt{\frac{1}{36} + 5.5 \times 10^{8}}$$ Solve for the concentration of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\): \(x = 9.975 \times 10^{-4} \ \text{M}\) Thus, the concentration of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(9.975 \times 10^{-4}\) M.

05

Solve for the equilibrium concentration of \(\mathrm{Ni}^{2+}\)

Using the change table, we find that the equilibrium concentration of \(\mathrm{Ni}^{2+}\) is also equal to the concentration of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\): Equilibrium concentration of \(\mathrm{Ni}^{2+} = x = 9.975 \times 10^{-4} \ \text{M}\) Thus, the concentration of \(\mathrm{Ni}^{2+}\) in the solution is \(9.975 \times 10^{-4}\) M.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!