The following equilibria were attained at \(298 \mathrm{~K}:\) $$\begin{array}{c} \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s) \quad K_{c}=5.6 \times 10^{9} \\ \mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q) \\ K_{c}=1.6 \times 10^{7}\end{array}$$ Based on these equilibria, calculate the equilibrium constant $\text { for } \mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)$ at \(298 \mathrm{~K}\).

Let the given equilibria be Reaction 1 and Reaction 2: Reaction 1: \(\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s), K_{c1}=5.6 \times 10^{9}\) Reaction 2: \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q), K_{c2}=1.6 \times 10^{7}\) Required equilibrium: \(\mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)\) To combine Reaction 1 and Reaction 2 to get the required equilibrium, we need to reverse Reaction 1 and multiply Reaction 2 by 1, since the stoichiometry of the reactions is maintained throughout the process. The new combined reaction will be"> Reverse Reaction 1: \(\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q) + \mathrm{Cl}^{-}(a q)\) Multiply Reaction 2 by 1: \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q)\) Combine these to get the required equilibrium: \(\mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)\) This new equilibrium is what we were asked to find, so now let's manipulate the equilibrium constants.

When we reverse a reaction, we must take the reciprocal of its equilibrium constant. Additionally, when we multiply a reaction by a factor, we raise its equilibrium constant to the power of the same factor. Since we reversed Reaction 1 and multiplied Reaction 2 by 1 (which doesn't affect the equilibrium constant), we get the new equilibrium constants: Reverse Reaction 1: \(K_{c1'} = \frac{1}{K_{c1}} = \frac{1}{5.6 \times 10^{9}}\) Now, multiply the two equilibrium constants together: Combined equilibrium constant: \(K_{c} = K_{c1'} \times K_{c2} = \frac{1}{5.6 \times 10^{9}} \times 1.6 \times 10^{7}\)

Finally, let's multiply the two constants together and find the combined equilibrium constant: \(K_c = \frac{1.6 \times 10^{7}}{5.6 \times 10^{9}}\) \(K_c = 2.857 \times 10^{-3}\) So, the equilibrium constant for the required equilibrium: \(\mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)\) at \(298 \mathrm{~K}\) is approximately \(2.857 \times 10^{-3}\).