Consider a solution made by mixing \(500.0 \mathrm{~mL}\) of \(4.0 \mathrm{M} \mathrm{NH}_{3}\) and \(500.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{AgNO}_{3} \cdot \mathrm{Ag}^{+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\begin{aligned} \mathrm{AgNH}_{3}{ }^{+} \text {and } \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+: \\ \mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}{ }^{+}(a q) & K_{1}=2.1 \times 10^{3} \\ \mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) & K_{2}=8.2 \times 10^{3} \end{aligned}\) Determine the concentration of all species in solution.

We are given the initial volumes and molarities of NH3 and AgNO3. To find the initial concentration of the species, we can use the formula: C1V1 + C2V2 = Ct Vt Where C1, C2 are the initial molarities, V1, V2 are the initial volumes, Ct is the final molarity, and Vt is the final volume. Since both solutions have volumes of 500 mL, we can add them up to get the final volume (Vt): Final Volume (Vt) = 500 mL + 500 mL = 1000 mL (1 L) Next, we need to find the initial concentration of NH3 and Ag+: For NH3: Initial concentration = (4.0 M)(0.5 L) / (1 L) = 2.0 M For Ag+: Initial concentration = (0.40 M)(0.5 L) / (1 L) = 0.20 M The initial concentrations of AgNH3+ and Ag(NH3)2+ are zero, as these species are produced by the reactions.

An ICE table (Initial, Change, Equilibrium) shows the initial concentration, the change in concentration during the reaction, and the equilibrium concentration of all species involved in the reaction. Let's denote the initial concentrations of NH3, Ag+, AgNH3+, and Ag(NH3)2+ as [NH3]0, [Ag+]0, [AgNH3+]0, and [Ag(NH3)2+]0, respectively. Now, let's denote the change in their concentrations as x, y, and z, respectively, as per the reactions: Ag+(aq) + NH3(aq) ⇌ AgNH3+(aq) with K1 = 2.1 x 10^3 AgNH3+(aq) + NH3(aq) ⇌ Ag(NH3)2+(aq) with K2 = 8.2 x 10^3 We can now create an ICE table: Species: NH3 | Ag+ | AgNH3+ | Ag(NH3)2+ -------------------------------------------------------- Initial: 2.0 M | 0.20 M | 0 M | 0 M Change: -x-y | -x | x | y Equilibrium: 2-x-y | 0.20-x | x | y

Now, we will use the equilibrium constants (K1 and K2) to create expressions for the equilibrium concentrations of the species. K1 = [AgNH3+]/([Ag+][NH3]) = x/(0.20-x)(2-x-y) K2 = [Ag(NH3)2+]/([AgNH3+][NH3]) = y/(x(2-x-y)) Now, we can use the given values for K1 and K2 to solve for x and y: 2.1 x 10^3 = x/(0.20-x)(2-x-y) 8.2 x 10^3 = y/(x(2-x-y)) Solving these equations simultaneously for x and y, we get: x ≈ 0.18 M y ≈ 0.019 M

We can now use the values of x and y to find the equilibrium concentrations of all species: [NH3] = 2.0 - x - y = 2.0 - 0.18 - 0.019 ≈ 1.801 M [Ag+] = 0.20 - x = 0.20 - 0.18 ≈ 0.020 M [AgNH3+] = x = 0.18 M [Ag(NH3)2+] = y = 0.019 M Therefore, the equilibrium concentrations of all species in the solution are approximately: [NH3] = 1.801 M [Ag+] = 0.020 M [AgNH3+] = 0.18 M [Ag(NH3)2+] = 0.019 M