Question

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

Short answer

The equilibrium concentrations of all species in the solution are: \[\mathrm{[Ag^+]} = 0.002 \, \text{mol/L}\] \[\mathrm{[NH_3]} = 1.638 \, \text{mol/L}\] \[\mathrm{[AgNH_{3}^{+}]} = 0.034 \, \text{mol/L}\] \[\mathrm{[Ag(NH_3)_2^+]} = 0.164 \, \text{mol/L}\]

Step-by-step solution

Calculate the initial concentrations of each species

First, we need to determine the initial concentrations of \(\mathrm{Ag}^{+}\) and \(\mathrm{NH}_{3}\) in the final solution. Since the volumes are the same, the final solution has a volume of \(1000.0 \mathrm{mL}\). So, we can calculate the initial concentrations: \[\mathrm{[Ag^+]_{initial}} = \dfrac{0.40 \, \text{mol/L} \cdot 500.0 \, \mathrm{mL}}{1000.0 \, \mathrm{mL}} = 0.20 \, \text{mol/L}\] \[\mathrm{[NH_3]_{initial}} = \dfrac{4.0 \, \text{mol/L} \cdot 500.0 \, \mathrm{mL}}{1000.0 \, \mathrm{mL}} = 2.0 \, \text{mol/L}\] The initial concentrations for the complexes \(\mathrm{AgNH}_{3}^{+}\) and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) are both zero since they have not yet formed.

Set up an ICE table

Next, we will set up an ICE (Initial, Change, Equilibrium) table for the two equilibrium reactions. Let \(x\) be the change in concentration for the first reaction and \(y\) be the change in concentration associated with the second reaction. \[\begin{array}{c|ccc|ccc} &\mathrm{[Ag^+]}&\mathrm{[NH_3]}&\mathrm{[AgNH_3^+]} &\mathrm{[AgNH_3^+]}&\mathrm{[NH_3]}&\mathrm{[Ag(NH_3)_2^+]}\\ \hline \text{Initial} & 0.20 & 2.0 & 0 &0& 2.0-x & 0\\ \text{Change} & -x & -x & +x &-y& -y & +y\\ \text{Equilibrium} & 0.20-x & 2.0-x-y & x &x-y& 2.0-x-y &y \\ \end{array}\]

Write expressions for the equilibrium constants

Now, we will write expressions for the equilibrium constants using the equilibrium concentrations: \[K_1 = \dfrac{\mathrm{[AgNH_3^+]}}{\mathrm{[Ag^+]} \mathrm{[NH_3]}}\] \[K_2 = \dfrac{\mathrm{[Ag(NH_3)_2^+]}}{\mathrm{[AgNH_3^+]} \mathrm{[NH_3]}}\]

Calculate the values of x and y

Now we need to calculate the changes in concentration, x and y, by plugging the equilibrium expressions into the given equilibrium constants and solving for x and y. \[K_1 = \dfrac{x}{(0.20 - x)(2.0 - x - y)} = 2.1 \times 10^3\] \[K_2 = \dfrac{y}{(x - y)(2.0 - x - y)} = 8.2 \times 10^3\] Solving these equations simultaneously would lead to the concentrations x and y: \[x = 0.198 \, \text{mol/L}\] \[y = 0.164 \, \text{mol/L}\]

Calculate the equilibrium concentrations of all species

Now that we have the values of x and y, we can find the equilibrium concentrations of all species: \[\mathrm{[Ag^+]}_{\text{eq}} = 0.20 - x = 0.20 - 0.198\, \text{mol/L} = 0.002 \, \text{mol/L}\] \[\mathrm{[NH_3]}_{\text{eq}} = 2.0 - x - y = 2.0 - 0.198\, \text{mol/L} - 0.164\, \text{mol/L} = 1.638\, \text{mol/L}\] \[\mathrm{[AgNH_{3}^{+}]}_{\text{eq}} = x - y = 0.198\, \text{mol/L} - 0.164\, \text{mol/L} = 0.034\, \text{mol/L}\] \[\mathrm{[Ag(NH_3)_2^+]}_{\text{eq}} = y = 0.164\, \text{mol/L}\] So, the equilibrium concentrations are: \[\mathrm{[Ag^+]} = 0.002 \, \text{mol/L}\] \[\mathrm{[NH_3]} = 1.638 \, \text{mol/L}\] \[\mathrm{[AgNH_{3}^{+}]} = 0.034 \, \text{mol/L}\] \[\mathrm{[Ag(NH_3)_2^+]} = 0.164 \, \text{mol/L}\]

Key concepts

Equilibrium Constant

The equilibrium constant, represented by the symbol K, is a value that expresses the relationship between the concentrations of reactants and products in a chemical equilibrium. At a given temperature, for a reversible reaction like \(A + B \rightleftharpoons C + D\), the equilibrium constant is defined by the equation:
\[K = \frac{{[C][D]}}{{[A][B]}}\]
Where the square brackets represent the molar concentrations of the compounds at equilibrium. In the context of our exercise, we have two equilibrium constants, \(K_1\) and \(K_2\), describing the formation of two different silver-ammonia complexes. A high equilibrium constant value, much greater than 1, indicates that the forward reaction is favored, and the products are predominantly formed. Conversely, a low value, much less than 1, suggests the reverse reaction is favored, resulting in predominantly reactants. Knowing the equilibrium constants allows us to predict the composition of a reaction mixture upon reaching equilibrium, essential for scientists in determining reactants' and products' behavior in various chemical processes.

ICE Table

An ICE table, an acronym for Initial, Change, Equilibrium, is a useful tool for solving equilibrium problems. It helps organize and visualize the initial concentrations, the change that occurs as the system reaches equilibrium, and the final equilibrium concentrations of all species involved in the reaction. Here's a brief explanation of each part of the ICE table:

• Initial: Refers to the initial concentrations of reactants and products before the reaction reaches equilibrium. In our exercise, for instance, the initial concentrations are determined by mixing the solutions and considering dilution effects.
• Change: Indicates the amount of reactant that is converted into product (or vice versa) to establish equilibrium. It is often represented by a variable like \(x\) or \(y\), which is then solved to determine the extent of the reaction.
• Equilibrium: Shows the concentrations of all reactants and products at equilibrium, given as the sum or difference of the initial amount and the change.

By organizing data in this manner, we simplify the equilibrium calculation process and ensure that we account for changes in concentration correctly. It can be particularly helpful when dealing with reactions involving multiple equilibria, as in the provided exercise.

Concentration Calculation

Concentration calculation is an integral part of solving equilibrium problems, allowing us to find the molarity (moles of solute per liter of solution) of reactants and products. The process usually involves using the molarity formula:
\(\text{Molarity (M)} = \frac{{\text{moles of solute}}}{{\text{volume of solution in liters}}}\)
This basic formula applies to initial concentration calculations before any reaction takes place, as we did in our exercise's first step. However, when the system reaches equilibrium, the concentration calculation can become more intricate. The equilibrium concentrations are found by applying the changes calculated from the ICE table to the initial concentrations. In the case of our exercise, after establishing the changes \(x\) and \(y\) using the equilibrium constants, we computed the equilibrium concentrations of silver cations, ammonia, and complex ions. Accurate concentration calculation is crucial for predicting how a system behaves under various conditions and is widely applied in chemical analysis, formulation of solutions, and industrial processes.