Question

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 \(\mathrm{AgNH}_{3}^{+}\) and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}:\) \(\mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) \quad K_{1}=2.1 \times 10^{3}\) \(\mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K_{2}=8.2 \times 10^{3}\)

Short answer

The equilibrium concentrations are: - [Ag+] = 0.034 M - [NH₃] = 1.834 M - [AgNH₃⁺] = 0.166 M - [Ag(NH₃)₂⁺] = 0.166 M

Step-by-step solution

Calculate initial concentrations

After mixing the two solutions, the total volume is 1000.0 mL. Use the dilution formula to find the initial concentrations of Ag+ (from AgNO3) and NH3 in the mixture. Initial concentration of Ag+: \(\frac{C_1V_1}{V_t} = \frac{(0.40~M)(500.0~mL)}{1000.0~mL} = 0.20~M\) Initial concentration of NH3: \(\frac{C_2V_2}{V_t} = \frac{(4.0~M)(500.0~mL)}{1000.0~mL} = 2.0~M\) The initial concentrations are: - [Ag+] = 0.20 M - [NH3] = 2.0 M Since there is no complex initially formed in the solution, the initial concentrations are: - [AgNH3+] = [Ag(NH3)2+] = 0 M

Set up the reaction quotient expressions and ICE table

Now, let's express the equilibrium constant for each reaction: \(K_1 = \frac{[AgNH_3^+]}{[Ag^+][NH_3]}\) \(K_2 = \frac{[Ag(NH_3)_2^+]}{[AgNH_3^+][NH_3]}\) Set up an ICE table to find the change in concentrations: \[ \begin{array}{c|c|c|c} & [Ag^+] & [NH_3] & [AgNH_3^+] & [Ag(NH_3)_2^+] \\ \hline \text{Initial} & 0.20 & 2.0 & 0 & 0 \\ \text{Change} &{-x} &{-x} & {+x} & {+x}\\ \text{Equilibrium} & 0.20-x & 2.0-x & x& x \\ \end{array} \]

Solve for the equilibrium concentrations

Now, plug the expressions from the ICE table back into the equilibrium constant expressions for each reaction: \(K_1 = \frac{x}{(0.20-x)(2.0-x)}\) \(K_2 = \frac{x}{x(2.0-x)}\) Solving the first reaction for x gives: \(x = \frac{K_1(0.20-x)(2.0 - x)}{1 + K_1(0.20-x)}\) Solving this equation with the given value of \(K_1\) for x, we get: \(x = 0.166~M\) Plug this value of x back into the second equilibrium expression and solve for the equilibrium concentrations: \([Ag^+]_{eq} = 0.20 - x = 0.034~M\) \([NH_3]_{eq} = 2.0 - x = 1.834~M\) \([AgNH_3^+]_{eq} = [Ag(NH_3)_2^+]_{eq} = x = 0.166~M\) So, the equilibrium concentrations are: - [Ag+] = 0.034 M - [NH3] = 1.834 M - [AgNH3+] = 0.166 M - [Ag(NH3)2+] = 0.166 M

Key concepts

Equilibrium Constant

Understanding the equilibrium constant (K) is crucial when studying chemical reactions that reach a state of balance. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the reactants and products remain constant over time. The equilibrium constant is a quantitative measure of the ratio of the concentrations of products to reactants, raised to the power of their stoichiometric coefficients.
For the reactions given in the exercise—where silver ions react with ammonia to form different complexes—the equilibrium constants are denoted as K1 and K2. These values indicate how far the reactions will proceed before reaching equilibrium. A larger K suggests a reaction that will favor the formation of products under standard conditions. In the example provided, both K1 and K2 are greater than 1, indicating the formations of \(AgNH_3^+\) and \(Ag(NH_3)_2^+\) are favored.
Here’s what the equilibrium constant expressions look like for our silver-ammonia system:

• For \(Ag^+ + NH_3 \rightleftharpoons AgNH_3^+\): \(K_1 = \frac{[AgNH_3^+]}{[Ag^+][NH_3]}\)
• For \(AgNH_3^+ + NH_3 \rightleftharpoons Ag(NH_3)_2^+\): \(K_2 = \frac{[Ag(NH_3)_2^+]}{[AgNH_3^+][NH_3]}\)

To predict the direction in which a reaction will proceed—forward or reverse—compare K to the reaction quotient, Q.

Reaction Quotient

To understand how a system not at equilibrium will shift, we use the reaction quotient (Q). It is calculated in the same way as the equilibrium constant but with initial, not equilibrium, concentrations. For a reaction \(aA + bB \rightleftharpoons cC + dD\), the reaction quotient is \(Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}\) where the concentration terms are those at a given point in time, not necessarily at equilibrium.
By comparing Q to the equilibrium constant K, we can predict the direction in which the reaction will proceed:

• If \(Q < K\), the reaction will proceed forward to produce more products.
• If \(Q > K\), the reaction will proceed in reverse to produce more reactants.
• If \(Q = K\), the system is at equilibrium, and no net change will occur.

In the exercise, since the initial concentrations of the complexes are zero, the reaction quotient Q would also start at zero. Given that \(K > 0\), the reaction will proceed forward to establish equilibrium.

ICE Table

An ICE (Initial, Change, Equilibrium) table is a systematic way to organize data about the concentration changes in a reaction over time. It helps in solving for equilibrium concentrations without requiring extensive algebra. To complete an ICE table, begin by filling in the initial concentrations (I) of reactants and products. Then determine the changes (C) that need to occur to reach equilibrium. Finally, calculate the equilibrium concentrations (E) by combining the initial amounts and changes for each species involved in the reaction.
In this exercise, an ICE table was used to calculate how much of the reactants will convert into products. The 'Change' row uses the variable \(x\) to represent the amount of reactants consumed and products formed. By applying the equilibrium constants, we solve for \(x\), giving us the concentrations of reactants and products at equilibrium. For the given reactions involving \(Ag^+\) and \(NH_3\), the ICE table helps us understand the stoichiometry and predicts that both \(AgNH_3^+\) and \(Ag(NH_3)_2^+\) will be formed in significant quantities, validating the equilibrium concepts presented throughout the solution.