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 \(\begin{aligned} \mathrm{AgNH}_{3}{ }^{+} \text {and } \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+: \\ \mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}{ }^{+}(a q) & K_{1}=2.1 \times 10^{3} \\ \mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) & K_{2}=8.2 \times 10^{3} \end{aligned}\) Determine the concentration of all species in solution.

Short answer

The equilibrium concentrations of the species in the solution are approximately: \[ \mathrm{[NH_{3}]} \approx 1.801 \mathrm{~M} \] \[ \mathrm{[Ag^{+}]} \approx 0.020 \mathrm{~M} \] \[ \mathrm{[AgNH_{3}^{+}]} \approx 0.18 \mathrm{~M} \] \[ \mathrm{[Ag(NH_{3})_{2}^{+}]} \approx 0.019 \mathrm{~M} \]

Step-by-step solution

Calculate the initial concentration of the species

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.

Create an ICE table

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

Write the expressions for the equilibrium constants and solve for x and 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

Calculate the equilibrium concentrations

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

Key concepts

Understanding ICE Tables

An ICE table is a handy tool used in chemistry to track changes in concentrations during a chemical reaction. It's called "ICE" because it stands for Initial, Change, and Equilibrium. Each part of the table helps you figure out what's happening in the reaction.

Here's how it works:

• **Initial:** Start with the concentrations of reactants and products before the reaction begins. In our example, the initial concentrations of \([NH_3]\) and \([Ag^+]\) are calculated based on given molarities and volumes.
• **Change:** This shows how much of each substance is used up or created. We represent these changes with variables like \(x\) and \(y\), which reflect the shifts due to the reaction process.
• **Equilibrium:** Finally, we determine the concentrations once the reaction has finished reaching equilibrium. By using the initial values and the changes, we can identify the amounts present at equilibrium.

Using the ICE table simplifies the understanding of complex reactions. It organizes your calculations and clarifies how equilibrium concentrations are achieved from initial conditions.

Molarity Calculations Made Easy

Molarity is a way to express the concentration of a solution. It's calculated by taking the number of moles of solute and dividing it by the volume of the solution in liters. Here's why molarity calculations are essential:

To find initial concentrations, you can use the formula: \[ C_i = \frac{{C_1 V_1 + C_2 V_2}}{{V_t}} \]
Where:

• \(C_1\), \(C_2\) are the initial molarities of your solutions.
• \(V_1\), \(V_2\) are the initial volumes.
• \(V_t\) is the total volume after mixing.

For example, in our exercise, we calculate the initial molarities of \(NH_3\) and \(Ag^+\) after mixing equal volumes of two solutions.

These calculations are crucial as they set up the values you need for the ICE table. Once you know the initial concentrations, you can easily follow the reaction's progress and find the equilibrium concentrations. Molarity calculations lay the groundwork for understanding chemical reactions by providing all the initial data you need!

Diving into Chemical Equilibrium

Chemical equilibrium is the point in a reaction when the rate of the forward reaction equals that of the reverse reaction. It might sound complex, but let's break it down into simpler terms.

In chemical equilibrium:

• **Balanced Rates:** The concentrations of reactants and products stay constant, not because the reactions have stopped, but because the forward and reverse reactions happen at the same rate.
• **Equilibrium Constants (K):** These constants (like \(K_1\) and \(K_2\) in our example) help quantify the ratios of products to reactants at equilibrium. Each constant is specific to a particular reaction at a given temperature.

For instance, when calculating the equilibrium concentrations in our exercise, we used given values for \(K_1\) and \(K_2\) to establish the relationships between different species in the reaction. This helped us find the concentrations of each species at equilibrium.

Understanding chemical equilibrium is key to predicting how a chemical system behaves over time, helping you solve many real-world chemical problems.