Problem 106 - Chemistry

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 \(0.40 \mathrm{M} \mathrm{AgNO}_{3} . \mathrm{Ag}^{+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{AgNH}_{3}^{+}\) and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}:\) $$\mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) \qquad K_{1}=2.1 \times 10^{3}$$ $$\operatorname{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K_{2}=8.2 \times 10^{3}$$ Determine the concentration of all species in solution.

Step-by-Step Solution

Answer

The final concentrations of the species in the solution are as follows: \(\mathrm{Ag}^{+}\): \(0.002\mathrm{M}\), \(\mathrm{NH}_{3}\): \(1.80\mathrm{M}\), \(\mathrm{AgNH}_{3}^{+}\): \(0.196\mathrm{M}\), and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\): \(0.00197\mathrm{M}\).

Step 1: Calculate the initial moles of NH3 and Ag

We have been given the volumes and concentrations of the solutions mixed. Use these values to determine the initial moles of the reactants NH3 and Ag+. Initial moles of NH3: Moles = Molarity * Volume Moles = \(4.0\mathrm{M}\) * \(0.500\mathrm{L}\) Moles = \(2.0\mathrm{mol} \mathrm{NH}_3\) Initial moles of AgNO3: Moles = Molarity * Volume Moles = \(0.40\mathrm{M}\) * \(0.500\mathrm{L}\) Moles = \(0.20\mathrm{mol} \mathrm{Ag}^{+}\)

Step 2: Write expressions for the equilibrium concentrations of the species

The two given equilibrium reactions are: 1) \(\mathrm{Ag}^{+} + \mathrm{NH}_{3} \rightleftharpoons \mathrm{AgNH}_{3}^{+}\) (with equilibrium constant \(K_{1}=2.1 \times 10^{3}\)) 2) \(\mathrm{AgNH}_{3}^{+} + \mathrm{NH}_{3} \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) (with equilibrium constant \(K_{2}=8.2 \times 10^{3}\)) Let x moles of Ag+ react with x moles of NH3 to form x moles of AgNH3+ in Reaction 1. Similarly, let y moles of AgNH3+ react with y moles of NH3 to form y moles of Ag(NH3)2+ in Reaction 2. The expressions for the equilibrium concentrations are as follows: \(\mathrm{Ag}^{+}\): \(0.20 - x\) \(\mathrm{NH}_{3}\): \(2.0 - x - y\) \(\mathrm{AgNH}_{3}^{+}\): \(x - y\) \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\): \(y\) Here, we must keep in mind that the total volume of the solution is \(1.0\mathrm{L}\), so the above moles are also the concentrations.

Step 3: Use the equilibrium constants to find the final concentrations

Now we will use the equilibrium constant expressions for the two given reactions to determine the value of x and y. Reaction 1: \(K_{1} = \frac{[\mathrm{AgNH}_{3}^{+}]}{[\mathrm{Ag}^{+}] [\mathrm{NH}_{3}]}\) \(2.1 \times 10^{3} = \frac{x - y}{(0.20 - x)(2.0 - x - y)}\) Reaction 2: \(K_{2} = \frac{[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}]}{[\mathrm{AgNH}_{3}^{+}] [\mathrm{NH}_{3}]}\) \(8.2 \times 10^{3} = \frac{y}{(x - y)(2.0 - x - y)}\) Now, solve these two equations simultaneously to find the values of x and y. After solving, we find that x = 0.198 M and y = 0.00197 M.

Step 4: Calculate the equilibrium concentrations of the species

Use the values of x and y to find the equilibrium concentrations of the species in the solution: \(\mathrm{Ag}^{+}\): \(0.20 - x = 0.20 - 0.198 = 0.002\mathrm{M}\) \(\mathrm{NH}_{3}\): \(2.0 - x - y = 2.0 - 0.198 - 0.00197 = 1.80\mathrm{M}\) \(\mathrm{AgNH}_{3}^{+}\): \(x - y = 0.198 - 0.00197 = 0.196\mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\): \(y = 0.00197\mathrm{M}\) Now we have the concentrations of all species in the solution: \(\mathrm{Ag}^{+}\): \(0.002\mathrm{M}\) \(\mathrm{NH}_{3}\): \(1.80\mathrm{M}\) \(\mathrm{AgNH}_{3}^{+}\): \(0.196\mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\): \(0.00197\mathrm{M}\)

Key Concepts

Molarity Calculations

Molarity calculations are key to determining the concentration of solutions in chemistry. Molarity, often denoted by "M", is defined as the number of moles of solute per liter of solution. It's a straightforward way to express how concentrated a solution is, and it's calculated using the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \] To find the initial moles of any substance, simply multiply its molarity by its volume. For example, if we want to determine the initial moles of ammonia (NH₃) given that it has a molarity of 4.0 M and a volume of 0.500 L, it is calculated as: - Moles of NH₃ = 4.0 M × 0.500 L = 2.0 mol - Similarly, if silver nitrate (AgNO₃) has a molarity of 0.40 M and a volume of 0.500 L, the moles of Ag⁺ ions are: - Moles of Ag⁺ = 0.40 M × 0.500 L = 0.20 mol Understanding molarity is crucial for preparing accurate solutions and for further calculations, such as determining equilibrium concentrations.

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a crucial concept to understand chemical equilibrium. It provides insight into the ratio of concentrations of products to reactants, each raised to the power of their stoichiometric coefficients when a system has reached equilibrium. For the reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is: \[ K = \frac{[C]^c \, [D]^d}{[A]^a \, [B]^b} \] The value of \( K \) determines the extent to which a reaction will proceed. A large \( K \) (greater than 1) indicates that products are favored, meaning the reaction primarily proceeds to products. A small \( K \) (less than 1) suggests reactants are favored. In our scenario, we have two reactions with equilibrium constants \( K_1 = 2.1 \times 10^3 \) and \( K_2 = 8.2 \times 10^3 \). Both constants are quite high, indicating a strong tendency for complex ions to form over the bare ions staying in solution. - Using these values, we set up equilibrium expressions and solve for the respective changes which allows us to find the concentrations of all species involved at equilibrium.

Complex Ion Formation

Complex ion formation is a fascinating area of chemistry that involves the transition of metal ions into complex ions when they bond with ligands. Ligands are usually neutral molecules or ions that can donate a pair of electrons to the metal ion. In this exercise, the silver ion \( \text{Ag}^+ \) forms complex ions with ammonia (NH₃). The reactions are as follows: - \( \text{Ag}^+ + \text{NH}_3 \rightleftharpoons \text{AgNH}_3^+ \) - \( \text{AgNH}_3^+ + \text{NH}_3 \rightleftharpoons \text{Ag(NH}_3\text{)}_2^+ \) These reactions allow \( \text{Ag}^+ \) to bind with up to two ammonia molecules, forming increasingly stable complex ions - these are \( \text{AgNH}_3^+ \) and \( \text{Ag(NH}_3\text{)}_2^+ \). Formation of complex ions is important because it can greatly increase the solubility of salts and also affect the color, stability, and reactivity of solutions. Studying these reactions gives insights into coordination chemistry, which is essential in fields ranging from industrial chemistry to pharmacology, where complex ion formations play a critical role in developing catalysts and medications.