Question

A solution is prepared by mixing \(100.0 \mathrm{mL}\) of \(1.0 \times 10^{-4} \mathrm{M}\) \(\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{mL}\) of \(8.0 M \mathrm{NaF}\). $$\mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}^{+}(a q) \quad K_{1}=7.9 \times 10^{4}$$ $$\mathrm{BeF}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{2}(a q) \quad K_{2}=5.8 \times 10^{3}$$ $$\operatorname{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{3}^{-}(a q) \quad K_{3}=6.1 \times 10^{2}$$ $$\mathrm{BeF}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{4}^{2-}(a q) \quad K_{4}=2.7 \times 10^{1}$$ Calculate the equilibrium concentrations of \(\mathrm{F}^{-}, \mathrm{Be}^{2+}, \mathrm{BeF}^{+}\) \(\mathrm{BeF}_{2}, \mathrm{BeF}_{3}^{-},\) and \(\mathrm{BeF}_{4}^{2-}\) in this solution.

Short answer

The equilibrium concentrations of the species are: [F⁻] ≈ 3.99987 M [Be²⁺] ≈ 0.490125 x 10⁻⁴ M [BeF⁺] ≈ 9.875 x 10⁻⁶ M [BeF₂] ≈ 8.415 x 10⁻⁹ M [BeF₃⁻] ≈ 3.212 x 10⁻¹¹ M [BeF₄²⁻] ≈ 8.673 x 10⁻¹³ M

Step-by-step solution

Write down the expressions for each ion complex formation reaction

The four equilibrium reactions are given by: \(1): \mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}^{+}(a q) \quad K_{1}=7.9 \times 10^{4}\) \(2): \mathrm{BeF}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{2}(a q) \quad K_{2}=5.8 \times 10^{3}\) \(3): \operatorname{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{3}^{-}(a q) \quad K_{3}=6.1 \times 10^{2}\) \(4): \mathrm{BeF}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{4}^{2-}(a q) \quad K_{4}=2.7 \times 10^{1}\)

Calculate the initial concentrations of the ions involved in the reactions

We are given the initial concentrations of Be(NO3)₂ (1.0 x 10⁻⁴ M) and NaF (8.0 M). Therefore, the initial concentrations of Be²⁺ and F⁻ are: [Be²⁺]₀ = 1.0 x 10⁻⁴ M [F⁻]₀ = 8.0 M Since the solution volume doubles, the concentrations will be halved: [Be²⁺]₀ = 0.5 x 10⁻⁴ M [F⁻]₀ = 4.0 M

Use the expressions and initial concentrations to solve for the equilibrium concentrations

Since the equilibria occur sequentially, the concentration of the successive anions will be lower than the previous anion. Let [F⁻] = [F⁻]₀ - x - 2y - 3z, where x, y, z are the concentrations of BeF⁺, BeF₂, and BeF₃⁻, respectively. The concentration of Be²⁺, BeF⁺, BeF₂, BeF₃⁻ and BeF₄²⁻ decrease successively and we can approximate: \[x << y, y << z, z << [F⁻]\] Using the approximation, we can find x, y, and z from the equilibrium constants: \(K_{1} = \frac{[BeF^{+}][F^{-}]}{[Be^{2+}]} = \frac{x(4 - x)}{0.5\times10^{-4}}\) \(K_{2} = \frac{[BeF_{2}][F^{-}]}{[BeF^{+}]} = \frac{y(4 - x)}{x}\) \(K_{3} = \frac{[BeF_{3}^{-}][F^{-}]}{[BeF_{2}]} = \frac{z(4 - x)}{y}\) From equation (1): x ≈ \(\frac{0.5\times10^{-4} \times K_1}{4}\) From equation (2): y ≈ \(\frac{x \times K_2}{4}\) From equation (3): z ≈ \(\frac{y \times K_3}{4}\) Now calculate x, y, and z: x ≈ 9.875 x 10⁻⁶ M y ≈ 8.415 x 10⁻⁹ M z ≈ 3.212 x 10⁻¹¹ M The equilibrium concentrations of each species are as follows: [F⁻] ≈ 4.0 M - x - 2y - 3z ≈ 3.99987 M [Be²⁺] ≈ 0 - x ≈ 0.5 x 10⁻⁴ M - 9.875 x 10⁻⁶ M ≈ 0.490125 x 10⁻⁴ M [BeF⁺] ≈ x ≈ 9.875 x 10⁻⁶ M [BeF₂] ≈ y ≈ 8.415 x 10⁻⁹ M [BeF₃⁻] ≈ z ≈ 3.212 x 10⁻¹¹ M [BeF₄²⁻] ≈ K_4 × z ≈ 8.673 x 10⁻¹³ M These are the equilibrium concentrations of the species in the solution.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is vital for interpreting reaction dynamics and predicting the outcome of a chemical process. In a chemical reaction, chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. At this point, the reaction proceeds in both directions at the same rate, and the concentrations of all species involved become constant, although not necessarily equal.

Having a clear grasp of this allows students to predict the stability of a reaction mixture over time. It's essential to know that reaching equilibrium doesn't imply that reactants have been fully converted to products or vice versa; instead, a specific ratio determined by the equilibrium constant is established. This knowledge is particularly useful when dealing with exercises like calculating the equilibrium concentrations of various species in a complex ion formation reaction.

Complex Ion Formation

Complex ion formation is a fascinating aspect of chemistry involving the combination of simple ions or molecules to form a larger, charged complex. This process is commonly observed in transition metal chemistry, where a central metal ion binds to several ligands creating a complex ion. The exercise provided involves the stepwise formation of beryllium fluoride complexes through sequential reactions.

Each step in the formation of complex ions is itself a reversible reaction that can achieve chemical equilibrium. As the beryllium ion reacts with fluoride ions in stages, progressively larger and more complex beryllium-fluoride entities are produced. Complex ion formation is a valuable topic to understand as it is pertinent to many areas of chemistry, including biochemistry, environmental chemistry, and industrial processes. By learning about this process, students gain insight into how metal ions in solution can impact various chemical properties and behaviors.

Equilibrium Constant

The equilibrium constant, denoted as K, quantifies the position of equilibrium for a reversible chemical reaction. This constant is a ratio of the concentration of products to reactants, each raised to the power of their respective coefficients in the balanced chemical equation, taken at the point of equilibrium.

For the complex ion formation reactions described in the exercise, an equilibrium constant (K) is associated with each step of the reaction sequence. These constants are vital for calculating the equilibrium concentrations of reactants and products. The exercise showcases successive equilibria, where the product of one reaction serves as a reactant for the next. The equilibrium constants help in making reasonable approximations to simplify the calculations (e.g., assuming the concentrations of intermediate complexes are negligible compared to the fluoride ion). By mastering this concept, students can determine the direction in which a reaction will proceed under certain conditions and predict the concentrations of different species at equilibrium.