Question

Given $$ \begin{aligned} \mathrm{HF}(a q) & \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) & K_{\mathrm{a}} &=6.9 \times 10^{-4} \\\ \mathrm{HF}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{HF}_{2}^{-}(a q) & K &=2.7 \end{aligned} $$ calculate \(K\) for the reaction $$ 2 \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HF}_{2}^{-}(a q) $$

Short answer

Question: Calculate the equilibrium constant for the reaction \(2 \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HF}_{2}^{-}(a q)\), given the following reactions and their respective equilibrium constants: 1. \(\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\) with \(K_{\mathrm{a}}=6.9 \times 10^{-4}\) 2. \(\mathrm{HF}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}_{2}^{-}(a q)\) with \(K=2.7\) Answer: The equilibrium constant for the given reaction is approximately 3913.

Step-by-step solution

Write down the given equations and constants

We are given the following equilibrium reactions and their respective constants: Equation 1: \(\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\), with \(K_{\mathrm{a}}=6.9 \times 10^{-4}\) Equation 2: \(\mathrm{HF}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}_{2}^{-}(a q)\), with \(K=2.7\) We need to find the equilibrium constant, \(K_{\text{new}}\), for the following reaction: Equation 3: \(2 \mathrm{HF}(a q) \rightleftharpoons\mathrm{H}^{+}(a q)+\mathrm{HF}_{2}^{-}(a q)\).

Combine equations 1 and 2 to form equation 3

To form equation 3, we will add equation 2 to equation 1. In order to have only one \(\mathrm{F}^{-}\) ion in the product, we will first invert equation 1 and then add it to equation 2. Inverted Equation 1: \(\mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}(a q)\), with \(K_{\mathrm{a}}^{-1}\) Sum of Equations: \(\mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) + \mathrm{HF}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}(a q) + \mathrm{HF}_{2}^{-}(a q)\) Now, cancel \(\mathrm{F}^{-}(a q)\) on both sides of the equation to get equation 3: Equation 3: \(2 \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HF}_{2}^{-}(a q)\)

Relate the equilibrium constants

Now that we have found a relationship between the reactions, we can find the corresponding relationship for the constants. When multiplying the inverted equation 1 by equation 2, we simply multiply the equilibrium constants, since they are the products of their respective concentrations: \(K_{\text{new}}= K_{\mathrm{a}}^{-1} \times K\)

Calculate \(K_{\text{new}}\)

Now, let's plug in the given values for \(K_{\mathrm{a}}\) and \(K\) to find the equilibrium constant for the new reaction: \(K_{\text{new}}= (6.9 \times 10^{-4})^{-1} \times 2.7\) \(K_{\text{new}}= \frac{2.7}{6.9 \times 10^{-4}}\) \(K_{\text{new}} \approx 3913\) The equilibrium constant for the reaction \(2 \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HF}_{2}^{-}(a q)\) is approximately 3913.

Key concepts

Chemical Equilibrium

In chemistry, chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of the reactants and products over time. It is a dynamic state where the concentrations of all reactants and products remain constant, but both reactions continue to occur.

Understanding chemical equilibrium is crucial for studying how reactions occur and predicting the outcomes. It also helps in determining the most efficient conditions for industrial processes, such as in the synthesis of ammonia in the Haber process. In educational exercises, the concept is often illustrated using reversible reactions, where the formation of reactants from products is just as feasible as the production of products from reactants. When students calculate equilibrium constants, they're applying this concept to find a numerical value that represents the ratio of product concentrations to reactant concentrations at equilibrium.

Equilibrium Constant

The equilibrium constant (K) quantifies the balance of reactants and products in a chemical reaction at equilibrium. It is a constant value that is dependent solely on temperature for a particular reaction. The equilibrium constant is derived from the law of mass action, which states that for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations raised to the power of their stoichiometric coefficients will always be constant.

The value of the equilibrium constant provides insights into the reaction mixture at equilibrium. A high K value indicates a greater amount of products, signifying the reaction favors the products. Conversely, a low K value means the reaction mixture contains mainly reactants. It's essential for students to know how to calculate the equilibrium constant from known concentrations of reactants and products as it's critical for predicting how far a reaction will proceed under specific conditions.

Acid Dissociation Constant

The acid dissociation constant (Ka) is a particular type of equilibrium constant that measures the strength of an acid in solution. It specifically applies to the dissociation (separation) of an acid into its ions. The smaller the value of Ka, the weaker the acid as it does not dissociate significantly in solution. In contrast, a larger Ka value denotes a stronger acid that dissociates to a greater extent.

For example, in the reaction HF ⟺ H+ + F-, Ka represents how much HF dissociates into H+ and F- at equilibrium. When students are working on exercises involving Ka, it is crucial for them to understand that it provides a measure of the tendency of an acid to lose a proton. The calculation of Ka can also involve the use of the pH of the solution and the concentrations of the acid and its ions, further emphasizing the importance of equilibrium principles in understanding acid-base chemistry.