Question

Given $$\begin{array}{ll} \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) & K_{\mathrm{a}}=6.9 \times 10^{-1} \\ \mathrm{HF}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{HF}_{2}-(a q) & K=2.7 \end{array}$$ calculate \(K\) for the reaction $$ 2 \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HF}_{2}^{-}(a q) $$

Short answer

Answer: The equilibrium constant for this reaction is approximately 1.863.

Step-by-step solution

Write down the given reactions and equilibrium constants

We have two reactions: 1. \(\mathrm{HF}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{F}^{-}(aq)\) with \(K_\mathrm{a} = 6.9 \times 10^{-1}\) 2. \(\mathrm{HF}(aq) + \mathrm{F}^{-}(aq) \rightleftharpoons \mathrm{HF}_{2}^{-}(aq)\) with \(K = 2.7\) Our goal is to find the equilibrium constant for the following reaction: 3. \(2\mathrm{HF}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HF}_{2}^{-}(aq)\)

Combine the given reactions to form the desired reaction

In order to obtain the desired reaction, we can add reactions 1 and 2: 1. \(\mathrm{HF}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{F}^{-}(aq)\) + 2. \(\mathrm{HF}(aq) + \mathrm{F}^{-}(aq) \rightleftharpoons \mathrm{HF}_{2}^{-}(aq)\) ———————————————————————————————————————————————— 3. \(2\mathrm{HF}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HF}_{2}^{-}(aq)\)

Calculate the equilibrium constant for the combined reaction

According to the laws of equilibrium, when two equilibrium reactions are added together, the equilibrium constants are multiplied. Therefore, the equilibrium constant for the combined reaction 3 (\(K_3\)) can be calculated as: \(K_3 = K_\mathrm{a} \times K\) Substituting the given values: \(K_3 = (6.9 \times 10^{-1}) \times 2.7\) \(K_3 = 1.863\) So, the equilibrium constant for the reaction \(2\mathrm{HF}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HF}_{2}^{-}(aq)\) is approximately 1.863.

Key concepts

Chemical Equilibrium

Understanding the concept of chemical equilibrium is fundamental when studying chemical reactions. It's a state in which the concentration of reactants and products in a chemical reaction is constant over time. This doesn't imply that the reactants and products are in equal concentrations but rather that their rates of formation are equal, leading to a stable ratio of products to reactants.

At equilibrium, a dynamic process occurs where the forward reaction, converting reactants to products, happens at the same rate as the reverse reaction, where products revert to reactants. Despite this ongoing process, there is no net change in the concentration of either reactants or products over time.

This concept is visually represented in the reaction between hydrogen fluoride (HF) and fluoride ions (F-). Initially, when HF is added to water, it starts dissociating into H+ and F- ions (the forward reaction). At the same time, H+ and F- ions can recombine to form HF (the reverse reaction). As the reaction proceeds, the rate of dissociation decreases while the rate of reformation increases, until they become equal, reaching what is called a chemical equilibrium.

Acid Dissociation Constant

The acid dissociation constant, abbreviated as Ka, is a specific type of equilibrium constant that measures the strength of an acid in solution. It is the ratio of the concentration of the dissociated (ionized) form of an acid to the concentration of the undissociated (non-ionized) form.

For the weak acid HF, we have the equilibrium expression involving its dissociation into hydrogen ions (H+) and fluoride ions (F-). The Ka value is then given by the following equation:\[\begin{equation}K_\text{a} = \frac{[H^+][F^-]}{[HF]}\end{equation}\]where square brackets denote concentration in molarity (M). The larger the value of Ka, the stronger the acid, meaning a higher tendency of the acid to lose its proton and form the corresponding anions in solution. In the context of the exercise, the Ka for HF indicates its tendency to dissociate which forms a part of the final equilibrium expression for the reaction under study.

Reaction Quotient

The reaction quotient, denoted as Q, is a measure that tells how far a reaction has proceeded towards equilibrium. It is calculated using the same formula as the equilibrium constant (K), but with the initial concentrations of the reactants and products instead of the concentrations at equilibrium.

For a general reaction where compounds A and B react to form C and D, the reaction quotient is expressed as:\[\begin{equation}Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\end{equation}\]where the lowercase letters represent the respective coefficients in the balanced chemical equation. If Q is less than K, the forward reaction is favored, and the system will proceed toward equilibrium by forming more products. If Q is greater than K, the reverse reaction is favored, and the system will shift toward equilibrium by forming more reactants.

In the step-by-step solutions provided for the exercise, we essentially calculate the equilibrium constant (K) for a new reaction formed by the combination of two others. Understanding how Q relates to K is critical because it allows us to determine the direction in which a reaction will need to shift to reach equilibrium under different initial conditions.