Question

Choose from the conjugate acid-base pairs \(\mathrm{F}^{-} / \mathrm{HF}, \mathrm{NH}_{3} / \mathrm{NH}_{4}^{+}\). and \(\mathrm{NO}_{3}^{-} / \mathrm{HNO}_{3}\), to complete the following equation with the pair that gives an equilibrium constant \(K_{c}>1\). $$ +\mathrm{H}_{2} \mathrm{CO}_{3} \longrightarrow \mathrm{HCO}_{3}^{-} $$

Short answer

The equation is completed with \( \mathrm{NO}_{3}^{-} / \mathrm{HNO}_{3} \).

Step-by-step solution

Identify the Equation Components

We have the equation: \[ + \mathrm{H}_{2} \mathrm{CO}_{3} \longrightarrow \mathrm{HCO}_{3}^{-} \]. We need to find a conjugate acid-base pair from the given options that can react with \( \mathrm{H}_{2} \mathrm{CO}_{3} \) to produce \( \mathrm{HCO}_{3}^{-} \) and another species, causing an equilibrium constant \( K_c > 1 \).

List Possible Conjugate Base Reactions

For each conjugate acid-base pair, determine the reaction that involves \( \mathrm{H}_{2} \mathrm{CO}_{3} \):1. \( \mathrm{F}^{-} + \mathrm{H}_{2} \mathrm{CO}_{3} \rightleftharpoons \mathrm{HF} + \mathrm{HCO}_{3}^{-} \) 2. \( \mathrm{NH}_{3} + \mathrm{H}_{2} \mathrm{CO}_{3} \rightleftharpoons \mathrm{NH}_{4}^{+} + \mathrm{HCO}_{3}^{-} \) 3. \( \mathrm{NO}_{3}^{-} + \mathrm{H}_{2} \mathrm{CO}_{3} \rightleftharpoons \mathrm{HNO}_{3} + \mathrm{HCO}_{3}^{-} \).

Assess Reaction Favorability

Evaluate which conjugate base is a stronger base compared to \( \mathrm{HCO}_{3}^{-} \), making a strong acid that mostly dissociates into the conjugate acid:- \( \mathrm{HF} \) is a weak acid, equilibrium favors reactants.- \( \mathrm{NH}_{4}^{+} \) approximates a weak acid, might favor products but less so.- \( \mathrm{HNO}_{3} \) is a strong acid, equilibrium greatly favors products, making \( K_c > 1 \).Therefore, \( \mathrm{NO}_{3}^{-} / \mathrm{HNO}_{3} \) is suitable.

Select the Appropriate Pair

Based on the assessment, the pair that produces a favorable equilibrium constant (\( K_c > 1 \)) by effectively donating a proton to \( \mathrm{H}_{2} \mathrm{CO}_{3} \) and resulting in \( \mathrm{HCO}_{3}^{-} \) is \( \mathrm{NO}_{3}^{-} / \mathrm{HNO}_{3} \). Thus the completed equation is: \[ \mathrm{NO}_{3}^{-} + \mathrm{H}_{2} \mathrm{CO}_{3} \longrightarrow \mathrm{HNO}_{3} + \mathrm{HCO}_{3}^{-} \].

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a number that provides insight into the position of equilibrium for a chemical reaction. It is calculated based on the concentrations of products and reactants at equilibrium. For a generalized reaction, \[ aA + bB \rightleftharpoons cC + dD \], the equilibrium constant is expressed as \[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]. The value of \( K_c \) can tell us whether the reaction tends to form more products or more reactants when equilibrium is reached.

• If \( K_c > 1 \), the products are favored at equilibrium, indicating that the reaction proceeds mostly to the right, forming more products than reactants.
• If \( K_c < 1 \), the reactants are favored, meaning the reaction does not proceed far, resulting in more reactants remaining at equilibrium.
• If \( K_c \approx 1 \), the concentrations of reactants and products are roughly equal.

In the context of acid-base reactions, a high \( K_c \) implies that the forward reaction, including proton transfer, is favorable.

Acid-Base Reaction

Acid-base reactions involve the transfer of protons from an acid to a base. According to the Brønsted-Lowry theory, an acid is a substance that can donate a proton (\( H^+ \)), while a base is a substance that can accept a proton. Here, conjugate acid-base pairs play a crucial role.

• An acid, upon donating a proton, transforms into its conjugate base.
• A base, after accepting a proton, becomes its conjugate acid.

Consider these given pairs: \( \mathrm{F}^{-} / \mathrm{HF} \), \( \mathrm{NH}_{3} / \mathrm{NH}_{4}^{+} \), and \( \mathrm{NO}_{3}^{-} / \mathrm{HNO}_{3} \). When these react with \( \mathrm{H}_{2} \mathrm{CO}_{3} \), they can form new acids and bases, potentially shifting the equilibrium as these new conjugate partners balance the reaction dynamics.

Proton Transfer

In an acid-base reaction, the concept of proton transfer is fundamental. During this process, a proton from an acid is transferred to a base. For example, in the reactions outlined:

• \( \mathrm{NO}_{3}^{-} + \mathrm{H}_{2} \mathrm{CO}_{3} \rightleftharpoons \mathrm{HNO}_{3} + \mathrm{HCO}_{3}^{-} \) involves \( \mathrm{H}_{2} \mathrm{CO}_{3} \), which donates a proton to \( \mathrm{NO}_{3}^{-} \).

The proton transfer direction will depend on which reacting pair forms a strong acid or base; strong acids like \( \mathrm{HNO}_{3} \) dissociate completely, making the reaction tend to form products (forward direction).

• This process is what makes \( K_c \) greater than 1, as the dissociation of a strong acid shifts equilibrium towards the right, favoring products.

Understanding proton transfer helps predict the behavior of reactants and products at equilibrium, especially in systems where conjugate acid-base pairs are involved.