Problem 126 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the following reactions. (i) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (ii) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)\) (b) We usually use single arrows for reactions when the forward reaction is appreciable ( \(K\) much greater than 1) or when products escape from the system, so that equilibrium is never established. If we follow this convention, which of these equilibria might be written with a single arrow?

Step-by-Step Solution

Answer

The equilibrium constants for the given reactions are found using the dissociation constants from Appendix D. For reaction (i), the equilibrium constant \(K_\mathrm{eq}\) can be calculated as \(K_\mathrm{eq} = \dfrac{K_\mathrm{a1}(\mathrm{HCO}_3^-)}{K_\mathrm{b}(\mathrm{OH}^-)}\). For reaction (ii), the equilibrium constant \(K_\mathrm{eq}\) can be calculated as \(K_\mathrm{eq} = \dfrac{K_\mathrm{a}(\mathrm{NH}_4^+)}{K_\mathrm{a2}(\mathrm{HCO}_3^-)}\). Comparing these equilibrium constants to the criteria mentioned in the problem statement (\(K \gg1\)), it can be determined if any of the equilibria can be represented using a single arrow.

Step 1: Determine the dissociation constants from Appendix D

You will need to find the dissociation constants for the relevant substances from Appendix D. You should have the following values: \(K_\mathrm{a1} (\mathrm{HCO}_3^-)\): This is the dissociation constant for bicarbonate ion. \(K_\mathrm{b} (\mathrm{OH}^-)\): This is the dissociation constant for hydroxide ion. \(K_\mathrm{a} (\mathrm{NH}_4^+)\): This is the dissociation constant for ammonium ion. \(K_\mathrm{a2} (\mathrm{HCO}_3^-)\): This is another dissociation constant for bicarbonate ion.

Step 2: Determine the equilibrium constant for reaction (i)

First, we will find the equilibrium constant for reaction (i): \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) The relevant dissociation constants are \(K_\mathrm{a1} (\mathrm{HCO}_3^-)\) and \(K_\mathrm{b} (\mathrm{OH}^-)\). To find the equilibrium constant for this reaction, we can use the relationship: \(K_\mathrm{eq} = \dfrac{K_\mathrm{a1}(\mathrm{HCO}_3^-)}{K_\mathrm{b}(\mathrm{OH}^-)}\) Using the dissociation constants obtained from Appendix D, perform the calculation and find the value of \(K_\mathrm{eq}\) for reaction (i).

Step 3: Determine the equilibrium constant for reaction (ii)

Next, we will find the equilibrium constant for reaction (ii): \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)\) The relevant dissociation constants are \(K_\mathrm{a} (\mathrm{NH}_4^+)\) and \(K_\mathrm{a2} (\mathrm{HCO}_3^-)\). To find the equilibrium constant for this reaction, we can use the relationship: \(K_\mathrm{eq} = \dfrac{K_\mathrm{a}(\mathrm{NH}_4^+)}{K_\mathrm{a2}(\mathrm{HCO}_3^-)}\) Using the dissociation constants obtained from Appendix D, perform the calculation and find the value of \(K_\mathrm{eq}\) for reaction (ii).

Step 4: Assess if any of the equilibria could be represented by a single arrow

We usually use single arrows for reactions when the forward reaction is appreciable (\(K \gg1\)) or when the products escape from the system. Compare the values of the equilibrium constants for reactions (i) and (ii) to these criteria. If any of the equilibrium constants are much greater than 1, then that equilibrium can be written using a single arrow. By following these steps, you should be able to determine the equilibrium constants for the given reactions and assess if they warrant the use of a single arrow for representation.

Key Concepts

Dissociation Constants

Dissociation constants are key to understanding how substances break down into their ions in a solution. They tell us the extent to which a compound will dissociate and are especially important in acid-base chemistry. When an acid dissociates, it releases hydrogen ions, and the dissociation constant (commonly referred to as \(K_a\) for acids or \(K_b\) for bases) quantifies this process.

For example, in our problem, we look at the dissociation constants \(K_{a1}\) and \(K_{a2}\) for bicarbonate ion \((\mathrm{HCO}_3^{-})\), in conjunction with \(K_b\) for hydroxide ion \((\mathrm{OH}^-)\) and \(K_a\) for ammonium ion \((\mathrm{NH}_4^+)\). By using these constants, we can calculate the equilibrium constant for the respective reactions. Understanding dissociation constants allows us to predict the position of the equilibrium and the strength of acids and bases.

Chemical Equilibrium

Chemical equilibrium refers to the state where the concentrations of reactants and products remain constant over time. At this point, the forward and reverse reactions occur at the same rate. It's crucial to grasp that equilibrium does not mean the reactants and products are present in equal amounts, but rather their rates of formation are balanced.

For reactions involving acids and bases, calculating the equilibrium constant \(K_{eq}\) is essential. It tells us which side of the reaction is favored. A large \(K_{eq}\) (much greater than 1) indicates that products are favored, possibly justifying the use of a single arrow in reaction equations.

• Reaction (i): For the bicarbonate and hydroxide reaction, \(K_{eq}\) is calculated using \(\frac{K_{a1}}{K_{b}}\)
• Reaction (ii): For the ammonium and carbonate reaction, \(K_{eq}\) is calculated using \(\frac{K_{a}}{K_{a2}}\)

Understanding these calculations helps in predicting whether a reaction is product- or reactant-favored.

Acid-Base Reactions

Acid-base reactions involve the transfer of protons between reactants. Recognizing whether a reaction tends toward the right (producing more products) or the left (favoring reactants) is vital. This tendency is linked to the strength of the acids and bases involved, as well as the calculated equilibrium constants.

In our exercises, we assess reactions such as \(\mathrm{NH}_4^+\) with \(\mathrm{CO}_3^{2-}\), where protons are transferred between the ammonium ion and bicarbonate. When calculating the equilibrium constants for these reactions, we use dissociation constants to assess which side is favored.

Stronger acids and bases dissociate more in solution, leading to larger dissociation constants. This influences the equilibrium position, indicating how much of each species is present at equilibrium. Understanding these dynamics helps in crafting balanced chemical reactions and predicting reaction behavior in different conditions.