Problem 126 - Chemistry The Central Science

Question

(a) Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the following reactions. \((\mathrm{i}) \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{co}_{3^{-}(a q)}+\mathrm{H}_{2} \mathrm{O}(l)\) (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 for- ward reaction is appreciable \((K\) much greater than 1\()\) 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 as follows: 1. For reaction (i), the equilibrium constant \(K_1 = 4.69 \times 10^3\). 2. For reaction (ii), the equilibrium constant \(K_2 = 11.84\). Among these reactions, only reaction (i) should be represented by a single arrow, as its equilibrium constant is much greater than 1, indicating a greater forward reaction compared to the reverse reaction.

Step 1: Find the dissociation constants

Refer to Appendix D to find the dissociation constants for the species involved in the reactions. Write down the values for later use. For reaction (i): \( K_{a1(HCO_3^{-})} \) = 4.69 x 10^-11 \( K_{b(OH^{-})} \) = 1.00 x 10^-14 For reaction (ii): \( K_{a1(NH_4^{+})} \) = 5.56 x 10^-10 \( K_{a1(HCO_3^{-})} \) = 4.69 x 10^-11 ##STEP 2: Calculate the equilibrium constants for each reaction##

Step 2: Calculate the equilibrium constants

Use the formula for calculating the equilibrium constant for each reaction and substitute the values obtained in Step 1: For reaction (i): \(K_1 = \frac{K_{a1(HCO_3^{-})}}{K_{b(OH^{-})}} = \frac{4.69 \times 10^{-11}}{1.00 \times 10^{-14}} = 4.69 \times 10^3 \) For reaction (ii): \(K_2 = \frac{K_{a1(NH_4^{+})}}{K_{a1(HCO_3^{-})}} = \frac{5.56 \times 10^{-10}}{4.69 \times 10^{-11}} = 11.84 \) ##STEP 3: Determine if any of the equilibria should be represented by a single arrow##

Step 3: Check for single arrow representation

Compare the equilibrium constants with 1 to determine if any of the equilibria should be represented by a single arrow: For reaction (i): As \(K_1 = 4.69 \times 10^3 \), which is much greater than 1, it indicates the forward reaction is much greater than the reverse reaction and thus a single arrow may be appropriate. For reaction (ii): As \(K_2 = 11.84 \), which is greater than 1 but not by a large amount, it indicates that the equilibrium is not too far towards the forward reaction, so a single arrow would likely not be appropriate in this case. To summarize the solutions: 1. Equilibrium constant for reaction (i) is \(4.69 \times 10^3\). 2. Equilibrium constant for reaction (ii) is 11.84. 3. Only reaction (i) should be represented by a single arrow, as its equilibrium constant is much greater than 1.

Key Concepts

Dissociation Constants

Dissociation constants are invaluable in understanding how compounds separate into their constituent ions in solution. These constants, often represented as Ka for acids and Kb for bases, provide insight into the strength of the acid or base. A strong acid or base will have a larger dissociation constant (Ka or Kb respectively) indicating it dissociates completely in water. Conversely, a weaker one has a smaller constant and does not dissociate completely.

When calculating equilibrium constants for acid-base reactions, these dissociation constants become crucial. As seen in the exercise, to find the equilibrium constant (K) for the reaction of bicarbonate with hydroxide, we consider the dissociation constants of bicarbonate (\(K_{a1(HCO_3^{-})}\)) and hydroxide (\(K_{b(OH^{-})}\)). Here, understanding that the hydroxide ion is a strong base (with a high Kb value) and bicarbonate is a relatively weaker acid aids in predicting the direction of the equilibrium and the magnitude of the equilibrium constant.

Chemical Equilibrium

Chemical equilibrium describes a state in which the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. It is important to note that being at equilibrium does not imply that reactants and products are present in equal concentrations, but rather that their rates of formation are balanced.

The equilibrium condition is dynamically achieved, as reactions still occur but compensate each other. This concept allows chemists to calculate the equilibrium constant, K, which provides a ratio of product concentrations to reactant concentrations at equilibrium, with each raised to the power of their stoichiometric coefficients. In the examples provided, the magnitude of these constants indicates the extent to which the reactions proceed towards the products.

Reaction Quotient

The reaction quotient, Q, is a measure of the relative amounts of products and reactants present during a reaction at any given point in time. It is calculated in the same way as the equilibrium constant, K, using the current concentrations of reactants and products. The value of Q is compared to K to predict which direction a reaction will proceed to reach equilibrium. If Q is less than K, the forward reaction will be favored to produce more products. Conversely, if Q is greater than K, the reaction will proceed in the reverse direction to produce more reactants. Understanding how to calculate and interpret Q can help assess the progress of a reaction and predict shifts in equilibrium in response to changes in concentration, pressure, or temperature.

Acid-Base Reactions

Acid-base reactions are a cornerstone of chemical equilibrium studies, typically involving the transfer of protons (H+) between reactant species. An acid will release H+ ions while a base will accept them. The strength of the acid or base in a given reaction influences the position of the equilibrium and is signified by its dissociation constant.

In our exercise, we analyzed acid-base equilibria involving bicarbonate and ammonium. Understanding the nature of these reactants, aided by their dissociation constants, we can calculate the equilibrium constants for their reactions. Notably, these reactions are crucial in many biological and environmental systems, including buffering systems in blood and soil pH management.