Question

Given \(K_{a}\) values of \(5.76 \times 10^{-10}\) and \(4.8 \times 10^{-10}\) for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{HCN}\) respectively. What is the equilibrium constant for the following reaction? $$ \mathrm{NH}_{4}^{+}(a q .)+\mathrm{CN}^{-}(a q .) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCN}(a q .) $$ (a) \(0.83\) (b) \(1.2\) (c) \(8.0 \times 10^{-11}\) (d) \(27.6 \times 10^{-10}\)

Short answer

The equilibrium constant for the reaction is 1.2 or answer (b).

Step-by-step solution

Understand the Reaction

The given reaction represents the reaction between ammonium ion (\text{NH}_{4}^{+}) and cyanide ion (\text{CN}^{-}) to produce ammonia (\text{NH}_{3}) and hydrogen cyanide (\text{HCN}). Both \text{NH}_{4}^{+} and \text{HCN} can act as acids in water, and their dissociation in water is characterized by their respective acid dissociation constants, \(K_{a}\).

Deduce the Relationship of the Equilibrium Constants

From the provided \(K_{a}\) values, we can infer that as \text{NH}_{4}^{+} donates a proton to become \text{NH}_{3}, \text{HCN} accepts a proton to form from \text{CN}^{-}. The equilibrium constant \(K\) for the overall reaction is the ratio of the \(K_{a}\) of \text{NH}_{4}^{+} to the \(K_{a}\) of \text{HCN}. Thus, \(K = K_{a}(\text{NH}_{4}^{+}) / K_{a}(\text{HCN})\).

Calculate the Equilibrium Constant for the Reaction

Using the values provided for \(K_{a}\) of \text{NH}_{4}^{+} and \text{HCN}, calculate the equilibrium constant \(K\): \[K = \frac{5.76 \times 10^{-10}}{4.8 \times 10^{-10}}\]

Perform the Calculation

Divide the \(K_{a}\) value of \text{NH}_{4}^{+} by the \(K_{a}\) value of \text{HCN} to find the equilibrium constant \(K\) for the reaction: \[K = \frac{5.76 \times 10^{-10}}{4.8 \times 10^{-10}} = 1.2\]

Key concepts

Acid Dissociation Constant

Understanding the acid dissociation constant (\(K_a\)) is crucial when studying chemistry, especially acid-base reactions. It indicates the strength of an acid in solution. Simply put, the acid dissociation constant refers to the equilibrium constant for the dissociation of an acid into its conjugate base and a proton (\text{H}^+). For example, ammonium ion (\text{NH}_4^+)\rightleftharpoons ammonia (\text{NH}_3) and a proton.

In a mathematical sense, if we have a generic acid denoted by HA, the dissociation can be written as HA \rightleftharpoons H^+ + A^-, and the acid dissociation constant would be expressed as \(K_a = \frac{[H^+][A^-]}{[HA]}\). Here, the concentrations are measured at equilibrium. A higher value of \(K_a\) suggests a stronger acid, because it implies that the acid dissociates more in solution, producing more protons and conjugate base.

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry, occurring when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of the reactants and products remain constant over time. It's important to recognize that this does not mean the reactants and products are in equal concentrations, but rather that their ratios do not change.

At equilibrium, the rate at which the products convert back into reactants is equal to the rate at which the reactants form the products. To quantify this state, chemists use an equilibrium constant (\(K\)), which relates to the concentration (or pressure, for gases) of the reactants and products. The equilibrium constant for a reaction like aA + bB \rightleftharpoons cC + dD would be formulated as \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\), where the letters represent the stoichiometric coefficients, and the square brackets denote the concentration of substances at equilibrium. The value of \(K\) is temperature-dependent and remains constant at a given temperature unless changes in the system occur.

Reaction Quotient

The reaction quotient (\(Q\)) is a measure that compares the relative amounts of products and reactants present during a reaction at any point in time, not just at equilibrium. It uses the same formula as the equilibrium constant but with the concentrations of the reactants and products at a particular moment. This means \(Q = \frac{[products]}{[reactants]}\), based on the stoichiometry of the balanced chemical equation.

If \(QK\), the reaction will proceed in the reverse direction to reach equilibrium, producing more reactants. When \(Q\) equals \(K\), the system is at equilibrium. Where the reaction quotient is essential in determining the direction in which a reaction must proceed to reach equilibrium, understanding \(Q\) helps chemists predict how changes in conditions (like concentration, pressure, or temperature) will affect the reaction progress.