Problem 58 - Chemistry The Central Science In Si Units

Question

The acid-dissociation constant for chlorous acid \(\left(\mathrm{HClO}_{2}\right)\) is \(1.1 \times 10^{-2} .\) Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{ClO}_{2}^{-}\), and \(\mathrm{HClO}_{2}\) at equilibrium if the initial concentration of \(\mathrm{HClO}_{2}\) is \(0.0200 \mathrm{M}\)

Step-by-Step Solution

Answer

\([\mathrm{H_3O}^+] = 0.0148 \, M, \quad [\mathrm{ClO_2}^-] = 0.0148 \, M, \quad [\mathrm{HClO}_2] = 0.0052 \, M\).

Step 1: Understand the Equilibrium Reaction

The dissociation of chlorous acid in water can be represented by the equation: \( \mathrm{HClO}_2 (aq) + \mathrm{H_2O} (l) \rightleftharpoons \mathrm{H_3O}^+ (aq) + \mathrm{ClO_2}^- (aq) \). The acid-dissociation constant, \( K_a \), for this reaction is given as \( 1.1 \times 10^{-2} \).

Step 2: Write the Expression for the Equilibrium Constant

The expression for the acid-dissociation constant \( K_a \) is: \[ K_a = \frac{[\mathrm{H_3O}^+][\mathrm{ClO_2}^-]}{[\mathrm{HClO}_2]} \] where \([\mathrm{H_3O}^+]\), \([\mathrm{ClO_2}^-]\), and \([\mathrm{HClO}_2]\) are the equilibrium concentrations of the respective species.

Step 3: Set Up the ICE Table

Set up the initial concentrations, the change in concentrations, and the equilibrium concentrations (ICE table) for the reaction. - Initial: \([\mathrm{HClO}_2] = 0.0200 \, M\), \([\mathrm{H_3O}^+] = 0\), \([\mathrm{ClO_2}^-] = 0\) - Change: \([\mathrm{HClO}_2] = -x\), \([\mathrm{H_3O}^+] = +x\), \([\mathrm{ClO_2}^-] = +x\) - Equilibrium: \([\mathrm{HClO}_2] = 0.0200 - x\), \([\mathrm{H_3O}^+] = x\), \([\mathrm{ClO_2}^-] = x\).

Step 4: Solve for x Using the Equilibrium Constant

Plug the equilibrium values from the ICE table into the \( K_a \) expression: \[ 1.1 \times 10^{-2} = \frac{x \cdot x}{0.0200 - x} \]Solve for \( x \) assuming \( x \) is small compared to the initial concentration, so \( 0.0200 - x \approx 0.0200\): \[ 1.1 \times 10^{-2} = \frac{x^2}{0.0200} \]\[ x^2 = 2.2 \times 10^{-4} \]Take the square root:\[ x = \sqrt{2.2 \times 10^{-4}} \approx 0.0148 \].

Step 5: Calculate Equilibrium Concentrations

Using \( x \approx 0.0148 \), calculate the equilibrium concentrations:- \([\mathrm{H_3O}^+] = x = 0.0148 \, M\)- \([\mathrm{ClO_2}^-] = x = 0.0148 \, M\)- \([\mathrm{HClO}_2] = 0.0200 - x = 0.0200 - 0.0148 = 0.0052 \, M\).

Key Concepts

Chlorous Acid

Chlorous acid, represented chemically as \( \mathrm{HClO}_2 \), is an oxoacid of chlorine. It is not a stable compound and is best known in solution. Chlorous acid is important in the context of acid-base chemistry, where it participates in dissociation reactions in water. In these reactions, acids donate protons to water, forming hydronium ions \( \mathrm{H_3O}^+ \). Understanding the behavior of chlorous acid in aqueous solution helps us predict how acidic or basic a solution will be when this acid is present.
Chlorous acid is a relatively strong acid compared to other oxoacids of chlorine, such as hypochlorous acid \( \mathrm{HClO} \). The strength of an acid in solution is often measured by its ability to donate protons, generally evaluated through its acid-dissociation constant \( K_a \). For chlorous acid, this constant has a value of \( 1.1 \times 10^{-2} \). This signifies that chlorous acid can significantly dissociate in water, yielding considerable concentrations of hydronium ions.

Equilibrium Concentrations

Equilibrium concentrations refer to the steady-state levels of all chemically relevant species in a reaction mixture when the rates of the forward and backward reactions are equal. In the case of chlorous acid's dissociation in water, it establishes an equilibrium among \( \mathrm{HClO}_2, \mathrm{H_3O}^+, \) and \( \mathrm{ClO}_2^- \).
When calculations are involved, such as with equilibrium constants, it's vital to compute the equilibrium concentrations of all species present using the known initial concentrations and changes defined by the reaction's stoichiometry. For chlorous acid, initially at a concentration of \( 0.0200 \, \mathrm{M} \), equilibrium is reached with concentrations distributed according to the reaction dynamics. By calculating these concentrations correctly, predictions can be made regarding the acidic properties of the solution.

ICE Table

An ICE table is an incredibly useful tool for equilibria calculations in chemistry. ICE stands for Initial, Change, and Equilibrium, which are the three stages of analysis when handling a dissociation reaction.
To set up an ICE table for the hydrolysis of chlorous acid, one should:

• Write down the initial concentrations: Before any reaction occurs, \( [\mathrm{HClO}_2] = 0.0200 \). Both products, \( [\mathrm{H_3O}^+] \) and \( [\mathrm{ClO}_2^-] \), have initial concentrations of 0.
• Define the change that occurs: Since \( x \) amount of \( \mathrm{HClO}_2 \) dissociates, the changes are noted as \( -x \) for \( \mathrm{HClO}_2 \) and \( +x \) for each of the products.
• Calculate the equilibrium concentrations: Apply the changes to the initial concentrations to find the equilibrium state.

ICE tables simplify tracking the changes in concentrations throughout the reaction process, making solving equilibrium problems much more systematic.

Equilibrium Constant

The equilibrium constant \( K_a \) is a crucial concept and represents the ratio of the concentrations of the products to the reactants, raised to the power of their respective coefficients in the balanced equation. For the dissociation of chlorous acid: \[ \mathrm{HClO}_2 \rightleftharpoons \mathrm{H_3O}^+ + \mathrm{ClO}_2^- \] The expression for the equilibrium constant is: \[ K_a = \frac{[\mathrm{H_3O}^+][\mathrm{ClO}_2^-]}{[\mathrm{HClO}_2]} \] This equation allows us to calculate the degree to which the acid dissociates. A larger \( K_a \) value means a greater extent of dissociation, indicating a stronger acid.
In practical terms, the equilibrium constant provides insights into the concentrations of products and reactants at equilibrium. By understanding this constant's role, we can predict the behavior of acids and bases in solution, such as their potential to affect pH or precipitate formation. In our process, \( K_a \) was used to find \( x \), the concentration at equilibrium of \( \mathrm{H_3O}^+ \) and \( \mathrm{ClO}_2^- \), which helped us determine that the solution was considerably acidic.