Problem 139 - General Chemistry

Question

Cyanic acid, HOCN, is a weak acid with a \(K_{a}\) value of \(3.5 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). In a \(0.293 M\) solution of the acid, the degree of ionization is \(3.5 \times 10^{-2}\). Calculate the degree of ionization in a \(0.293 \mathrm{M}\) solution to which sufficient \(\mathrm{HCl}\) is added to make it \(4.19 \times 10^{-2} M \mathrm{HCl}\) in the given volume.

Step-by-Step Solution

Answer

The degree of ionization decreases due to the common ion effect from added HCl.

Step 1: Understanding the Problem

The problem provides us with the degree of ionization of cyanic acid, HOCN, in a 0.293 M solution. It asks us to determine how this degree of ionization changes when HCl is added to the solution, increasing the concentration of HCl to 0.0419 M.

Step 2: Calculate Initial H⁺ Concentration

First, we need to find the initial concentration of H⁺ ions in the solution before adding HCl. Given the ionization degree is 0.035, and the initial concentration is 0.293 M, the initial concentration of H⁺ from HOCN ionization is: \[ [H^+]_{initial} = 0.293 imes 0.035 = 0.010255 ext{ M} \]

Step 3: New H⁺ Concentration After HCl Addition

Adding HCl will increase the concentration of H⁺ ions. The concentration of H⁺ ions after adding HCl is the sum of the initial H⁺ concentration from the ionization and the concentration of H⁺ from HCl:\[ [H^+]_{new} = 0.010255 ext{ M} + 0.0419 ext{ M} = 0.052155 ext{ M} \]

Step 4: Apply the Ionization Equation With Added HCl

Use the equilibrium expression for the ionization of cyanic acid involving the acid dissociation constant \( K_a \):\[ K_a = \frac{[H^+][OCN^-]}{[HOCN]} \]Substitute the expressions after adding HCl:\[ 3.5 \times 10^{-4} = \frac{[HOCN]_{ionized} imes 0.052155}{0.293 - [HOCN]_{ionized}} \]

Step 5: Solve for Degree of Ionization

Assuming the ionization due to the added HCl is negligible, calculate the ionization due to HOCN using the adjusted H⁺ concentration. Rearrange:\[ [HOCN]_{ionized} = \frac{3.5 \times 10^{-4} imes (0.293 - [HOCN]_{ionized})}{0.052155} \]Let \( x = [HOCN]_{ionized} \), and solve this equation to find the new degree of ionization.

Key Concepts

Weak Acids

Weak acids are fascinating compounds with unique properties that set them apart from strong acids. Unlike strong acids, which completely dissociate into ions in water, weak acids only partially ionize. This means that when a weak acid is dissolved in water, only a small fraction of its molecules turn into hydrogen ions (\( H^+ \) or hydronium ions) and anions. The rest remain as complete molecules.
Understanding the behavior of weak acids involves recognizing that they exist in equilibrium. This balance is influenced by factors like concentration and temperature. Adding more acid to the solution doesn't linearly increase the concentration of ions, because the equilibrium constant of the reaction governs the extent of ionization. It's important to know that the degree of ionization does not exceed 100%, which is a key point differentiating weak acids from their strong counterparts.
Examples of weak acids include acetic acid (\( CH_3COOH \)) and cyanic acid (\( HOCN \)). These acids are usually described by their ability to donate protons (hydrogen ions) to a base, shown by their acid dissociation constant, or \( K_a \). This constant is a measure of the strength of the weak acid, indicating how well it can donate protons.

Degree of Ionization

The degree of ionization is an important concept when studying weak acids. It tells us the fraction of the total acid molecules that have ionized in solution. This is vital for calculating the pH of the solution or determining how certain changes, like adding another acid, will affect the system.
In a solution, the degree of ionization can be calculated as the ratio of ionized acid concentration to the initial acid concentration, usually expressed as a percentage. For instance, if \( x \) moles of an acid initially present in \( 1 M \) solution ionizes into its constituent ions, the degree of ionization is \( \frac{x}{1} \times 100\% \) .
Consider cyanic acid from our exercise. Initially, it has a degree of ionization calculated based on its \( H^+ \) concentration before any HCl is added. The presence of additional HCl introduces more hydrogen ions into the solution, altering the dynamics of ionization. As a rule, when more protons are present in the solution, the degree of ionization typically decreases because the equilibrium shifts to favor the non-ionized form of the acid, thanks to Le Chatelier's principle.

Equilibrium Constants

Equilibrium constants such as \( K_a \) are essential for understanding the chemistry of weak acids. The acid dissociation constant, \( K_a \), directly measures a weak acid's strength by indicating the extent to which it ionizes in solution. The greater the \( K_a \) value, the stronger the acid, meaning it's more likely to donate \( H^+ \) ions.
The value of \( K_a \) is calculated based on the concentrations of the products and reactants at equilibrium. In reactions involving weak acids, it takes into account both the dissociated ions and the undissociated acid molecule. For cyanic acid, the expression for its \( K_a \) is: \[ K_a = \frac{[H^+][OCN^-]}{[HOCN]} \]
This formula helps explain the relationship between the \( H^+ \) concentration in the solution and the degree of ionization. While standard calculations often assume \( H^+ \) ions are solely from the weak acid, added acids like HCl complicate the equilibrium by adding extra \( H^+ \). Thus, careful calculation is necessary to correctly interpret the scenario.
Learning how equilibrium constants function is crucial for solving problems in acid-base chemistry and predicting how a system will react under various conditions.