Question

Ionization constant of acetic acid is \(1.8 \times 10^{-5}\) The concentration of \(\mathrm{H}^{+}\) ions in \(0.1 \mathrm{M}\) solution is: (a) \(1.8 \times 10^{-3} \mathrm{M}\) (b) \(1.8 \times 10^{-5} \mathrm{M}\) (c) \(1.3 \times 10^{-3} \mathrm{M}\) (d) \(1.34 \times 10^{-3} \mathrm{M}\)

Short answer

The concentration of H⁺ ions is approximately 1.34 × 10⁻³ M, choice (d).

Step-by-step solution

Write the Ionization Equation

Acetic acid (CH₃COOH) ionizes in water to form acetate ions (CH₃COO⁻) and hydrogen ions (H⁺). The chemical equation for this equilibrium process is: \[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

Write the Expression for the Ionization Constant

The ionization constant (Ka) expression for acetic acid is given as:\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \] where \([\text{CH}_3\text{COO}^-]\) and \([\text{H}^+]\) are the concentrations of the acetate and hydrogen ions, respectively, and \([\text{CH}_3\text{COOH}]\) is the concentration of acetic acid.

Substitute Initial Concentration and Changes

Let the initial concentration of acetic acid be 0.1 M, and let the concentration that ionizes be \(x\). Thus, at equilibrium: \[ [\text{CH}_3\text{COOH}] = 0.1 - x \] \[ [\text{CH}_3\text{COO}^-] = x \]\[ [\text{H}^+] = x \] Substituting these into the Ka expression gives: \[ 1.8 \times 10^{-5} = \frac{x \cdot x}{0.1 - x} \approx \frac{x^2}{0.1} \] because \(x\) is very small compared to 0.1.

Simplify and Solve for x

Now, solve for \(x\): \[ 1.8 \times 10^{-5} \times 0.1 = x^2 \]\[ x^2 = 1.8 \times 10^{-6} \]Taking the square root of both sides gives: \[ x = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \] Thus, the concentration of \(\text{H}^+\) ions is approximately \(1.34 \times 10^{-3} \text{ M} \).

Key concepts

Acetic Acid Ionization

When acetic acid, known chemically as \( \text{CH}_3\text{COOH} \), is dissolved in water, it undergoes ionization. This process breaks it down into acetate ions \( \text{CH}_3\text{COO}^- \) and hydrogen ions \( \text{H}^+ \). Although this reaction seems simple, not all acetic acid molecules will ionize. The extent of ionization relies heavily on the acid's inherent property, defined by its ionization constant, and on the solution's conditions, such as concentration. An important point to grasp here is that acetic acid is a weak acid, meaning it doesn't fully ionize in solution—unlike strong acids such as hydrochloric acid. This partial ionization can be represented by a reversible reaction:

• \( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \)

This equation captures the dynamic nature of the ionization, where the forward and reverse reactions occur concurrently, reaching a balance we call equilibrium.

Equilibrium Expressions

To delve deeper into acetic acid ionization, we use equilibrium expressions, which are mathematical representations of the chemical equilibrium. For acetic acid, the equilibrium or ionization constant \( K_a \) is critical. It quantifies the ratio of the concentration of the products to the reactants at equilibrium:

• \( K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \)

Here:

• \([\text{CH}_3\text{COO}^-]\) is the concentration of the acetate ions.
• \([\text{H}^+]\) is the concentration of the hydrogen ions.
• \([\text{CH}_3\text{COOH}]\) is the concentration of un-ionized acetic acid.

The \( K_a \) value for acetic acid is \(1.8 \times 10^{-5}\), informing us about the degree to which the acid ionizes under standard conditions. It's through this precise mathematical approach that we interpret the behavior of acetic acid in solutions.

Concentration Calculations

Understanding how to calculate concentrations in an ionization reaction requires a grasp of how each component’s concentration changes as the reaction proceeds to equilibrium. Starting with an initial concentration of acetic acid, consider that a small fraction \( x \) of this initial amount ionizes. The equilibrium concentrations become:

• \([\text{CH}_3\text{COOH}] = 0.1 - x \)
• \([\text{CH}_3\text{COO}^-] = x \)
• \([\text{H}^+] = x \)

We apply these expressions back into the ionization constant formula:

• \( K_a = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1} \)

Since \( x \) is much smaller than \( 0.1 \) due to acetic acid's weak ionization, the reaction simplifies. This simplification allows easier calculation of \( x \), which equals the concentration of \( \text{H}^+ \) ions. Hence, \( x \approx 1.34 \times 10^{-3} \text{ M} \). Performing these calculations reveals the proportion of ionization, crucial for various practical applications, including buffer solutions and acidity adjustments.

Chemical Equilibrium

Chemical equilibrium plays a pivotal role in understanding reactions like acetic acid ionization. It's the point at which the rate of the forward reaction equals the rate of the reverse reaction, resulting in constant concentrations of the different species involved. In the equation:

• \(\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+\),

the system reaches equilibrium, meaning both the reactants and products are present, yet their concentrations remain unchanged. This dynamic equilibrium doesn't represent a static situation but rather a balance of processes.
Factors such as concentration, temperature, and pressure influence the position of equilibrium according to Le Chatelier's principle. For weak acids like acetic acid, achieving equilibrium is vital for predicting how changes in conditions will affect the reaction composition. Understanding equilibrium concepts ensures accurate predictions about how solutions behave under different conditions, a common challenge in chemical processes and industrial applications.