Q.The ionization constant of acetic acid is 1.74 × 10⁻⁵. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Concept: Weak Acid Ionization — Acetic acid ($\text{CH}_3\text{COOH}$) is a weak acid, so its dissociation is small and we can use the approximation $\alpha \ll 1$.

Step 1: Set up the equilibrium.

For $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$, initial concentration $c = 0.05$ M.

At equilibrium: $[\text{CH}_3\text{COOH}] = c(1-\alpha)$, $[\text{CH}_3\text{COO}^-] = [\text{H}^+] = c\alpha$.

Step 2: Write the ionization constant.

Given $K_a = 1.74 \times 10^{-5}$ and $c = 0.05$ M.

Step 3: Apply the approximation $\alpha \ll 1$ so $1-\alpha \approx 1$.

Then $K_a \approx c\alpha^2$, so

$\alpha = \sqrt{\frac{K_a}{c}} = \sqrt{\frac{1.74 \times 10^{-5}}{0.05}} = \sqrt{3.48 \times 10^{-4}} = 1.866 \times 10^{-2}$ …

The Concept: Weak Acid Ionization

Acetic acid ($CH_3COOH$) is a classic weak acid. Unlike a strong acid, it does not fully dissociate in water. Instead, it establishes an equilibrium:

The strength of this acid is quantified by its ionization constant, $K_a$. For a weak acid, $K_a$ is small (here, $1.74 \times 10^{-5}$), meaning the equilibrium lies far to the left — most of the acid remains undissociated.

The degree of dissociation, $\alpha$, is the fraction of the original acid molecules that have ionized. If we start with a concentration $C$ (here, 0.05 M), then at equilibrium:

• $[CH_3COOH] = C(1 - \alpha)$
• $[H^+] = C\alpha$
• $[CH_3COO^-] = C\alpha$

The key insight: because $K_a$ is very small, $\alpha$ will also be small (much less than 1). This lets us use a simplifying approximation that avoids solving a quadratic equation.

Step-by-Step Solution

1. Write the equilibrium expression.

From the reaction, the ionization constant is:

Substituting the equilibrium concentrations in terms of $C$ and $\alpha$:

This is the exact expression. It is a quadratic in $\alpha$.

2. Apply the weak acid approximation.

Since $K_a$ is very small, $\alpha \ll 1$, so $1 - \alpha \approx 1$. This simplifies the expression to:

This is the Ostwald dilution law for weak electrolytes. It tells us that for a given $K_a$, the degree of dissociation increases as the solution becomes more dilute (as $C$ decreases).

3. Calculate $\alpha$.

Rearrange the approximate formula:

Plug in the values: $K_a = 1.74 \times 10^{-5}$, $C = 0.05$ M.

So, $\alpha \approx 0.01866$.

Check the approximation: $0.01866 \times 100\% = 1.87\%$, which is well under 5%. The approximation is valid. …