Q.Calculate the degree of ionization of 0.05M acetic acid if its pK a value is 4.74. How is the degree of dissociation affected when its solution also contains

The key idea is Weak Acid Ionization: acetic acid ($CH_3COOH$) is a weak acid, so its ionization is suppressed in the presence of a common ion ($H^+$ from HCl).

Step 1: Degree of ionization in pure water

For a weak acid, $K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2$ (since $\alpha$ is small).

Given $pK_a = 4.74$, so $K_a = 10^{-4.74} = 1.82 \times 10^{-5}$.

With $C = 0.05$ M:

Step 2: Effect of added HCl (common ion effect)

In presence of $x$ M HCl, $[H^+] \approx x$ (from strong acid dominates).

From $K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \approx \frac{x \cdot (C\alpha')}{C}$, we get:

$\alpha' = \frac{K_a}{x}$ …

The key here is understanding that acetic acid is a weak acid — it only partially ionizes in water. The degree of ionization $\alpha$ tells us what fraction of the acid molecules have actually dissociated into $H^+$ and $CH_3COO^-$. When we add a strong acid like HCl, which provides a high concentration of $H^+$ ions, the equilibrium shifts left (Le Chatelier’s principle), suppressing the ionization of the weak acid. This is the common ion effect in action.

Let’s work through it step by step.

1. Find $K_a$ from $pK_a$ The relationship is $pK_a = -\log K_a$, so $K_a = 10^{-pK_a}$.

This is a small number, confirming acetic acid is weak.

2. Set up the ionization equilibrium for pure 0.05 M acetic acid Let initial concentration be $C = 0.05$ M. If $\alpha$ is the degree of ionization, then:

At equilibrium:

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

The acid dissociation constant is:

3. Solve for $\alpha$ in pure solution Since $K_a$ is small and $C$ is moderate, $\alpha$ will be small, so we can approximate $1-\alpha \approx 1$. Then:

That’s about 1.91% ionization. Let’s check if the approximation was valid: $1-\alpha = 0.9809$, close to 1, so it’s fine.

4. Now add HCl — the common ion effect HCl is a strong acid, fully dissociated. So in a solution containing both acetic acid and HCl, the total $[H^+]$ comes mostly from HCl. Let $C_{HCl}$ be the concentration of added HCl. The equilibrium now has initial $[H^+] = C_{HCl}$ (from HCl) plus a tiny amount from the weak acid. Let $\alpha'$ be the new degree of ionization of acetic acid. Then:

The $K_a$ expression becomes:

5. Simplify using the fact that $C\alpha'$ is tiny compared to $C_{HCl}$ Because HCl suppresses ionization, $\alpha'$ will be even smaller than before. So $C\alpha' \ll C_{HCl}$ (we’ll verify after). Also $1-\alpha' \approx 1$. Then:

6. Case (a): 0.01 M HCl $C_{HCl} = 0.01 \, \text{M}$ …