Question

At 298 K, 0.1 M solution of acetic acid is 1.34 % ionized. What is the dissociation constant of acetic acid?

• A. $1.4 \times 10^{-3}$
• B. $1.8 \times 10^{-5}$
• C. $1.6 \times 10^{-3}$
• D. $1.34 \times 10^{-5}$

Hint:

To quickly approximate $(1.34)^2$ without painful manual multiplication, notice that $13^2 = 169$ and $14^2 = 196$. Therefore, $1.34^2$ must lie between $1.69$ and $1.96$, pointing directly toward $1.8$ among the options!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem states that a weak acid (acetic acid) has a concentration of $0.1\ \text{M}$ and is $1.34\ \%$ ionized at a given temperature. We need to find the acid dissociation constant ($K_a$).

Step 2: Key Formula or Approach:
For a weak monobasic acid, Ostwald's Dilution Law relates the dissociation constant ($K_a$) to the initial molar concentration ($C$) and the degree of dissociation ($\alpha$) via:
$$K_a = \frac{C\alpha^2}{1-\alpha}$$ Since the degree of ionization is very small ($\alpha \ll 1$), the denominator $(1-\alpha)$ can be simplified to approximately 1. This gives the simplified expression:
$$K_a = \alpha^2 C$$

Step 3: Detailed Explanation:
Let's list our given values and convert the percentage to a regular decimal fraction:
Molar concentration ($C$) = $0.1\ \text{M} = 10^{-1}\ \text{M}$
Percentage Ionization = $1.34\ \%$
Degree of dissociation ($\alpha$) = $\frac{1.34}{100} = 1.34 \times 10^{-2}$
Now, substitute these parameters into the expression for $K_a$:
$$K_a = (1.34 \times 10^{-2})^2 \times 0.1$$ Square the terms inside the parentheses:
$$(1.34)^2 = 1.7956$$ $$(10^{-2})^2 = 10^{-4}$$ $$K_a = 1.7956 \times 10^{-4} \times 10^{-1}$$ $$K_a = 1.7956 \times 10^{-5} \approx 1.8 \times 10^{-5}$$ This matches option (B).

Step 4: Final Answer:
The dissociation constant of the acid is $1.8 \times 10^{-5}$, which matches option (B).