Question

Acetic acid is 5% ionised in its decimolar solution. Calculate the dissociation constant of acetic acid.

Hint:

"Decimolar" means \( 0.1 \, M \), "Centimolar" means \( 0.01 \, M \), and "Millimolar" means \( 0.001 \, M \). Knowing these terms is essential for solving concentration problems.

Answer 1

Step 1: Understanding the Concept:
Acetic acid is a weak electrolyte. Its dissociation constant (\( K_a \)) is related to its concentration (\( C \)) and degree of dissociation (\( \alpha \)) by Ostwald's dilution law.
Step 2: Key Formula or Approach:
For a weak acid:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]
If \( \alpha \) is very small (\(<5% \)), the formula can be simplified to:
\[ K_a = C \alpha^2 \]
Step 3: Detailed Explanation:
Given:
Degree of ionisation, \( \alpha = 5% = \frac{5}{100} = 0.05 \).
Concentration, \( C = \text{decimolar} = 0.1 \, M = 10^{-1} \, M \).
Using the formula:
\[ K_a = 0.1 \times (0.05)^2 \]
\[ K_a = 10^{-1} \times 0.0025 \]
\[ K_a = 10^{-1} \times 2.5 \times 10^{-3} \]
\[ K_a = 2.5 \times 10^{-4} \]
Step 4: Final Answer:
The dissociation constant (\( K_a \)) of acetic acid is \( 2.5 \times 10^{-4} \).