Question

A monobasic weak acid dissociates to 1.2 % in its 0.01 M solution at 298 K. Calculate the dissociation constant of it.

• A. \( 1.04 \times 10^{-8} \)
• B. \( 1.44 \times 10^{-6} \)
• C. \( 1.30 \times 10^{-6} \)
• D. \( 1.18 \times 10^{-5} \)

Hint:

For weak acids, the dissociation constant can be calculated using the concentration of dissociated and undissociated acid, with dissociation percentage.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Given the degree of dissociation as a percentage and molar concentration, we need to find the acid dissociation constant \( \text{K}_\text{a} \).
Step 2: Key Formula or Approach:
For a weak acid: \( \text{K}_\text{a} = c \alpha^2 \) (where \( \alpha \) is small).
Step 3: Detailed Explanation:
Given:
Concentration \( c = 0.01 \text{ M} = 10^{-2} \text{ M} \)
Percentage dissociation = 1.2%
\( \alpha = 1.2 / 100 = 0.012 = 1.2 \times 10^{-2} \)
Calculate \( \text{K}_\text{a} \):
\[ \text{K}_\text{a} = c \alpha^2 = 10^{-2} \times (1.2 \times 10^{-2})^2 \] \[ \text{K}_\text{a} = 10^{-2} \times 1.44 \times 10^{-4} \] \[ \text{K}_\text{a} = 1.44 \times 10^{-6} \] Step 4: Final Answer:
The dissociation constant is \( 1.44 \times 10^{-6} \).