Question

At \( 27^\circ C \), the degree of dissociation of weak acid (HA) in its 0.5M aqueous solution is 1%. Its \( K_a \) value is approximately:

• A. \( 5 \times 10^{-4} \)
• B. \( 5 \times 10^{-5} \)
• C. \( 5 \times 10^{-6} \)
• D. \( 5 \times 10^{-8} \)

Hint:

For weak acids, use the formula \( K_a = C \alpha^2 \), where \( \alpha \) is the degree of dissociation and \( C \) is the concentration of the acid.

Answer 1

The Correct Option is B

Step 1: Define the given values
Degree of dissociation \( \alpha = 1\% = 0.01 \)
Initial concentration \( C = 0.5M \)
Step 2: Use the expression for \( K_a \)
\[ K_a = C \alpha^2 \] Substituting the values: \[ K_a = (0.5) \times (0.01)^2 \] \[ K_a = 5 \times 10^{-5} \] Thus, the correct answer is \( 5 \times 10^{-5} \).