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 a weak monobasic acid, the dissociation constant is concentration times the degree of dissociation squared.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For \(\text{HA}\rightleftharpoons\text{H}^++\text{A}^-\) with small \(\alpha\), \(K_a=C\alpha^2\).

Step 2: Substitute
\(C=0.01\ \text{M}\) and \(\alpha=1.2\%=0.012\).

Step 3: Calculation
\(K_a=0.01\times(0.012)^2=0.01\times1.44\times10^{-4}=1.44\times10^{-6}\)

Step 4: Compare the options
The second option, \(1.44\times10^{-6}\), is correct.

Final Answer:
The dissociation constant is \(1.44\times10^{-6}\). \[ \boxed{\text{(B) }1.44\times10^{-6}} \]