Question

A monobasic acid with a concentration of 0.04 M is dissociated to the extent of 0.5% in an aqueous solution. What is the dissociation constant of the acid?

• A. \(10^{-6}\)
• B. \(10^{-5}\)
• C. \(10^{-4}\)
• D. \(10^{-7}\)

Hint:

Use \(K_a = C\alpha^2\) since \(\alpha\) is small here.

Answer 1

The Correct Option is A

Step 1: The degree of dissociation is given as a percentage. So \(\alpha = 0.5\% = 0.005\).

Step 2: For a weak monobasic acid \(HA \rightleftharpoons H^+ + A^-\), the dissociation constant is \(K_a = \dfrac{C\alpha^2}{1-\alpha}\), where C is the initial concentration.

Step 3: Since \(\alpha\) is very small, \(1-\alpha \approx 1\), so \(K_a \approx C\alpha^2\).

Step 4: Substitute the values: \(K_a = 0.04 \times (0.005)^2 = 0.04 \times 0.000025 = 1 \times 10^{-6}\). \[\boxed{K_a = 10^{-6}}\]