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: Set up an ICE table. Let initial concentration C = 0.04 M and degree of dissociation $\alpha = 0.005$.

Step 2: At equilibrium, $[HA] = C(1-\alpha) = 0.04 \times 0.995 = 0.0398$ M, and $[H^+] = [A^-] = C\alpha = 0.04 \times 0.005 = 2 \times 10^{-4}$ M.

Step 3: Apply the equilibrium expression: $K_a = \dfrac{[H^+][A^-]}{[HA]} = \dfrac{(2\times10^{-4})^2}{0.0398} = \dfrac{4\times10^{-8}}{0.0398}$

Step 4: $K_a \approx 1.005\times10^{-6}$, which rounds to $10^{-6}$. \[\boxed{K_a = 10^{-6}}\]