A monobasic weak acid dissociates $2 \%$ in its 0.002 M solution. Calculate the dissociation constant of weak acid.

Dissociation constant calculation for weak acids

For a weak acid HA with dissociation equilibrium $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$, the acid dissociation constant is given by $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$.

Given initial concentration $C = 0.002\ \text{M}$ and percentage dissociation 2%, the degree of dissociation is $\alpha = 0.02$.

From the equilibrium concentrations $[\text{H}^+] = [\text{A}^-] = C\alpha$ and $[\text{HA}] = C(1-\alpha)$, the dissociation constant becomes $K_a = \frac{C\alpha^2}{1-\alpha}$.

Since $\alpha = 0.02$ is sufficiently small, we apply the approximation $1-\alpha \approx 1$, yielding $K_a \approx C\alpha^2$.

Substituting values: $K_a \approx (2 \times 10^{-3}) \times (2 \times 10^{-2})^2 = 8 \times 10^{-7}$.

The calculated value matches option B.