Question

The dissociation constant of a weak monobasic acid is 3.2 x 104. Calculate the degree of dissociation in its 0.04 M solution.

Answer 1

Degree of dissociation ≈ 0.086 (8.6%).

Answer 2

For a weak acid HA dissociating as

$$\text{HA} \leftrightarrow \text{H}^+ + \text{A}^-$$

let the initial concentration be $C = 0.04\,M$ and the degree of dissociation be $\alpha$. At equilibrium:

$$[\text{H}^+] = [\text{A}^-] = C\alpha, \quad [\text{HA}] = C(1-\alpha)$$

The acid dissociation constant is given by

$$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{C^2\alpha^2}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$$

Given $K_a = 3.2\times10^{-4}$, we have

$$3.2\times10^{-4} = \frac{0.04\,\alpha^2}{1-\alpha}$$

Rearrange to form a quadratic in $\alpha$:

$$0.04\,\alpha^2 = 3.2\times10^{-4}(1-\alpha)$$ $$0.04\,\alpha^2 + 3.2\times10^{-4}\,\alpha - 3.2\times10^{-4} = 0$$

Divide through by $0.04$ to simplify:

$$\alpha^2 + 0.008\,\alpha - 0.008 = 0$$

Using the quadratic formula:

$$\alpha = \frac{-0.008 \pm \sqrt{(0.008)^2 + 4\times0.008}}{2}$$ $$\alpha = \frac{-0.008 \pm \sqrt{0.000064 + 0.032}}{2} = \frac{-0.008 \pm \sqrt{0.032064}}{2}$$ $$\sqrt{0.032064} \approx 0.1791$$ $$\alpha = \frac{-0.008 + 0.1791}{2} \approx \frac{0.1711}{2} \approx 0.0856$$

Thus, the degree of dissociation is approximately 0.086 (or 8.6%).