Question

(vii) Which formula co-relates degree of dissociation and concentration of electrolyte?

• A. $c = \sqrt{\frac{K_a}{\alpha}}$
• B. $\alpha = \sqrt{\frac{K_a}{c}}$
• C. $c = \sqrt{K_a \alpha}$
• D. $c = \sqrt{\frac{\alpha}{K_a}}$

Answer 1

Answer: (B)

$\alpha = \sqrt{\frac{K_a}{c}}$

Answer 2

The relationship between the degree of dissociation ($\alpha$), concentration of the electrolyte ($c$), and the dissociation constant ($K_a$) is derived from the equilibrium expression for a weak electrolyte.

Consider a weak acid, HA, dissociating in water:

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

Initial concentrations:

$c \quad 0 \quad 0$

At equilibrium:

$c(1-\alpha) \quad c\alpha \quad c\alpha$

The acid dissociation constant, $K_a$, is given by:

$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$

Substitute the equilibrium concentrations:

$K_a = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

$K_a = \frac{c^2\alpha^2}{c(1-\alpha)}$

$K_a = \frac{c\alpha^2}{1-\alpha}$

For weak electrolytes, the degree of dissociation ($\alpha$) is very small (i.e., $\alpha << 1$). Therefore, $(1-\alpha)$ can be approximated as 1.

So, the equation simplifies to:

$K_a \approx c\alpha^2$

Rearranging this equation to solve for $\alpha$:

$\alpha^2 = \frac{K_a}{c}$

Taking the square root of both sides:

$\alpha = \sqrt{\frac{K_a}{c}}$