Question

Oswalds dilution law is applicable for both weak electrolytes and strong electrolytes- True or False?

• A. True
• B. False

Answer 1

The correct option is B

False

For a particular temperature, consider an aqueous solution of a weak electrolyte HA. In aqueous solution, HA remains in equilibrium with its ions.

HA+ $H_{2}O$ $\leftrightharpoons$ $H^{+}$+ $A^{-}$

$K_{eq}$ = $\frac{\left[H^{+}\right]\left[A\right]}{\left[HA\right]\left[H_{2}O\right]}$

$K_a$ = $K_{eq}$ [$H_{2}O$] = $\frac{\left[H^{+}\right]\left[A\right]}{\left[H_{2}O\right]}$

If the initial concentration of HA is c M, and the degree of dissociation be $\alpha$, then at equilibrium,

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

Throughout this chapter, whenever weak electrolytes are used, the value of α is very small. In this case, it can be approximated as 1- $\alpha$ $\approx$1

Hence K $\alpha$ = c ${\alpha }^2$ or $\alpha$ = $\sqrt{\frac{K\alpha }{c}}$

This is known as the Oswald's dilution law. It is strictly applicable only for weak electrolytes. You can't apply the above logic to strong electrolytes like HCl, HI, $HCl{O}_4$, NaOH, KOH etc

At infinite dilution, c $\to$ 0 and 1 - $\alpha$$\to$ 0 or $\alpha$ $\to$ 1

We can say that at infinite dilution, even weak electrolytes behave like strong ones - they dissociate almost completely. Experimental verifications reveal that this Oswald's law is applicable only for weak electrolytes.