If alpha is the degree of ionization,C is the | vedclass

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Question

(C) For a weak acid $HA$ dissociating as $HA \rightleftharpoons H^+ + A^-$,the equilibrium constant $K_a$ is given by $K_a = \frac{[H^+][A^-]}{[HA]}$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For a weak acid $HA$ dissociating as $HA \rightleftharpoons H^+ + A^-$,the equilibrium constant $K_a$ is given by $K_a = \frac{[H^+][A^-]}{[HA]}$.At equilibrium,$[H^+] = C\alpha$,$[A^-] = C\alpha$,and $[HA] = C(1 - \alpha)$.Substituting these,$K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha}$.Since the electrolyte is weak,$\alpha \ll 1$,so $(1 - \alpha) \approx 1$.Thus,$K_a \approx C\alpha^2$,which simplifies to $\alpha^2 = \frac{K_a}{C}$ or $\alpha = \sqrt{\frac{K_a}{C}}$.

If $\alpha$ is the degree of ionization,$C$ is the concentration of a weak electrolyte,and $K_a$ is the acid ionization constant,then the correct relationship between $\alpha$,$C$,and $K_a$ is:

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