. What is the acid dissociation constant for an acid at equilibrium, HX H ++ X – ?

Break Down the Problem

1. We need to determine the acid dissociation constant ($K_a$) for the equilibrium reaction: $HX \rightleftharpoons H^+ + X^-$

Relevant Concepts

2. The acid dissociation constant is defined by the following expression: $K_a = \frac{[H^+][X^-]}{[HX]}$ where:
   • $[H^+]$ is the concentration of hydrogen ions at equilibrium,
   • $[X^-]$ is the concentration of the conjugate base at equilibrium,
   • $[HX]$ is the concentration of the undissociated acid at equilibrium.

Analysis and Detail

3. At equilibrium, let:

   • The concentration of $HX$ at the start be $[HX]_0$,
   • The change in concentration be $x$.

   Then, at equilibrium:

   • $[HX] = [HX]_0 - x$,
   • $[H^+] = x$,
   • $[X^-] = x$.

   Substituting these into the expression for $K_a$: $K_a = \frac{(x)(x)}{[HX]_0 - x} = \frac{x^2}{[HX]_0 - x}$

Verify and Summarize

4. The $K_a$ expression shows that the acid dissociation constant depends on the change in concentration of the products and the remaining reactants at equilibrium. To compute a specific value for $K_a$, values for $[H^+]$, $[X^-]$, and $[HX]$ need to be determined from experimental data.

Final Answer

The acid dissociation constant $K_a$ for the equilibrium reaction is given by: $K_a = \frac{[H^+][X^-]}{[HX]}$