To find the correct equation for the acid dissociation constant, [tex]\(K_a\)[/tex], of hydrofluoric acid (HF) when it dissolves in water, we need to look at the equilibrium reaction:
[tex]\[ HF(aq) + H_2O(l) \rightleftarrows H_3O^+(aq) + F^-(aq) \][/tex]
The general definition of the acid dissociation constant, [tex]\( K_a \)[/tex], for an acid [tex]\( HA \)[/tex] dissociating in water can be expressed as:
[tex]\[ HA(aq) + H_2O(l) \rightleftarrows H_3O^+(aq) + A^-(aq) \][/tex]
Here, the acid dissociation constant [tex]\( K_a \)[/tex] is given by:
[tex]\[ K_a = \frac{[H_3O^+][A^-]}{[HA]} \][/tex]
Applying this to our specific reaction, HF dissociating in water:
[tex]\[ HF(aq) + H_2O(l) \rightleftarrows H_3O^+(aq) + F^-(aq) \][/tex]
The [tex]\( K_a \)[/tex] expression for this reaction would be:
[tex]\[ K_a = \frac{[H_3O^+][F^-]}{[HF]} \][/tex]
Given the choices:
A. [tex]\( K_a = \frac{[H_3O^+][F^-]}{[HF]} \)[/tex]
B. [tex]\( K_a = \frac{[H_3O^+][F^-]}{[HF][H_2O]} \)[/tex]
C. [tex]\( K_a = \frac{[HF]}{[H_3O^+][F^-]} \)[/tex]
D. [tex]\( K_a = \frac{[HF][H_2O]}{[H_3O^+][F^-]} \)[/tex]
We can see that the correct choice corresponding to the equation for the acid dissociation constant, [tex]\( K_a \)[/tex], of HF is:
[tex]\[ K_a = \frac{[H_3O^+][F^-]}{[HF]} \][/tex]
Thus, the correct answer is:
A. [tex]\( K_a = \frac{[H_3O^+][F^-]}{[HF]} \)[/tex]
