Will someone help me with this chem question ?

Question

a 0.500m solution of a weak acid,HX, is only partially ionized.the [h+] was found to be 4.02x 10^-3m find the dissociation constant for this acid?

Answer 1

Assume the dissociation equilibrium HX ↔ H+ + X-

The expression for the dissociation constant is

Ka = [H+] [X-] / [HX]

By the reaction equation, if [H+] = 4.02x10^-3 m, then [X-] = [H+] = 4.02x10^-3 m, and [HX] = [HX]o - [H+] = 0.500 m - 4.02x10^-3 m. Just substitute into the equilibrium expression to get:

Ka = (4.02x10^-3 m) (4.02x 0^-3 m) / (0.500 m - 4.02x10^-3 m)

Doing the math on my calculator,

Ka = 3.26x10^-5, or pKa = 4.49.