what is the acid ionization/equilibrium....?

Question

Calculate the acid ionization constant, ka, of a weak monoprotic acid a 1.0 M solution of which has an hydronium ion concentration of 4.18x10^-3 M.

does ka=(4.18x10^-3) (4.18x10^-3) / (what do i put here?)


Can anyone help me with this and please explain the process? thanks!

Answer 1

The equilibrium is

HA <> H+ + A-
at equilibrium

[ H+ ] = [ A- ] = 4.18 x 10^-3

[HA] = 1- 4.18x10^-3 = 0.996

Ka = [H+][A-] / [HA ] = (4.18x10^-3)^2 / 0.996 =1.75 x 10^-5

Answer 2

Ka = Ka^2/Kw