HF acid ionization constant (Ka) is 3.5 X 10-4. In a solution of HF acid @ equilibrium [H+] = 0.090 M. What is the concentration of F-?
HF><H+ + a=x^2/.090
solve for x
To find the concentration of F- in the given solution, we need to use the ionization constant (Ka) of HF and the equilibrium concentration of H+.
The ionization constant (Ka) represents the extent to which a acid reacts with water to form H+ ions. The equation for the ionization of HF is as follows:
HF + H2O ⇌ H3O+ + F-
The equilibrium constant expression for this reaction is given by:
Ka = [H3O+][F-] / [HF]
We are given the value of Ka as 3.5 × 10^-4 and the equilibrium concentration of [H+] as 0.090 M.
[H+] = [H3O+]
[H+] = 0.090 M
Since HF is a weak acid, we can assume that the concentration of [HF] will be approximately equal to the initial concentration of HF. Therefore, we can assume that [HF] ≈ [HF]initial.
Assuming that the initial concentration of HF is represented as [HF]initial, we can write the equilibrium concentration of HF as [HF] = [HF]initial - [H+].
Substituting these values into the equilibrium constant expression, we get:
Ka = (0.090) × [F-] / ([HF]initial - 0.090)
Now, rearranging the equation to solve for [F-]:
[F-] = (Ka × ([HF]initial - 0.090)) / 0.090
Substituting the value of Ka as 3.5 × 10^-4, we can calculate the concentration of [F-]:
[F-] = (3.5 × 10^-4 × ([HF]initial - 0.090)) / 0.090
Please provide the initial concentration of HF ([HF]initial) to obtain the final concentration of [F-].