Final answer:
The equilibrium fluoride ion concentration can be calculated by setting up the equilibrium expression using the given Ka (acid dissociation constant) for hydrofluoric acid (HF), and solving for x, which represents the change in concentration during the dissociation of HF in water.
Step-by-step explanation:
This question involves the calculation of the equilibrium fluoride ion concentration in a solution composed of hydrofluoric acid (HF) and hydrochloric acid (HCl). To answer this question, we will use the concept of chemical equilibrium, where we use the acid dissociation constant (Ka) given for HF.
For HF, the dissociation reaction in water is: HF <=> H+ + F-, with a given Ka of 6.8 x 10-4. We initially have 0.13 M HF, which partially dissociates into H+ and F-. Let’s assume x is the change in concentration, our equilibrium concentrations would be: [HF] = 0.13 - x, [H+] = x, [F-] = x.
Applying the Ka expression for HF, we get: Ka = [H+][F-] / [HF] = x*x / (0.13 - x) = 6.8 x 10⁻⁴. Solving this equation for x, should give us the equilibrium concentration of F-, expressed in mM. Because Ka is very small, we can make the assumption that x will be very small compared to the initial concentration of HF, thus negligible, which simplifies the equation a lot.
Note: The presence of HCl, a strong acid, does not interfere with this calculation because it entirely dissociates to H+ and Cl-, and only contributes to the overall [H+] of the solution, but does not play any role in the dissociation equilibrium of HF.
Learn more about Chemical Equilibrium