Final answer:
The student's question involves calculating the equilibrium concentrations of hydronium, fluoride, and hydroxide ions in a hydrofluoric acid solution using an ICE table and the acid dissociation constant (Ka). Approaching the problem, you would set up the equilibrium expression with Ka and solve for the variables, assuming minimal dissociation of HF due to it being a weak acid.
Step-by-step explanation:
The question relates to the dissociation of hydrofluoric acid (HF) and the calculation of the concentrations of hydronium ions ([H3O+]), fluoride ions ([F-]), and hydroxide ions ([OH-]) in a 0.580 M HF solution given its Ka value of 6.8 x 10^-4. To find these concentrations, you would set up an ICE table (Initial, Change, Equilibrium) and use the acid dissociation constant (Ka) to solve for the equilibrium concentrations based on the reaction HF (aq) + H2O (l) ⇌ H3O+ (aq) + F- (aq).
Assuming complete dissociation, which is not the case for a weak acid, but allows approximation:
- [H3O+] = [F-] = x
- [HF] = 0.580 - x ≈ 0.580 M (since x will be small compared to 0.580 M)
- Ka = [H3O+][F-] / [HF] = x^2 / (0.580 - x) ≈ x^2 / 0.580 M
- Solve for x to find [H3O+] and [F-]
Since HF is a weak acid, its dissociation in water is not complete. However, the addition of fluoride from sources like sodium fluoride will suppress the dissociation of HF further, and therefore, the approximation that [HF] stays close to 0.580 M is reasonable.
Finally, the concentration of hydroxide ions, [OH-], can be calculated using the ion-product constant for water (Kw), via the relationship Kw = [H3O+][OH-]. Given [H3O+] from the previous calculation, solve for [OH-]. Keep in mind the charge balance and the relationship between the acid dissociation constant (Ka) of HF and the base dissociation constant (Kb) of F-.