Final answer:
The pH of a 0.10 M hydrofluoric acid solution with a Ka of 6.77 x 10^-4 is approximately 2.12, calculated by finding the concentration of H3O+ ions using an ICE table and applying the Ka value.
Step-by-step explanation:
The dissociation of hydrofluoric acid (HF) in water can be represented by the following equilibrium reaction:
- HF (aq) + H2O (l) ⇌ H3O+ (aq) + F- (aq)
To calculate the pH of the HF solution, we must first find the concentration of H3O+ ions at equilibrium. We use the Ka value provided and set up an ICE table. For a 0.10 M HF solution, the initial concentrations are [HF] = 0.10 M, and [H3O+] and [F-] are both essentially 0 M.
At equilibrium, a small amount of HF has dissociated, which we can represent by the variable x, altering the concentrations to [HF] = 0.10 - x, and [H3O+] = [F-] = x. Applying the given Ka of 6.77 x 10^-4:
Ka = [H3O+][F-]/[HF] = x^2 / (0.10 - x)
Solving for x, we obtain the concentration of H3O+ ions. The pH is then determined using the formula:
pH = -log[H3O+]
After computing the pH, we arrive at the answer that the pH of the 0.10 M HF solution is approximately 2.12.