Final answer:
The initial pH of the 0.20 M HF solution before the addition of any 0.10 M KOH is calculated using the dissociation of HF and the acid-ionization constant (Ka). The concentration of H+ is determined and then used to calculate the pH, which is approximately 2.06.
Step-by-step explanation:
The question requires calculating the pH before any 0.10 M KOH is added to a solution of 0.20 M HF. Using the acid-ionization constant (Ka) for HF which is 6.8 × 10^-4, we can set up an equation based on the dissociation of HF in water: HF(aq) → H+(aq) + F-(aq)
At equilibrium, we have:
[H+] = [F-] = x
Ka = [H+][F-] / [HF] = x^2 / (0.20 - x)
Assuming x is small compared to 0.20, the equation simplifies to Ka ≈ x^2 / 0.20. Solving for x gives us the concentration of H+ in the solution. We can then use the relation pH = -log[H+] to find the pH.
The calculation will look like this:
Ka = 6.8 × 10^-4 = x^2 / 0.20
x^2 = (6.8 × 10^-4)(0.20)
x = √(6.8 × 10^-4 × 0.20)
x = [H+] ≈ 0.008687
pH = -log(0.008687) ≈ 2.06
Therefore, the initial pH of the solution before the addition of KOH is approximately 2.06.