To find the pH of a 0.100 M solution of hydrofluoric acid (HF), we first need to set up the dissociation equation for HF. HF is a weak acid and partially dissociates in water:
HF ⇌ H+ + F–
Using the given Ka value of 6.3 x 10-4, we can establish the following equilibrium expression:
Ka = \\frac{[H^+][F^-]}{[HF]}
Let x be the concentration of H+ ions at equilibrium. Since HF dissociates into one H+ and one F–, we have:
- [H+] = x
- [F–] = x
- [HF] = 0.100 – x
Substituting these values into the Ka expression gives:
6.3 x 10-4 = \\frac{x imes x}{0.100 – x}
Assuming that x is much smaller than 0.100, we can simplify the equation to:
6.3 x 10-4 = \\frac{x^2}{0.100}
By rearranging this equation to solve for x, we get:
x2 = 6.3 x 10-4 * 0.100
x2 = 6.3 x 10-5
x = √(6.3 x 10-5)
x ≈ 0.00794 M
This x value represents the concentration of H+ ions in the solution. To find the pH, we use the formula:
pH = -log[H+]
Substituting in our value for x:
pH ≈ -log(0.00794) ≈ 2.10
Therefore, the pH of a 0.100 M solution of hydrofluoric acid (HF) is approximately 2.10.