Final answer:
To determine the pH of a 0.10 M NH4F solution, we compare the ionization constants of NH4+ (Ka = 5.6 × 10⁻¹⁰) and F- (Kb calculated from Kw and the Ka of HF to be 1.47 × 10⁻¹ⁱ). Since Ka > Kb, the solution is acidic with a pH of less than 7.
Step-by-step explanation:
The pH of a 0.10 M solution of NH4F can be determined by considering the ionization of both NH4+ and F-. NH4+ is the conjugate acid of NH3 and F- is the conjugate base of HF. According to the given Ka values, the ionization constant for NH4+, K2, is 5.6 × 10⁻¹⁰, which suggests that NH4+ can dissociate to form H3O+ and NH3. The ionization constant for F-, K1, would be taking K_w (1.0 × 10⁻) and dividing it by the Ka of HF (6.8 × 10⁻⁴), giving 1.47 × 10⁻¹ⁱ (K_w / Ka of HF = Kw/Ka).
Given that the Ka for NH4+ is greater than the Kb for F- (K1), we conclude that the solution is acidic. For a more precise pH value, one would need to set up equations to use for the hydrolysis of NH4+ and the reaction of F- with water to form HF and OH-. However, based on the given K values, we can say that the solution has a pH of less than 7.