Final answer:
To determine the pH of a 0.22 M NaF solution, we need to consider the dissociation of NaF. The F- ions react with water to form HF and OH- ions. The concentration of HF is used to calculate the concentration of H+ ions, which allows us to determine the pH of the solution.
Step-by-step explanation:
To determine the pH of a 0.22 M NaF solution at 25°C, we need to consider the dissociation of NaF. NaF is a salt composed of sodium cations (Na+) and fluoride anions (F-). When NaF dissolves in water, it will dissociate into Na+ ions and F- ions.
The F- ions will react with water to form HF (hydrofluoric acid) and OH- (hydroxide ions). The HF will then dissociate partially into H+ (protons) and F- ions. Since the F- ions are consumed in the reaction, the concentration of F- will be lower than the initial concentration of NaF.
To calculate the pH of the solution, we need to determine the concentrations of H+ and OH- ions. We can use the Ka of HF to calculate the concentration of H+ ions, and the concentration of OH- ions will be equal to the concentration of HF. Finally, we can calculate the pH using the equation pH = -log[H+].
First, let's write the dissociation reactions:
NaF(s) → Na+(aq) + F-(aq)
F-(aq) + H2O(l) → HF(aq) + OH-(aq)
HF(aq) → H+(aq) + F-(aq)
Let's calculate the concentrations step-by-step:
- The initial concentration of F- is 0.22 M, which is also the concentration of HF.
- Using the Ka of HF (Ka = 3.5 x 10^-5), we can calculate the concentration of H+ ions. Since HF is a weak acid, we can assume that most of the HF will remain undissociated. Therefore, [H+] ≈ x, where x is the concentration of H+ ions.
- The concentration of OH- ions is equal to the concentration of HF, which is 0.22 M.
- Finally, we can calculate the pH using the equation pH = -log[H+].
Let's plug in the values:
- [HF] = [H2O] = 0.22 M
- Ka = 3.5 x 10^-5
Now let's calculate:
3.5 x 10^-5 = x * x / 0.22 => x = sqrt(0.22 * 3.5 x 10^-5) ≈ 1.66 x 10^-3 M
The concentration of H+ ions is approximately 1.66 x 10^-3 M. Therefore, the pH of the solution is -log(1.66 x 10^-3) ≈ 2.78.