Given information:
Concentration of NaF = 0.10 M
Ka of HF = 6.8*10⁻⁴
To determine:
pH of 0.1 M NaF
Explanation:
NaF (aq) ↔ Na+ (aq) + F-(aq)
[Na+] = [F-] = 0.10 M
F- will then react with water in the solution as follows:
F- + H2O ↔ HF + OH-
Kb = [OH-][HF]/[F-]
Kw/Ka = [OH-][HF]/[F-]
At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x
10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x
x = [OH-] = 1.21*10⁻⁶ M
pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92
pH = 14 - pOH = 14-5.92 = 8.08
Ans: (b)
pH of 0.10 M NaF is 8.08
