What is the pH of 0.10 M NaF (aq). The Ka of HF is 6.8 x 10-4

2.08

8.08

5.92

14.92

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