Determine the pH of a 0.22 M NaF solution at 25°C.The K_a of HF is 3.5 × 10^-5. A) 10.20 B) 5.10 C) 8.90 D) 11.44 E) 2.56

Question

Quizplus · Accepted Answer

NaF is a salt of a weak base (F-) and a strong acid (NaOH). In water, it will hydrolyze and form a basic solution.

The hydrolysis reaction is: 
F- + H2O ⇌ HF + OH-

The equilibrium constant for this reaction can be expressed as:
Kb = [HF][OH-]/[F-]

We can use the relationship between Ka and Kb for the conjugates:

Ka x Kb = Kw = 1.0 x 10^-14

Ka for HF is 3.5 x 10^-4. Therefore, Kb for F- can be found:

Kb = Kw/Ka = (1.0 x 10^-14)/(3.5 x 10^-4) = 2.9 x 10^-11

pKb = -log(Kb) = -log(2.9 x 10^-11) = 10.54

pOH = 1/2 (pKw - pKb) = 1/2 (14.00 - 10.54) = 1.73

pH = 14.00 - pOH = 14.00 - 1.73 = 12.27

However, we need to consider the effect of the Na+ ions, which will slightly decrease the pH. Therefore, the final pH is expected to be slightly lower than 12.27.

The correct answer is therefore (C) 8.90.

Determine the PH of a 0