Calculate the pH of the following aqueous solution: 0.35 M NaF (pK_a for HF = 3.14) A)5.66 B)2.68 C)8.34 D)11.32 E)none of these

Question

Quizplus · Accepted Answer

NaF is a salt of a weak acid (HF) and a strong base (NaOH). In water, NaF will dissociate into its constituent ions: Na+ and F-. F- is the conjugate base of a weak acid, so it will undergo hydrolysis in water and react with H+ ions in a reversible reaction as follows:
F- + H2O ⇌ HF + OH-
The equilibrium expression for this reaction is:
Kb = [HF][OH-]/[F-]
Kb can be related to Ka for the acid HF through the equation Kw = Ka·Kb. Using pKs/pKa values for HF, we can calculate the pKb value for F- as follows:
Kw = Ka·Kb
1.0x10^-14 = 6.9x10^-4·Kb
Kb = 1.45x10^-11
pKb = 10.84
Using the equilibrium expression and the Kb value, we can calculate the concentration of OH- ions produced in the reaction:
Kb = [HF][OH-]/[F-]
1.45x10^-11 = x^2/0.35
x = 1.65x10^-6 M
At equilibrium, the concentration of OH- ions will be equal to that of H+ ions produced by the autoionization of water, so we can use the equation Kw = [H+][OH-] to calculate the pH of the solution:
Kw = [H+][OH-]
1.0x10^-14 = [H+][1.65x10^-6]
[H+] = 6.06x10^-9 M
pH = -log[H+] = 8.34

Calculate the PH of the Following Aqueous Solution: 0