Question

Find the pH of a 0.245 M NaF solution. (The Ka of hydrofluoric acid, HF, is 3.5×10−4.)   

Answer 1

Ans. Being salt of weak acid (HF) and strong base, NaF is a strong electrolyte. It completely dissociates into constituent ions Na⁺ and F^- when dissolved in water.

The fluoride ion (F^-) is a conjugate base of weak acid. Such anions turn the solution basic by accepting a proton from water and thus increasing [OH^-].

Na⁺ simply acts as a spectator ion.

# Following complete dissociation of NaF in water, the initial [F^-] = 0.245M

Since F^- acts as conjugate base, we need to calculate base dissociation (Kb) constant from the given acid dissociation constant (Ka).

            Ka x Kb = 10^-14

            Or, Kb = 10^-14 / Ka = 10^-14 / (3.5 x 10^‑4) = 2.8571 x 10^-11

# Create an ICE table with initial [F^-] = 0.245 M as shown in figure.

Now,

Base dissociation constant, Kb = [HF] [OH^-] / [F^-]           - all concentrations at equilibrium

            Or, 2.8571 x 10^-11 = (X) (X) / (0.245 - X)

            Or, 2.8571 x 10^-11 = X² / (0.245)                 ; assuming 0.245 >> X

            Or, 2.8571 x 10^-11 x 0.245 = X²

            Or, X = (7.00 x 10^-12)^½

            Hence, X = 1.323 x 10^-6

# Now, [OH^-] at equilibrium = X = 1.323 x 10^-6

pOH os solution = - log [OH^-] = - log (1.323 x 10^-6) = 5.88

And, using pH + pOH = 14.00

            pH = 14.00 – pOH = 14.00 – 5.88 = 8.12

Hence, pH of NaF solution (0.245 M) = 8.12