what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻ + H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF, the [ OH ⁻] from the given pH, and finally the mass needed to produce that OH⁻ concentration.

The equilibrium constant, Kb, can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is 6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14 ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information:

F⁻ HF OH⁻

Equilibrium X 2.51 x 10⁻⁶ 2.51 x 10⁻⁶

(2.51 x 10⁻⁶) ² / X = 1.5 x 10⁻¹¹ ⇒ X = (2.51 x 10⁻⁶) ² / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL (0.35 L) we need to add:

0.41 mol HF / 1 L * 0.35 L = 0.144 mol

and finally the mass will be:

0.144 mol NaF * 42.0 g/mol NaF = 6.03 g NaF