Answer:
Step-by-step explanation:
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