Calculate the pH of a 0.370 M Lithium Fluoride (LiF) solution?

Refer to: https://depts.washington.edu/eooptic/links/acidstrength.html
Ka(HF) = 6.6 × 10⁻⁴
Hence, Kb(F⁻) = Kw / Ka(HF) = (1.0 × 10⁻¹⁴) / (6.6 × 10⁻⁴) = 1.52 × 10⁻¹¹
(Ka value from different sources may be slightly different.)

Consider the dissociation of F⁻.
                      F⁻(aq) + H₂O(ℓ) ⇌ HF(aq) + OH⁻(aq)      Kb = 1.52 × 10⁻¹¹
Initial (M):       0.370       ---             0             0
Change (M):      -y          ---           +y           +y
Eqm (M):     0.0370 - y    ---            y             y
               ≈ 0.370 (as Kb is very small)

At equilibrium:
Kb = [HF] [OH⁻] / [F⁻]
1.52 × 10⁻¹¹ = y² / 0.370
y = √(0.370 × 1.52 × 10⁻¹¹)
[OH⁻] = y M = 2.37 × 10⁻⁶ M

pH = Kw - pOH = 14.00 + log(2.37 × 10⁻⁶) = 8.37

HF, a weak acid, has a Ka of 6.6 x 10^-4.  So [H+][F-] is 6.6x10^-4.  If we assume that [F-] >> [H+] then plug in 0.37 M for F- and calculate H+.  Convert to pH.