pOH = 3.08
NX3 + H2O <----> NHX3+ + OH-
Kb = 4.0 x 10^-6
Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).
c(NH₄⁺) = c(OH⁻) = x.
x² = Kb · c(NH₃)
x² = 4.0 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.
x = c(OH⁻) = √(7.0 × 10⁻⁷)
= 8.367 × 10⁻⁴
pOH = -log(c(OH⁻))
=- log ( 8.367 × 10⁻⁴)
= 3.08
