Question

If kb for nx3 is 1.5×10−6, what is the poh of a 0.175 m aqueous solution of nx3?

Answer 1

Answer is: pOH = 3,29.
Kb(NH₃) = 1,5·10⁻⁶.
c(NH₃) = 0,175M.
pOH = ?
Chemical reaction: NH₃ + H₂O ⇄ NH₄⁺ + OH⁻.
Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).
c(NH₄⁺) = c(OH⁻) = x.
x² = Kb · c(NH₃) 
x² = 1,5·10⁻⁶ · 0,175 = 2,625 ·10⁻⁷.
x = c(OH⁻) = √2,625 ·10⁻⁷ = 5,12 · 10⁻⁴.
pOH = -log(c(OH⁻)) = 3,29.