If kb for nx3 is 4.0×10−6, what is the poh of a 0.175 m aqueous solution of nx3?

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

answer; talc and gypsum;