If Kb for NX3 is 8.5×10−6, what is the pOH of a 0.175 M aqueous solution of NX3?

Answer 1

NX₃ + H₂O ==> HNX₃⁺ + OH^-

Kb = 8.5x10^-6 = [HNX₃⁺]{OH^-] / [NX₃]

8.5x10^-6 = (x)(x) / 0.175 - x (and assuming x is small we can ignore it in the denominator)

x² = 1.49x10^-6

x = 1.2x10^-3 = [OH^-] (note: this is small compared to 0.175 so above assumption was valid)

pOH = -log [OH^-] = -log 1.2x10^-3

pOH = 2.9