Kb(NH₂OH) = 1,8·10⁻⁵.
c₀(NH₂OH) = 0,0500 M =
0,05 mol/L.
c(NH₂⁺) = c(OH⁻) = x.
c(NH₂OH) = 0,05 mol/L - x.
Kb = c(NH₂⁺) · c(OH⁻) / c(NH₂OH).
0,0000000066 = x² / (0,05 mol/L - x).
solve quadratic equation: x = c(OH⁻) = 0,000018 mol/L.
pOH = -log(0,000018 mol/L) = 4,74.
pH = 14 - 4,74 = 9,23.
