Respuesta :

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

pOH=-lg[OH⁻]
pOH = 14-pH

14-pH = -lg[OH⁻]

[OH⁻]=10^(pH-14)

[OH⁻]=10^(11.8-14)=0.00631 = 6.31·10⁻³ mol/L