a. daca baza este total ionizata,, [OH⁻]=c, mol/l=2x0⁻²mol/l
pOH= -log[OH⁻]= 2 - lg2
pH+pOH=14------> pH= 12+log2
sau [H⁺][OH⁻]⁻= 10⁻¹⁴mol²/l²-----> [H⁺]= 10⁻¹⁴/2x10⁻²=>5x10⁻¹³mol/l
pH= 13 -log5
b. se deduce [OH⁻]= √c Kb=> √8 x10⁻⁴mol/l=2,83x10⁻²mol/l
pOH= 2-log 2,83
pH=12+ log 2,83