Odpowiedź :
[OH-]=10^(-2) mol/dm3
pOH = -log[OH-]
pOH =-log(10^(-2))=-(-2)=2
pH +pOH = 14
pH = 14 -pOH
pH = 14 -2 =12
pOH = -log[OH-]
pOH =-log(10^(-2))=-(-2)=2
pH +pOH = 14
pH = 14 -pOH
pH = 14 -2 =12