trebuie stiut: pH=-lg [H3O⁺]; pOH=-lg[OH⁻]; [H3O⁺][OH⁻]=10⁻¹⁴mol²/ l² ;pH+pOH=14;
Ka= [H3O⁺]²/C,acid
a) pH=3-->[H3O⁺]=10⁻³mol/l
c=[H3O⁺]²/Ka----> c= 10⁻⁶/1,8x10⁻⁵===== calculeaza !!!
b)pH= 13--> pOH=1---> [OH⁻]= 10⁻¹mol/l; deoarece KOH este o baza tare,
[OH⁻]=[KOH]= 0,1M
