pH.initial = 2 , Vs = 200 mL = 0,2 L
stim ca HNO3 = acid tare el va disocia complet in solutie apoasa
=> pH = -lg[H3O⁺] => [H3O⁺] = Ca = 10⁻² mol/L, Ca = Cm acid
stim Cm = niu/Vs
=> niu = CmxVs = 10⁻²x0,2 = 0,002 moli
Vs.final = Vs1 - 50 = 200 - 50 = 150 mL = 0,15 L
=> Cm.final = 0,002/0,15 = 0,013(3)
=> pH = -lg(0,013(3)) = -(-1,875) = 1,88
