1. [HO⁻] = [NaOH] = 5ₓ10⁻² [H⁺] = 10⁻¹⁴ : (5ₓ10⁻2) = 2ₓ10⁻¹³ mol/l
pH = - lg(2ₓ10⁻¹³) = - 0,3 + 13 = 12,7
2. acid : [H⁺] = 10⁻³ Vs = 2ₓ10⁻³ l n acid = 2ₓ10⁻⁶ moli
baza :pOH = 4 [HO⁻] = 10⁻⁴ Vs = 3ₓ10⁻³ l n baza = 3ₓ10⁻⁷ moli
reactioneaza : 3ₓ10⁻⁷ moli acid + 3ₓ10⁻⁷ moli baza
solutia finala : Vs = 5ₓ10⁻³ l nH⁺ = 2ₓ10⁻⁶ - 3ₓ10⁻⁷ = 1,7ₓ10⁻⁶
[H⁺] = 1,7ₓ10⁻⁶/5ₓ10⁻³ = 0,34ₓ10⁻³ = 34ₓ10⁻⁵ mol/l
pH = - 1,53 +5 = 3,47
3. [H⁺] = 10⁻pH = 10⁻⁹,⁶⁷ lg[H⁺] = -9,67 [H⁺] = 2,138ₓ10⁻10