7. pH = 3 ⇔ [H⁺] = 10⁻³ ⇒ [HO⁻] = 10 ⁻¹¹ mol/ l
c = n/Vs n = c·Vs = 10⁻¹¹ ·10⁻¹ = 10⁻¹²
8. [H3O⁺] = 10⁻⁶/10⁻² = 10⁻⁴moli/ l pH = 4
9. n KOH = 112/56 = 2mmoli = 0,002 moli Vs = 0.2 l
[HO⁻] = [KOH] = 0,002/0,2 = 0,01 moli/l = 10⁻²
[H⁺] = 10⁻¹² pH = 12
10. [HO⁻] = c = 10⁻¹ [H⁺] = 10⁻¹³ pH = 13
