Kwas mocny + woda
HA + H20 —>H3Oł + A , C0l A- = C0
H20 + H20 <=> H3Oł + OH-
[H301 = C0 - [OH ] oraz Kw = [H30*][0H]
- dwa równania
Kw = [H30*]([H30*] - C0)
[H30+][H30+] - [H3O+]C0 - K, = 0 A = C02 + 4K,
[H30+] = (C0 + oC02 - 4KJ* )/2 gdy C02 » Kw +o [H30+] = C0 gdy C0 —^0, to [H301=Kw^ ^pH^7