Oblicz pH roztworu zawierającego w 1 dm³ 0.020 M Na₂SO₄ i 0.020 M HCl.

c_Na+ = 2 *cNa₂SO₄ = 2*0.020 M = 0.040 M

cSO₄^2- = 0.020 M

cCl^- = 0.020 M

I = ½ (0.040*12 + 0.020*22 + 0.020*12 +0.020*12) =0.080

Dla jonu H+: a*B = 3.0, A = 0.51

ogniwa: K(+)-red | A(-)-utl elektro: K(-)-red | A(+)-utl

→(+)e.joniz.,powin.e^-(-)char.metal. ↓(+)char.metal.(-)e.joniz.,powin.e^-,elektrouj.

[H3O^+]=10^-pH	[HCOOH]	[H3O^+]	[HCOO^−]
pocz.	0.10	0	0
zmiana	−4.2 × 10^-3	+4.2 × 10^-3	+4.2 × 10^−3
równow.	0.10 − 4.2 × 10^−3 = 0.0958 = 0.10	4.2 × 10^−3	4.2 × 10^−3

---