Odpowiedź:
Wyjaśnienie:
Dane: Szukane:
C=0,01m/dm3 [H+] =?
K = 1,75*10⁻⁵ pH =?
K = α²*C/(1 - α)
CH3COOH to słaby elektrolit:
K = α²*C; (1 - α) ≅ 1
α² = K/C
α = √K/C
α = √1,75*10⁻²/10⁻²
α = 4,18*10⁻²
α = Czdys/C = [H+]/C
[H+] =α *C
[H+] = 4,18*10⁻² * 10⁻²
[H+] = 4,18*10⁻⁴ m/dm3
pH = -log[H+]
pH = -log(4,18*10⁻⁴)
pH = 3,38