Odpowiedź:
Wyjaśnienie:
Dane: Szukane:
K = 5*10⁻⁴ pH =?
Rozwiązanie:
Ca(OH)₂ ⇔ Ca²⁺ + 2OH⁻
Ks =[Ca²⁺]*[OH⁻]²
{Ca²⁺] = S
2[OH⁻]= 2S
Ks = S*(2S)² = 4S³
S = ∛Ks / 4 = ∛5*10⁻⁴/4 = ∛1,25*10⁻⁴ = 0,05mol/dm³
[OH⁻] = 2S = 0,1 mol/dm³
pOH = -log[OH⁻] = -log(10⁻¹) = 1
pH = 14 - pOH = 14 - 1 = 13
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Odpowiedź:
Wyjaśnienie:
Dane: Szukane:
K = 5*10⁻⁴ pH =?
Rozwiązanie:
Ca(OH)₂ ⇔ Ca²⁺ + 2OH⁻
Ks =[Ca²⁺]*[OH⁻]²
{Ca²⁺] = S
2[OH⁻]= 2S
Ks = S*(2S)² = 4S³
S = ∛Ks / 4 = ∛5*10⁻⁴/4 = ∛1,25*10⁻⁴ = 0,05mol/dm³
[OH⁻] = 2S = 0,1 mol/dm³
pOH = -log[OH⁻] = -log(10⁻¹) = 1
pH = 14 - pOH = 14 - 1 = 13