Odpowiedź:
Wyjaśnienie:
Dane: Szukane:
Csoli= 0,1mol/dm³ pH =?
Kb= 10⁻⁵
NH₄Cl ⇄ NH₄⁺ + Cl⁻
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺ (sól mocnego kwasu)
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Kb=[NH₄⁺]*[OH⁻] / [NH₃]
[NH₄⁺] = Csoli
[NH₃] = [H₃O⁺] = x
Kb = Cs *[OH⁻]/x
[OH⁻] wyznaczymy z iloczynu jonowego wody, Kw:
Kw = [OH⁻]*[H⁺] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴/x
wtedy Kb wynosi:
Kb = (Cs* 10⁻¹⁴/x) / x
Kb = Cs*10⁻¹⁴/x²
x² = Cs*10⁻¹⁴/10⁻⁵
x = √0,1*10⁻¹⁴/10⁻⁵
x = √10⁻¹⁰
x = 10⁻⁵
[H₃O⁺] = 10⁻⁵
pH = -log(10⁻⁵)
pH = 5
