Di dalam 200 mL larutan terlarut 5,35 gram NH4Cl. Jika Kb = 10^ -5 , Ar N = 14, H = 1, Cl = 35,5. Hitung pH larutan tersebut.
Dzaky111
M = [NH₄Cl] = g/Mr x 1000/V = 5,35/53,5 x 1000/200 = 0,1 x 5 = 0,5 M Kb = 10⁻⁵ pH = .........? Penyelesaian: Hidrolisis (asam kuat basa lemah)
= 5,35/53,5 x 1000/200
= 0,1 x 5
= 0,5 M
Kb = 10⁻⁵
pH = .........?
Penyelesaian:
Hidrolisis (asam kuat basa lemah)