Berapakah PH larutan Ba(OH)2 0,01M

mapel : kimia kelaa xi
"larutan asam basa "

semoga bermanfaat ya.

Mapel: Kimia
Kelas: XI
Materi: Asam Basa

Jawab

Ba(OH)2 → Ba²+ + 2OH-

[OH-] = a×Ma
[OH-] = 2×0.01
[OH-] = 0.02
[OH-] = 2×10-²

poH= -log[OH-]
poH= -log2×10-²
poH= 2-log2

pH=14- [OH-]
pH= 14-(2-log2)
pH= 12+log2
pH= 12+0,3
pH= 12,3