Question

PH larutan Ba(OH)2 0,1 MOLAR ADALAH

Answer 1

[Ba(OH)2] = 0,1 M
[OH-] = Xb x M
= 2 x 0,1 = 0,2 M

pOH = - log [OH-] = - log (0,2)
= 1 - log 2

pH = 13 + log 2

Answer 2

Ba(OH)₂ = 0,1M
[OH⁻] = a . [Ba(OH)₂]
[OH⁻] = 2 . 1 x 10⁻¹
[OH⁻] = 2 x 10⁻¹
pOH = -log 2 x 10⁻¹
pOH = 1 - log 2
pH = 14 - (1 - log 2)
pH = 13 + log 2