Question

Tentukan pH larutan:
0,05 mol NH₃ dalam 500 ml larutan (Kb = 1 x 10⁻5)

Answer 1

NH3 0,05 mol (Kb = 1 × 10^-5)
V = 500 mL = 0,5 L

M = n/V
M = 0,05 / 0,5
M = 0,1

[OH^-] = √Kb × Mb
[OH^-] = √1 × 10^-5 × 0,1
[OH^-] = √10^-6
[OH^-] = 10^-3

pOH = - log [OH^-]
pOH = - log 10^-3
pOH = 3

pH = 14 - pOH
pH = 14 - 3
pH = 11