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= 5 mmol
[NH₃OH] = mol / volume
= 5 / 50
= 0,1 M
[OH⁻] = √Kb.M
= √(10⁻⁵)(0,1)
= 10⁻³ M
pOH = -log [OH⁻]
pOH = 3
pH + pOH = 14
pH = 11
SOLVED :)