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= 1,8 × 10^-5 × (120 ÷ 90)
= 2,4 × 10^-5
pOH = 5 - log 2,4 ⇒ pH = 9 + log 2,4
b. Volume total = 200 + 300 + 500 = 1000 mL
mol NH3 = 120 ÷ 1000 = 0,12
mol NH4OH = 90 ÷ 1000 = 0,09
OH^- = Kb × (mol basa ÷ mol garam)
= 1,8 × 10^-5 × (0,12 ÷ 0,09)
= 2,4 × 10^-5
pOH = 5 - log 2,4 ⇒ pH = 9 + log 2,4