Soal:
Jawaban:
Materi : Larutan Penyangga / Buffer
Kelas : 11
Kelas : 11
n NH3 = M × V
n NH3 = 0,1 M × 100 mL
n NH3 = 10 mmol
n NH4NO3 = M × V
n NH4NO3 = 0,2 M × 200 mL
n NH4NO3 = 40 mmol
Kb NH3 = 1,8 × 10^-5
OH- = Kb × n basa / n garam
OH- = 1,8 × 10^-5 × 10 mmol / 40 mmol
OH- = 4,5 × 10^-6
pOH = – log OH-
pOH = – log (4,5 × 10^-6)
pOH = 6 – log 4,5
pH = pKw – pOH
pH = 14 – (6 – log 4,5)
pH = 8 + log 4,5
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