Tentukan harga ph larutan nh4oh 0,9m (kb=9x10^-6

Mapel: Kimia
Kelas: XI
Materi: Asam Basa

Jawab

NH4OH
[OH-] = √kb×M
[OH-] = √9×10^-6 × 9×10-¹
[OH-] = √81×10^-7
[OH-] = 9×10^-3,5

pOH = -log[OH-]
pOH = -log9×10^-3,5
pOH = 3,5-log9

pH= 14-(3,5-log9)
pH= 10,5+log9
pH= 10,5+log3²
pH= 10,5+2log3