jika harga kb NH3=2.10^-5.tentukan ph larutan NH3 0.2 M

Mapel: Kimia
Kelas: XI
Materi: Asam Basa
Kata Kunci: pH, pOH

Jawab

NH3
basa lemah

[OH-] = √kb×M
[OH-] = √2×10^-5×2×10-¹
[OH-] = √2×10^-6
[OH] = 2×10-³

pOH= -log[OH]
pOH= -log(2×10-³)
pOH= 3-log2

pH= pKw - pOH
pH= 14-(3-log2)
pH= 11+log2

OH- = akar kb x M = akar 2x10^-5 x 0.2
OH- = 2x10^-3
pOH = 3-log 2
pH = 14-(3-log2) = 11+log 2