Diketahui Kb NH3=10^-5 maka besarnya pH NH3 0,2m adalah …..

Jawaban:

[OH-] = √kb. Mb

= √10^-5 . 2 × 10^-1

= √2 . 10^-6

= √2. 10^-3

POH = -(log √2+log 10^-3)

= -(log √2+(-3))

= -(log √2-3)

= -log √2 + 3

= 3 – log √2

PH = 14-(3-log √2)

= 11+log√2

Penjelasan:

maaf klo salah:')