PH larutan asam amonium hidroksida 0,5 M (Kb = 1,8 × 10-5
Jawaban:
11 + log 3
Penjelasan:
[OH-] = √Kb.Mb
=> √1,8×10^-5.5×10^-1
=> 3×10^-3 M
pOH = -log[OH-]
=> -log 3×10^-3
=> 3 – log 3
pH = 14 – pOH
=> 14 – ( 3 – log 3 )
=> 11 + log 3
PH larutan asam amonium hidroksida 0,5 M (Kb = 1,8 × 10-5
Jawaban:
11 + log 3
Penjelasan:
[OH-] = √Kb.Mb
=> √1,8×10^-5.5×10^-1
=> 3×10^-3 M
pOH = -log[OH-]
=> -log 3×10^-3
=> 3 – log 3
pH = 14 – pOH
=> 14 – ( 3 – log 3 )
=> 11 + log 3