Larutan buffer dibuat dengan mencampurkan 400 mL NH3 0.1 dengan 100 mL NH4CL 0.2M. Jika Kb NH3 = 1,76×10 ^ -5, hitung pH larutan buffer
Tolong dibantu yaa terimakasih banyak ( no 2 ) kalau mau nomor 3 yaa gpp
Penjelasan:
2. mmol NH3 = 400 x 0,1 = 40 mmol
mmol NH4Cl = 100 x 0,2 = 20 mmol
[OH-] = Kb x ( mmol basa / mmol garam )
[OH-] = 1,76×10^-5 ( 40 / 20 )
[OH-] = 3,52×10^-5 M
pOH = – log [OH-]
pOH = – log 3,52×10^-5
pOH = 5 – log 3,52
pH = 14 – pOH
pH = 14 – ( 5 – log 3,52 )
pH = 9 + log 3,52