Suatu larutan penyangga dibuat dengan mencampurkan 400 ml NH3 0,1 dan 100 ml NH4Cl 0,2 M. Jika diketahui kb NH3=1,76x10^-5 hitunglah larutan pH penyangga tersebut.
Alsha08
Dik : n NH3 (Basa lemah) = 400.0,1 = 40 mmol n NH4Cl(Garam) = 100.0,2 = 20 mmol n NH4 ( Asam Konjugasi ) = 1 x n NH4Cl = 1.20 = 20 mmol Kb = 1,76 x 10^-5 Dit : pH ? Dij : [ OH-] = Kb. n basa lemah/ n asam konjugasi = 1,76x10^-5 . 40/20 = 1.76 x 10^-5 . 2 = 3,52 x 10^-5 pOH = -log [OH-] = - log [3,52 x 10^-5] = 5 - log 3,52
mmol nh4cl = 100. 0,2 = 20
nh4cl --> nh4+ cl -
20 20
oh- = kb. mmol nh3 : mmol nh4+
= 3.53 x 10 -5
n NH4Cl(Garam) = 100.0,2 = 20 mmol
n NH4 ( Asam Konjugasi ) = 1 x n NH4Cl = 1.20 = 20 mmol
Kb = 1,76 x 10^-5
Dit : pH ?
Dij : [ OH-] = Kb. n basa lemah/ n asam konjugasi
= 1,76x10^-5 . 40/20
= 1.76 x 10^-5 . 2
= 3,52 x 10^-5
pOH = -log [OH-]
= - log [3,52 x 10^-5]
= 5 - log 3,52
pH = 14 - pOH
= 14 - (5-log 3,52)
= 9 + log 3,52