Page 31 - modul penyangga

+
2-
(NH4)2SO4 → 2 NH4 + SO4
+
mol NH4 = 2 . mol (NH4)2SO4 = 2 . 0,005 = 0,01 M

-
[OH ] = Kb. [ ℎ]
[ ]
= Kb. [NH 4 OH]
+
ൣ ൧
4

-5
= 2 . 10 . (0,005)
(0,01)
= 10
-5
- -5
pOH = -log [OH ] = -log 10 = 5

pH = 14 – pOH = 14 – 5 = 9

b. Bila 50 mL larutan NH4OH 0,2 M dicampurkan ke dalam 50 mL

larutan HCl 0,1 M. Hitung pH campuran yang terjadi jika Kb= 2. 10 ?
-5

Jawab:

n NH4OH = M . V = 0,2 M . 50 mL = 10 mmol

n HCl = M . V = 0,1 M . 50 mL = 5 mmol

NH4OH(aq) + HCl(aq) → NH4Cl(aq) + H2O(l)

n mula-mula : 10 mmol 5 mmol - -

n berekasi : -5 mmol -5 mmol +5 mmol +5 mmol

n sisa : 5 mmol - 5 mmol 5 mmol

-
[OH ] = Kb . ℎ

= Kb . 4
4
-5
= 2 . 10 . (5 )
(5 )
= 2 . 10
-5
-5
pOH = -log [OH ] = -log 2 . 10 = 5 – log 2
-
pH = 14 – pOH = 14 – (5 - log 2) = 9 + log 2

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