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Kb = tetapan basa
Contoh Soal :
1. Hitung PH larutan berikut !
a. H2SO4 0,1 M
-6
b. HF 0,01 M (Ka = 10 )
c. KOH 0,001 M
-5
d. NH4OH 0,4 M (Kb = 10 )
Penyelesaian :

a. H2SO4 → Asam Kuat
a = 2, Ma = 0,1 M
+
[H ] = a x Ma
= 2 x 0,1
= 2 x 10
-1
+
PH = - log [H ]
= - log 2 x 10 -1
= 1 -log 2
b. HF → Asam Lemah
-6
Ma = 0,01 M, Ka = 10
+
[H ] = √Ka x Ma
= √10-6 x 0,01
-8
= √10
= 10
-4
+
PH = - log [H ]
= - log 10 = 4
-4
c. KOH → Basa Kuat
b = 1, Mb = 0,001 M
-
[OH ] = b x Mb
= 1 x 0,001
-3
= 10
-
POH = - log [OH ]
-3
= - log 10
= 3
PH + POH = 14
PH + 3 = 14
PH = 11
d. NH4OH → Basa Lemah
-
[OH ] = √Kb x Mb
-5
= √10 x 0,4
-6
= √4 x 10
= 2 x 10 -3
-
POH = - log [OH ]
= - log 2 x 10 -3
= 3 – log 2
PH = 14 – (3 – log 2)
= 11 + log 2
2. Suatu asam kuat H2X mempunyai PH = 3 – log 2. Hitung molaritas
H2X !
Penyelesaian :
PH = 3 – log 2
-3
+
[H ] = 2 x 10
-3
a x Ma = 2 x 10
2 x Ma = 2 x 10
-3
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