Page 260 - BAHAN AJAR KAPITA SELEKTA KIMIA BERBASIS CASE METHOD DAN PROJECT

pH = - log [H ]
+
pH = - log 1,34 x 10 -3
pH = 3-log 1,34

pH = 3-0,12

pH = 2,88
3. Hitunglah pH dari larutan basa kalsium hidroksida (Ca(OH)2) 0,05 M = 5 x 10 .
-2
Pembahasan :

Mg(OH)2 (aq) → Mg 2+ (aq) + 2OH (aq)
-
–
[OH ] = α . Mbasa
-2
–
[OH ] = 5×10 . 2
–
-2
[OH ] = 10×10
-1
–
[OH ] = 10
pOH = -log [OH ]
–
pOH = -log 10 -1
pOH = 1

pH = 14-pOH
pH = 14-1

pH = 13

–5
4. Tentukan pH dari larutan NH4OH 0,4 M (Kb = 10 )
Pembahasan :

+ -
NH4OH (aq) → NH4 (aq) + OH (aq)

[OH ] = √ Kb . M basa
-

[OH ] = √ 10 . 4 x 10
-
−5
−1

[OH ] = √ 4 x 10
−6
-
-3
-
[OH ] = 2 x 10
–
pOH = -log [OH ]
-3
pOH = -log 2 x 10
pOH = 3- log 2
pOH = 3-0,3
pOH = 2,7

pH = 14-pOH
pH = 14-2,7

pH = 11,3

236 | Berbasis Case Method & Project