Page 12 - BAB VII - LARUTAN

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
–
[OH ] = 5×10 . 2
–
-2
[OH ] = 10×10 -2
–
[OH ] = 10 -1
–
pOH = -log [OH ]
–
pOH = -log 10 -1

pOH = 1
pH = 14-pOH

pH = 14-1

pH = 13
4. Tentukan pH dari larutan NH4OH 0,4 M (Kb = 10 )
–5
Pembahasan :

NH4OH (aq) → NH4 (aq) + OH (aq)
+
-
[OH ] = √ Kb . M basa
-
[OH ] = √ 10 . 4 x 10
−5
-
−1
[OH ] = √ 4 x 10
-
−6
[OH ] = 2 x 10 -3
-
pOH = -log [OH ]
–
pOH = -log 2 x 10
-3
pOH = 3- log 2

pOH = 3-0,3
pOH = 2,7

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

pH = 11,3