Page 64 - Modul Asam Basa

Contoh Soal

= 0,2 M

Langkah 2:

Tulis persamaan reaksi ionisasi dari Mg(OH)2

2+
–
Mg(OH)2 → Mg + 2 OH

–
[OH ] = x. Mb

–
[OH ] = 2 x 0,2 M

–
[OH ] = 0,4 M

pOH = –log 0,4

= –log 4 x 10–1

= 1 – log 4

= 1 – 0,602 = 0,398

Langkah 3:

Tentukan harga pH

pH + pOH= 14

pH = 14 – 0,398

pH = 13,602

4. Tentukan pOH larutan KOH 0,05 M!

Jawab:

-
+
Reaksi ionisasi: KOH(aq) → K (aq) + OH (aq)

–
[OH ] = x. Mb

–
[OH ] = 1 x 0,05 M

-2
–
[OH ] = 5 x 10 M

-
pOH = - log [OH ]

pOH = - log 5 x 10 -2

pOH = 2 – log 5

pH + pOH= 14

pH = 14 – (2 – log 5)

pH = 12 + log 5 = 12 + 0,699 = 12, 699

E-MODUL KIMIA ASAM BASA 52