Question

diketahui 0,6 gram asam cuka perdagangan dilarutkan dalam 1 liter air. Jika:
harga Ka = [tex]{10}^{ - 6}[/tex] ,
Ar H = 1,
Ar c = 12,
Ar o = 16
Tentukan:
1) pH​

Answer 1

Verified answer

Penjelasan:

CH3COOH 0,6 gr

v = 1 liter

Mr = 60

mol CH3COOH

= m/Mr

= 0,6/60

= 0,01 mol

M CH3COOH

= n x 1000/v (ml)

= 0,01 x 1000/1000

= 0,01 M

[H+]

= √(Ka . M)

= √(0,01 . 10^-6)

= √10^-8

= 10^-4

PH

= -log H+

= -log 10^-4

= 4