Question

Suatu larutan mengandung 0,58 gram asam asetat dilarutkan dalam 10 liter air. Jika Ka = maka pH larutan adalah...

Answer 1

Mr CH3COOH = 60
n = massa : Mr
n = 0,58 : 60
n = 0,009 M
M = n : V
M = 0,001 : 10 L
M = 0,0009
H+ = √Ka× M
H+ = √ ×
H+ = √
H+ =
pH = 4,5 log 3