Question

1. Berapa pH larutan yang terjadi jika dalam 500 mL larutan terdapat 12 gram CH₃COOH dan 8,2 gram CH₃COONa? (Ka=1,8x10⁻⁵)

Answer 1

Mol CH3COOH = 500 . 12 = 6000
mol CH3COONA = 8,2 . V

CH3COOH = CH3COONA
6000 = 8,2 V
V = 6000/8,2 = 731 , 70732

CH3COOH α = √Ka.M =  1,8X10^-5.6000 = √1,08 X 10^-1
pH = -log H
     = - log √1,08 X 10^-1
    
kb = 14/5 = 1x10^9

OH = √1x10^9 . 731,7032

pH = 14 - log OH