Question

Di dalam 500ml larutan terdapat 4,1 gram CH3COONa yang terlarut. Jika nilai Ka CH3COOH = 10^-5 dan Ar Na = 23, C =12, O = 16, maka pH larutan tersebut adalah

Answer 1

Mr CH3COONa = 82
mol Ch3COONa = gram/mr = 4,1 / 82 = 0,05
[CH3COONa] = mol/liter = 0,05 / 0,5 = 0,1

[OH-] = √kw/ka . [CH3COONa]
         = √10^-14 / 10^-5 . 10^-1
         =√10^-10
[OH-]  =10^-5

POH = 5
PH = 14 - 5
     =9