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PH= 4-log 5
Ka= 10^-5
DITANYA : α (derajat ionisasi) ?
DIJAWAB :
PH = - log [H+]
4-log5= - log [H+]
[H+]= 5.10^-4
[H+] = √Ma.Ka
5.10^-4 = √Ma x 10^-5
(5.10^-4)²= Ma x 10^-5
Ma = 25.10^-8 : 10^-5
Ma = 25.10^-3 M
α = √Ka : Ma
α = √10^-5 : 25.10^-3
α = √4.10^-4
α = 2.10^-2