Kz= (H3O+)e.(F-)e/(HF)e HF + H2O ⇔ F- + H3O+
n0 Cz 0 0
Δn -x +x +x
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ne Cz-x x x
Als x klein is mag men Cz-x=Cz of Kz< 10-4
Kz= x.x/Cz of x2=Kz.Cz x=(Kz.Cz )½
(H3O+)=x pH=-log(H3O+)= -log(Kz.Cz )½=-½log Kz-½log Cz= ½pKz -½log Cz
vb 0,02 mol/l HF pH? pH= 0,5.3,17+0,5.2=2,58