Izračunajte koncetraciju H+ jona u rastvoru koji sadrži 0,3 mol/dm3 sirćetne kiseline i 0,3 mol/dm3 natrijum-acetata.Vrednost K(CH3COOH) = 1,8 x 10-5.Ako neko može da pomogne,malo sam zapeo.
c(CH3COONa)=0.3 mol/l a c(CH3COOH)=0.3 mol/l
CH3COOH------------->CH3COO- + H+
CH3COONa------------->CH3COO - + Na+
Ka={CH3COO-}*{H+}/{CH3COOH}
{CH3COO-}={CH3COONa}=0.3 mol/l
{H+}={CH3COOH}*Ka/{CH3COO-}=0.3*1.8*10(-5)/0.3=1.8*10(-5) mol/l
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H+
c(CH3COONa)=0.3 mol/l a c(CH3COOH)=0.3 mol/l
CH3COOH------------->CH3COO- + H+
CH3COONa------------->CH3COO - + Na+
Ka={CH3COO-}*{H+}/{CH3COOH}
{CH3COO-}={CH3COONa}=0.3 mol/l
{H+}={CH3COOH}*Ka/{CH3COO-}=0.3*1.8*10(-5)/0.3=1.8*10(-5) mol/l
E super veliko hvala
E super veliko hvala micoloknico