Izracunajte konstantu disocijacije amonijaka ,ako je u rastvoru koncetracije 0.1 mol/dm3 1.3% molekula NH3 disusovano na jone.
NH3+H2O------------->NH4+ + OH-
K={NH4+}*{OH-}/{NH3}
{NH3}=0.1 mol/l, {NH4+}={OH-}=0.0013 mol/l
{NH4+}={OH-}=0.013*{NH3}=0.013*0.1=0.0013 mol/l
K=0.0013*0.0013/0.1=1.69*10(-5)
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NH3+H2O------------->NH4+ +
NH3+H2O------------->NH4+ + OH-
K={NH4+}*{OH-}/{NH3}
{NH3}=0.1 mol/l, {NH4+}={OH-}=0.0013 mol/l
{NH4+}={OH-}=0.013*{NH3}=0.013*0.1=0.0013 mol/l
K=0.0013*0.0013/0.1=1.69*10(-5)