Amonijum jon

Submitted by MD on Mon, 12/09/2013 - 18:54
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Koliko se molova NH4+ jona nalazi u 200cm^3 rastvora amonijaka, čija je koncetracija 0,12 mol/dm^3? Kb = 1,8*10^-5 mol/dm^3

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milicaradenkovic - Mon, 12/09/2013 - 19:01

NH3+H2O----------->NH4+ + OH-

Kb={NH4+}*{OH-}/{NH3}

{NH3]=0.12*0.2=0.024 mol/l

{NH4+}={OH-}=x

x*2=1.8*10(-5)*0.024

x*2=4.32*10(-7)

x=(koren(4.32*10(-7)))

x=6.57*10(-4)

{NH4+}=6.57*10(-4) mol/l

6.57*10(-4) mol------------------1 l

x mol------------------------0.2 l

x=6.57*10(-40*0.2/1=1.314*10(-4)

{NH4+}=1.314*10(-4) mol/l