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Submitted by banejeca943 on Sun, 01/05/2014 - 17:52
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Izracunati pH rastvora nastalog mesanjem 300cm3 rastvora NH4Cl pH=5 i 200cm3 rastvora NaOH pH=13
Kb(NH3)=1.8*10^-5 mol/dm3
Izracunati pH rastvora nastalog mesanjem 300cm3 rastvora NH4Cl pH=5 i 200cm3 rastvora NaOH pH=13
Kb(NH3)=1.8*10^-5 mol/dm3
n(NaOH)={NaOH}*V=0.1*0.2=0.02
n(NaOH)={NaOH}*V=0.1*0.2=0.02 mol
{NaOH}={OH-}=0.1 mol/l jer je pH=13 a pOH=1
Ka={NH3}*{H+}/{NH4+}
Ka=Kw/Kb=1*10(-14)/1.8*10(-5)=5.55*10(-10)
5.55*10(-10)=(1*10(-5))*2/{NH4+}
{NH4+}=0.18 mol/l
{NH4Cl}={NH4+}=0.18 mol/l
n{NH4Cl}={NH4Cl}*V=0.18*0.3=0.054 mol
n(NH4Cl)1=n(NaOH)=0.02 mol
n(NH4Cl koji je preostao)=n(NH4Cl)-n(NH4Cl)1=0.054-0.02=0.034 mol
pOH=pKb+log(n(baze)/n(kiseline))
pOH=4.74+log(0.02/0.034)=4.74+0.23=4.97
pH=14-pOH=14-4.97=9.03
Hvala.
Hvala.