Koliko se Oh i H+ jona nalazi u 500cm3 rastvora NaBrO,ako se u ovoj zapremini rastvora nalazi 14.85g ove soli? Kk(HBrO)=2,5x10^-9mol/dm3resenje: 3,01x10^12H+ 3.o1x10^20 OH-
NaBrO+H2O--------------->HBrO+NaOH
NaBrO-------------->Na+ + BrO-
Kb=Kw/Kk=1*10(-14)/2.5*10(-9)=4*10(-6)
BrO- + H2O------------>HBrO + OH-
n=14.85/119=0.125 mol
c=n/V=0.125/0.5=0.25 mol/l NaBrO
c(NaBrO)=c(BrO-)=0.25 mol/l
Kb={HBrO}*{OH-}/{BrO-}
{HBrO}={OH-}=x
4*10(-6)=x*2/0.25
x=(koren(4*10(-6)*0.25)}=1*10(-3) mol/l
0.001 mol--------------------1 l
x mol-------------------------0.5 l
x=0.5*0.001/1=5*10(-4) mol OH- jona
1 mol------------------------6*10(23)
5*10(-4) mol-----------------x
x=6*10(23)*5*10(-4)/1=3*10(20) jona OH-
HBrO+H2O------------->BrO- + H+
Kk={H+}*{BrO-}/{HBrO}
2.5*10(-9)={H+}*0.25/1*10(-3)
{H+}=2.5*10(-9)*1*10(-3)/0.25=1*10(-11) mol/l
1*10(-11) mol----------------------1 l
x mol------------------------------------0.5 l
x=1*10(-11)*0.5/1=5*10(-12) mol jona H+
1 mol--------------------6*10(23)
5*10(-12) mol----------x
x=5*10(-12)*6*10(23)/1=3*10(12) jona H+
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H+ OH-
NaBrO+H2O--------------->HBrO+NaOH
NaBrO-------------->Na+ + BrO-
Kb=Kw/Kk=1*10(-14)/2.5*10(-9)=4*10(-6)
BrO- + H2O------------>HBrO + OH-
n=14.85/119=0.125 mol
c=n/V=0.125/0.5=0.25 mol/l NaBrO
c(NaBrO)=c(BrO-)=0.25 mol/l
Kb={HBrO}*{OH-}/{BrO-}
{HBrO}={OH-}=x
4*10(-6)=x*2/0.25
x=(koren(4*10(-6)*0.25)}=1*10(-3) mol/l
0.001 mol--------------------1 l
x mol-------------------------0.5 l
x=0.5*0.001/1=5*10(-4) mol OH- jona
1 mol------------------------6*10(23)
5*10(-4) mol-----------------x
x=6*10(23)*5*10(-4)/1=3*10(20) jona OH-
HBrO+H2O------------->BrO- + H+
Kk={H+}*{BrO-}/{HBrO}
2.5*10(-9)={H+}*0.25/1*10(-3)
{H+}=2.5*10(-9)*1*10(-3)/0.25=1*10(-11) mol/l
1*10(-11) mol----------------------1 l
x mol------------------------------------0.5 l
x=1*10(-11)*0.5/1=5*10(-12) mol jona H+
1 mol--------------------6*10(23)
5*10(-12) mol----------x
x=5*10(-12)*6*10(23)/1=3*10(12) jona H+