Hidroliza

Submitted by Makicaaa on Mon, 12/23/2013 - 08:16
Makicaaa's picture
Forums: 
Opšta diskusija

Help 

Koliko grama kalijum acetata treba rastvoriti u 600 cm3 vode da bi pH dobijenog rastvora iznosila 8?

Resenje:0.106

milicaradenkovic's picture

milicaradenkovic - Mon, 12/23/2013 - 15:25

CH3COOH------------->CH3COO- + H+

Ka={CH3COO-}*{H+}/{CH3COOH}

Kb=Kw/Ka=1*10(-14)/1.8*10(-5)=5.55*10(-10)

CH3COO- + H2O-------------->OH- + CH3COOH

Kb={OH-}*{CH3COOH}/{CH3COO-}

{OH-}={CH3COO-}=1*10(-6) mol/l

pOH=14-pH=14-8=6

5.5*10(-10)=(1*10(-6))*2/{CH3COO-}

{CH3COO-}=1*10(-12)/5.5*10(-10)=0.00182 mol/l

{CH3COOK}={CH3COO-}=0.00182 mol/l

n(CH3COOK)={CH3COOK}*V=0.00182*0.6=0.00109 mol

m(CH3COOK)=n(CH3COOK)*Mr=0.00109*98=0.1068 g

 

milicaradenkovic's picture

milicaradenkovic - Mon, 12/23/2013 - 15:26

Izvini greska je kod {OH-}={CH3COOH}=1*10(-6) mol/l a ne {OH-}={CH3COO-}=1*10(-6) mol/l