koncentracija

Submitted by Tina Trezic on Thu, 06/26/2014 - 19:00
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Izracunati koncentraciju OH- jonova ( u mol/dm3) u vodenom rastvoru CH3COOH, cija je koncentracija 0,05 mol/dm3 . (Ka=1,0* 10-5) resenje je 1,054 * 10-11

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milicaradenkovic - Thu, 06/26/2014 - 20:08

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

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

(CH3COO-)=(H+)=x

1.8*10(-5)=x2/0.05

x=(koren(9*10(-7)))=9.5*10(-4) 

{CH3COO-}=x=9.5*10(-4) mol/l

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

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

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

5.55*10(-10)=x*0.05/9.5*10(-4)

x=5.225*10(-13)/0.05=1.045*10(-11)

{OH-}=x=1.045*10(-11) mol/l Ka=1.8*10(-5) za CH3COOH a ne 1*10(-5)