pH

Submitted by Marijana maya on Thu, 01/02/2014 - 20:22
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Moze li pomoc oko ovog zadatka??

Izracunati pH rastvora sumporne kiseline koncentracije 0,050 mol dm-3 na 25 stepeni C.

Ka,1(H2SO4,25 stepeni C) JE 103

Ka,2(H2SO4,25 stepeni C) JE 1,2*10-2

Resenje je 1,23.

Unapred zahvalna

milicaradenkovic's picture

milicaradenkovic - Thu, 01/02/2014 - 20:54

H2SO4+H2O--------------->H3O+ + HSO4-

{H+}={HSO4-}=0.05 mol/l

HSO4- +H2O------------->H3O+ + SO42-

{HSO4-}=0.05-x

{H+}=x

{SO42-}=0.05+x

K={SO42-}*{H+}/{HSO4-}

1.2*10(-2)*(0.05-x)=x*(0.05+x}

6*10(-4)-1.2*10(-2)*x=x*2+0.05*x

x*2+0.062*x-6*10(-4)=0

x=0.0085

{H3O+}2=0.0085 mol/l

{H3O+}1=0.05 mol/l

{H3O+}u.={H3O+}1+{H3O+}2=0.05+0.0085=0.0585 mol/l i pH=1.23

Marijana maya's picture

Marijana maya - Thu, 01/02/2014 - 21:03

Savrseno jasno.Hvala puno!!!