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
H2SO4+H2O--------------->H3O+
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
Savrseno jasno.Hvala puno!!!
Savrseno jasno.Hvala puno!!!