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Submitted by zzz555 on Sat, 02/08/2014 - 13:30

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U 100 cm3 rastvora NaH2PO4 koncentracije 0,1 mol/dm3 dodato je 10 cm3 rastvora NaOH koncentracije 0,1 mol/dm3.
Izracunati pH rastvora.
Ka( H3PO4)= 1x 10 ( -2)
Ka ( H2PO4-)=1x 10 (-7)
Ka ( HPO4 -) = 1x 10 (-12)
NaH2PO4+NaOH-----------------
NaH2PO4+NaOH------------------->Na2HPO4+H2O
H2PO4- + H2O---------------------HPO42- + H3O+
Ka={HPO42-}*{H3O+}/{H2PO4-}
NaH2PO4--------------->Na+ + HPO42-
{HPO42-}=c{NaH2PO4}=0.0818 mol
n(NaH2PO4)=c*V=0.1*0.1=0.001 mol u visku je so
n(NaOH)=c*V=0.01*0.1=0.001 mol
n1(NaH2PO4)=n(NaOH)=0.001 mol
n2(NaH2PO4)=n(NaH2PO4)-n1(NaH2PO4)=0.01-0.001=0.009 mol
Vu=V2+V1=100+10=110 cm3
c(NaH2PO4)=n2(NaH2PO4)/Vu=0.009/0.11=0.0818 mol/l
c(Na2HPO4)=c1(NaOH)
c1(NaOH)=n(NaOH)/Vu=0.001/0.11=0.0091 mol/l
{HPO42-}=c1(NaOH)=0.0091 mol/l
Ka={H3O+}*{HPO42-}/{H2PO4-}
{H3O+}=Ka*{H2PO4-}/{HPO42-}
{H3O+}=1*10(-7)*0.0818/0.0091=8.99*10(-7) mol/l
pH= - log{H3O+}= -log{8.99*10(-7)}=6.05
Bas ti hvala!!!
Bas ti hvala!!!