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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)

milicaradenkovic's picture

milicaradenkovic - Sat, 02/08/2014 - 13:54

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

 

 

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zzz555 - Sat, 02/08/2014 - 14:58

Bas ti hvala!!!