Neorganska hemija

jovana96's picture

 Dati su vodeni rastvori Ba3(PO4)2 Ksp= 1,3x10^-29, BaCrO4 Ksp=1,17x10^-10, i BaF2  Ksp=1,84x10^-7.Koji rastvor sadrži najviše jona Ba2+ u mg/l?

milicaradenkovic's picture

BaCrO4------------------>Ba2+ + CrO42-

Ksp={Ba2+}*{CrO42-}

Ksp=R*2

{Ba2+}=R i {CrO42-}=R

1.17*10(-10)=R*2

R=(koren(1.17*10(-10)))=1.08*10(-5)

{Ba2+}=1.08*10(-5) mol/l

m(Ba2+)=(Ba2+)*Mr=1.08*10(-5)*137=1.48 mg/l

BaF2------------------>Ba2+ + 2F-

Ksp={Ba2+}*{2*F-}*2

{Ba2+}=R i {F-}=R

Ksp=4*R*3

1.84*10(-7)=4*R*3

R=(3koren(1.84*10(-7)/4))=3.6*10(-3)

{Ba2+}=3.6*10(-3) mol/l

m(Ba2+)=(Ba2+)*Mr=3.6*10(-3)*137=493.2 mg/l

Ba3(PO4)2------------------->3Ba2+ + 2PO43-

Ksp={3*Ba2+}*3*{2*PO43-}*2

Ksp=108*R*5

{Ba2+}=R i {PO43-}=R

1.3*10(-29)=108*R*5

R=(5koren(1.3*10(-29)/108))=6.6*10(-7)

{Ba2+}=3*R=3*6.6*10(-7)=1.98*10(-6) mol/l

m(Ba2+)={Ba2+}*Mr=1.98*10(-6)*137=0.271 mg/l 

Najvise jona sadrzi BaF2.