
In
reactions involving aqueous ionic compounds, the dissolved ionic compounds break
into their respective ions and float around in the solution independent of each
other but in the ratio according to their chemical formula.
For example, the following balanced equation: Na2SO4(aq)
+ Pb(NO3)2(aq) à 2NaNO3(aq) + PbSO4(s)
would actually have Na2SO4(aq) exist as Na+ and
SO42- ions floating around separately but in a 2 Na+
: 1 SO42- ratio. When
aqueous ionic equations are written out with the ions separated the equation is
called the “Total Ionic Equation.” For example note how the (aq) compounds are split up
2Na+(aq) + SO42-(aq) + Pb2+(aq)
+ 2NO3- (aq) à 2Na+(aq) + 2NO3-
(aq) + PbSO4(s)
Notice
that PbSO4(s) is not split up into ions. Can you think why this is the case?
In
the total ionic equation, notice that there are two ions that are exactly in the
same state and with the same charge on both sides of the arrow.
These are the Na+(aq) and NO3-(aq) ions.
Since these ions don’t change in the course of the reaction, they are
called “spectator ions.” In
the “Net Ionic Equation” the spectator ions are removed and only the
components of the rxn that change remain. Ex: SO42-(aq) + Pb2+(aq) à PbSO4(s)
Exercise:
For the following reactions, balance the equations, write
the total ionic equation, circle the spectator ions and write the
net ionic equation.
1.
KBr(aq) + F2(g)
à
KF(aq) + Br2(l)
2.
Zn(s) + HCl(aq)
à
ZnCl2(aq) +
H2(g)
3.
FeCl3(aq) +
Na2CO3(aq) à
Fe2(CO3)3(s) +
NaCl(aq)
4.
CaCl2(aq) +
Na3PO4(aq) à
NaCl(aq) + Ca3(PO4)2(s)
5.
Al(s) + CuCl2(aq)
à
AlCl3(aq) +
Cu(s)
Questions: