Chemistry Reference
In-Depth Information
Let's take the example that we started discussing and set it up as a
problem, as shown below:
Example 1
A student mixes aqueous solutions of lead (II) chlorate and sodium
iodide and finds that a white precipitate forms.
For this chemical reaction:
A. Write a word equation.
B. Write a balanced chemical reaction, with subscripts
indicating state or phase.
C. Write a net ionic equation.
D. Identify the precipitate.
Most of the first two steps should be easy to you, if you studied Chapter
5. You may want to try these steps on your own, for additional practice. I
will jump ahead and complete step A and most of step B. Then we will
discuss how you determine the subscripts indicating the state or phase of
each substance.
A.
Write a word equation.
lead (II) chlorate + sodium iodide
lead (II) iodide and sodium chlorate
B.
Write a balanced chemical reaction, with subscripts indicating state or
phase.
PbI
2
+ 2NaClO
4
Note that I included the “state” subscript of (aq) for the reactants but
not for the products. We know that the reactants are in aqueous form,
because that is mentioned in the original question. To figure out if the
products are aqueous or not, we need to consult either the solubility rules
or Figure 6-3a. The rules tell us that chlorate salts are essentially all soluble,
so we will add the (aq) subscript to the sodium chlorate. We also note that
Figure 6-3a states that lead (II) iodide is insoluble, which means that it is
probably the precipitate that appeared in the bottom of the test tube. We
use the (s) subscript to indicate a precipitate.
Adding this information to the formula that we wrote in step B, we
would get the following chemical equation:
Pb(ClO
4
)
2(aq)
+ 2NaI
(aq)
Pb(ClO
4
)
2(aq)
+ 2NaI
(aq)
PbI
2(s)
+ 2NaClO
4(aq)
When we say that the “product” called sodium chlorate is aqueous, we
mean that it doesn't really form. It exists as free ions in solution. The lead (II)
