4.8
Chemical Analysis and Titration Are Applications of Solution Stoichiometry
163
A solution of AgNO
3
is added
to a solution of CaCl
2
,
producing a precipitate of AgCl.
(Andy Washnik.)
Simple reasoning works well
here. Since the coefficients of Ag
1
and Cl
2
are the same, the
numbers of moles that react must
be equal.
To solve the problem, the first step will be to calculate the molarities of the ions in the so
lutions being mixed. Next, using the volume and molarity of the Cl
2
solution, we will calculate
the moles of Cl
2
available. Then we’ll use the coefficients of the equation to find the moles
of Ag
1
that react. Finally, we’ll use the molarity of the Ag
1
solution to determine the volume of
the 0.100
M
AgNO
3
solution needed.
SOLUTION:
We begin by finding the concentrations of the ions in the reacting solutions:
0.100
M
AgNO
3
contains 0.100
M
Ag
1
and 0.100
M
NO
3
2
0.400
M
CaCl
2
contains 0.400
M
Ca
2
1
and 0.800
M
Cl
2
We’re only interested in the Ag
1
and Cl
2
; the Ca
2
1
and NO
3
2
are spectator ions. For our pur
poses, then, the solution concentrations are 0.100
M
Ag
1
and 0.800
M
Cl
2
. Having these val
ues, we can now restate the problem: How many milliliters of 0.100
M
Ag
1
solution are needed
to react completely with 25.0 mL of 0.800
M
Cl
2
solution?
The moles of Cl
2
available for reaction are obtained from the molarity and volume
(0.0250 L) of the Cl
2
solution.
The coefficients of the equation tell us the Ag
1
and Cl
2
combine in a 1to1 mole ratio, so
. Finally, we calculate the volume of the Ag
1
solution us
ing its molarity as a conversion factor. As we’ve done earlier, we use the factor that makes the
units cancel correctly.
Our calculations tell us that we must use 0.200 L or 200 mL of the AgNO
3
solution. We could
also have used the following chain calculation, of course.
0.0200
mol Ag
+
*
1.00 L Ag
+
soln
0.100
mol Ag
+
=
0.200 L Ag
+
soln
0.0200 mol Cl

3
0.0200 mol Ag
+
0.0250
L Cl

soln
*
0.800 mol Cl

1.00
L Cl

soln
=
0.0200 mol Cl

25.0 mL Cl

soln
3
? mL Ag
+
soln
0.0250
L Cl

soln
*
0.800
mol Cl

1.00
L Cl

soln
*
1
mol Ag
+
1
mol Cl

*
1.00 L Ag
+
soln
0.100
mol Ag
+
=
0.200 L Ag
+
soln
IS THE ANSWER REASONABLE?
The silver ion concentration is oneeighth as large as the
chloride ion concentration. Since the ions react oneforone, we will need eight times as much
silver ion solution as chloride solution. Eight times the amount of chloride solution, 25 mL, is
200 mL, which is the answer we obtained. Therefore, the answer appears to be correct.
Practice Exercise 35:
Suppose 18.4 mL of 0.100
M
AgNO
3
solution was needed to
react completely with 20.5 mL of CaCl
2
solution. What is the molarity of the CaCl
2
solu
tion? Use the net ionic equation in the preceding example to work the problem. (Hint: How
can you calculate molarity from moles and volume, and how can you calculate the molarity
of the CaCl
2
solution from the molarity of Cl
2
?)
Practice Exercise 36:
How many milliliters of 0.500
M
KOH are needed to react
completely with 60.0 mL of 0.250
M
FeCl
2
solution to precipitate Fe(OH)
2
? The net ionic
equation is
4.8
C
HEMICAL ANALYSIS AND TITRATION ARE
APPLICATIONS OF SOLUTION STOICHIOMETRY
Chemical analyses fall into two categories. In a
qualitative analysis
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 Spring '10
 Dr.KimberlyEdwards
 Chemistry, Stoichiometry, Chemical reaction

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