Page 174 - Modern Analytical Chemistry

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Chapter 6 Equilibrium Chemistry 157

6G.2 A More Complex Problem: The Common Ion Effect
Calculating the solubility of Pb(IO 3 ) 2 in distilled water is a straightforward prob-
–
lem since the dissolution of the solid is the only source of Pb 2+ or IO 3 . How is the
solubility of Pb(IO 3 ) 2 affected if we add Pb(IO 3 ) 2 to a solution of 0.10 M
Pb(NO 3 ) 2 ? Before we set up and solve the problem algebraically, think about the
chemistry occurring in this system, and decide whether the solubility of Pb(IO 3 ) 2
will increase, decrease, or remain the same. This is a good habit to develop.
Knowing what answers are reasonable will help you spot errors in your calcula-
tions and give you more confidence that your solution to a problem is correct.
We begin by setting up a table to help us keep track of the concentrations of
2+
–
Pb and IO 3 in this system.
PbI 2 (s) t Pb (aq) + 2IO 3 (aq)
2+
–
Initial concentration solid 0.10 0
Change in concentration solid +x +2x
Equilibrium concentration solid 0.10 + x = x 0+2x =2x

Substituting the equilibrium concentrations into the solubility product expression
(equation 6.33)
2
(0.10 + x)(2x) = 2.5 ´10 –13
and multiplying out the terms on the left leaves us with
2
3
4x + 0.40x = 2.5 ´10 –13 6.34
This is a more difficult equation to solve than that for the solubility of Pb(IO 3 ) 2 in dis-
tilled water, and its solution is not immediately obvious. A rigorous solution to equa-
tion 6.34 can be found using available computer software packages and spreadsheets.
How might we solve equation 6.34 if we do not have access to a computer? One
possibility is that we can apply our understanding of chemistry to simplify the algebra.
2+
From Le Châtelier’s principle, we expect that the large initial concentration of Pb will
significantly decrease the solubility of Pb(IO 3 ) 2 . In this case we can reasonably expect the
2+
equilibrium concentration of Pb to be very close to its initial concentration; thus, the
2+
following approximation for the equilibrium concentration of Pb seems reasonable
2+
[Pb ] = 0.10 + x »0.10 M
Substituting into equation 6.34
2
(0.10)(2x) = 2.5 ´10 –13
and solving for x gives
2
0.40x = 2.5 ´10 –13
x = 7.91 ´10 –7
Before accepting this answer, we check to see if our approximation was reasonable.
In this case the approximation 0.10 + x » 0.10 seems reasonable since the difference
between the two values is negligible. The equilibrium concentrations of Pb 2+ and
–
IO 3 , therefore, are
2+
[Pb ] = 0.10 + x » 0.10 M
–6
–
[I ]=2x = 1.6 ´10 M