Page 173 - Modern Analytical Chemistry

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156 Modern Analytical Chemistry

6G Solving Equilibrium Problems
UO 2 2+ Ladder diagrams are a useful tool for evaluating chemical reactivity, usually provid-
ing a reasonable approximation of a chemical system’s composition at equilibrium.
+0.327 V (pH = 0) When we need a more exact quantitative description of the equilibrium condition, a
ladder diagram may not be sufficient. In this case we can find an algebraic solution.
Perhaps you recall solving equilibrium problems in your earlier coursework in
chemistry. In this section we will learn how to set up and solve equilibrium prob-
E lems. We will start with a simple problem and work toward more complex ones.
+0.209 V (pH = 1)
6G.1 A Simple Problem: Solubility of Pb(IO 3) 2 in Water
When an insoluble compound such as Pb(IO 3 ) 2 is added to a solution a small por-
+0.090 V (pH = 2) tion of the solid dissolves. Equilibrium is achieved when the concentrations of Pb 2+
–
and IO 3 are sufficient to satisfy the solubility product for Pb(IO 3 ) 2 . At equilibrium
U 4+ the solution is saturated with Pb(IO 3 ) 2 . How can we determine the concentrations
–
of Pb 2+ and IO 3 , and the solubility of Pb(IO 3 ) 2 in a saturated solution prepared by
adding Pb(IO 3 ) 2 to distilled water?
Figure 6.10 We begin by writing the equilibrium reaction
Ladder diagram showing the effect of a 2+ –
change in pH on the areas of predominance Pb(IO 3 ) 2 (s) t Pb (aq) + 2IO 3 (aq)
2+
for the UO 2 /U 4+ half-reaction.
and its equilibrium constant
– 2
2+
K sp = [Pb ][IO 3 ] = 2.5 ´10 –13 6.33
2+
–
As equilibrium is established, two IO 3 ions are produced for each ion of Pb . If we
2+
assume that the molar concentration of Pb at equilibrium is x then the molar con-
–
centration of IO 3 is 2x. To help keep track of these relationships, we can use the
following table.
2+
–
PbI 2 (s) t Pb (aq) + 2IO 3 (aq)
Initial concentration solid 0 0
Change in concentration solid +x +2x
Equilibrium concentration solid 0 + x = x 0+2x =2x

Substituting the equilibrium concentrations into equation 6.33
2
(x)(2x) = 2.5 ´10 –13
and solving gives
3
4x = 2.5 ´10 –13

x = 3.97 ´10 –5
2+
–
The equilibrium concentrations of Pb and IO 3 , therefore, are
–5
2+
[Pb ]= x = 4.0 ´10 M
–
–5
[I ]=2x = 7.9 ´10 M
2+
Since one mole of Pb(IO 3 ) 2 contains one mole of Pb , the solubility of Pb(IO 3 ) 2
2+
is the same as the concentration of Pb ; thus, the solubility of Pb(IO 3 ) 2 is
–5
4.0 ´10 M.