Page 175 - Modern Analytical Chemistry

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158 Modern Analytical Chemistry

The solubility of Pb(IO 3 ) 2 is equal to the additional concentration of Pb 2+ in solu-
tion, or 7.9 ´10 –7 mol/L. As expected, the solubility of Pb(IO 3 ) 2 decreases in the
presence of a solution that already contains one of its ions. This is known as the
common ion effect common ion effect.
The solubility of an insoluble salt As outlined in the following example, the process of making and evaluating ap-
decreases when it is placed in a solution proximations can be extended if the first approximation leads to an unacceptably
already containing one of the salt’s ions.
large error.

EXAMPLE 6. 9
–4
Calculate the solubility of Pb(IO 3 ) 2 in 1.0 ´10 M Pb(NO 3 ) 2 .
SOLUTION
2+
Letting x equal the change in the concentration of Pb , the equilibrium
concentrations are
–4
–
2+
[Pb ] = 1.0 ´10 + x [IO 3 ]=2x
and
–4
2
(1.0 ´10 + x)(2x) = 2.5 ´10 –13
We start by assuming that
–4
2+
–4
[Pb ] = 1.0 ´10 + x ≈ 1.0 ´10 M
–5
and solve for x, obtaining a value of 2.50 ´10 . Substituting back gives the
2+
calculated concentration of Pb at equilibrium as
2+
–4
–4
–5
[Pb ] = 1.0 ´10 + 2.50 ´10 = 1.25 ´10 M
a value that differs by 25% from our approximation that the equilibrium
–4
concentration is 1.0 ´10 M. This error seems unreasonably large. Rather than
shouting in frustration, we make a new assumption. Our first assumption that
the concentration of Pb 2+ is 1.0 ´10 –4 M was too small. The calculated
concentration of 1.25 ´10 –4 M, therefore, is probably a little too large. Let us
assume that
–4
2+
–4
[Pb ] = 1.0 ´10 + x ≈1.2 ´10 M
Substituting into the solubility product equation and solving for x gives us
x = 2.28 ´10 –5
2+
or a concentration of Pb at equilibrium of
–5
–4
2+
–4
[Pb ] = 1.0 ´10 + (2.28 ´10 ) = 1.23 ´10 M
which differs from our assumed concentration of 1.2 ´10 –4 M by 2.5%. This
seems to be a reasonable error since the original concentration of Pb 2+ is
given to only two significant figures. Our final solution, to two significant
figures, is
–5
2+
–4
–
[Pb ] = 1.2 ´10 M [IO 3 ] = 4.6 ´10 M
and the solubility of Pb(IO 3 ) 2 is 2.3 ´10 –5 mol/L. This iterative approach to
solving an equation is known as the method of successive approximations.