Solubility-Product Constants (Chemical Equilibrium)
Solubility-Product Constants
** Most, but
not all, sparingly soluble salts are essentially completely dissociated in
saturated aqueous solution.
** For example, when an excess of barium iodate is equilibrated with water, the dissociation process is adequately described by the equation:
** For example, when an excess of barium iodate is equilibrated with water, the dissociation process is adequately described by the equation:
Using
Equation:
we
write:
** The
denominator represents the molar concentration of Ba(IO3)2
in the solid, which is a phase that is separate from but in contact with the
saturated solution. The concentration of a compound in its solid state is,
however, constant. In other words, the number of moles of Ba(IO3)2
divided by the volume of the solid Ba(IO3)2 is constant
no matter how much excess solid is present. Therefore, the previous equation
can be rewritten in the form:
where the new
constant is called the solubility-product constant or the solubility product.
It is important to appreciate that Equation above shows that the position of
this equilibrium is independent of the amount of Ba(IO3)2
as long as some solid is present. In other words, it does not matter whether
the amount is a few milligrams or several grams.
Solubility-product constants at 25 °C inorganic salts
** The examples
that follow demonstrate some typical uses of solubility product expressions.
The Solubility of a Precipitate in Pure Water
With the
solubility-product expression, we can calculate the solubility of a sparingly soluble
substance that ionizes completely in water.
Example (1): What
mass (in grams) of Ba(IO3)2 (487 g/mol) can be dissolved
in 500 mL of water at 25 °C?
Solution
** The
solubility-product constant for Ba(IO3)2 is 1.57 × 10-9
(see table above). The equilibrium between the solid and its ions in solution
is described by the equation:
** The equation
describing the equilibrium reveals that 1 mol of Ba2+ is formed for
each mole of Ba(IO3)2 that dissolves. Therefore,
** Since two
moles of iodate are produced for each mole of barium ion, the iodate concentration
is twice the barium ion concentration:
** Substituting
this last equation into the equilibrium-constant expression gives:
** Since 1 mol
Ba2+ is produced for every mole of Ba(IO3)2,
** To compute
the number of millimoles of Ba(IO3)2 dissolved in 500 mL
of solution, we write
** The mass of
Ba(IO3)2 in 500 mL is given by:
The Effect of a Common Ion on the Solubility of a Precipitate
The common-ion
effect is a mass-action effect predicted from Le Châtelier’s principle
and is
demonstrated by the following examples.
Example (2): Calculate
the molar solubility of Ba(IO3)2 in a solution that is
0.0200 M in Ba(NO3)2.
Solution
** The solubility is not equal to [Ba2+] in this case because Ba(NO3)2 is also a source of barium ions. We know, however, that the solubility is related to [IO3-]:
** There are
two sources of barium ions: Ba(NO3)2 and Ba(IO3)2.
The contribution from the nitrate is 0.0200 M, and that from the iodate is
equal to the molar solubility,or ½ [IO3-]. Thus,
** By
substituting these quantities into the solubility-product expression, we find
that
** Since this
is a cubic equation, we would like to make an assumption that would simplify
the algebra required to find [IO3-]. The small numerical
value of Ksp suggests that the solubility of Ba(IO3)2
is quite small, and this finding is confirmed by the result obtained in Example
(1). Also, barium ion from Ba(NO3)2 will further suppress
the limited solubility of Ba(IO3)2. Therefore, it seems
reasonable to assume that 0.0200 is large with respect to ½ [IO3-]
in order to find a provisional answer to the problem. That is, we assume that ½
[IO3-] << 0.0200, so
** The original
equation then simplifies to:
** The
assumption that (0.0200 + ½ × 2.80 × 10-4) ~ 0.0200 causes minimal
error because the second term, representing the amount of Ba2+
arising from the dissociation of Ba(IO3)2, is only about
0.7% of 0.0200. Usually, we consider an assumption of this type to be
satisfactory if the discrepancy is less than 10%. Finally, then,
** If we
compare this result with the solubility of barium iodate in pure water (Example
1), we see that the presence of a small concentration of the common ion has
decreased the molar solubility of Ba(IO3)2 by a factor of
about 5.
Example (3): Calculate
the solubility of Ba(IO3)2 in a solution prepared by mixing 200 mL of 0.0100 M Ba(IO3)2 with 100 mL of 0.100 M NaIO3.
Solution
** First, establish
whether either reactant is present in excess at equilibrium. The amounts taken are:
** If the
formation of Ba(IO3)2 is complete:
** Thus,
** As in
Example (1),
** In this case, however,
** where 2[Ba2+]
represents the iodate contributed by the sparingly soluble Ba(IO3)2.
We find a provisional answer after making the assumption that [IO3-]
~ 0.0200. Therefore,
** Since the
provisional answer is nearly four orders of magnitude less than 0.0200 M, our
approximation is justified, and the solution does not need further refinement.
** Notice that
the results from the last two examples demonstrate that an excess of iodate
ions is more effective in decreasing the solubility of Ba(IO3)2
than is the same excess of barium ions.
Reference: Fundamentals of analytical chemistry / Douglas A. Skoog, Donald M. West, F. James Holler, Stanley R. Crouch. (ninth edition) , 2014 . USA
Reference: Fundamentals of analytical chemistry / Douglas A. Skoog, Donald M. West, F. James Holler, Stanley R. Crouch. (ninth edition) , 2014 . USA
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