# 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(IO

_{3})_{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(IO_{3})_{2}divided by the volume of the solid Ba(IO_{3})_{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(IO

_{3})_{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.
** 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(IO**

_{3})_{2}(487 g/mol) can be dissolved in 500 mL of water at 25 °C?**Solution**

** The
solubility-product constant for Ba(IO

_{3})_{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 Ba

^{2+}is formed for each mole of Ba(IO_{3})_{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
Ba

^{2+}is produced for every mole of Ba(IO_{3})_{2},
** To compute
the number of millimoles of Ba(IO

_{3})_{2}dissolved in 500 mL of solution, we write
** The mass of
Ba(IO

####

_{3})_{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(IO**

_{3})_{2}in a solution that is 0.0200 M in Ba(NO_{3})_{2}.**Solution**

** The solubility is not equal to [Ba

^{2+}] in this case because Ba(NO

_{3})

_{2}is also a source of barium ions. We know, however, that the solubility is related to [IO

_{3}

^{-}]:

** There are
two sources of barium ions: Ba(NO

_{3})_{2}and Ba(IO_{3})_{2}. The contribution from the nitrate is 0.0200 M, and that from the iodate is equal to the molar solubility,or ½ [IO_{3}^{-}]. 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 [IO

_{3}^{-}]. The small numerical value of K_{sp}suggests that the solubility of Ba(IO_{3})_{2}is quite small, and this finding is confirmed by the result obtained in Example (1). Also, barium ion from Ba(NO_{3})_{2}will further suppress the limited solubility of Ba(IO_{3})_{2}. Therefore, it seems reasonable to assume that 0.0200 is large with respect to ½ [IO_{3}^{-}] in order to find a provisional answer to the problem. That is, we assume that ½ [IO_{3}^{-}] << 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 Ba^{2+}arising from the dissociation of Ba(IO_{3})_{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(IO

_{3})_{2}by a factor of about 5.**Example (3): Calculate the solubility of Ba(IO**

_{3})_{2}in a solution prepared by mixing 200 mL of 0.0100 M Ba(IO_{3})_{2}with 100 mL of 0.100 M NaIO

_{3}**.**

**Solution**

** First, establish
whether either reactant is present in excess at equilibrium. The amounts taken are:

** If the
formation of Ba(IO

_{3})_{2}is complete:
** Thus,

** As in
Example (1),

** In this case, however,

** where 2[Ba

^{2+}] represents the iodate contributed by the sparingly soluble Ba(IO_{3})_{2}. We find a provisional answer after making the assumption that [IO_{3}^{-}] ~ 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(IO

_{3})_{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*

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