CK-12-Chemistry Intermediate

19.3. Solubility Equilibrium http://www.ck12.org

The moles of each ion from the original solutions are calculated by multiplying the molarity by the volume in liters.

mol Ba^2 += 0.0050 M×0.010 L = 5.0× 10 −^5 mol Ba^2 +

mol SO 42 −= 0.0020 M×0.020 L = 4.0× 10 −^5 mol SO 42 −

The concentration of each ion after dilution is then calculated by dividing the moles by the final solution volume of
0.030 L.

[Ba^2 +] =

5. 0 × 10 −^5 mol

0 .030 L

= 1. 7 × 10 −^3 M

[SO^24 −] =

4. 0 × 10 −^5 mol

0 .030 L

= 1. 3 × 10 −^3 M

Now, the ion product is calculated.

Ba^2 + [SO 42 −] = (1.7× 10 −^3 )(1.3× 10 −^3 ) = 2.2× 10 −^6

Since the ion product is greater than the Ks p, a precipitate of barium sulfate will form.

Step 3: Think about your result.

Two significant figures are appropriate for the calculated value of the ion product.

Practice Problem

2. Calculate the ion product for calcium hydroxide when 20.0 mL of 0.010 M CaCl 2 is mixed with 30.0 mL of
   0.0040 M KOH. Decide if a precipitate will form.

The Common Ion Effect

In a saturated solution of calcium sulfate, an equilibrium exists between the solid calcium sulfate and its ions in
solution.

CaSO 4 (s)⇀↽Ca^2 +(aq)+SO^24 −(aq) Ksp= 2. 4 × 10 −^5

Suppose that some calcium nitrate were added to this saturated solution. Immediately, the concentration of the
calcium ion in the solution would increase. As a result, the ion product [Ca^2 +][SO 42 −] would increase to a value
that is greater than the Ks p. According to Le Châtelier’s principle, the equilibrium above would shift to the left in
order to relieve the stress of the added calcium ion. Additional calcium sulfate would precipitate out of the solution
until the ion product, once again, becomes equal to Ks p. Note that in the new equilibrium, the concentrations of the
calcium ion and the sulfate ion would no longer be equal to each other. The calcium ion concentration would be
larger than the sulfate ion concentration.

This situation describes the common ion effect. Acommon ionis an ion that is common to more than one salt in
a solution. In the above example, the common ion is Ca^2 +. Thecommon ion effectis a decrease in the solubility
of an ionic compound as a result of the addition of a common ion. Adding calcium ions to a saturated solution of
calcium sulfate causes additional CaSO 4 to precipitate from the solution, lowering its solubility. The addition of a
solution containing sulfate ion, such as potassium sulfate, would result in the same common ion effect.