### Common Ion Effect on Solubility

```Common Ion Effect on Solubility
How is the solubility of a solid affected by other ion species in solution?
Why?
The solubility product (K sp) for a salt allows chemists to predict the concentration of dissolved ions in
solution. At this point you have learned to calculate the concentration of ions in water when an insoluble
salt is dissolved in pure water. But what if the beaker already has some ions dissolved in it? Can the salt
still dissolve to the same extent? Is the solubility of the salt changed?
Model 1 – Dissolving a Salt in 500 mL of Solution
Note: Proceed to Question 1. Do not fill in these tables until instructed to do so.
Ca(OH)2 (s) ↔ Ca2+ (aq) + 2OH1− (aq) K sp = 5.5 × 10 −6
Beaker 1
Initial
Beaker 2
Ca2+
OH1−
0M
0M
Initial
Change
Change
Equilibrium
Equilibrium
Beaker 3
Initial
Ca2+
OH1−
0M
0.200 M
Ca2+
OH1−
0M
0.500 M
Beaker 4
Ca2+
OH1−
0.200 M
0M
Initial
Change
Change
Equilibrium
Equilibrium
1. The solubility of what salt is being investigated in the four beakers in Model 1?
Common Ion Effect on Solubility
1
2. Why does solid Ca(OH)2 not appear in the ICE tables in Model 1?
3. Match each of the following descriptions with one of the beakers in Model 1. In each case,
assume the change in volume as the solid(s) are added is minimal.
a. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of water in beaker
.
b. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.500 M sodium
hydroxide solution in beaker .
c. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.200 M sodium
hydroxide solution in beaker .
d. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.200 M calcium
nitrate solution in beaker .
4. Based on the solubility product constant, K sp, for calcium hydroxide given in Model 1, do you
expect most of the 1.00 mole sample of solid to dissolve in any of the four beakers?
5. Three of the beakers in Model 1 contain a common ion in solution.
a. Which beakers in Model 1 contain a common ion?
b. Consider the source of the common ion in each beaker. Does the common ion come from the
insoluble salt? If no, explain the source of the common ion.
6. Predict what will be in Beakers 1 and 2 when the systems in those beakers reach equilibrium. Use
particulate drawings below to communicate your predictions. You may omit water molecules for
clarity.
7. Write the equation for the solubility product constant, K sp, for this salt.
8. Divide the work among group members to fill in each of the ICE tables in Model 1. Use the
solubility product constant to find the increase in Ca2+ ion concentration, or x value, in each
beaker. Add this to Model 1. (Work space for these calculations has been provided in Model 1.)
2POGIL™ Activities for AP* Chemistry
9. What does a large x value mean? That is, do large x values mean more solid has dissolved?
10. Does the presence of a common ion increase or decrease the solubility of the insoluble salt? Support your answer with evidence from Model 1.
11. According to Model 1, as the concentration of the common ion is increased, how does the solubility of the insoluble salt change? Support your answer with evidence from Model 1.
12. Are the data in Model 1 consistent with LeChâtelier’s Principle? Support your answer with specific examples from Model 1.
13. Consider the four beakers below. One-gram samples of barium sulfate, BaSO4, are added to each
beaker. Rank the beakers in terms of solubility of the barium sulfate from least soluble to most
soluble. The solubility product constant for barium sulfate is K sp = 1.1 × 10−10.
Beaker 1
Beaker 2
Beaker 3
Beaker 4
Water0.100 M0.100 M0.200 M
NaClNa2SO4
Ba(NO3 )2
Common Ion Effect on Solubility
3
14. The solubility product constant, K sp, for silver carbonate, Ag 2CO3, is 8.10 × 10 −12. Compare the
solubility of silver carbonate in water to its solubility in a 0.100 M solution of silver nitrate.
Support your comparison with calculations similar to those in Model 1.
4POGIL™ Activities for AP* Chemistry
Extension Questions
15. In a precipitation reaction (otherwise known as a double replacement or metathesis reaction) the
extent to which a precipitate is formed is dependent on the solubility product constant of the
precipitate. A species with a large K sp (highly soluble) may not precipitate until very high concentrations are reached. A species with a low K sp (mostly insoluble) will precipitate readily. Use the
ideas explored in this activity to explain why the formation of a precipitate may increase when
non-stoichiometric amounts of reactants are used (one reactant will limit, while the other will be
in excess).
16. Would the four beakers in Model 1 have the same reading on a pH meter? Explain your reasoning.
17. Consider any of the beakers in Model 1. Predict what might happen to the solubility of the
insoluble salt if a 1.0 M acid solution was added dropwise to the beaker. Hint: Which ion species
in the beaker will react with the acid? How will this reaction alter the concentration of the common ion?)
Common Ion Effect on Solubility
5
```