Lesson 4: Solubility Product
Remember from Lesson 3 that a saturated solution is in a state of dynamic equilibrium. It there is solid at the bottom of the solution, the concentration of the saturated solution stays the same but some solid is continually dissolving while an equal amount of solute is precipitating out of solution. The concentrations of ions in saturated solutions have a relationship to one another somewhat like the relationship between the concentration of H3O+ and OH- in water in that if the concentration of one ion increases, the concentration of the other ion will decrease to maintain the equilibrium relationship.
Sodium Chloride
Consider saturated sodium chloride solution. Quite a bit of sodium chloride can be dissolved in water, about 6 moles in one liter. That makes the concentration of both the sodium ion and the chloride ion about 6 M.
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If nothing happened, we would still have 6M Na+ and a higher concentration, perhaps, 8M, of Cl-. But something does happen. Crystals of NaCl form from the reaction of some of the extra Cl- with some of the Na+ that was in the solution. The concentration of Na+ goes down to around 5 M as the concentration of Cl- increases to somewhere around 7M. |
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As the concentration of one ion increases, the concentration of the other ion decreases. As you may have suspected, there is an equation that relates the concentrations of the dissociated ions of sodium chloride.
Let me draw the parallel.
Water ionizes to form H3O+ and OH-. |
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The reaction is reversible. |
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The concentrations of H3O+ and OH- are related by this equation: |
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When multiplied together, the concentrations of H3O+ and OH- give a fairly constant value called the ionization constant of water, or Kw. |
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H3O+ + OH-
H3O+ + OH-
Now sodium chloride.
Sodium chloride dissolves and dissociates in water to Na+ and Cl-. |
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The reaction is reversible: |
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The concentrations of Na+ and Cl- are related by this equation: |
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When multiplied together, the concentrations of Na+ and Cl- give a fairly constant value called the solubility product constant, or Ksp. For sodium chloride, the value of Ksp is about 36. |
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Practice
See if you can figure out what the concentration of Na+ would be if we were able to increase the concentration of Cl- up to 10 M. Take a moment to figure that out.
Answer
You should have calculated about 3.6 M for the sodium ion concentration. I got that value by saying that the concentration of Na+ times the concentration of Cl- is equal to 36 (the Ksp value for sodium chloride). If the concentration of Cl- is going to be 10 M, then the concentration of of Na+ has to be 36 divided by 10. That comes at to 3.6 M. |
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Silver Chloride
The same line of reasoning can be used with any salt that dissolves in water, even if it dissolves only a very small amount.
Silver chloride will dissolve somewhat in water. However, it reaches saturation very quickly--that is, when the concentrations of silver and chloride ions are about 1.3 x 10-5M. |
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Still we can write a solubility product equation for it. |
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The value for the Ksp of silver chloride, however, is about 1.8 x 10-10, a very small number. |
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The smaller the Ksp, the less soluble the salt; the larger the Ksp, the more soluble the salt. Sodium chloride is very soluble in water and has a large value of Ksp (36), but silver chloride (a salt that is listed as "insoluble" by the Solubility Rules) has a very small Ksp.
Practice
Try using the information from above to calculate the Ag+ concentration if the chloride ion concentration were 3.0 M.
Answer
In this case the answer turns out to be a very small number, which can be calculated using the process shown here. Use the value for Ksp that we determined in the previous example and plug in 3.0 M for the concentration of the chloride ion. Then solve for the silver ion concentration. |
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Lead(II) Chloride
When the formula for a salt contains more than just one of each ion, the solubility product equation gets a little more complicated. (Example 11 in your workbook.)
Let's use PbCl2 as an example. When PbCl2 dissolved in water, two ions of Cl- are released for every one ion of Pb2+. |
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When saturation is reached we show that the reaction is reversible. |
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The equation for the solubility product is: |
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The concentration of Cl- is squared because the balanced equation for the reaction shows a 2 as the coefficient in front of Cl-. |
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Looking at the reaction in this way might help you to remember that: |
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Example - Determining concentrations at equilibrium
Here is an example of how to calculate the equilibrium concentration of one substance given the Ksp and the equilibrium concentrations of the other substance. (Also shown in example 12 in your workbook.)
For the reaction |
Ksp = [Pb+2] x [Cl-]2 | |
| 2.0 x 10-5 = [Pb+2] x (2.0 x 10-3)2 | ||
| 2.0 x 10-5 = [Pb+2] x 4.0 x 10-6 | ||
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| 5.0 M = [Pb+2] |
Practice Problems: Equilibrium Concentrations
Try your hand at answering the following questions (also shown in exercise 13 in your
workbook.) Check your answers below and then move on to the Wrap-Up.
a. |
The Ksp for AgCl is 1.8 x 10-10. If Ag+ and Cl- are both in solution and in equilibrium with AgCl. What is [Ag+] if [Cl-] = .020 M? |
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If Ag+ and Cl- were both present at 0.0001 M, would a precipitate occur? |
c. |
What concentration of Ag+ would be necessary to bring the concentration of Cl- to 1.0 x 10-6 M or lower? |
Answers: Equilibrium Concentrations
Here are the answers to the questions above.
a. |
The Ksp for AgCl is 1.8 x 10-10. If Ag+ and Cl- are both in solution and in equilibrium with AgCl. What is [Ag+] if [Cl-] = .020 M? |
[Ag+] = 9.0 x 10-9M
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If Ag+ and Cl- were both present at 0.0001 M, would a precipitate occur? |
Yes, a precipitate would occur because these concentrations together are higher than what the Ksp allows.
c. |
What concentration of Ag+ would be necessary to bring the concentration of Cl- to 1.0 x 10-6 M or lower? |
[Ag+] = 1.8 x 10-4M or higher