Common Ion Effect

Emily Wilson
7 min read
Study Guide Overview
This study guide covers the common ion effect on solubility equilibria. It explains how adding a common ion decreases solubility, using Le Chatelier's Principle. It includes practice problems involving Ksp calculations, ICE tables, and molar solubility with and without common ions. The guide also emphasizes the importance of this topic for the AP exam.
#Solubility Equilibria and the Common Ion Effect
Hey there, future AP Chem master! Let's dive into how solubility changes when we're not just dealing with pure water. It's all about the common ion effect, and it's a game-changer! 💡
This topic is crucial because it connects solubility, equilibrium, and Le Chatelier's principle—all big hitters on the AP exam. Expect to see it in both multiple-choice and free-response questions.
#The Common Ion Effect: What is it?
The common ion effect is a fancy way of saying that if you add a salt that shares an ion with another salt already in solution, the solubility of the second salt goes down. It's like adding more of one ingredient to a recipe – it throws off the balance. Think of it as a crowded dance floor; if there are already a lot of dancers of one type, it's harder for more of that type to join in!
#Key Points
- A common ion is an ion that is present in two or more different compounds.
- Adding a common ion to a solution decreases the solubility of a slightly soluble salt.
- This effect is a direct result of Le Chatelier's Principle.
The common ion effect is a direct application of Le Chatelier's Principle. Remember, systems at equilibrium will shift to relieve stress. Adding a common ion is a stress that the system counteracts by reducing the solubility of the salt.
#How Does it Work? 🤔
Let's say we have a salt, , that dissolves like this:
The solubility product, , is:
Now, if we add a salt like , which also has ions, we're increasing the concentration of in the solution. According to Le Chatelier's Principle, the equilibrium will shift to the left, favoring the formation of the solid and reducing the solubility of .
Think of it like a seesaw. If you add weight to one side (the common ion), the seesaw tips the other way (less of the salt dissolves).
#Common Ion Effect Practice Problem ✏️
Let's work through an example to make this crystal clear. We'll calculate the molar solubility of () in pure water and then in a solution of .
#Molar Solubility in Pure Water
First, the equilibrium:
Set up an ICE table:
Initial | - | ||
Change | - | ||
Equilibrium | - |
So,
Solving for x, we get:
#Molar Solubility in
Now, we have an initial concentration of of from . Our ICE table changes:
Initial | - | ||
Change | - | ||
Equilibrium | - |
Our expression becomes:
We can ignore the '+x' because x will be very small compared to 0.0010. Solving for x:
Notice that the solubility of is much lower in the solution than in pure water! That's the common ion effect in action.
When solving for solubility with a common ion, you can often ignore the '+x' in the equilibrium expression if the initial concentration of the common ion is significantly larger than the Ksp value.
#Justifying the Common Ion Effect with Le Chatelier's Principle
Let's break down why this happens using Le Chatelier's Principle. Remember, Le Chatelier's Principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.
In our example, the stress is the addition of the common ion (). The system relieves this stress by shifting the equilibrium backwards, towards the solid . This means less dissolves, and the molar solubility decreases.
When explaining the common ion effect on the AP exam, always refer to Le Chatelier's Principle. It's your go-to explanation for equilibrium shifts.
#Another Example
Consider the dissolution of ():
If we dissolve in a solution containing , the presence of the common ion, , will shift the equilibrium to the left, reducing the solubility of .
Don't forget to consider the stoichiometry when dealing with polyatomic ions. For example, if you have , the concentration of will be 3x, and the concentration of will be 2x.
#Final Exam Focus
- Master the ICE table: It's your best friend for solving solubility problems.
- Le Chatelier's Principle is key: Always use it to explain equilibrium shifts.
- Common ion effect: Understand how it impacts solubility.
- Don't forget : It's the foundation for all solubility calculations.
When tackling free-response questions, show all your work. Even if you make a calculation error, you can still earn points for setting up the problem correctly.
# Practice Questions
Practice Question
#Multiple Choice Questions
-
The molar solubility of in pure water is M. What is the molar solubility of in a M solution? (A) M (B) M (C) M (D) M
-
Which of the following will decrease the solubility of in a saturated solution? (A) Adding (B) Adding (C) Adding water (D) Increasing the temperature
#Free Response Question
Consider the dissolution of in water. The for is .
(a) Write the balanced chemical equation for the dissolution of in water.
(b) Calculate the molar solubility of in pure water.
(c) Calculate the molar solubility of in a M solution of .
(d) Explain why the solubility of is different in pure water versus in the solution, using principles of equilibrium.
#Scoring Breakdown
(a) (1 point)
(b) (2 points)
- Setting up the ICE table correctly (1 point)
- Correctly solving for x (1 point)
(c) (2 points)
- Setting up the ICE table correctly with the common ion (1 point)
- Correctly solving for x (1 point)
(d) (2 points)
- Mentioning the common ion effect (1 point)
- Explaining that the presence of shifts the equilibrium to the left, decreasing solubility (1 point)
#Answers
- (D)
- (B)
You've got this! Remember to breathe, stay focused, and apply these concepts. You're going to do great! 💪
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