Introduction to Solubility Equilibria

Caleb Thomas
7 min read
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Study Guide Overview
This study guide covers solubility equilibria, focusing on the solubility product constant (Ksp). It explains how to calculate Ksp from solubility, and how to determine molar solubility using Ksp and RICE tables. Finally, it shows how to predict precipitation using the ion product (Qsp) compared to Ksp.
Solubility Equilibria: Your Ultimate Guide ๐
Hey there, future AP Chem master! Let's dive into solubility equilibria, a topic that might seem tricky but is totally conquerable with the right approach. Think of this as your pre-exam cheat sheet, designed to make everything click.
Introduction to Solubility Equilibria
What is Solubility Anyway? ๐ค
We often talk about substances being "soluble" or "insoluble," but the truth is, everything dissolves to some extent. It's all about how much dissolves.
Solubility is essentially an equilibrium process, where a solid dissolves into its ions in a solution.
The Dissolution Reaction
For example, let's look at lead iodide:
Even though we call "insoluble," this reaction does occur, just with a tiny equilibrium constant (K). This special K for dissolution is called the solubility product constant, or Ksp.
Ksp is the product of the ion concentrations raised to their stoichiometric coefficients. Solids are NOT included in the Ksp expression.
Ksp: The Magic Number
- High Ksp: Indicates a highly soluble compound (like NaCl or KOH). The reaction goes far to the right.
- Low Ksp: Indicates a poorly soluble compound (like PbI2). The reaction barely proceeds to the right.
Think of Ksp like a measure of how much a solid 'likes' to dissolve. A big Ksp means it's a social butterfly that loves to dissolve, while a small Ksp means it's a wallflower that prefers to stay solid.

Image from Purdue University: Visualizing the dynamic equilibrium of dissolution.
Calculating Ksp: Step-by-Step
Grams per Liter vs. Molarity
Solubility is often given in grams per liter (g/L) or moles per liter (mol/L, or molarity, M).
When calculating Ksp, always convert solubility to molarity (mol/L) first!
Example Problem: Lead Chromate
Let's calculate the Ksp of lead chromate ():
The solubility of is 4.5 \times 10^{-5}
g/L. What is its solubility product?
-
Write the dissolution reaction:
-
Convert g/L to mol/L (Molarity):
-
Plug into the Ksp expression:
Don't forget to raise ion concentrations to the power of their stoichiometric coefficients in the Ksp expression!
Calculating Molar Solubility
Molar solubility is the concentration of the metal cation in a saturated solution. It is essentially solving for equilibrium concentrations, but we don't include the solid reactant in our calculations. Time to bring back our trusty RICE table!
Example Problem: Calcium Carbonate
Calculate the molar solubility of ().
-
Write the dissolution reaction:
-
Set up the RICE table:
Reaction Initial ---- 0.00 M 0.00 M Change ---- +x +x Equilibrium ---- x x -
Plug into the Ksp expression and solve:
The molar solubility of is 5.8 \times 10^{-5}
M.
Relating Qsp and Ksp
Just like with regular equilibrium, we can compare the ion product (Qsp) to Ksp to predict if a precipitate will form:
- Qsp > Ksp: Precipitation occurs; the solution is supersaturated.
- Qsp < Ksp: No precipitation; the solution is unsaturated.
- Qsp = Ksp: The solution is at equilibrium (saturated).
Think of Qsp as the current state of the solution, and Ksp as the solution's 'ideal' state. If Qsp is too high, the solution needs to precipitate to get back to Ksp.
Final Exam Focus ๐ฏ
High-Priority Topics
- Calculating Ksp from solubility data.
- Calculating molar solubility using Ksp and RICE tables.
- Predicting precipitation using Qsp and Ksp.
- Understanding the relationship between solubility, Ksp, and equilibrium.
Common Question Types
- Multiple-choice questions involving Ksp calculations and comparisons.
- Free-response questions requiring you to set up RICE tables and calculate molar solubility.
- Questions that combine solubility with other equilibrium concepts.
Last-Minute Tips
- Time Management: Start with the questions you know best. Don't get bogged down on one problem.
- Common Pitfalls: Pay close attention to stoichiometric coefficients and units. Double-check your calculations.
- Challenging Questions: Break down complex problems into smaller steps. Use RICE tables to organize your thoughts.
Practice Question
Practice Questions
Multiple Choice Questions
-
The solubility of is
1.3 \times 10^{-5}
M. What is the value of the solubility product constant, , for ? (A)1.3 \times 10^{-5}
(B)2.6 \times 10^{-5}
(C)1.7 \times 10^{-10}
(D)1.3 \times 10^{-10}
-
The for is
3.9 \times 10^{-11}
. What is the molar solubility of ? (A)1.95 \times 10^{-11}
(B)3.39 \times 10^{-6}
(C)2.14 \times 10^{-4}
(D)4.26 \times 10^{-4}
Free Response Question
A 50.0
mL sample of 0.020
M is mixed with 50.0
mL of 0.10
M . A precipitate of forms. The of is 1.6 \times 10^{-5}
.
(a) Write the balanced net ionic equation for the precipitation reaction. (b) Calculate the initial concentrations of and ions in the mixture. (c) Calculate the value of the ion product, , for the initial conditions. (d) Does a precipitate form? Justify your answer. (e) Calculate the equilibrium concentration of ions in the solution after the precipitate forms.
FRQ Scoring Breakdown:
(a) (1 point)
(b) (2 points)
- (1 point)
- (1 point)
(c) (1 point)
- (1 point)
(d) (1 point)
- Yes, a precipitate forms because (1 point)
(e) (3 points)
-
Set up ICE table:
2Cl^-
Initial 0.010 0.050 Change -x -2x Equil 0.010-x 0.050-2x -
(1 point)
-
Since is small, assume x is small:
1.6 \times 10^{-5} = (0.010)(0.050)^2
(1 point) -
Solve for x: . (1 point)
You've got this! Go ace that exam! ๐ช

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Question 1 of 8
Solubility is best described as which type of process? ๐ค
Irreversible
Equilibrium
Spontaneous
Non-spontaneous