17.4 Solubility and Ksp
PRE-HEALTH PREP
Ksp and Molar Solubility Calculations
General Chemistry 17.4
The solubility product constant (Ksp) is the equilibrium constant for an ionic solid dissolving into its constituent ions in water. ICE tables provide a systematic method for converting between Ksp and molar solubility, with worked examples covering 1:1 salts like AgCl, multi-ion compounds like Ag2S and BiI3, and the algebraic handling of stoichiometric coefficients. When comparing the solubility of different salts, direct Ksp comparison is valid only when the salts produce the same number of ions; otherwise, molar solubility must be calculated for each compound.
What Is Ksp? The Solubility Product Constant Explained
This lesson introduces solubility equilibrium, a new application of the equilibrium principles you have already studied. In previous chapters, you encountered special equilibrium constants for specific types of reactions: Ka for weak acid dissociation, Kb for weak base dissociation, and Kw for the autoionization of water. Each of these describes a particular substance dissociating in aqueous solution.
Now we add another to the collection. When an ionic solid dissolves in water, the equilibrium constant for that process is called Ksp, where "sp" stands for solubility product. You may also see it referred to as the solubility product constant. Despite the new name, Ksp is simply an equilibrium constant for a very specific type of reaction: an ionic solid dissociating into its constituent ions in water.
Key Point: Ksp is just another equilibrium constant. The only thing that makes it special is the type of reaction it describes — an ionic solid dissolving into its ions in aqueous solution.
How to Write Ksp Expressions in 2 Steps
Step 1: Write the Dissolution Reaction
Every Ksp problem begins with the same type of reaction: a solid ionic compound on the reactant side dissociating into its individual ions. Consider silver chloride as an example:
Example: Dissolution of AgCl
AgCl(s) → Ag+(aq) + Cl−(aq)
Solid AgCl dissociates into one silver ion and one chloride ion — one of each because there is one of each in the formula.
Step 2: Write the Ksp Expression
The Ksp expression follows the standard equilibrium expression rules. However, because the reactant is always a pure solid, it has an activity of one and does not appear in the expression. This means Ksp expressions never have a denominator:
Example: Ksp Expression for AgCl
Ksp = [Ag+][Cl−]
The solid AgCl does not appear. This is why Ksp is called a solubility product — you are literally multiplying the product ion concentrations together, with no denominator.
Why No Denominator? In any equilibrium expression, pure solids and pure liquids are excluded (activity = 1). Since the reactant in a Ksp reaction is always a solid, Ksp expressions consist entirely of product ion concentrations. You will never have a denominator in a Ksp expression.
The ICE Table Method for Molar Solubility
To solve Ksp problems quantitatively, you set up an ICE (Initial, Change, Equilibrium) table just as you would for any other equilibrium calculation. For the dissolution of AgCl:
| AgCl(s) | Ag+(aq) | Cl−(aq) | |
|---|---|---|---|
| Initial | — | 0 | 0 |
| Change | −x | +x | +x |
| Equilibrium | — | x | x |
Before any AgCl dissolves, the concentrations of Ag+ and Cl− in pure water are both zero. As the solid dissolves, each ion increases by +x, giving equilibrium concentrations of x for both ions.
The variable x in the ICE table represents the molar concentration of the solid that dissolves. This quantity has a special name: the molar solubility. Regardless of the ion ratio in the formula, x always equals the molar solubility — it is the molar concentration of the ionic solid that dissolves.
The 2 Most Common Ksp Questions: You will almost always be asked one of two things: (1) given Ksp, calculate the molar solubility, or (2) given the molar solubility, calculate Ksp. Master these two directions and you have the foundation for this entire topic.
Calculating Molar Solubility from Ksp: 2 Worked Examples
Example: AgCl (1:1 Dissociation)
Problem: What is the molar solubility of AgCl? (Ksp = 1.8 × 10−10)
Setup: From the ICE table, [Ag+] = x and [Cl−] = x, so:
Ksp = (x)(x) = x2
Solve:
1.8 × 10−10 = x2
x = √(1.8 × 10−10) = 1.3 × 10−5 M
The molar solubility of AgCl is 1.3 × 10−5 M. In this case, the concentration of each ion also equals 1.3 × 10−5 M, since both have 1:1 coefficients.
Example: Ag2S (3-Ion Dissociation)
Problem: What is the molar solubility of Ag2S? (Ksp = 6.0 × 10−51)
Dissolution reaction:
Ag2S(s) → 2 Ag+(aq) + S2−(aq)
ICE table: Starting at zero, [Ag+] = 2x and [S2−] = x at equilibrium.
Ksp expression:
Ksp = [Ag+]2[S2−] = (2x)2(x) = 4x3
Solve:
6.0 × 10−51 = 4x3
x3 = 1.5 × 10−51
x = (1.5 × 10−51)1/3 = 1.1 × 10−17 M
Common Mistake: When squaring (2x), students often square the x but forget to square the 2. Remember: (2x)2 = 4x2, not 2x2. The coefficient shows up in two places — as the multiplier of x in the ICE table and as the exponent in the Ksp expression.
Notice the Ksp for Ag2S (10−51) is vastly smaller than that of AgCl (10−10). In general, a smaller Ksp means less dissolved solid and a lower molar solubility. A larger Ksp indicates greater solubility. However, direct Ksp comparisons require caution when the salts produce different numbers of ions — more on that later in this lesson.
Ion Concentrations vs. Molar Solubility
Individual ion concentrations are not always equal to the molar solubility. For Ag2S with x = 1.1 × 10−17 M:
- Sulfide ion concentration: [S2−] = x = 1.1 × 10−17 M
- Silver ion concentration: [Ag+] = 2x = 2.2 × 10−17 M
Don't Confuse Concentration with the Ksp Formula: A common student error is squaring the silver ion concentration when asked for [Ag+]. You only square it when plugging it into the Ksp expression. The actual concentration of Ag+ in solution is simply 2x — no exponent needed.
Calculating Ksp from Molar Solubility
Example: BiI3 (Bismuth Iodide)
Problem: The molar solubility of BiI3 is 1.32 × 10−5 M. Calculate Ksp.
Dissolution reaction:
BiI3(s) → Bi3+(aq) + 3 I−(aq)
ICE table: [Bi3+] = x and [I−] = 3x at equilibrium.
Ksp expression:
Ksp = [Bi3+][I−]3 = (x)(3x)3 = 27x4
Solve:
Ksp = 27(1.32 × 10−5)4 = 8.20 × 10−19
Terminology Alert: Don't confuse molar solubility (the variable x — how much solid dissolves) with the solubility product constant (Ksp — the equilibrium constant). Both contain the word "solubility," but they are different quantities.
Shortcut: Writing Ksp Directly from the Formula
When no common ion is present, you can write the Ksp expression directly from the formula without setting up a full ICE table. Look at the subscripts in the compound's formula:
- For BiI3: one Bi gives (1x)1, three I gives (3x)3
- Ksp = (1x)1(3x)3 = 27x4
Each subscript in the formula becomes both the coefficient of x and the exponent in the Ksp expression. This shortcut produces the same result as the full ICE table approach and saves time on exams.
Limitation: This shortcut applies only when no common ion is already present in solution. For common ion and precipitation problems, you must use the full ICE table. Those situations are covered in the next lesson.
Comparing Solubility of Different Salts
When asked which salt has the highest solubility in water, you might be tempted to simply pick the one with the largest Ksp. Sometimes that works — but not always.
| Ksp | |
|---|---|
| AgCl | 1.8×10−10 |
| AgBr | 5.0×10−13 |
| AgI | 8.3×10−17 |
| Ag2S | 6.0×10−51 |
| Co(OH)2 | 1.3×10−15 |
Practice Problem: Which of the following salts has the highest solubility in water?
a) AgCl
b) AgBr
c) AgI
d) Ag2S
e) Co(OH)2
When You Can Compare Ksp Directly
If two salts produce the same number of ions when they dissolve, their Ksp values relate to molar solubility through the same mathematical expression. In that case, the salt with the larger Ksp is genuinely more soluble.
Example: Comparing Silver Halides
Consider AgCl, AgBr, and AgI. Each dissolves into exactly two ions (one Ag+ and one halide), so for all three: Ksp = x2. Because they share the same Ksp-to-x relationship, whichever has the largest Ksp has the largest molar solubility. Among these three, AgCl wins.
When You Must Calculate x
When salts produce different numbers of ions, the relationship between Ksp and x differs for each salt. A salt with a smaller Ksp could potentially have a larger molar solubility if its Ksp expression involves a higher power of x.
Example: AgCl vs. Co(OH)2
For AgCl (Ksp = 1.8 × 10−10): Ksp = x2, so x = 1.3 × 10−5 M
For Co(OH)2 (Ksp = 1.3 × 10−15): Ksp = 4x3, so x = (Ksp/4)1/3 = 6.9 × 10−6 M
Even though Co(OH)2 has a much smaller Ksp, its molar solubility (6.9 × 10−6 M) is only about half that of AgCl (1.3 × 10−5 M), not orders of magnitude less. The different Ksp-to-x relationships make direct Ksp comparison unreliable across groups.
3-Step Strategy for Comparing Solubilities
When comparing multiple salts, follow this approach:
- Group the salts by the number of ions they produce upon dissolving.
- Within each group, the salt with the largest Ksp is the most soluble — no calculation needed.
- Across groups, calculate the actual molar solubility (x) for the top candidate in each group, then compare.
Exam Tip: Professors are fond of choosing examples where the "just pick the highest Ksp" shortcut gives the wrong answer. Always check whether the salts produce the same or different numbers of ions before comparing Ksp values directly.
7 Key Takeaways
- Ksp is the equilibrium constant for an ionic solid dissolving into its ions in water. Ksp expressions have no denominator because the reactant is always a pure solid.
- Molar solubility (x) is the molar concentration of the solid that dissolves. It always corresponds to x in the ICE table, regardless of the ion ratio.
- For a 1:1 salt like AgCl, Ksp = x2. For salts with unequal ion ratios, include the coefficients as both multipliers and exponents — for example, Ag2S gives Ksp = 4x3.
- When calculating (2x)2 or (3x)3, always raise the coefficient to the power as well: (2x)2 = 4x2, not 2x2.
- Individual ion concentrations may differ from the molar solubility by a factor of the stoichiometric coefficient. Do not apply the Ksp exponent when reporting a concentration.
- You can compare Ksp values directly only when the salts produce the same number of ions. Otherwise, calculate the actual molar solubility for each salt before comparing.
- To compare solubilities across different salt types, group by ion count, eliminate within each group using Ksp, then calculate x for the remaining candidates.