What Is Solubility Equilibria?
In AP Chemistry, we spend a good deal of time discussing solubility, that is, whether or not a solute can dissolve in a solvent, the latter almost always being water. However, after discussing equilibrium, we can apply these ideas to the dissolution of various compounds that we previously classified as insoluble.
In reality, everything dissolves to some extent; we are just describing to what extent dissolution occurs. For example, NaCl is able to dissolve much easier than a substance like PbI2, which we would usually call insoluble. However, we can still write out the reaction for the dissolution of PbI2:
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
You may think that this reaction simply does not occur because we call PbI2 "insoluble," but instead, this reaction just has a tiny K value. In fact, for the dissolution of lead iodide, the K value is 4.41 x 10^−9. When dealing with the equilibrium of dissolution, we call this value Ksp, where sp is short for solubility product. We call it the solubility product because when finding Ksp, we simply have to multiply [Pb2+][I-]^2 (or generally the product of the reactants raised to their stoichiometric coefficients) because we do not include solids in equilibrium calculations.
The beauty of Ksp values is that we can use them to predict which salts are more or less soluble - simply compare their Ksp values! For salts with the same ion ratio (like 1:1 salts such as AgCl and AgBr), the one with the larger Ksp is more soluble. For example, if AgCl has Ksp = 1.8 × 10^-10 and AgBr has Ksp = 5.0 × 10^-13, then AgCl is more soluble than AgBr.
For extremely soluble compounds such as NaCl or KOH, our Ksp will be well above 1, further reflecting that its dissolution goes far forward. In fact, there's a direct connection to those solubility rules you learned earlier - remember how we said all sodium, potassium, and nitrate salts are soluble? Well, these "soluble" salts according to our rules have Ksp values greater than 1! Meanwhile, those "insoluble" salts from our rules (like carbonates, sulfides, and hydroxides of most metals) have Ksp values much less than 1. When dealing with Ksps of this magnitude (>1), we often think of the reaction as going in one direction because, in many cases, most, if not all, of the solute dissolves. However, we cannot think that when Ksp is very small, such as in the case of lead iodide.
Calculating Ksp
When dealing with solubility, we often see two major measures of how soluble a compound is. The first is grams per liter, and the other is moles per liter or molarity. When doing calculations involving calculating Ksp or finding solubility, you want to be sure to convert from grams per liter to molarity.
Let’s look at a sample problem involving converting and calculating our solubility product:
The solubility of PbCrO4 is 4.5 * 10^-5 g/L. What is its solubility product?
Let’s start by writing out the expression for the dissolution of lead chromate:
PbCrO4(s) ⇌ Pb2+(aq) + CrO42-(aq)
Next, let’s convert from g/L to mol/L:
4.5 * 10^-5 g/L * 1mol/323.2g = 1.4 * 10^-7 mol/L.
Finally, we can plug into our formula for solubility product:
Ksp = [1.4 * 10^-7][1.4 * 10^-7] = 1.9 * 10^-14.
Remember that if you have coefficients, you also need to adjust your equilibrium concentrations before solving for Ksp!
Calculating Molar Solubility
Calculating molar solubility is essentially the same as solving for equilibrium concentrations as we did in Unit 7.7. However, we need to again keep in mind that our solid reactant is not included in concentrations or equilibrium calculations. However, we can return to our best friend, the (R)ICE Table!! Let’s apply these new concepts to a practice problem:
Calculate the molar solubility of CaCO3 (Ksp = 3.36 x 10-9).
The equation for the molar solubility of CaCO3 is as follows:
CaCO3 (s) ⇌ Ca2+ (aq) + CO32- (aq)
Let’s fill out an ICE Box and solve for concentrations!
| Reaction | CaCO₃ | Ca²⁺ | CO₃²⁻ |
|---|---|---|---|
| Initial | --- | 0.00 M | 0.00 M |
| Change | --- | +x | +x |
| Equilibrium | --- | x | x |
| Finally, plugging into Ksp and solving, we find: x^2 = 3.36 x 10^(-9) ⇒ x = 5.8*10^(-5). |
Relating Qsp and Ksp
Like regular equilibrium, we can also relate Ksp to the reaction quotient to see if a precipitation reaction will produce more or less precipitate to adjust to equilibrium. Like before, if Q > K, the reaction will produce more reactant. If Q < K, the reaction will produce more product. And if Q = K, the reaction is at equilibrium.
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| balanced chemical equation | A chemical equation where the number of atoms of each element is equal on both the reactant and product sides. |
| dissolution | The process by which a solute dissolves in a solvent to form a solution, involving the breaking of bonds or interactions in the solute and formation of new interactions with the solvent. |
| equilibrium system | A system in which the forward and reverse reactions occur at equal rates, resulting in constant concentrations of reactants and products. |
| Ksp | The solubility product constant; the equilibrium constant for the dissolution of a sparingly soluble salt into its ions. |
| molar solubility | The solubility of a substance expressed as the number of moles of solute that dissolve per liter of solution in a saturated solution. |
| salt | An ionic compound formed from the reaction of an acid and a base. |
| saturated solution | A solution that contains the maximum amount of dissolved solute at a given temperature, in equilibrium with undissolved solute. |
| solubility | The maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature, typically expressed in moles per liter (molarity) or grams per 100 mL of solvent. |
| solubility rules | Guidelines that predict whether an ionic compound is soluble or insoluble in water based on the identity of the cation and anion. |
| soluble salts | Ionic compounds that dissolve readily in water, typically corresponding to Ksp values greater than 1. |
| stoichiometry | The quantitative relationship between reactants and products in a balanced chemical equation that determines the rates of change of their concentrations. |
Frequently Asked Questions
What is Ksp and how does it relate to solubility?
Ksp is the solubility product constant for a slightly soluble salt—a mass-action equilibrium expression that uses the concentrations of the dissociated ions in a saturated solution. For a salt that dissolves as: AgCl(s) ⇌ Ag+(aq) + Cl−(aq), Ksp = [Ag+][Cl−]. Ksp tells you how much of the solid can dissolve (molar solubility) at equilibrium: you set the solubility = s, write ion concentrations from the dissolution stoichiometry, plug into the Ksp expression, and solve (often with an ICE table). The relationship between Ksp and solubility depends on stoichiometry (e.g., MxAy gives Ksp = [M]x[A]y). If Ksp > 1 the salt is generally soluble (CED 7.11.A.3). Use Ksp and the ion product Q to predict precipitation (Q > Ksp → precipitate). For practice and worked examples, see the Topic 7.11 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng), the Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7), and thousands of practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do I calculate the solubility of a salt if I know its Ksp value?
Do the dissolution stoichiometry, make an ICE table, write the Ksp expression, substitute concentrations in terms of molar solubility s, then solve for s. Steps (quick): 1. Write balanced dissolution. Example: AgCl(s) ⇌ Ag+(aq) + Cl−(aq). 2. Let molar solubility = s (mol·L−1). At equilibrium [Ag+] = s, [Cl−] = s. 3. Write Ksp: Ksp = [Ag+][Cl−] = s·s = s^2. 4. Solve: s = sqrt(Ksp). So if Ksp(AgCl) = 1.8×10−10, s = √(1.8×10−10) ≈ 1.34×10−5 M. Different stoichiometry: CaF2(s) ⇌ Ca2+ + 2F− → [Ca2+] = s, [F−] = 2s → Ksp = s(2s)^2 = 4s^3 → s = (Ksp/4)^(1/3). Use an ICE table when initial concentrations or common ions are present (common-ion lowers solubility). Show algebra and units on the AP free-response; use sig figs. More examples and practice: see the Topic 7.11 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng) and Unit 7 resources (https://library.fiveable.me/ap-chemistry/unit-7). For extra practice, try the AP problem set (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about the difference between solubility and Ksp - can someone explain?
Solubility is how much of a solid salt will dissolve in a solvent (usually reported as molar solubility: mol L-1 that dissolves to make a saturated solution). Ksp (the solubility product) is the equilibrium constant for that dissolution reaction—it’s a number you get from the mass-action expression using the concentrations of the dissolved ions at saturation (solids aren’t included). How they connect: write the balanced dissolution (stoichiometry matters), express ion concentrations in terms of the molar solubility (s), plug into Ksp = [ions]^coefficients, and solve for s. Example: AgCl(s) ⇌ Ag+ + Cl–, Ksp = [Ag+][Cl–] = s^2. For CaF2(s) ⇌ Ca2+ + 2F–, Ksp = [Ca2+][F–]^2 = s(2s)^2 = 4s^3. Use ICE tables, watch common-ion effects, and compare Q to Ksp to predict precipitation. (Per the CED, Ksp > 1 generally means a salt is soluble.) For step-by-step examples, see the Topic 7.11 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng). For more practice, try unit problems (https://library.fiveable.me/ap-chemistry/unit-7) or the 1000+ practice questions (https://library.fiveable.me/practice/ap-chemistry).
Why do some salts dissolve more than others in water?
Some salts dissolve more than others because of two competing factors that set the dissolution equilibrium: how strongly the ions are held in the solid (lattice energy) vs. how much water stabilizes those ions (hydration). Those combined effects show up in the solubility product, Ksp. A larger Ksp means a salt produces more ions at equilibrium (and Ksp > 1 typically indicates solubility per the CED). Molar solubility depends on the dissolution stoichiometry (use an ICE table and the mass-action expression to relate Ksp to the concentration of ions). Other important influences: a common ion reduces solubility (common-ion effect), pH can increase or decrease solubility for salts with basic or acidic ions, and temperature changes Ksp. For AP practice, make sure you can (1) write the balanced dissolution, (2) set up Ksp = [ions]n with stoichiometric powers, and (3) solve for molar solubility or Ksp using an ICE table (see the Topic 7.11 study guide on Fiveable for worked examples: https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng). For more problems, try the unit review (https://library.fiveable.me/ap-chemistry/unit-7) and Fiveable practice problems (https://library.fiveable.me/practice/ap-chemistry).
What does it mean when a solution is saturated in equilibrium?
A saturated solution in equilibrium means the solid salt and its dissolved ions are in dynamic balance: the rate a tiny bit of solid dissolves equals the rate ions recombine and precipitate, so overall concentrations stay constant. At saturation the ion product (Q) equals the solubility product constant Ksp for that dissolution reaction—if Q < Ksp more solid can dissolve, if Q > Ksp precipitation occurs. You can describe saturation with molar solubility (the molarity of dissolved ions at equilibrium) and use the dissolution stoichiometry in the Ksp expression to calculate it (CED 7.11.A.1–A.4). Common-ion or added/removed ions shift that equilibrium and change molar solubility without changing Ksp. For worked examples and practice calculating solubility from Ksp, see the Topic 7.11 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do I know if the stoichiometry affects the Ksp calculation?
Yes—stoichiometry matters. Ksp is written from the balanced dissolution equation, so the coefficients become exponents and multipliers when you express Ksp in terms of the molar solubility s. Quick rules: - Write the balanced dissolution: e.g., MX(s) ⇌ M+ + X− - Let solubility = s → [M+] = s, [X−] = s → Ksp = [M+][X−] = s·s = s^2 - For MX2(s) ⇌ M2+ + 2 X− - [M2+] = s, [X−] = 2s → Ksp = [M2+][X−]^2 = s·(2s)^2 = 4s^3 - For general aA_bB ⇌ aA^{b+} + bB^{a−}: Ksp = [A]^{a}[B]^{b} and use [A]=a·s, [B]=b·s with the proper coefficients. On the AP exam you should set up an ICE table (CED keywords: dissolution stoichiometry, molar solubility, ICE table) and solve algebraically for s. For practice and walkthroughs, see the Topic 7.11 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
What's the relationship between the solubility rules we learned earlier and Ksp values?
Solubility rules give a quick, qualitative answer (this salt is generally soluble or insoluble). Ksp gives the quantitative picture behind those rules. If a salt’s Ksp > 1 it’s considered soluble (CED 7.11.A.3), but more often Ksp values are <<1 for slightly soluble salts. To connect them you model dissolution as an equilibrium, write the balanced dissolution (stoichiometry matters), make an ICE table, and use the mass-action expression Ksp = [products]^(coefficients). Example: AgCl(s) ⇌ Ag+ + Cl− so Ksp = [Ag+][Cl−] = s^2; s = sqrt(Ksp). For CaF2(s) ⇌ Ca2+ + 2F−, Ksp = [Ca2+][F−]^2 = (s)(2s)^2 = 4s^3. Remember common-ion effect and Q (ion product)—if Q > Ksp precipitation occurs; if Q < Ksp more dissolves. Practice computing molar solubilities and using ICE tables for the AP LO (7.11.A)—see the topic study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
If Ksp is greater than 1, does that automatically mean the salt is soluble?
Short answer: not automatically—but the CED does note that Ksp values > 1 generally correspond to soluble salts (7.11.A.3). That statement is a useful quick rule, but you should be careful. Why: Ksp is the equilibrium constant for a heterogeneous dissolution reaction, and the numeric Ksp depends on the balanced dissolution stoichiometry (7.11.A.2). To decide if a salt is “soluble” for AP work you should: - write the dissolution equation and Ksp expression, - solve for the molar solubility (s)—that gives a direct, quantitative measure of how many moles dissolve (7.11.A.1, 7.11.A.4), - remember other factors (common-ion effect, pH, complex formation) can lower or raise solubility even if Ksp is large. So use Ksp>1 as a quick flag, but always compute molar solubility from the Ksp and reaction stoichiometry for a definitive AP-level answer. See the Topic 7.11 study guide for worked examples (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do I set up the ICE table for a dissolution equilibrium reaction?
Start by writing the balanced dissolution equation and the Ksp expression. Pick a variable (usually x) for the molar solubility (mol L–1) of the salt or for one ion, then fill an ICE table with concentrations (not solids). Example (AgCl): AgCl(s) ⇌ Ag+(aq) + Cl−(aq) ICE: - Initial: [Ag+] = 0 (or given if common ion), [Cl−] = 0 (or given) - Change: +x, +x - Equilibrium: [Ag+] = x, [Cl−] = x Plug into Ksp: Ksp = [Ag+][Cl−] = x·x = x^2 → solve for x. For different stoichiometry, adjust changes. Example CaF2(s) ⇌ Ca2+ + 2F−: - Change: +x, +2x - Ksp = [Ca2+][F−]^2 = x·(2x)^2 = 4x^3. If a common ion or initial concentration exists, put that in the Initial row and change accordingly (e.g., initial [Cl−] = 0.10 M → equilibrium [Cl−] = 0.10 + x). Use approximations (ignore +x if x << initial) and check validity. This procedure matches the AP CED objective to compute solubility from Ksp (7.11.A). For worked examples and more practice, see the Topic 7.11 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng) and lots of practice problems (https://library.fiveable.me/practice/ap-chemistry).
I don't understand how to go from molar solubility back to calculating Ksp - can someone walk me through it?
Start with the balanced dissolution equation, set molar solubility = s, write concentrations from stoichiometry, then plug into the Ksp expression. Quick steps + two examples: - AgCl(s) ⇌ Ag+(aq) + Cl−(aq). If molar solubility = s (mol·L−1), [Ag+] = s, [Cl−] = s so Ksp = [Ag+][Cl−] = s·s = s^2. - If s = 1.3×10−5 M, Ksp = (1.3×10−5)^2 = 1.69×10−10. - CaF2(s) ⇌ Ca2+(aq) + 2 F−(aq). If molar solubility = s, [Ca2+] = s, [F−] = 2s so Ksp = [Ca2+][F−]^2 = s·(2s)^2 = 4s^3. - If s = 1.0×10−3 M, Ksp = 4(1.0×10−3)^3 = 4.0×10−9. Notes: keep units mol·L−1, use ICE tables for more complex cases, and remember common-ion or complexation changes the [ions] (so you can’t just use s directly then). This is exactly the skill tested in Topic 7.11 (use dissolution stoichiometry, ICE, mass-action)—see the Topic 7.11 study guide for more worked examples (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng). For extra practice, try problems at https://library.fiveable.me/practice/ap-chemistry.
Why do we model salt dissolution as an equilibrium instead of just a regular reaction?
We model salt dissolution as an equilibrium because dissolution is reversible: ions leave the solid into solution and ions in solution can recombine and re-form solid. In a saturated solution those two rates balance (dynamic equilibrium), so concentrations stop changing even though particles keep moving. That lets us use the mass-action law and define Ksp (solubility product) with the solid’s activity treated as 1 (heterogeneous equilibrium). Ksp links molar solubility to ion concentrations and tells you whether a salt will stay dissolved (Q < Ksp), precipitate (Q > Ksp), or be at equilibrium (Q = Ksp). This framework also explains the common-ion effect, selective precipitation, and supersaturation—topics you’ll need for AP tasks that ask you to calculate solubility or predict precipitation (see CED EKs 7.11.A.1–4). For a focused review and practice problems, check the Topic 7.11 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng) and unit resources (https://library.fiveable.me/ap-chemistry/unit-7).
What's the difference between calculating Ksp for something like NaCl versus Ca(OH)2?
Short answer: the math depends on the dissolution stoichiometry. For NaCl (1:1) the dissolution is NaCl(s) ⇌ Na+(aq) + Cl−(aq). If molar solubility = s, [Na+] = s and [Cl−] = s, so Ksp = [Na+][Cl−] = s^2. For Ca(OH)2 (1:2) the dissolution is Ca(OH)2(s) ⇌ Ca2+(aq) + 2 OH−(aq). If molar solubility = s, [Ca2+] = s and [OH−] = 2s, so Ksp = [Ca2+][OH−]^2 = s(2s)^2 = 4s^3. Key points from the CED: always write the balanced dissolution equation, set up an ICE table, express ion concentrations in terms of s using stoichiometric coefficients, then substitute into the mass-action Ksp expression (Topic 7.11). Charges and coefficients change the algebra (s^2 vs 4s^3 here), so salts with the same Ksp can have very different molar solubilities. For more practice and examples, see the Topic 7.11 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng) and Unit 7 review (https://library.fiveable.me/ap-chemistry/unit-7).
How do I know when to use x, 2x, or 3x in my Ksp expression?
You use x, 2x, 3x (or 1x, 2x, etc.) based on the stoichiometry of the balanced dissolution equation—the coefficient for each ion equals how many times the molar solubility appears in its concentration. Steps: write the balanced dissolution, let the molar solubility = s (or x), assign each ion concentration from stoichiometry, then plug into Ksp. Examples: - AgCl(s) ⇌ Ag+ + Cl–. If solubility = s, [Ag+] = s, [Cl–] = s → Ksp = [Ag+][Cl–] = s·s = s^2. - CaF2(s) ⇌ Ca2+ + 2 F–. If solubility = s, [Ca2+] = s, [F–] = 2s → Ksp = s·(2s)^2 = 4s^3. - Al2(SO4)3(s) ⇌ 2 Al3+ + 3 SO4^2–. If solubility = s, [Al3+] = 2s, [SO4^2–]=3s → Ksp = (2s)^2(3s)^3 = 108 s^5. Use an ICE table when initial ions/ common ions are present (common-ion effect). This is exactly the dissolution-stoichiometry idea in the CED (7.11.A.2); review the Topic 7.11 study guide for worked examples (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng) and more practice at the Unit 7 page (https://library.fiveable.me/ap-chemistry/unit-7) or the practice problem bank (https://library.fiveable.me/practice/ap-chemistry).
Can you explain why the math is different for salts with different formulas like AB versus AB2?
It's different because the dissolution stoichiometry fixes how many ions are produced from each formula unit, and Ksp uses those ion concentrations. Write the balanced dissolution, let s = molar solubility (mol L⁻¹ of formula units that dissolve), then express Ksp in terms of s. - For AB(s) ⇌ A⁺ + B⁻: [A⁺] = s, [B⁻] = s so Ksp = [A⁺][B⁻] = s·s = s² → s = sqrt(Ksp). - For AB2(s) ⇌ A²⁺ + 2B⁻: [A²⁺] = s, [B⁻] = 2s so Ksp = [A²⁺][B⁻]² = s·(2s)² = 4s³ → s = (Ksp/4)^(1/3). So the exponents and numeric factors come straight from the coefficients in the balanced equation (CED 7.11.A.2). For more practice converting Ksp ↔ molar solubility and using ICE tables, check the Topic 7.11 study guide (https://library.fiveable.me/ap-chemistry/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng) and extra problems at the Unit 7 page (https://library.fiveable.me/ap-chemistry/unit-7) or the practice set (https://library.fiveable.me/practice/ap-chemistry).