Molar Solubility from Ksp — Compute molar solubility from Ksp - Chemistry 2

SECTION 1

Historical Context & Motivation

For centuries, chemists noticed that some salts dissolve easily in water while others barely dissolve at all. Early alchemists and apothecaries cataloged substances as "soluble" or "insoluble," but these labels were imprecise. A compound described as insoluble actually does dissolve—just in incredibly tiny amounts. The real question was: how much of a "barely soluble" substance actually enters the solution? Answering that question required the development of equilibrium chemistry and quantitative methods that took decades to mature.

1864 • Law of Mass Action

Norwegian chemists Cato Guldberg and Peter Waage proposed that the rate of a reaction is proportional to the concentrations of the reactants. This law laid the mathematical groundwork for all equilibrium expressions, including K_sp.

1884 • Le Chatelier's Principle

Henri Le Chatelier described how equilibrium systems respond to disturbances. His principle helps predict how changes in ion concentration affect solubility.

1889 • Nernst & Solubility Product

Walther Nernst formalized the solubility product constant (K_sp), giving chemists a single number to describe how far the dissolution of a sparingly soluble salt proceeds at equilibrium.

1923 • Debye–Hückel Theory

Peter Debye and Erich Hückel showed that interionic attractions in solution affect effective concentrations (activities), refining solubility predictions for more concentrated solutions.

The central question this lesson addresses is straightforward: given the K_sp value for a sparingly soluble ionic compound, how do we calculate its molar solubility—the number of moles that dissolve per liter of solution? This skill connects equilibrium theory to real-world applications like water purification, kidney stone formation, and mineral extraction.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need a solid grasp of several key ideas. A sparingly soluble (or slightly soluble) salt is one that dissolves only to a very small extent—think of silver chloride or barium sulfate. When such a salt is placed in water, some of it dissolves and produces ions until the solution reaches dynamic equilibrium: the rate of dissolution equals the rate of precipitation. At that point, the ion concentrations remain constant, and the solution is said to be saturated.

Solubility Product (Ksp)

The equilibrium constant for the dissolution of a sparingly soluble ionic solid. It equals the product of the ion concentrations, each raised to the power of its coefficient in the balanced equation. A smaller K_sp means the compound is less soluble.

Molar Solubility (s)

The number of moles of solute that dissolve per liter of saturated solution, expressed in mol/L (or M). This is the quantity we calculate from K_sp. It directly tells you "how much dissolves."

ICE Table Approach

We organize dissolution problems using an ICE table (Initial, Change, Equilibrium). For pure water, initial ion concentrations are 0, and the change is expressed in terms of the variable s.

Stoichiometric Ratios Matter

When a salt like Ca₃(PO₄)₂ dissolves, it produces ions in specific mole ratios from the balanced equation. The ion concentrations in the K_sp expression are multiples of s, not simply equal to s.

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KEY TAKEAWAY

Think of K_sp like the capacity of a parking lot. K_sp tells you the maximum "product" of ion concentrations the solution can hold. Molar solubility is like figuring out how many cars actually fit—you have to account for how many parking spaces each car takes up (the stoichiometric coefficients).

SECTION 3

Visual Explanation — Dissolution at Equilibrium

This diagram shows a beaker of water containing undissolved AgCl solid at the bottom. Violet circles represent dissolved Ag⁺ ions and cyan circles represent Cl⁻ ions in solution. At equilibrium, the solid continues dissolving and precipitating at equal rates, so the ion concentrations stay constant.

In the diagram above, notice that both Ag⁺ and Cl⁻ ions are present in equal numbers because AgCl produces a 1:1 ratio when it dissolves. The equilibrium expression is K_sp = [Ag⁺][Cl⁻]. If we call the molar solubility s, then both ion concentrations equal s, and K_sp = s². This is the simplest type of K_sp problem, and it's the foundation for all more complex cases.

SECTION 4

Mathematical Framework

The strategy for computing molar solubility always follows the same pattern. First, write the balanced dissolution equation. Next, define s as the molar solubility and express each ion concentration in terms of s using the stoichiometric coefficients. Then substitute into the K_sp expression and solve for s. The algebraic complexity depends on the salt type.

Case 1: 1:1 Salts (e.g., AgCl, BaSO₄)

DISSOLUTION EQUATION

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Each formula unit produces one cation and one anion. Both concentrations equal s.

Ksp EXPRESSION — 1:1 SALT

Ksp = [Ag⁺][Cl⁻] = (s)(s) = s²

Solving: s = √(K_sp)

Case 2: 1:2 or 2:1 Salts (e.g., PbCl₂, Ag₂CrO₄)

DISSOLUTION EQUATION — 1:2 SALT

PbCl₂(s) ⇌ Pb²⁺(aq) + 2 Cl⁻(aq)

If s mol/L dissolves: [Pb²⁺] = s, [Cl⁻] = 2s.

Ksp EXPRESSION — 1:2 SALT

Ksp = [Pb²⁺][Cl⁻]² = (s)(2s)² = 4s³

Solving: s = ³√(K_sp / 4)

General Formula

GENERAL DISSOLUTION

AₘBₙ(s) ⇌ m A^(n+)(aq) + n B^(m−)(aq)

K_sp = (ms)ᵐ × (ns)ⁿ = mᵐ × nⁿ × $s^{m+n}$ . Solve for s by taking the (m + n)th root.

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Common Mistake Alert

Students often forget to include the coefficient in the concentration. For PbCl₂, [Cl⁻] = 2s, not s. The coefficient also gets raised to the exponent in the K_sp expression: (2s)² = 4s², not 2s².

SECTION 5

Detailed Breakdown by Salt Type

Different ionic compounds break apart into different numbers of ions. The table below organizes the most common salt types you'll encounter, showing how the K_sp expression and the formula for s change depending on the stoichiometry. Study this table carefully—it is your roadmap for solving any molar solubility problem.

Salt Type	Example	Ion Concentrations	Ksp in Terms of s	Solve for s
AB (1:1)	AgCl, BaSO₄	[A⁺] = s, [B⁻] = s	s²	s = √Ksp
AB₂ (1:2)	PbCl₂, CaF₂	[A²⁺] = s, [B⁻] = 2s	4s³	s = ³√(Ksp/4)
A₂B (2:1)	Ag₂CrO₄	[A⁺] = 2s, [B²⁻] = s	4s³	s = ³√(Ksp/4)
AB₃ (1:3)	AlF₃, Fe(OH)₃	[A³⁺] = s, [B⁻] = 3s	27s⁴	s = ⁴√(Ksp/27)
A₃B₂ (3:2)	Ca₃(PO₄)₂	[A²⁺] = 3s, [B³⁻] = 2s	108s⁵	s = ⁵√(Ksp/108)

Summary of Ksp–molar solubility relationships by salt type

Three side-by-side ICE tables show how stoichiometric coefficients affect the equilibrium concentrations and the resulting K_sp expression. Notice that both 1:2 and 2:1 salts produce K_sp = 4s³.

The diagram reinforces a critical insight: the stoichiometric coefficients from the balanced equation become both the multipliers of s in the equilibrium concentrations and the exponents in the K_sp expression. Mastering this connection is the single most important step in becoming fluent with these problems.

SECTION 6

Worked Example

Let's work through a full example to see the method in action. We'll tackle a 1:2 salt to illustrate the role of stoichiometric coefficients.

Find the Molar Solubility of PbCl₂

1
Step 1 — Write the Balanced Dissolution Equation

PbCl₂(s) ⇌ Pb²⁺(aq) + 2 Cl⁻(aq). One formula unit of PbCl₂ produces one Pb²⁺ ion and two Cl⁻ ions.
2
Step 2 — Define s and Build the ICE Table

Let s = molar solubility of PbCl₂ in mol/L. Then at equilibrium: [Pb²⁺] = s and [Cl⁻] = 2s. The factor of 2 comes from the coefficient in front of Cl⁻.
3
Step 3 — Write the Ksp Expression

K_sp = [Pb²⁺][Cl⁻]² = (s)(2s)² = (s)(4s²) = 4s³.
4
Step 4 — Substitute Ksp and Solve for s

The K_sp of PbCl₂ is 1.7 × 10⁻⁵ at 25 °C. Substituting: 1.7 × 10⁻⁵ = 4s³. Divide both sides by 4: s³ = 4.25 × 10⁻⁶. Take the cube root: s = $(4.25 × 10⁻⁶)^{1/3}$ .

s ≈ 1.62 × 10⁻² mol/L
5
Step 5 — Verify (Optional Check)

Plug back in: [Pb²⁺] = 0.0162 M, [Cl⁻] = 0.0324 M. K_sp = (0.0162)(0.0324)² = (0.0162)(1.05 × 10⁻³) ≈ 1.7 × 10⁻⁵. ✓ This matches the given K_sp.

SECTION 7

Strengths & Limitations of the Ksp Method

The K_sp method is a powerful tool, but it comes with assumptions and limitations that you should understand. Real solutions can behave differently from the ideal model, especially when additional ions are present or when the dissolved salt participates in other equilibria such as acid-base reactions.

Strengths	Limitations
Simple algebra gives a quick estimate of solubility.	Assumes ideal behavior—ignores interionic attractions in solution.
Works well for sparingly soluble salts at low concentrations.	Less accurate for moderately soluble salts where ion concentrations are high.
Allows comparison of solubilities across different salt types.	Cannot directly compare Ksp values of salts with different stoichiometries (e.g., AgCl vs. PbCl₂).
Provides a foundation for common-ion effect and precipitation calculations.	Does not account for complex-ion formation or pH effects on solubility.
Ksp values are tabulated for hundreds of compounds at 25 °C.	Ksp changes with temperature; tabulated values apply only at the stated temperature.

Strengths and limitations of computing molar solubility from Ksp

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KEY TAKEAWAY

The K_sp calculation gives you a theoretical baseline for molar solubility—like a weather forecast that works perfectly on clear days. When extra factors come into play (common ions, complex ions, pH), you'll need additional tools from later chapters, just as a forecaster needs extra data to predict storms.

SECTION 8

Connection to Advanced Concepts

Once you master computing molar solubility in pure water, you're ready for more complex situations. In AP Chemistry and college courses, you'll encounter scenarios where the solution already contains a common ion, where pH affects one of the ions, or where a dissolved metal ion forms a complex with another species. Each of these modifies the simple K_sp calculation, but they all build on the same ICE-table foundation.

This Lesson	Advanced Extension
Dissolving in pure water (initial ion concentrations = 0)	Common-ion effect: a shared ion is already in solution, reducing molar solubility
Ksp expression involves only ion concentrations	Ion-product Q vs. Ksp: predicting whether precipitation will occur
Assumes no side reactions of the ions	pH-dependent solubility: hydroxide or basic-anion salts are more soluble in acidic solutions
Uses concentrations (molarity)	Activity coefficients: more accurate predictions via thermodynamic Ksp

How this lesson's method extends into more advanced aqueous equilibria

Think of molar solubility in pure water as the "base case." Every advanced solubility problem starts here and then adds one or more complications. Mastering this base case ensures that the more challenging problems are manageable—each new factor is just one additional layer on top of what you already know.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Compound X has K_sp = 4.0 × 10⁻¹⁰ and compound Y has K_sp = 2.0 × 10⁻⁸. Both are 1:1 salts (AB type). Without calculating, which salt is more soluble and why?

Show answer

Compound Y is more soluble. Because both salts share the same AB stoichiometry, their K_sp values can be directly compared. Since K_sp = s² for 1:1 salts, a larger K_sp corresponds to a larger molar solubility. Y's K_sp is about 50 times larger than X's, so Y dissolves more per liter. Important: you can only compare K_sp values directly when the salts have the same ion ratio.

Question 2: BASIC CALCULATION

Calculate the molar solubility of BaSO₄ in pure water at 25 °C. K_sp = 1.1 × 10⁻¹⁰.

Show answer

BaSO₄ is a 1:1 salt: BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq). Let s = molar solubility. Then [Ba²⁺] = s and [SO₄²⁻] = s. K_sp = s² = 1.1 × 10⁻¹⁰. Taking the square root: s = √(1.1 × 10⁻¹⁰) = 1.05 × 10⁻⁵ mol/L. This means only about 0.0000105 moles of BaSO₄ dissolve per liter—a very small amount, consistent with barium sulfate being considered insoluble in general chemistry.

Question 3: INTERMEDIATE

Calculate the molar solubility of CaF₂ in pure water. K_sp = 3.9 × 10⁻¹¹. Then determine the equilibrium concentration of F⁻ ions.

Show answer

CaF₂ is a 1:2 salt: CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq). Let s = molar solubility. Then [Ca²⁺] = s and [F⁻] = 2s. K_sp = (s)(2s)² = 4s³. Substituting: 3.9 × 10⁻¹¹ = 4s³, so s³ = 9.75 × 10⁻¹², and s =

(9.75 × 10⁻¹²)^{1/3}

≈ 2.13 × 10⁻⁴ mol/L. The F⁻ concentration is 2s = 2(2.13 × 10⁻⁴) = 4.27 × 10⁻⁴ M. Notice that the fluoride concentration is twice the molar solubility because each formula unit releases two fluoride ions.

Question 4: APPLIED

Silver chromate (Ag₂CrO₄) is used as an indicator in the Mohr titration for chloride analysis. Given K_sp = 1.1 × 10⁻¹², find (a) the molar solubility in pure water and (b) the mass (in grams) of Ag₂CrO₄ that dissolves in 500 mL of water. (Molar mass of Ag₂CrO₄ = 331.73 g/mol)

Show answer

Ag₂CrO₄(s) ⇌ 2 Ag⁺(aq) + CrO₄²⁻(aq). This is a 2:1 salt. Let s = molar solubility. Then [Ag⁺] = 2s and [CrO₄²⁻] = s. K_sp = (2s)²(s) = 4s³. Substituting: 1.1 × 10⁻¹² = 4s³, so s³ = 2.75 × 10⁻¹³, and s =

(2.75 × 10⁻¹³)^{1/3}

≈ 6.50 × 10⁻⁵ mol/L. For part (b): moles dissolved in 0.500 L = (6.50 × 10⁻⁵)(0.500) = 3.25 × 10⁻⁵ mol. Mass = (3.25 × 10⁻⁵ mol)(331.73 g/mol) ≈ 0.0108 g or about 10.8 mg. This tiny mass illustrates why Ag₂CrO₄ precipitates so readily in the Mohr titration.

Question 5: CRITICAL THINKING

Consider two salts: Salt A is AgBr (K_sp = 5.0 × 10⁻¹³, a 1:1 salt) and Salt B is Ag₂CrO₄ (K_sp = 1.1 × 10⁻¹²). Salt B has a larger K_sp. Does Salt B necessarily have a larger molar solubility? Calculate both and explain your finding.

Show answer

This is a classic trap! For AgBr (1:1): K_sp = s², so s = √(5.0 × 10⁻¹³) = 7.07 × 10⁻⁷ mol/L. For Ag₂CrO₄ (2:1): K_sp = 4s³, so s = ³√(1.1 × 10⁻¹² / 4) = ³√(2.75 × 10⁻¹³) ≈ 6.50 × 10⁻⁵ mol/L. Salt B actually has a molar solubility nearly 100 times larger than Salt A, despite their K_sp values being close. However, the reason isn't just the larger K_sp—it's that the mathematical relationship is different (s² vs. 4s³). This demonstrates that you cannot rank solubilities of salts with different stoichiometries by comparing their K_sp values alone. You must calculate s.

SUMMARY

Lesson Summary

Summary

The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. To find molar solubility (s), write the balanced dissolution equation, express each ion concentration as a multiple of s using the stoichiometric coefficients, substitute into the Ksp expression, and solve algebraically. For a 1:1 salt (AB), K_sp = s². For a 1:2 or 2:1 salt, K_sp = 4s³. The coefficients serve as both multipliers and exponents.

Remember that Ksp values cannot be directly compared across different salt types to rank solubilities—you must compute s first. The ICE table method provides a reliable, systematic approach for every problem. This skill is the foundation for predicting precipitation, the common-ion effect, and pH-dependent solubility in more advanced aqueous equilibria topics.