Comparing Solubilities — Compare solubilities using Ksp and stoichiometry - Chemistry 2

SECTION 1

Historical Context & Motivation

For centuries, chemists noticed that different salts dissolve to very different extents in water. Some, like table salt, dissolve readily, while others, like limestone, barely dissolve at all. The challenge was finding a quantitative way to describe and predict these differences. Early approaches relied on qualitative labels—"soluble" versus "insoluble"—but these categories couldn't answer more nuanced questions, such as which of two sparingly soluble salts contributes more ions to a solution.

1864 • Guldberg & Waage — Law of Mass Action

Norwegian chemists Cato Guldberg and Peter Waage formalized the relationship between concentrations and the position of chemical equilibrium, setting the stage for all equilibrium constants.

1889 • Nernst's Solubility Product Concept

Walther Nernst proposed the solubility product constant (K_sp) as a special equilibrium constant for the dissolution of sparingly soluble ionic solids.

1923 • Debye–Hückel Theory

Peter Debye and Erich Hückel developed a model for ion interactions in solution, explaining why measured solubilities sometimes deviate from simple K_sp predictions.

1950s–Today • Quantitative Analytical Chemistry

Modern analytical techniques and comprehensive K_sp tables allow chemists to compare solubilities with precision, guiding applications from water treatment to pharmaceutical design.

These advances left chemists with a powerful but sometimes tricky tool. Because K_sp expressions depend on the stoichiometry of each salt's dissolution reaction, you cannot always tell which salt is more soluble just by glancing at K_sp values. That is the central question this lesson addresses: how do you correctly compare the solubilities of salts that produce different numbers of ions?

SECTION 2

Core Principles & Definitions

Before diving into comparisons, make sure the following foundational ideas are clear. Each one plays a role when you convert a K_sp value into an actual molar solubility.

Molar Solubility (s)

The number of moles of a salt that dissolve per liter of solution to form a saturated solution. It is the bridge between K_sp and how much actually dissolves.

Solubility Product (Ksp)

The equilibrium constant for the dissolution of a sparingly soluble ionic compound. For A_mB_n, K_sp = [A]^m[B]ⁿ.

Stoichiometric Coefficients

The mole ratios from the balanced dissolution equation determine how ion concentrations relate to the molar solubility s. These exponents make K_sp expressions different for 1:1, 1:2, and 2:3 salts.

ICE Table Approach

An ICE (Initial–Change–Equilibrium) table tracks ion concentrations as the salt dissolves. Each ion's equilibrium concentration is a multiple of s dictated by stoichiometry.

✦

KEY TAKEAWAY

Think of K_sp like a final exam score that combines several test sections. Two students can have the same total score (K_sp) but different numbers of sections. To compare their per-section averages (molar solubility), you must account for how many sections each student took. That's exactly what stoichiometry does when you convert K_sp to s.

SECTION 3

Visual Explanation — From Ksp to Molar Solubility

Three salt types shown side-by-side. Notice how the formula linking K_sp to s changes depending on the number of ions produced. A 1:2 salt uses a cube root, while a 2:3 salt requires a fifth root. This is why you cannot compare K_sp values directly when stoichiometries differ.

The diagram above highlights the central challenge. When comparing AgCl (a 1:1 salt) and CaF₂ (a 1:2 salt), you might expect AgCl to be more soluble because its K_sp (1.8 × 10⁻¹⁰) is larger than CaF₂'s K_sp (3.9 × 10⁻¹¹). However, once you solve for molar solubility, CaF₂ turns out to be about 16 times more soluble (2.1 × 10⁻⁴ M vs. 1.3 × 10⁻⁵ M). The cube-root relationship for CaF₂ effectively "stretches" its K_sp into a much larger molar solubility than the square-root relationship for AgCl.

💡

When CAN You Compare Ksp Directly?

If two salts have the same ion ratio (for example, both are 1:1 salts like AgCl and AgBr, or both are 1:2 salts like CaF₂ and BaF₂), then the one with the larger K_sp IS the more soluble salt. The root function is the same for both, so the ranking is preserved.

SECTION 4

Mathematical Framework

The general procedure for comparing solubilities involves three steps: write the dissolution equation, express K_sp in terms of the molar solubility s, and then solve for s. Below are the key equations for the most common salt types you will encounter.

1:1 SALT (e.g., AgCl, BaSO₄)

Ksp = s² → s = √Ksp

The salt AB dissolves as AB → A⁺ + B⁻. Each ion has concentration s, so K_sp = s × s = s².

1:2 SALT (e.g., CaF₂, PbCl₂)

Ksp = (s)(2s)² = 4s³ → s = ∛(Ksp / 4)

The salt AB₂ dissolves as AB₂ → A²⁺ + 2B⁻. The cation concentration is s and the anion concentration is 2s.

2:1 SALT (e.g., Ag₂CrO₄)

Ksp = (2s)²(s) = 4s³ → s = ∛(Ksp / 4)

The salt A₂B dissolves as A₂B → 2A⁺ + B²⁻. Notice this gives the same algebraic form, 4s³, as the 1:2 salt—the math is symmetric.

GENERAL FORMULA FOR AmBn

Ksp = (ms)^m × (ns)^n = m^m × n^n × s^(m+n)

For the general salt A_mB_n: solve s = [(K_sp) / (m^m × nⁿ)]^1/(m+n). This is the master formula.

⚠

Common Mistake Alert

Students often forget to include the stoichiometric coefficient inside the parentheses before raising to a power. For CaF₂, the fluoride concentration is 2s, not s. Writing K_sp = s³ instead of 4s³ will give the wrong molar solubility.

SECTION 5

Detailed Breakdown — Salt Type Comparison

The table below gathers several sparingly soluble salts with their K_sp values, salt types, and calculated molar solubilities. Study the rankings carefully—the order by K_sp is not the same as the order by molar solubility whenever the stoichiometries differ.

Salt	Type	Ksp	Ksp → s formula	Molar Solubility (s)
AgCl	1:1	1.8 × 10⁻¹⁰	s = √Ksp	1.3 × 10⁻⁵ M
CaF₂	1:2	3.9 × 10⁻¹¹	s = ∛(Ksp / 4)	2.1 × 10⁻⁴ M
Ag₂CrO₄	2:1	1.1 × 10⁻¹²	s = ∛(Ksp / 4)	6.5 × 10⁻⁵ M
PbCl₂	1:2	1.7 × 10⁻⁵	s = ∛(Ksp / 4)	1.6 × 10⁻² M
BaSO₄	1:1	1.1 × 10⁻¹⁰	s = √Ksp	1.0 × 10⁻⁵ M

Ksp values and calculated molar solubilities at 25 °C

The left column ranks the five salts by K_sp value, while the right column ranks them by calculated molar solubility. Red dashed lines show salts that dropped in ranking; blue dashed lines show salts that climbed. The multi-ion salts (CaF₂ and Ag₂CrO₄) both moved up.

This comparison diagram drives home the lesson's main point: the Ksp ranking and the molar solubility ranking only match when all salts share the same stoichiometric type. When the types are mixed, you must calculate s for each salt individually before drawing any conclusions.

SECTION 6

Worked Example

Which is more soluble in water at 25 °C: silver chloride (AgCl, K_sp = 1.8 × 10⁻¹⁰) or calcium fluoride (CaF₂, K_sp = 3.9 × 10⁻¹¹)?

Comparing Solubilities of AgCl and CaF₂

1
Step 1 — Write the Dissolution Equations

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). This is a 1:1 salt. CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq). This is a 1:2 salt.
2
Step 2 — Express Ksp in Terms of s for AgCl

Let s₁ = molar solubility of AgCl. Then [Ag⁺] = s₁ and [Cl⁻] = s₁. K_sp = s₁² = 1.8 × 10⁻¹⁰.

s₁ = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M
3
Step 3 — Express Ksp in Terms of s for CaF₂

Let s₂ = molar solubility of CaF₂. Then [Ca²⁺] = s₂ and [F⁻] = 2s₂. K_sp = (s₂)(2s₂)² = 4s₂³ = 3.9 × 10⁻¹¹.

s₂ = ∛(3.9 × 10⁻¹¹ / 4) = ∛(9.75 × 10⁻¹²) = 2.1 × 10⁻⁴ M
4
Step 4 — Compare the Molar Solubilities

s₂ (CaF₂) = 2.1 × 10⁻⁴ M is much greater than s₁ (AgCl) = 1.3 × 10⁻⁵ M, even though AgCl has the larger K_sp.

CaF₂ is more soluble than AgCl.

SECTION 7

Strengths, Limitations & When to Compare Directly

Comparison Scenario	Method	Why It Works / Doesn't
Same stoichiometric type (e.g., two 1:1 salts)	Compare Ksp directly	Both salts use the same root (e.g., √Ksp), so the larger Ksp always gives the larger s.
Different stoichiometric types (e.g., 1:1 vs. 1:2)	Calculate s for each, then compare	Different roots (√ vs. ∛) mean a smaller Ksp can still produce a larger s. Direct Ksp comparison is unreliable.
Very different magnitudes of Ksp (many orders apart)	Quick estimation may work	If one Ksp is several orders of magnitude larger, it's likely more soluble—but always verify with calculation for certainty.
Presence of common ion or complex ion	Full equilibrium calculation needed	Common-ion effect suppresses solubility; complex-ion formation can increase it. Simple Ksp-to-s conversion is insufficient.

Decision guide: how to compare solubilities under different conditions

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THE BOTTOM LINE

Think of K_sp as a "compressed" measure of solubility. The stoichiometry acts like the compression ratio. Two MP3 files can be the same size on disk (same K_sp) but different lengths when you play them (different molar solubility) because one was compressed more aggressively than the other. You always need to "decompress" K_sp by solving for s when the stoichiometries differ.

SECTION 8

Connection to Advanced Topics

The simple K_sp-to-s calculation assumes ideal behavior: no ion–ion interactions, no side reactions, and pure water as the solvent. In reality, more advanced factors can affect solubility, and understanding these connections prepares you for college chemistry and beyond.

This Lesson's Approach	Advanced Consideration
Concentrations used in Ksp expression	Activities (effective concentrations) account for ion–ion interactions via activity coefficients (γ).
Pure water assumed	Common-ion effect: adding a shared ion shifts equilibrium left, decreasing solubility.
Temperature held constant at 25 °C	Van 't Hoff equation relates Ksp changes to temperature, since dissolution can be endothermic or exothermic.
Only dissolution equilibrium	Complex-ion formation (e.g., AgCl + Cl⁻ → AgCl₂⁻) can dramatically increase apparent solubility.

Simple model vs. advanced considerations

For now, mastering the stoichiometric conversion from K_sp to molar solubility gives you a reliable tool for comparing sparingly soluble salts in pure water. As you move into AP Chemistry or college-level courses, you will build on this foundation by incorporating activity coefficients, common-ion effects, and complex-ion equilibria to make even more accurate predictions.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Two salts, AX and AX₂, both have a K_sp of exactly 1.0 × 10⁻⁸. Without calculating, which salt do you expect to have the greater molar solubility, and why?

Show answer

AX₂ will have the greater molar solubility. For AX (a 1:1 salt), s = √(K_sp) = √(1.0 × 10⁻⁸) = 1.0 × 10⁻⁴ M. For AX₂ (a 1:2 salt), s = ∛(K_sp / 4) = ∛(2.5 × 10⁻⁹) ≈ 1.36 × 10⁻³ M. Even with the same K_sp, AX₂ is about 14 times more soluble because the cube root "stretches" the K_sp into a larger molar solubility than the square root does.

Question 2: BASIC CALCULATION

Calculate the molar solubility of PbI₂ given that its K_sp = 9.8 × 10⁻⁹ at 25 °C.

Show answer

PbI₂ dissolves as PbI₂ → Pb²⁺ + 2 I⁻. This is a 1:2 salt, so K_sp = (s)(2s)² = 4s³. Solving: s³ = (9.8 × 10⁻⁹) / 4 = 2.45 × 10⁻⁹. Taking the cube root: s = ∛(2.45 × 10⁻⁹) ≈ 1.35 × 10⁻³ M. Thus, approximately 1.35 × 10⁻³ mol of PbI₂ dissolves per liter of saturated solution.

Question 3: INTERMEDIATE

Which is more soluble in pure water: BaCrO₄ (K_sp = 1.2 × 10⁻¹⁰) or Ag₂CrO₄ (K_sp = 1.1 × 10⁻¹²)? Show your work.

Show answer

BaCrO₄ is a 1:1 salt: s = √(1.2 × 10⁻¹⁰) = 1.1 × 10⁻⁵ M. Ag₂CrO₄ is a 2:1 salt: K_sp = (2s)²(s) = 4s³, so s = ∛(1.1 × 10⁻¹² / 4) = ∛(2.75 × 10⁻¹³) ≈ 6.5 × 10⁻⁵ M. Despite having a K_sp about 100 times smaller, Ag₂CrO₄ is roughly 6 times more soluble than BaCrO₄ because the cube root relationship significantly amplifies the molar solubility.

Question 4: APPLIED

A water treatment facility needs to remove either fluoride or sulfate ions from drinking water by precipitation. They are considering using CaF₂ (K_sp = 3.9 × 10⁻¹¹) or BaSO₄ (K_sp = 1.1 × 10⁻¹⁰). Which salt leaves fewer dissolved ions in solution? Explain using molar solubility.

Show answer

For BaSO₄ (1:1 salt): s = √(1.1 × 10⁻¹⁰) = 1.0 × 10⁻⁵ M, producing 2 × 1.0 × 10⁻⁵ = 2.0 × 10⁻⁵ M total ions. For CaF₂ (1:2 salt): s = ∛(3.9 × 10⁻¹¹ / 4) = 2.1 × 10⁻⁴ M, producing s + 2s = 3 × 2.1 × 10⁻⁴ = 6.3 × 10⁻⁴ M total ions. BaSO₄ leaves far fewer dissolved ions in solution (roughly 30 times fewer), making it the better choice if the goal is minimizing dissolved ions. Even though CaF₂ has a smaller K_sp, its 1:2 stoichiometry results in a much higher molar solubility.

Question 5: CRITICAL THINKING

A student claims: "If Salt A has a K_sp of 10⁻²⁰ and Salt B has a K_sp of 10⁻²², then Salt A is always more soluble, no matter what." Construct a counterexample using specific salt formulas and K_sp expressions to prove or disprove this claim.

Show answer

The claim is false. Let Salt A be a 1:1 salt with K_sp = 10⁻²⁰: s = √(10⁻²⁰) = 10⁻¹⁰ M. Now let Salt B be a 2:3 salt (like A₂B₃) with K_sp = 10⁻²²: K_sp = 108s⁵, so s =

(10⁻²² / 108)^{1/5}

≈

(9.26 × 10⁻²⁵)^{0.2}

≈ 3.9 × 10⁻⁵ M. Salt B has a smaller K_sp but is roughly 100,000 times more soluble! The fifth root in the 2:3 expression amplifies the molar solubility enormously. This proves that K_sp magnitude alone cannot determine relative solubility when stoichiometries differ.

SUMMARY

Lesson Summary

Summary

The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound, but it does not directly tell you how soluble one salt is compared to another. To compare solubilities, you must convert K_sp into molar solubility (s) using the stoichiometry of the dissolution reaction. For a 1:1 salt, s = √K_sp; for a 1:2 or 2:1 salt, s = ∛(K_sp / 4); and for a 2:3 salt, s = ⁵√(K_sp / 108).

The key rule: you may only compare Ksp values directly when the salts share the same stoichiometric type. When they differ, a salt with a smaller K_sp can still be more soluble because higher-order roots amplify small K_sp values into larger molar solubilities. Always write the balanced equation, set up the ICE table, and solve for s before drawing any solubility comparison.