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

SECTION 1

Historical Context & Motivation

The question of why certain salts dissolve readily while others remain stubbornly insoluble has occupied chemists for centuries. Early alchemists catalogued substances as "soluble" or "insoluble" based on qualitative observation, but quantifying the degree of solubility required a thermodynamic framework that did not emerge until the late nineteenth and early twentieth centuries. The development of the solubility product constant (K_sp) gave chemists a single equilibrium number for each sparingly soluble salt, but comparing K_sp values across salts of different stoichiometries introduced a subtle and often misunderstood complication.

1864 • Law of Mass Action

Cato Guldberg and Peter Waage formalize the relationship between equilibrium concentrations and a constant expression, laying the groundwork for all equilibrium constants including K_sp.

1889 • Nernst's Solubility Product

Walther Nernst applies the mass-action law specifically to dissolution equilibria of sparingly soluble salts, defining the solubility product as the product of ion concentrations each raised to its stoichiometric coefficient.

1923 • Debye–Hückel Theory

Peter Debye and Erich Hückel develop a model for ion–ion interactions in solution, explaining why measured solubilities deviate from ideal K_sp predictions and motivating the use of activity coefficients.

1950s–present • Modern Tabulations & Computational Chemistry

Extensive K_sp databases and computational approaches allow chemists to predict precipitation and compare solubilities across thousands of compounds under varying conditions.

A central question persists from Nernst's era to the modern classroom: if salt A has a K_sp of 10⁻⁸ and salt B has a K_sp of 10⁻¹⁰, is salt A necessarily more soluble? The answer depends entirely on whether the two salts share the same dissolution stoichiometry. This lesson clarifies when a direct K_sp comparison is valid, and when you must convert K_sp into molar solubility (s) before drawing conclusions.

SECTION 2

Core Principles & Definitions

Before comparing solubilities, it is essential to distinguish between two related but distinct quantities. The solubility product constant (K_sp) is a thermodynamic equilibrium constant that characterizes the dissolution of a sparingly soluble ionic compound at a given temperature. The molar solubility (s) is the number of moles of the compound that dissolve per liter of saturated solution. While K_sp is a fixed constant for each salt at a given temperature, s is the experimentally measurable concentration that tells us how much actually dissolves. Because the mathematical relationship between K_sp and s depends on the stoichiometry of the dissolution reaction, comparing K_sp values directly across different salt types can be misleading.

Ksp Is an Equilibrium Constant

K_sp equals the product of the equilibrium ion concentrations, each raised to the power of its stoichiometric coefficient. It is temperature-dependent and independent of how much excess solid is present.

Molar Solubility (s) Is the Observable Quantity

Molar solubility s represents moles of solid dissolved per liter. It is derived from K_sp via the ICE table and reflects how much solute the solution can hold at equilibrium.

Stoichiometry Determines the Ksp ↔ s Relationship

For a 1:1 salt like AgCl, K_sp = s². For a 1:2 salt like CaF₂, K_sp = 4s³. These different exponents mean that Ksp values with different stoichiometries cannot be compared directly.

Same-Type Salts: Direct Ksp Comparison Works

When two salts share identical stoichiometry (e.g., both are AB-type), the salt with the larger K_sp is guaranteed to have the larger molar solubility. The monotonic mapping s = (K_sp)^(1/n) preserves the ordering.

Different-Type Salts: Convert to s First

When comparing AB vs. AB₂ vs. A₂B₃ salts, you must calculate s for each compound individually and then compare the resulting molar solubilities. A smaller K_sp can yield a larger s when the stoichiometric factor differs.

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KEY TAKEAWAY

Think of K_sp as a "compressed" representation of solubility, where the compression algorithm depends on stoichiometry. Just as you cannot directly compare the file sizes of a JPEG and a RAW photo to determine which image has higher resolution, you cannot directly compare K_sp values across different salt types to determine which salt is more soluble. You must "decompress" each K_sp back to molar solubility using the appropriate stoichiometric formula before making a valid comparison.

SECTION 3

Visual Explanation — Stoichiometric Types at a Glance

Three common stoichiometric types and their K_sp–to–s conversion formulas. The bottom panel illustrates how CaF₂ (AB₂ type) has a smaller K_sp yet a larger molar solubility than AgCl (AB type).

The diagram above organizes the three most frequently encountered stoichiometric types for sparingly soluble salts. Notice that as the total number of ions produced per formula unit increases, the exponent on s in the K_sp expression grows, and a numerical coefficient (4 for AB₂, 108 for A₂B₃) appears. This coefficient and exponent act as a "packing factor" that compresses a relatively large molar solubility into a deceptively small K_sp. The counterintuitive comparison between AgCl and CaF₂ in the lower panel is the classic example used to demonstrate that a direct K_sp comparison fails when stoichiometries differ.

SECTION 4

Mathematical Framework — From Ksp to Molar Solubility

Consider a generic sparingly soluble salt M_aX_b that dissolves according to the equilibrium M_aX_b(s) ⇌ aMⁿ⁺(aq) + bX^m−(aq). If s is the molar solubility, then [Mⁿ⁺] = as and [X^m−] = bs at equilibrium. The general K_sp expression and its solution for s are derived below.

GENERAL Ksp EXPRESSION

Ksp = [M^(n+)]^a × [X^(m−)]^b = (as)^a × (bs)^b = a^a × b^b × s^(a+b)

s = molar solubility (mol L⁻¹); a, b = stoichiometric coefficients; a + b = total number of ions per formula unit.

SOLVING FOR MOLAR SOLUBILITY

s = (Ksp / a^a × b^b)^(1/(a+b))

This is the master equation for converting K_sp into molar solubility. The stoichiometric prefactor a^ab^b and the exponent (a + b) both depend on the dissolution stoichiometry.

Common Special Cases

AB TYPE (1:1)

Ksp = s² → s = √Ksp

Examples: AgCl, BaSO₄, PbSO₄. The coefficient a^ab^b = 1¹ × 1¹ = 1.

AB₂ TYPE (1:2)

Ksp = 4s³ → s = ∛(Ksp / 4)

Examples: CaF₂, PbCl₂, Mg(OH)₂. The coefficient = 1¹ × 2² = 4.

SECTION 5

Detailed Breakdown — When Can You Compare Ksp Directly?

A systematic decision process helps avoid errors when comparing solubilities. The flowchart below encodes the logic: first identify the stoichiometric type of each salt, then decide whether a direct K_sp comparison suffices or whether you must compute s for each compound.

Decision flowchart for comparing solubilities. When the stoichiometric types match (left branch), K_sp rankings and solubility rankings coincide. When they differ (right branch), you must convert each K_sp to molar solubility before comparing.

Salt Type	General Formula	Ksp in Terms of s	Solve for s
AB (1:1)	AgCl, BaSO₄	s²	s = √K_sp
AB₂ or A₂B (1:2)	CaF₂, PbCl₂, Ag₂CrO₄	4s³	s = ∛(K_sp / 4)
AB₃ (1:3)	Al(OH)₃, Fe(OH)₃	27s⁴	s = ⁴√(K_sp / 27)
A₂B₃ (2:3)	Bi₂S₃, Ca₃(PO₄)₂	108s⁵	s = ⁵√(K_sp / 108)

Summary of stoichiometric types with their Ksp expressions and molar solubility formulas

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Common Pitfall

Students frequently assume that a smaller K_sp always means lower solubility. This is only guaranteed when the salts being compared produce the same total number of ions in the same ratio. When in doubt, always convert to molar solubility.

SECTION 6

Worked Example — Ranking Three Salts by Solubility

Rank the following three salts from most soluble to least soluble in pure water at 25 °C: AgBr (K_sp = 5.0 × 10⁻¹³), CaF₂ (K_sp = 3.9 × 10⁻¹¹), and Ag₂CrO₄ (K_sp = 1.1 × 10⁻¹²).

Ranking Three Salts of Different Stoichiometric Types

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Step 1 — Identify the Stoichiometric Type of Each Salt

AgBr is an AB-type salt: AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq), so K_sp = s². CaF₂ is an AB₂-type salt: CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), so K_sp = 4s³. Ag₂CrO₄ is an A₂B-type salt: Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq), so K_sp = 4s³.

All three salts have different stoichiometric types (AB vs. AB₂ vs. A₂B), so direct K_sp comparison is invalid.
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Step 2 — Calculate Molar Solubility for AgBr

For the AB-type: s = √K_sp = √(5.0 × 10⁻¹³) = 7.1 × 10⁻⁷ M.

s(AgBr) = 7.1 × 10⁻⁷ M
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Step 3 — Calculate Molar Solubility for CaF₂

For the AB₂-type: s = ∛(K_sp / 4) = ∛(3.9 × 10⁻¹¹ / 4) = ∛(9.75 × 10⁻¹²) = 2.1 × 10⁻⁴ M.

s(CaF₂) = 2.1 × 10⁻⁴ M
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Step 4 — Calculate Molar Solubility for Ag₂CrO₄

For the A₂B-type: K_sp = (2s)²(s) = 4s³, so s = ∛(K_sp / 4) = ∛(1.1 × 10⁻¹² / 4) = ∛(2.75 × 10⁻¹³) = 6.5 × 10⁻⁵ M.

s(Ag₂CrO₄) = 6.5 × 10⁻⁵ M
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Step 5 — Rank by Molar Solubility

Comparing the three molar solubilities: s(CaF₂) = 2.1 × 10⁻⁴ M > s(Ag₂CrO₄) = 6.5 × 10⁻⁵ M > s(AgBr) = 7.1 × 10⁻⁷ M.

CaF₂ > Ag₂CrO₄ > AgBr (most to least soluble)

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Note on This Result

If you had ranked by K_sp alone, you would have placed CaF₂ (3.9 × 10⁻¹¹) as the least soluble—exactly the opposite of the correct answer. The A₂B and AB₂ stoichiometries compress larger solubilities into smaller K_sp values.

SECTION 7

Strengths, Limitations & Assumptions

The K_sp-to-s conversion framework is elegant and widely applicable, but it rests on several assumptions that may break down in real systems. Understanding these limitations is essential for correctly interpreting solubility data in laboratory or industrial contexts.

Aspect	Strength	Limitation
Simplicity	Quick ranking of solubilities with a single formula per salt type	Only valid in pure water with no common ions, no competing equilibria
Ideal solution assumption	Accurate for very dilute solutions where activity coefficients ≈ 1	Breaks down at higher ionic strengths; activity corrections (Debye–Hückel) needed
Temperature dependence	Ksp values are tabulated at standard temperature (25 °C) for easy lookup	Comparisons at other temperatures require recalculation via van 't Hoff equation
Common-ion effect	The same Ksp framework handles common-ion problems by modifying the ICE table	Solubility rankings may change in solutions containing common ions
Ion pairing / complexation	Works well for salts that fully dissociate into simple ions	If ions form complexes (e.g., Pb²⁺ + Cl⁻ → PbCl⁺), effective solubility exceeds Ksp prediction

Strengths and limitations of the Ksp-based solubility comparison method

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KEY TAKEAWAY

The K_sp-to-s conversion is analogous to converting between units in engineering: the conversion factor (the stoichiometric prefactor and exponent) is exact and reliable under the stated assumptions, but the physical quantity it represents—actual dissolved concentration—can be affected by environmental factors (ionic strength, pH, temperature) just as a bridge's load capacity depends on conditions not captured by the blueprints alone.

SECTION 8

Connections to Advanced Solubility Theory

The approach outlined in this lesson treats K_sp as a concentration-based equilibrium constant, which is the standard approximation for general chemistry. More rigorous treatments replace concentrations with thermodynamic activities and incorporate additional equilibria such as hydrolysis, complex formation, and redox reactions. The table below connects the introductory framework to its advanced counterparts, foreshadowing topics encountered in analytical chemistry, geochemistry, and environmental science.

Introductory Concept	Advanced Extension	Why It Matters
Ksp with concentrations	Thermodynamic Ksp with activities (a = γ × c)	Accounts for inter-ionic attractions; essential in seawater, biological fluids, and industrial brines
Pure water assumption	Common-ion & diverse-ion effects, ion pairing	Real solutions always contain other electrolytes that shift equilibrium positions
Single equilibrium (dissolution)	Simultaneous equilibria (Ksp + Kb + Kf)	Hydrolysis of basic anions (e.g., CO₃²⁻, S²⁻) significantly increases solubility beyond simple Ksp prediction
Molar solubility comparison	Conditional solubility product (Ksp′) at a given pH	pH-dependent solubility is critical for metal hydroxide precipitation and selective separation schemes

Connecting the introductory Ksp framework to advanced solubility concepts

In a subsequent course on analytical chemistry, you will encounter situations where the "simple" molar solubility comparison must be modified for pH, complex-ion formation, or simultaneous precipitation of multiple salts. The foundational skill of converting K_sp to s using correct stoichiometry—the focus of this lesson—remains the essential first step in all of these more advanced analyses.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why two salts with different stoichiometric types cannot be ranked in order of solubility solely by comparing their K_sp values. Use the mathematical relationship between K_sp and s to support your answer.

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The relationship K_sp = a^ab^b × s^(a+b) shows that both the numerical prefactor (a^ab^b) and the exponent (a + b) depend on stoichiometry. When these differ between two salts, the same s value maps to very different K_sp values, and conversely, the same K_sp can correspond to different s values. The mapping from K_sp to s is monotonic only within a single stoichiometric class, not across classes.

Question 2: BASIC CALCULATION

Which is more soluble in pure water: BaSO₄ (K_sp = 1.1 × 10⁻¹⁰) or AgCl (K_sp = 1.8 × 10⁻¹⁰)? Justify your answer.

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Both BaSO₄ and AgCl are AB-type (1:1) salts, so for each K_sp = s². Since they share the same stoichiometric type, direct comparison is valid: the salt with the larger K_sp is more soluble. AgCl (1.8 × 10⁻¹⁰) > BaSO₄ (1.1 × 10⁻¹⁰), so AgCl is more soluble. Numerically: s(AgCl) = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M and s(BaSO₄) = √(1.1 × 10⁻¹⁰) = 1.0 × 10⁻⁵ M, confirming the ranking.

Question 3: INTERMEDIATE

Determine which compound has greater molar solubility in pure water: PbCl₂ (K_sp = 1.7 × 10⁻⁵) or PbSO₄ (K_sp = 1.6 × 10⁻⁸). Show your work.

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PbCl₂ is an AB₂-type salt: PbCl₂(s) ⇌ Pb²⁺ + 2Cl⁻, K_sp = 4s³. Thus s = ∛(1.7 × 10⁻⁵ / 4) = ∛(4.25 × 10⁻⁶) = 1.6 × 10⁻² M. PbSO₄ is an AB-type salt: K_sp = s², so s = √(1.6 × 10⁻⁸) = 1.3 × 10⁻⁴ M. Despite PbSO₄ having a larger K_sp than many 1:1 salts, PbCl₂ is far more soluble (1.6 × 10⁻² M vs. 1.3 × 10⁻⁴ M) because the cube-root relationship for the 1:2 type yields a higher s from its larger K_sp.

Question 4: APPLIED

A wastewater treatment plant must choose between precipitating Pb²⁺ as PbCO₃ (K_sp = 7.4 × 10⁻¹⁴) or Pb₃(PO₄)₂ (K_sp = 1.0 × 10⁻⁵⁴). Which precipitate would leave the lowest residual Pb²⁺ concentration in solution? Calculate s for each.

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PbCO₃ is AB-type: s = √(7.4 × 10⁻¹⁴) = 2.7 × 10⁻⁷ M. Pb₃(PO₄)₂ is A₃B₂-type: Pb₃(PO₄)₂(s) ⇌ 3Pb²⁺ + 2PO₄³⁻, so K_sp = (3s)³(2s)² = 108s⁵. Then s = ⁵√(1.0 × 10⁻⁵⁴ / 108) = ⁵√(9.26 × 10⁻⁵⁷) ≈ 3.9 × 10⁻¹² M. Despite K_sp for Pb₃(PO₄)₂ being astronomically small, the fifth-root extraction is the key operation. The residual [Pb²⁺] = 3s = 1.2 × 10⁻¹¹ M for the phosphate vs. 2.7 × 10⁻⁷ M for the carbonate. Precipitating as Pb₃(PO₄)₂ leaves far less lead in solution, making it the superior choice for remediation.

Question 5: CRITICAL THINKING

Consider two hypothetical salts: MX with K_sp = 1.0 × 10⁻⁶ and MY₃ with K_sp = 1.0 × 10⁻⁶. Both have the same numerical K_sp. (a) Calculate s for each. (b) Derive a general condition (in terms of stoichiometric coefficients a and b) under which a salt with identical K_sp to a 1:1 salt will nonetheless be more soluble.

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(a) For MX (AB-type): s = √(1.0 × 10⁻⁶) = 1.0 × 10⁻³ M. For MY₃ (AB₃-type): K_sp = (s)(3s)³ = 27s⁴, so s = ⁴√(1.0 × 10⁻⁶ / 27) = ⁴√(3.70 × 10⁻⁸) = 4.4 × 10⁻² M. Thus MY₃ is about 44 times more soluble despite having the same K_sp. (b) For a general salt M_aX_b with the same K_sp as a 1:1 salt, the solubility ratio is s(M_aX_b) / s(MX) = (K_sp / a^ab^b)^(1/(a+b)) / K_sp^1/2. This ratio exceeds 1 whenever a + b > 2 and K_sp < a^ab^b, which is almost always satisfied for sparingly soluble salts (K_sp ≪ 1). In essence, for typical K_sp values, any salt producing more than 2 ions per formula unit will be more soluble than a 1:1 salt with the same K_sp.

SUMMARY

Lesson Summary

Summary

Comparing the solubilities of sparingly soluble salts requires careful attention to dissolution stoichiometry. The solubility product constant (K_sp) encodes both the molar solubility and the stoichiometric type of the salt into a single number. For salts of the same stoichiometric type (e.g., all AB or all AB₂), a direct K_sp comparison correctly identifies the more soluble compound. For salts of different stoichiometric types, you must convert each K_sp to molar solubility using the general formula s = (K_sp / a^ab^b)^1/(a+b) before comparing.

Key results to remember: AB-type salts give K_sp = s², AB₂/A₂B-type salts give K_sp = 4s³, and A₂B₃-type salts give K_sp = 108s⁵. Salts producing more ions per formula unit "compress" a larger molar solubility into a smaller K_sp, which is precisely why the classic CaF₂ vs. AgCl comparison yields the counterintuitive result that CaF₂ is more soluble despite its smaller K_sp. Always check stoichiometry before comparing.