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

SECTION 1

Historical Context & Motivation

The question of how much of a "hard-to-dissolve" salt actually enters solution has occupied chemists since the earliest days of quantitative analysis. Before thermodynamic equilibrium concepts matured, analysts relied on empirical solubility tables compiled through painstaking gravimetric experiments. The intellectual leap came when researchers recognized that even a precipitate sitting at the bottom of a beaker coexists with trace concentrations of its constituent ions in solution, and that the product of those ion concentrations is a thermodynamic constant at a given temperature. This insight transformed solubility from a descriptive catalog into a predictive, calculable quantity.

1864 • Law of Mass Action

Cato Guldberg and Peter Waage published the law of mass action, establishing that the rate of a reaction is proportional to the product of the concentrations of the reactants. This framework later provided the algebraic skeleton on which solubility equilibrium expressions are built.

1889 • Nernst's Solubility Product

Walther Nernst formalized the solubility product concept, arguing that for a sparingly soluble salt, the product of the ion concentrations raised to their stoichiometric powers is constant at equilibrium. This was the birth of the K_sp expression.

1923 • Debye–Hückel Theory

Peter Debye and Erich Hückel developed a model for ion–ion interactions in solution, introducing activity coefficients that refine K_sp calculations for real solutions, especially at higher ionic strengths.

1960s • Modern Analytical Tables

Large-scale compilations of K_sp values at 25 °C became standard references in the CRC Handbook and NIST databases, allowing chemists to predict precipitation, design separations, and control water quality with a single equilibrium constant.

The central question that K_sp answers is deceptively simple: If I add excess solid to pure water and wait for equilibrium, what molar concentration of dissolved salt will I find? Converting K_sp into that concentration—the molar solubility—requires nothing more than careful ICE-table algebra, but mastering the stoichiometric bookkeeping is essential for problems ranging from qualitative-analysis separations to environmental ion-removal calculations.

SECTION 2

Core Principles & Definitions

Before diving into calculations, it is crucial to anchor a few foundational ideas. A sparingly soluble salt is an ionic compound whose saturated solution contains very low concentrations of its constituent ions—typically below roughly 0.01 M. When such a salt is placed in water, a dynamic equilibrium is established between the undissolved solid and the dissolved ions. The equilibrium constant for this dissolution, written with activities approximated as molar concentrations, is called the solubility product constant, K_sp. Because the solid phase has an activity of 1, it does not appear in the expression.

Molar Solubility (s)

The number of moles of salt that dissolve per liter of solution to form a saturated solution at a given temperature. Typically expressed in mol·L⁻¹. This is the quantity we ultimately solve for.

Ksp Expression

For a generic salt M_aX_b, K_sp = [M^b+]^a [X^a−]^b. Each concentration is raised to its stoichiometric coefficient.

ICE Table Method

An ICE (Initial–Change–Equilibrium) table tracks how concentrations shift as the solid dissolves. For dissolution in pure water, ion concentrations start at 0 and increase by stoichiometric multiples of s.

Stoichiometric Ratio

The stoichiometric coefficients in the balanced dissolution equation dictate the relationship between s and each ion's equilibrium concentration. For example, if one mole of salt produces two moles of an ion, that ion's concentration equals 2s.

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KEY TAKEAWAY

Think of K_sp as a fixed "ion budget" that the solution cannot exceed at a given temperature. If one ion's concentration rises (say, through a common-ion effect), the other must fall to keep the product within budget. Molar solubility is simply the maximum number of formula units per liter the budget allows when no outside ions are present—analogous to finding the maximum number of identical kits you can assemble when each kit requires specific parts and your total inventory of each part type is capped by the budget.

SECTION 3

Visual Explanation — From Dissolution to Ksp

A saturated solution of MX₂ in water. The solid (purple block) dissolves until the ion concentrations satisfy the K_sp expression. Amber circles represent M²⁺ cations; pink circles represent X⁻ anions. The ICE table (upper left) and resulting K_sp polynomial (upper right) summarize the algebra.

The diagram above illustrates the key relationship between the macroscopic solid phase and the microscopic ionic species in solution. When one formula unit of MX₂ dissolves, it releases one M²⁺ ion and two X⁻ ions. If we define s as the molar solubility, then [M²⁺] = s and [X⁻] = 2s at equilibrium. Substituting these into the K_sp expression gives K_sp = (s)(2s)² = 4s³. Solving this polynomial for s yields the molar solubility. Notice that the stoichiometric coefficients—here 1 and 2—produce the numerical prefactor (4) and the overall power of s (3). Different salt types generate different algebraic forms, a point we explore next.

SECTION 4

Mathematical Framework

The general dissolution of a salt M_aX_b can be written as M_aX_b(s) ⇌ aM^b+(aq) + bX^a−(aq). The K_sp expression follows directly from the law of mass action, with the solid's activity set to unity. Below are the key equations that connect K_sp to molar solubility s.

GENERAL KSP EXPRESSION

Ksp = [M^(b+)]^a × [X^(a−)]^b

Each ion concentration is raised to the power of its coefficient in the balanced dissolution equation. The solid does not appear because its activity is 1.

SUBSTITUTION IN TERMS OF s

Ksp = (as)^a × (bs)^b = a^a × b^b × s^(a+b)

If s moles of M_aX_b dissolve per liter, then [M^b+] = as and [X^a−] = bs. The exponent (a + b) is the total number of ions produced per formula unit.

SOLVING FOR MOLAR SOLUBILITY

s = [ Ksp / (a^a × b^b) ] ^ (1/(a+b))

Isolate s by dividing K_sp by the stoichiometric prefactor and taking the (a + b)th root. This is the master formula; every specific salt type (1:1, 1:2, 2:3, etc.) is a special case.

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Common Salt Types at a Glance

1:1 salts (e.g., AgCl): K_sp = s² → s = √K_sp. | 1:2 salts (e.g., CaF₂): K_sp = 4s³ → s = (K_sp/4)^(1/3). | 2:3 salts (e.g., Bi₂S₃): K_sp = 108s⁵ → s = (K_sp/108)^(1/5).

SECTION 5

Detailed Breakdown — Common Salt Stoichiometries

The algebraic form of the K_sp–solubility relationship changes with every new stoichiometric ratio. The table below catalogs the most common patterns you will encounter in a general chemistry course, along with the resulting polynomial and its solution for s. Recognizing these patterns instantly saves significant time on examinations.

Salt Type	Example	Dissolution	Ksp in terms of s	Solve for s
1 : 1	AgCl	AgCl ⇌ Ag⁺ + Cl⁻	s²	s = √Ksp
1 : 2	CaF₂	CaF₂ ⇌ Ca²⁺ + 2 F⁻	4s³	s = (Ksp / 4)^(1/3)
2 : 1	Ag₂CrO₄	Ag₂CrO₄ ⇌ 2 Ag⁺ + CrO₄²⁻	4s³	s = (Ksp / 4)^(1/3)
1 : 3	AlF₃	AlF₃ ⇌ Al³⁺ + 3 F⁻	27s⁴	s = (Ksp / 27)^(1/4)
2 : 3	Bi₂S₃	Bi₂S₃ ⇌ 2 Bi³⁺ + 3 S²⁻	108s⁵	s = (Ksp / 108)^(1/5)

Ksp–molar-solubility relationships for the most common salt stoichiometries

A five-step algorithmic flowchart for computing molar solubility from K_sp. Begin with the balanced dissolution equation and work systematically downward through the ICE table to the final polynomial solve.

An important observation from the table and flowchart: two salt types that share the same stoichiometric sum (a + b) can produce different prefactors. For instance, both 1:2 and 2:1 salts yield K_sp = 4s³, because the prefactor a^a × b^b depends on both values. You cannot directly compare K_sp values of salts with different stoichiometries to rank solubilities—you must first convert each K_sp to s and then compare the s values.

SECTION 6

Worked Example — Molar Solubility of Lead(II) Iodide

Lead(II) iodide, PbI₂, is a bright yellow sparingly soluble salt with K_sp = 9.8 × 10⁻⁹ at 25 °C. Let us compute its molar solubility in pure water.

Molar Solubility of PbI₂

1
Step 1 — Write the Balanced Dissolution Equation

PbI₂(s) ⇌ Pb²⁺(aq) + 2 I⁻(aq). This is a 1:2 salt, producing one cation and two anions per formula unit dissolved.
2
Step 2 — Write the Ksp Expression

K_sp = [Pb²⁺][I⁻]². The solid PbI₂ is omitted because its activity is 1.
3
Step 3 — Build the ICE Table

Let s = molar solubility of PbI₂. In pure water, initial ion concentrations are 0. As s moles dissolve: [Pb²⁺] changes by +s, [I⁻] changes by +2s. At equilibrium: [Pb²⁺] = s, [I⁻] = 2s.
4
Step 4 — Substitute into Ksp and Simplify

K_sp = (s)(2s)² = (s)(4s²) = 4s³. This confirms the 1:2 pattern from the table.

9.8 × 10⁻⁹ = 4s³
5
Step 5 — Solve for s

Divide both sides by 4: s³ = (9.8 × 10⁻⁹) / 4 = 2.45 × 10⁻⁹. Take the cube root: s = $(2.45 × 10⁻⁹)^{1/3}$ . Using a calculator, s = 1.35 × 10⁻³ M. This means approximately 1.35 millimoles of PbI₂ dissolve per liter of pure water at 25 °C.

s = 1.35 × 10⁻³ mol·L⁻¹
6
Step 6 — Verify (Optional but Recommended)

Back-substitute: [Pb²⁺] = 1.35 × 10⁻³ M, [I⁻] = 2(1.35 × 10⁻³) = 2.70 × 10⁻³ M. Check: (1.35 × 10⁻³)(2.70 × 10⁻³)² = (1.35 × 10⁻³)(7.29 × 10⁻⁶) = 9.8 × 10⁻⁹ ✓. The result matches the given K_sp.

SECTION 7

Strengths, Limitations & Common Pitfalls

Strengths	Limitations
Simple, algebraic route from a single tabulated constant to the dissolved concentration.	Assumes ideal behavior—activities ≈ concentrations—which breaks down above ~0.01 M ionic strength.
Applicable to any sparingly soluble salt regardless of stoichiometry.	Ignores ion-pairing and complex-ion formation that can increase apparent solubility.
Easily extended to include common-ion and pH effects via modified ICE tables.	Does not account for hydrolysis of basic anions (e.g., CO₃²⁻, S²⁻), which can make the calculated s too low.
Provides a clear yes/no precipitation prediction when combined with Q (ion product) analysis.	Temperature dependence of Ksp is often neglected; most tabulated values are for 25 °C only.

Strengths and limitations of the Ksp → molar solubility approach

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Common Pitfall: Comparing Ksp Values Directly

A frequent exam error is to assume that a larger K_sp always means greater molar solubility. This is true only when the salts share the same stoichiometry. For example, K_sp(AgCl) = 1.8 × 10⁻¹⁰ and K_sp(Ag₂CrO₄) = 1.1 × 10⁻¹² look like AgCl is more soluble by two orders of magnitude, but computing s gives AgCl: 1.3 × 10⁻⁵ M vs. Ag₂CrO₄: 6.5 × 10⁻⁵ M—Ag₂CrO₄ is actually about five times MORE soluble.

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KEY TAKEAWAY

The K_sp → s calculation is a powerful first-order model, but it lives in the ideal-solution approximation. When ionic strength is non-negligible, when pH couples to the anion's basicity, or when complex ions form, additional equilibria must be layered on top. Treating K_sp solubility as the "ground-floor" model—accurate for dilute, simple systems—rather than a universal truth is the mark of a mature understanding of aqueous equilibria.

SECTION 8

Connection to Advanced Theory

The molar-solubility calculation you have learned is the gateway to several more sophisticated topics in aqueous equilibria. The table below maps the basic K_sp approach to its advanced extensions. Understanding where the simple model ends and the advanced model begins will prepare you for upper-division analytical chemistry and environmental engineering courses.

Basic Ksp Model	Advanced Extension	When It Matters
Pure water, no other ions	Common-ion effect: solve ICE with initial [ion] ≠ 0	Precipitation in mixed-salt solutions; selective precipitation separations
Concentrations ≈ activities	Activity coefficients via Debye–Hückel; use Ksp = Ksp(thermo) / γ products	Ionic strength > 0.01 M; seawater or biological fluids
Anion does not react with water	pH-dependent solubility: couple Ksp with Ka/Kb for the anion	Salts of weak-acid anions (CaCO₃, FeS, Ca₃(PO₄)₂)
No complex-ion formation	Couple Ksp with Kf (formation constant) for metal–ligand complexes	AgCl dissolving in excess NH₃; Cu(OH)₂ in concentrated NaOH

From the basic Ksp model to its advanced extensions

Thermodynamically, K_sp is related to the standard Gibbs energy of dissolution by ΔG° = −RT ln K_sp. Because dissolution of most salts is endothermic (positive ΔH°), K_sp generally increases with temperature—a quantitative relationship described by the van 't Hoff equation. Exploring these connections in upper-division courses will deepen your appreciation for why a single number at 25 °C captures so much predictive power.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the solid salt does not appear in the K_sp expression. If you added more solid to a saturated solution at constant temperature, what would happen to the molar solubility?

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The activity of a pure solid is defined as 1, so it contributes a factor of 1 to the equilibrium expression and can be omitted. Adding more solid to an already saturated solution does not change the ion concentrations because the system is at equilibrium; the extra solid simply sits undissolved at the bottom. The molar solubility—a property of the saturated solution—remains unchanged as long as temperature is constant, because K_sp depends only on temperature.

Question 2: BASIC CALCULATION

The K_sp of silver bromide (AgBr) is 5.4 × 10⁻¹³ at 25 °C. Calculate the molar solubility of AgBr in pure water.

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AgBr is a 1:1 salt: AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq). Letting s = molar solubility, we have K_sp = s × s = s². Therefore s = √(5.4 × 10⁻¹³) = 7.35 × 10⁻⁷ M. This means roughly 0.735 micromoles of AgBr dissolve per liter of water, which corresponds to about 0.14 mg L⁻¹—extremely insoluble.

Question 3: INTERMEDIATE

Calculate the molar solubility of calcium fluoride (CaF₂) in pure water given K_sp = 3.45 × 10⁻¹¹. Then determine the equilibrium concentrations of Ca²⁺ and F⁻.

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CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq). Let s = molar solubility. From the ICE table: [Ca²⁺] = s, [F⁻] = 2s. K_sp = (s)(2s)² = 4s³. Solving: s³ = 3.45 × 10⁻¹¹ / 4 = 8.625 × 10⁻¹², so s =

(8.625 × 10⁻¹²)^{1/3}

= 2.05 × 10⁻⁴ M. Therefore [Ca²⁺] = 2.05 × 10⁻⁴ M and [F⁻] = 2 × 2.05 × 10⁻⁴ = 4.10 × 10⁻⁴ M. Verification: (2.05 × 10⁻⁴)(4.10 × 10⁻⁴)² = 3.45 × 10⁻¹¹ ✓.

Question 4: APPLIED

An environmental engineer needs to know whether a wastewater stream will precipitate lead(II) chloride (PbCl₂). The K_sp of PbCl₂ is 1.7 × 10⁻⁵ at 25 °C. (a) Compute the molar solubility of PbCl₂ in pure water. (b) If the wastewater contains [Pb²⁺] = 0.020 M and [Cl⁻] = 0.030 M, will precipitation occur?

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(a) PbCl₂(s) ⇌ Pb²⁺(aq) + 2 Cl⁻(aq). K_sp = [Pb²⁺][Cl⁻]² = (s)(2s)² = 4s³. s³ = 1.7 × 10⁻⁵ / 4 = 4.25 × 10⁻⁶; s =

(4.25 × 10⁻⁶)^{1/3}

= 1.62 × 10⁻² M. (b) Compute the ion product Q = [Pb²⁺][Cl⁻]² = (0.020)(0.030)² = (0.020)(9.0 × 10⁻⁴) = 1.8 × 10⁻⁵. Since Q = 1.8 × 10⁻⁵ > K_sp = 1.7 × 10⁻⁵, the solution is slightly supersaturated and precipitation of PbCl₂ will occur.

Question 5: CRITICAL THINKING

Consider three salts: A (1:1, K_sp = 4.0 × 10⁻⁸), B (1:2, K_sp = 2.0 × 10⁻¹⁰), and C (1:3, K_sp = 1.0 × 10⁻¹²). Without using a calculator, rank these three salts from least to most soluble and justify your reasoning. Then verify with approximate computations.

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First compute s for each. Salt A (1:1): s = √(4 × 10⁻⁸) = 2 × 10⁻⁴ M. Salt B (1:2): s = (K_sp/4)^(1/3) =

(5 × 10⁻¹¹)^{1/3}

≈ 3.7 × 10⁻⁴ M. Salt C (1:3): s = (K_sp/27)^(1/4) =

(3.7 × 10⁻¹⁴)^{1/4}

≈ 2.5 × 10⁻⁴ M. Ranking from least to most soluble: A (2.0 × 10⁻⁴) < C (2.5 × 10⁻⁴) < B (3.7 × 10⁻⁴). This problem illustrates the critical lesson that K_sp magnitudes alone are misleading: Salt B has the smallest K_sp of salts A and B, yet it has the largest molar solubility because its stoichiometry spreads the s dependence across a cube root rather than a square root.

SUMMARY

Lesson Summary

Summary

The solubility product constant (Ksp) quantifies the equilibrium between a sparingly soluble ionic solid and its dissolved ions. To convert K_sp into molar solubility (s), write the balanced dissolution equation, construct an ICE table expressing each ion concentration in terms of s, substitute into the K_sp expression, and solve the resulting polynomial for s. The specific algebraic form—s², 4s³, 27s⁴, 108s⁵—depends on the stoichiometric coefficients in the dissolution reaction.

Keep in mind that direct comparison of Ksp values only ranks solubilities when salts share the same stoichiometry. For salts with different ion ratios, you must compute s for each salt individually. This foundational technique extends naturally to common-ion problems, pH-dependent solubility, and selective precipitation analyses—all of which build on the same ICE-table framework presented here.