What is the difference between solubility and solubility product constant ($K_{sp}$)?

Solubility ($s$) refers to 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 (molar solubility) or grams per liter. It's a measure of the concentration of the dissolved species. The solubility product constant ($K_{sp}$), on the other hand, is an equilibrium constant for the dissolution of a sparingly soluble ionic compound. It represents the product of the concentrations of the constituent ions, each raised to its stoichiometric coefficient, in a saturated solution. $K_{sp}$ is a constant at a given temperature, while solubility ($s$) can be influenced by factors like the common ion effect or pH, even if $K_{sp}$ remains constant.

How does the common ion effect influence the solubility of a sparingly soluble salt?

The common ion effect states that the solubility of a sparingly soluble salt decreases when a soluble salt containing a common ion is added to the solution. This is a direct consequence of Le Chatelier's Principle. For example, if you have a saturated solution of AgCl ($AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$), adding NaCl (which provides $Cl^-$ ions) will increase the concentration of $Cl^-$ ions. To relieve this stress, the equilibrium shifts to the left, causing more AgCl to precipitate out of the solution, thereby reducing the concentration of $Ag^+$ ions and thus the solubility of AgCl.

Why is the concentration of the solid not included in the $K_{sp}$ expression?

In the equilibrium expression for the dissolution of a sparingly soluble solid, the concentration of the pure solid is omitted because its concentration is essentially constant. A pure solid's concentration is determined by its density and molar mass, both of which are fixed values at a given temperature. As long as some solid is present, its 'concentration' (or activity) does not change, even if the amount of solid changes. Therefore, it is incorporated into the value of the equilibrium constant itself, simplifying the expression to only include the concentrations of the dissolved ions.

How can pH affect the solubility of certain salts?

pH significantly affects the solubility of salts whose anions are conjugate bases of weak acids (e.g., $CO_3^{2-}$, $S^{2-}$, $F^-$) or whose cations are acidic. For example, consider $CaCO_3$. In an acidic solution, $H^+$ ions react with $CO_3^{2-}$ ions to form $HCO_3^-$ and then $H_2CO_3$. This removes $CO_3^{2-}$ from the solution, shifting the $CaCO_3$ dissolution equilibrium ($CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$) to the right, thus increasing its solubility. Conversely, increasing pH (making it more basic) would decrease the solubility of such salts. Salts with basic anions are generally more soluble in acidic solutions.

What is the significance of comparing the ion product ($Q_{sp}$) with $K_{sp}$?

Comparing the ion product ($Q_{sp}$) with the solubility product constant ($K_{sp}$) allows us to predict whether a precipitate will form or if a solution is unsaturated. $Q_{sp}$ is calculated using the current (non-equilibrium) concentrations of ions, while $K_{sp}$ is the value at equilibrium. If $Q_{sp} K_{sp}$, the solution is supersaturated, and precipitation will occur until the ion concentrations decrease to the point where $Q_{sp} = K_{sp}$, establishing equilibrium.

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Solubility Equilibria of Sparingly Soluble Salts

Chemistry

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Version 1Updated 22 Mar 2026

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Solubility equilibrium for sparingly soluble salts describes the dynamic state reached when the rate of dissolution of a solid ionic compound into its constituent ions in a solvent equals the rate of precipitation of those ions back into the solid form. This equilibrium is quantitatively characterized by the solubility product constant, $K_{sp}$ , which is the product of the molar concentrations of…

Quick Summary

Solubility equilibria deal with the dynamic balance between a sparingly soluble ionic solid and its dissolved ions in a saturated solution. A sparingly soluble salt dissolves only to a small extent, establishing an equilibrium where the rate of dissolution equals the rate of precipitation.

This equilibrium is quantified by the solubility product constant, $K_{sp}$ , which is the product of ion concentrations, each raised to its stoichiometric coefficient. For a salt $A_x B_y$ , $K_{sp} = [A^{y+}]^x [B^{x-}]^y$ .

Molar solubility ( $s$ ) can be calculated from $K_{sp}$ and vice versa, with the relationship depending on the salt's stoichiometry (e.g., $s = \sqrt{K_{sp}}$ for AB type, $s = \sqrt[3]{K_{sp}/4}$ for $AB_2$ type).

The ion product ( $Q_{sp}$ ) is used to predict precipitation: if $Q_{sp} > K_{sp}$ , precipitation occurs. Factors like the common ion effect (decreases solubility), pH (increases solubility for salts with basic anions in acidic media), and complex ion formation (increases solubility) significantly influence solubility, all explainable by Le Chatelier's Principle.

These concepts are crucial for understanding chemical separations and environmental processes.

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Key Concepts

Calculating

K_{sp}

from Solubility and vice-versa

The relationship between molar solubility ( $s$ ) and $K_{sp}$ depends critically on the stoichiometry of the…

Common Ion Effect on Solubility

When a soluble salt containing an ion common to a sparingly soluble salt is added to the solution, the…

Effect of pH on Solubility

The solubility of salts containing basic anions (conjugate bases of weak acids) increases in acidic…

Equilibrium: —

A_x B_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)

$K_{sp}$ expression: —

K_{sp} = [A^{y+}]^x [B^{x-}]^y

s

K_{sp}

relations:**

- AB type: $K_{sp} = s^2 \implies s = \sqrt{K_{sp}}$ - $AB_2$ or $A_2B$ type: $K_{sp} = 4s^3 \implies s = \sqrt[3]{K_{sp}/4}$ - $A_x B_y$ type: $K_{sp} = x^x y^y s^{(x+y)}$

Precipitation condition:

- $Q_{sp} < K_{sp}$ : Unsaturated, no precipitation - $Q_{sp} = K_{sp}$ : Saturated, equilibrium - $Q_{sp} > K_{sp}$ : Supersaturated, precipitation occurs

Common Ion Effect: — Decreases solubility (

s

K_{sp}

remains constant.

pH Effect: — Increases solubility for salts with basic anions in acidic solutions (e.g.,

Mg(OH)_2

CaCO_3

PbS

Complex Ion Effect: — Increases solubility if one ion forms a stable complex (e.g., AgCl in

NH_3

SPARINGLY SOLUBLE SALTS: Shift PH, Add Reagents, Ion Numbers, Get Lower Yields.

Shift PH: Solubility changes with pH for salts with basic anions.

Add Reagents: Common ion effect (adding a common ion) decreases solubility.

Ion Numbers: Stoichiometry is crucial for

K_{sp}

and

s

relationship (e.g.,

s^2

4s^3

Get Lower Yields:

Q_{sp} > K_{sp}

means precipitation, reducing ion yield in solution.

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