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17.2: Relationship Between Solubility and Ksp

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    • Quantitatively related \(K_{sp}\) to solubility

    Considering the relation between solubility and \(K_{sq}\) is important when describing the solubility of slightly ionic compounds. However, this article discusses ionic compounds that are difficult to dissolve; they are considered "slightly soluble" or "almost insoluble." Solubility product constants (\(K_{sq}\)) are given to those solutes, and these constants can be used to find the molar solubility of the compounds that make the solute. This relationship also facilitates finding the \(K_{sq}\) of a slightly soluble solute from its solubility.


    Recall that the definition of solubility is the maximum possible concentration of a solute in a solution at a given temperature and pressure. We can determine the solubility product of a slightly soluble solid from that measure of its solubility at a given temperature and pressure, provided that the only significant reaction that occurs when the solid dissolves is its dissociation into solvated ions, that is, the only equilibrium involved is:


    In this case, we calculate the solubility product by taking the solid’s solubility expressed in units of moles per liter (mol/L), known as its molar solubility.

    Example \(\PageIndex{1}\): Calculation of Ksp from Equilibrium Concentrations

    We began the chapter with an informal discussion of how the mineral fluorite is formed. Fluorite, CaF2, is a slightly soluble solid that dissolves according to the equation:

    \[\ce{CaF2}(s)⇌\ce{Ca^2+}(aq)+\ce{2F-}(aq) \nonumber \]

    The concentration of Ca2+ in a saturated solution of CaF2 is 2.1 × 10–4 M; therefore, that of F is 4.2 × 10–4 M, that is, twice the concentration of Ca2+. What is the solubility product of fluorite?


    First, write out the Ksp expression, then substitute in concentrations and solve for Ksp:

    \[\ce{CaF2}(s)⇌\ce{Ca^2+}(aq)+\ce{2F-}(aq) \nonumber\]

    A saturated solution is a solution at equilibrium with the solid. Thus:

    \[K_\ce{sp}=\ce{[Ca^2+][F-]^2}=(2.1×10^{−4})(4.2×10^{−4})^2=3.7×10^{−11} \nonumber\]

    As with other equilibrium constants, we do not include units with Ksp.

    Exercise \(\PageIndex{1}\)

    In a saturated solution that is in contact with solid Mg(OH)2, the concentration of Mg2+ is 3.7 × 10–5 M. What is the solubility product for Mg(OH)2?

    \[\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}(aq)+\ce{2OH-}(aq) \nonumber\]


    2.0 × 10–13

    Example \(\PageIndex{2}\): Determination of Molar Solubility from Ksp

    The Ksp of copper(I) bromide, CuBr, is 6.3 × 10–9. Calculate the molar solubility of copper bromide.


    The solubility product constant of copper(I) bromide is 6.3 × 10–9.

    The reaction is:


    First, write out the solubility product equilibrium constant expression:

    \[K_\ce{sp}=\ce{[Cu+][Br- ]}\nonumber\]

    Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the Ksp:

    At equilibrium:

    \(K_\ce{sp}=\ce{[Cu+][Br- ]}\)

    Therefore, the molar solubility of CuBr is 7.9 × 10–5 M.


    Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. Equilibrium is the state at which the concentrations of products and reactant are constant after the reaction has taken place. The solubility product constant (\(K_{sp}\)) describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. The higher the \(K_{sp}\), the more soluble the compound is. \(K_{sq}\) is defined in terms of activity rather than concentration because it is a measure of a concentration that depends on certain conditions such as temperature, pressure, and composition. It is influenced by surroundings. \(K_{sp}\) is used to describe the saturated solution of ionic compounds. (A saturated solution is in a state of equilibrium between the dissolved, dissociated, undissolved solid, and the ionic compound.)