Skip to main content
Chemistry LibreTexts

18.4: Solubility Rules

  • Page ID
    288841
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

     

    To determine if a salt is soluble or to judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds will be insoluble in aqueous solutions. For this, we use solubility rules, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble, or insoluble). Table \(\PageIndex{1}\) lists some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules. If a compound is soluble, we use the (aq) label with it, indicating that it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected.

    Table \(\PageIndex{1}\): Some Useful Solubility Rules (soluble)
    These compounds generally dissolve in water (are soluble): Exceptions:
    All compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+ None
    All compounds of NO3 and C2H3O2 None
    Compounds of Cl, Br, I Ag+, Hg22+, Pb2+
    Compounds of SO42 Hg22+, Pb2+, Sr2+, Ba2+
    Table \(\PageIndex{2}\): Some Useful Solubility Rules (insoluble)
    These compounds generally do not dissolve in water (are insoluble): Exceptions:
    Compounds of CO32− and PO43− Compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+
    Compounds of OH Compounds of Li+, Na+, K+, Rb+, Cs+, NH4+, Sr2+, and Ba2+

    For example, consider the possible double-replacement reaction between Na2SO4 and SrCl2. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble, except for Ag+, Hg22+, and Pb2+, which are not being considered here. Therefore, Na2SO4 and SrCl2 are both soluble. The possible double-replacement reaction products are NaCl and SrSO4. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO4? Compounds of the sulfate ion are generally soluble, but Sr2+ is an exception: we expect it to be insoluble—a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be:

    \[\ce{Na2SO4(aq) + SrCl2(aq) → 2NaCl(aq) + SrSO4(s)}\nonumber \]

    You would expect to see a visual change corresponding to SrSO4 precipitating out of solution (Figure \(\PageIndex{2}\)).

    4856feea642ad7543b114732b2b5fc1e.jpg
    Figure \(\PageIndex{2}\): Double-Replacement Reactions. Some double-replacement reactions are obvious because you can see a solid precipitate coming out of solution. Source: Photo courtesy of Choij, http://commons.wikimedia.org/wiki/File:Copper_solution.jpg.

    Example \(\PageIndex{4}\):

    Will a double-replacement reaction occur? If so, identify the products.

    1. Ca(NO3)2 + KBr → ?
    2. NaOH + FeCl2 → ?

    Solution

    1. According to the solubility rules, both Ca(NO3)2 and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)—namely, CaBr2 and KNO3. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case.
    2. According to the solubility rules, both NaOH and FeCl2 are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH)2. NaCl is soluble, but, according to the solubility rules, Fe(OH)2 is not. Therefore, a reaction would occur, and Fe(OH)2(s) would precipitate out of solution. The balanced chemical equation is \[\ce{2NaOH(aq) + FeCl2(aq) → 2NaCl(aq) + Fe(OH)2(s)}\nonumber \]

    Exercise \(\PageIndex{4}\)

    \[\ce{Sr(NO3)2 + KCl → }\nonumber \]

    Answer

    No reaction; all possible products are soluble.

    Key Takeaways

    • Solubility rules are used to predict whether some double-replacement reactions will occur.

    18.4: Solubility Rules is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.