Skip to main content
Chemistry LibreTexts

2.3: Writing Complete and Net Ionic Reactions

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

    Identifying ions in aqueous compounds

    Aqueous ionic compounds, acids, and bases will split up into their respective ions in solution. Below is a summary of ions that could be generated from these three classes of compounds. Keep in mind that polyatomic ions stay as one unit in solution and do not split up into their individual atoms. 

    • Aqueous ionic compounds will split into their cation and anion in solution. The charge of most ions can be predicted from the periodic table, or from the identity of the polyatomic ion. For those cations that form more than one charge, the charge should be determined from the ions paired anion(s).
    • Strong acids will split up into H+ ion(s) and an anion (either a nonmetal ion or a polyatomic ion).
    • Strong bases will typically split up into OH- ion(s) and a cation.


    Example \(\PageIndex{1}\)

    What ions exist in a lead (II) nitrate, Pb(NO3)2, solution?


    Soluble ionic compounds will separate into the cation and anion.

    How do we know NO3- doesn't separate?

    How do we determine the 2+ charge on the Pb2+ ion? Lead is a metal that is capable of forming ions with various charges. To determine the charge here, we need to look 

    Exercise \(\PageIndex{1}\)

    What ions exist in each of the following solutions?

    a. LiCl

    b. HCl

    c. Ca(OH)2

    Answer a

    LiCl will separate into Li+ and Cl- ions.

    Answer b

    HCl will separate into H+ and Cl- ions.

    Answer c

    Ca(OH)2 will separate into Ca2+ and OH- ions.


    You'll use this same process to "take apart" the reactants in a chemical reaction. These ions will be available to interact with the other ions in the solution.

    Example \(\PageIndex{1}\)

    Add example text here.


    Add example text here.


    Forming products (“switching partners”)

    When two aqueous compounds are mixed together, their respective ions will be able to interact and “switch partners.” When one of the possible products is a solid (s), liquid (l), or gas (g), a reaction has occurred. The example below shows what occurs when Pb(NO3)2 (aq) is mixed with LiCl (aq). The Pb2+ ion is now able to react with the Cl- ion and the Li+ ion is now able to react with NO3- ion. We will discuss phases in the next section.

    Ions in Solution.png
    Figure \(\PageIndex{1}\): When Pb(NO3)2 and LiCl solutions are combined, the ions can interact.


    When the compounds “switch partners,” the products that are formed will depend on the charges of the ions. Keep in mind that cations will react with anions in a ratio that will cancel out their charges. When H+ (aq) reacts with OH- (aq), it produces H2O (l). For the following problems, indicate which ions will react with each other and write the newly formed compounds as products in the reaction.

    2.3: Writing Complete and Net Ionic Reactions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?