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11.2: Precipitation Reactions

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    \[\ce{Ba^{2+}(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)}\label{1}\]

    can be written to describe what happens, and such an equation is useful in making chemical calculations. However, Equation \(\ref{1}\) does not really represent the microscopic particles (that is, the ions) present in the solution. Thus we might write

    \[\ce{Ba^{2+}(aq) + 2Cl^{-}(aq) + 2Na^{+}(aq) + SO4^{2-}(aq) -> BaSO4(s) + 2Na^{+} + 2Cl^{-}(aq)}\label{2}\]


    Figure \(\PageIndex{1}\): The figure on the left shows all ions present in the solution before the reaction occurs. After the reaction occurs, Na+ and Cl- ions remain present in the solution, un-reacted (spectator ions). Ba2+ and SO42- ions however, react to form the insoluble, white BaSO4 (s), which precipitates out of the solution.

    Equation \(\ref{2}\) is rather cumbersome and includes so many different ions that it may be confusing. In any case, we are often interested in the independent behavior of ions, not the specific compound from which they came. A precipitate of BaSO4(s) will form when any solution containing Ba2+(aq) is mixed with any solution containing SO42–(aq) (provided concentrations are not extremely small). This happens independently of the Cl(aq) and Na+(aq) ions in Eq. \(\ref{2}\) . These ions are called spectator ions because they do not participate in the reaction (see the figure above). When we want to emphasize the independent behavior of ions, a net ionic equation is written, omitting the spectator ions. For precipitation of BaSO4 the net ionic equation is

    \[\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO{4}(s)}\label{3} \]

    Example \(\PageIndex{1}\): Precipitation Reaction

    When a solution of AgNO3 is added to a solution of CaCl2, insoluble AgCl precipitates. Write three equations to describe this process.


     Both AgNO3 and CaCl2 are soluble ionic compounds, and so they are strong electrolytes. The three equations are


    \(\ce{2Ag^+(aq) + 2NO3^{-}(aq) + Ca^{2+}(aq) + Cl^{-}(aq) -> 2AgCl(s) + Ca^{2+}(aq) + 2NO3^{-}(aq)}\)      


    The occurrence or nonoccurrence of precipitates can be used to detect the presence or absence of various species in solution. BaCl2 solution, for instance, is often used as a test for SO42–(aq) ion. There are several insoluble salts of Ba, but they all dissolve in dilute acid except for BaSO4. Thus, if BaCl2 solution is added to an unknown solution which has previously been acidified, the occurrence of a white precipitate is proof of the presence of the SO42– ion. AgNO3 solution is often used in a similar way to test for halide ion.

    If AgNO3 solution is added to an acidified unknown solution, a white precipitate indicates the presence of Cl ions, a cream-colored precipitate indicates the presence of Br ions, and a yellow precipitate indicates the presence of I ions. Further tests can then be made to see whether perhaps a mixture of these ions is present. When AgNO3 is added to tap water, a white precipitate is almost always formed. The Cl ions in tap water usually come from the Cl2 which is added to municipal water supplies to kill microorganisms.


    Example \(\PageIndex{2}\): Concentration

    When a solution of 0.1 M AgNO3is added to 50.0 cm3 of a CaCl2 solution of unknown concentration, 2.073 g AgCl precipitates. Calculate the concentration of the unknown solution.


    We know the volume of the unknown solution, and so only the amount of solute is needed to calculate the concentration. This can be found using Eq. (2a) in Example 1. From the equation the stoichiometric ratio S(CaCl2/AgCl) may be obtained. A road map to the solution of the problem is

    \(m_{AgCl} \underset{\text{M_{AgCl}}}{\mathop{\rightarrow}}\, n_{AgCl} \underset{\text{M_{CaCL_2}}}{\mathop{\rightarrow}}\, n_{CaCl_2}\)

    \(n_{CaCl-2} = 2.073 g AgCl * \frac{1mol AgCl}{143.32 g AgCl} * \frac{1 mol CaCl_2}{2 mol AgCl} = 7.23 * 10^{-3} \text{mol CaCl}_2\)

    \(c_{CaCl_2} = \frac{n_{CaCl_2}}{V_{soln}} = \frac{7.23 * 10^{-3} mol CaCl_2}{50.0 cm^3} * \frac{10^3 cm^3}{1 dm^3} = 0.145 \frac{mol}{dm^3}\)

    Thus the concentration of the unknown solution is 0.145 M.

    Because of the general utility of precipitates in chemistry, it is worth having at least a rough idea of which common classes of compounds can be precipitated from solution and which cannot. Table \(\PageIndex{1}\) gives a list of rules which enable us to predict the solubility of the most commonly encountered substances. Use of this table is illustrated in the following example.

    Table \(\PageIndex{1}\): Solubility. Rules
    Soluble in Water Important Exceptions (insoluble)
    All Group IA and NH4+ salts CaSO4, BaSO4, SrSO4, PbSO4
    All nitrates, chlorates, perchlorates and acetates AgX, Hg2X2, PbX2 (X= Cl, Br, or I)
    All sulfates  
    All chlorides, bromides, and iodides  
    Slightly Soluble in Water Important Exceptions (soluble)
    All carbonates and phosphates Group IA and NH4+ salts
    All hydroxides Group IA and NH4+ salts; Ba2+, Sr2+, Ca2+ sparingly soluble
    All sulfides Group IA, IIA and NH4+ salts; MgS, CaS, BaS sparingly soluble
    All oxalates Group IA and NH4+ salts
    The following electrolytes are of only moderate solubility in water: CH3COOAg, Ag2SO4, KClO4  They will precipitate only if rather concentrated solutions are used.

    Example \(\PageIndex{3}\): Net Ionic Equation

    Write balanced net ionic equations to describe any reactions which occur when the following solutions are mixed:

    1. 0.1 M Na2SO4 + 0.1 M NH4I
    2. 0.1 M K2CO3 + 0.1 M SrCl2
    3. 0.1 M FeSO4 + 0.1 M Ba(OH)2


    a) If any precipitate forms, it will be either a combination of Na+ ions and I ions, namely, NaI, or a combination of ammonium ions, NH4+, and sulfate ions, SO42–, namely, (NH4)2SO4. From Table 11.2 we find that NaI and (NH4)2SO4 are both soluble. Thus no precipitation reaction will occur, and there is no equation to write.

    b) Possible precipitates are KCl and SrCO3. From Table 11.2 we find that SrCO3 is insoluble. Accordingly we write the net ionic equation as


    omitting the spectator ions K+ and Cl.

    c) Possible precipitates are Fe(OH)2 and BaSO4. Both are insoluble. The net ionic equation is thus

    \(\ce{Fe^{2+}(aq) + SO4^{2-}(aq) + Ba^{2+}(aq) + 2OH^{-}(aq) -> Fe(OH)2(s) + BaSO4(s)}\)