Precipitation Reactions

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:15

The independent behavior of each type of ionAn atom or covalently bonded set of atoms that carries an overall net charge. in solutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. was illustrated in Chemical Bonding by means of precipitation reactions. Precipitation is a process in which a soluteThe substance added to a solvent to make a solution. separates from a supersaturatedDescribes a solution that has a greater concentration of a solute than under equilibrium conditions. solution. In a chemical laboratory it usually refers to a solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. crystallizing from a liquidA state of matter in which the atomic-scale particles remain close together but are able to change their positions so that the matter takes the shape of its container solution, but in weather reports it applies to liquid or solid water separating from supersaturated air.

A typical precipitation reaction occurs when an aqueous solution of barium chloride is mixed with one containing sodium sulfate. The equation

BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)      (1a)

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

Ba2+(aq) + 2Cl(aq) + 2Na+(aq) + SO42–(aq) → BaSO4(s) + 2Na+(aq) + Cl(aq)      (1b)

Equation (1b) 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 compoundA substance made up of two or more elements and having those elements present in definite proportions; a compound can be decomposed into two or more different substances. 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. (1b). These ions are called spectator ions because they do not participate in the reaction. When we want to emphasize the independent behavior of ions, a net ionic equationThe equation for a reaction in which strong electrolytes are written as their separate component ions and in which species that are unchanged are omitted. is written, omitting the spectator ions. For precipitation of BaSO4 the net ionic equation is

Ba2+(aq) + SO42–(aq) → BaSO4(s)      (1c)

EXAMPLE 1 When a solution of AgNO3 is added to a solution of CaCl2, insolubleUnable to dissolve appreciably in a solvent. AgCl precipitates. Write three equations to describe this process.

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

2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq)      (2a)

2Ag+(aq) + 2NO3(aq) + Ca2+(aq) + 2Cl(aq) → 2AgCl(s) + Ca2+(aq) + 2NO3(aq)      (2b)

Ag+(aq) + Cl(aq) → AgCl(s)      (2c)

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 acidIn Arrhenius theory, a substance that produces hydrogen ions (hydronium ions) in aqueous solution. In Bronsted-Lowry theory, a hydrogen-ion (proton) donor. In Lewis theory, a species that accepts a pair of electrons to form a covalent bond. 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 mixtureA combination of two or more substances in which the substances retain their chemical identity. 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.

Precipitates are also used for quantitative analysis of solutions, that is, to determine the amount of solute or the massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object. of solute in a given solution. For this purpose it is often convenient to use the first of the three types of equations described above. Then the rules of stoichiometry may be applied.

EXAMPLE 2 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.

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_{\text{AgCl}}\xrightarrow{M_{\text{AgCl}}}n_{\text{AgCl}}\xrightarrow{S\text{(CaCl}_{\text{2}}\text{/AgCl)}}n_{\text{CaCl}_{\text{2}}} \ 
 &  \ 
 & n_{\text{CaCl}_{\text{2}}}=\text{2}\text{.073 g AgCl }\times \text{ }\frac{\text{1 mol AgCl}}{\text{143}\text{.32 g AgCl}}\text{ }\times \text{ }\frac{\text{1 mol CaCl}_{\text{2}}}{\text{2 mol AgCl}}=\text{7}\text{.23 }\times \text{ 10}^{-\text{3}}\text{ mol CaCl}_{\text{2}} \ 
 &  \ 
 & c_{\text{CaCl}_{\text{2}}}=\frac{n_{\text{CaCl}_{\text{2}}}}{V_{\text{soln}}}=\frac{\text{7.23 }\times \text{ 10}^{-\text{3}}\text{ mol CaCl}_{\text{2}}}{\text{50}\text{.0 cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{10}^{\text{3}}\text{ cm}^{\text{3}}}{\text{1 dm}^{\text{3}}}=\text{0}\text{.145 mol dm}^{-\text{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 1 gives a list of rules which enable us to predict the solubilityThe extent to which a solute dissolves in a solvent; often expressed as the mass of a substance that will dissolve in 100 mL of solvent. of the most commonly encountered substances. Use of this table is illustrated in the following example.

TABLE 1 Solubility. Rules

Soluble in Water Important Exceptions (insoluble)
All Group IA and NH4+ salts
All nitrates, chlorates, perchlorates and acetates
All sulfates CaSO4, BaSO4, SrSO4, PbSO4
All chlorides, bromides, and iodides AgX, Hg2X2, PbX2 (X= Cl, Br, or I)
Sparingly 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:
They will precipitate only if rather concentrated solutions are used

EXAMPLE 3 Write balanced net ionic equations to describe any reactions which occur when the following solutions are mixed:

a) 0.1 M Na2SO4 + 0.1 M NH4I

b) 0.1 M K2CO3 + 0.1 M SrCl2

c) 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

Sr2+(aq) + CO32–(aq) → SrCO3(s)

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

Fe2+(aq) + SO42–(aq) + Ba2+(aq) + 2OH(aq) → Fe(OH)2(s) + BaSO4(s)