A reaction of two water soluble salts in which cation (positively charged ion) and anion (negatively charged ion) partners are "traded" is called a metathesis or double displacement reaction. Such a reaction is represented schematically below, where A and B are two different cations, X and Y two different anions.

Equation 1

AX  +  BY => AY +  BX                                        

For such a reaction to proceed to the right, there must be a thermodynamic driving force. The driving force is provided, according to Le Chatelier's principle, by removing one of the products from solution. This will occur if one of the products precipitates (i.e., leaves solution as a solid), is a weak electrolvte (i.e., is essentially unionized in solution), or decomposes to form a gas, which escapes from solution. Reaction of two salts to form solid, weak electrolyte, or gas is spontaneous. The free energy change, AG, for such a process is negative. This experiment deals with reactions that (potentially) form solids.

The formation of a precipitate is illustrated in equation form below. Three equations, of increasing simplicity, are shown.

Total Equation (TE)

-AgNO₃(aq)  +  NaCl(aq)  =>  AgCl(s)  +  NaNO₃(aq)    

Ionic Equation (IE)

Ag⁺(aq) + N0₃^-(aq) + Na⁺(aq) + Cl^-(aq) =>AgCl(s) + Na⁺(aq) + N0₃^-(aq)

Net Ionic Equation (NIE)

Ag⁺(aq) + Cl^-(aq) -> AgCl(s)                                                 

These are called, respectively, the total equation (TE), the ionic equation (IE), and the net ionic equation (NIE). The TE shows what reagents to obtain from the stockroom to carry out a given reaction. An IE is obtained by writing all soluble strong electrolytes from the TE in dissociated form; it bridges the total equation and the net ionic equation. The net ionic equation is obtained by omitting ions that appear on both sides of the ionic equation. Canceled ions are called spectator ions because they do not take part in the chemical reaction. Therefore, the NIE shows in simplest form what species take part in the reaction.

Solubility

Solubilitv is the amount of solute (in g or moles) that dissolves in a given amount of solvent (in mL, L, or g). If quite a bit of a solid dissolves in a solvent, the solid is soluble in that solvent. If only a small amount of a solid dissolves in the solvent, the solid is insoluble, or sparinglv soluble, in the solvent. A solubility of 1.0 gram solute per 100 mL solvent is considered the border line between soluble and insoluble. For example, 74.5 g of calcium chloride, CaCl₂, dissolves in 100 mL H₂0; thus CaCl₂ is soluble in water. In contrast, only 2.9 x l0^-4 g silver chloride, AgCl, dissolves in 100 mL water, so it is insoluble in water. Formation of a soluble salt from its ions has G⁰> 0; formation of an insoluble salt from its ions has G⁰<0.

A spontaneous process is one that takes place without assistance from outside. Examples are melting of ice at room temperature, rusting of iron, and combustion of natural gas in a furnace. A process is spontaneous either because it is exothermic (has H < 0) or because it increases disorder (has s> 0) or both. The effects of these two quantities on spontaneity are neatly summarized in terms of a single quantity, called the free energy, which is related to enthalpy (H) and entropy (S) by equation 2.

Equation 2

G=H -TS                                                             

Processes for which G <0 are spontaneous; those for which G> 0 are not.

It is useful for chemists to tabulate the standard free energies of formation, Gf⁰, for many chemical substances. AGf⁰ is the change in free energy which accompanies the formation (subscript t) of one mole of a substance in its standard state (superscript 0) from its elements in their standard states. For example, the formation of one mole of CaCO₃(s) from its elements is shown below:

Ca(s) + C(s) + 3/2 0₂(g) CaCO₃(s)

The free energy change which accompanies this process is, by definition, the standard free energy of formation of CaCO₃. G⁰ for any chemical reaction may be calculated from the Gf⁰ values of reactants and products using equation 3. 

Equation 3

G⁰ for reaction = ∑G_f⁰(products) - G_f⁰(reactants)     

To calculate the free energy change for a chemical reaction under conditions other than standard,  correct G⁰ for the non-standard conditions using equation 4.

Equation 4

G=G⁰ + RTInQ                                                                

Here R is the gas constant (8.312 x 10^-3 kJ/mole-K), T is the Kelvin temperature, and Q is the mass action quotient.

Example:

Predict whether the following chemical reaction is spontaneous: 

Ag⁺(aq) + Cl^-(aq) -> AgCl(s)       

We can readily calculate G for any net ionic equation from ion concentrations and Gf⁰ values for reactants and products. Table 1 gives standard free energies of formation, Gf⁰, for 1 M solutions of several cations (first and second columns) and anions (third and fourth columns). The fifth and sixth columns show Gf⁰ for several solid salts formed from combinations of the listed cations and anions. Note that two of the solids crystallize with waters of hydration (e.g., CaSO₄*2H₂0). Equations for formation of these solids must include the appropriate number of water molecules on the left side. Gf⁰ for water is -237.18 kJ/mole.

Table 1

Gf⁰ Values in kJ/mol for Cations, Anions, and Solids

(Data from the CRC Handbook of Chemistry and Physics, 63rd Edition, CRC Press, Inc., Boca Raton, FL, 1982)

Cations	Gf 0 (kJ/mol)	Anions	Gf 0 (kJ/mol)	Selected Solids	Gf 0 (kJ/mol)
Ag+	77.12	Cl-	-131.26	AgCl	-109.80
Ba2+	-560.74	I-	-51.59	Ag2SO4	-618.48
Ca2+	-553.54	SO42-	-744.63	AgI	-66.19
Pb2+	-24.39	Cr042-	-727.85	BaCrO4	-1345.3
A13+	-485.34	0H-	-157.29	CaCI2	-748.1
				Pb(OH)2	-452.29
				BaCl2*2H2O	-1296.5
				Al2(S04)3	-3100.1
				PbSO4	-813.2
				CaSO4*2H2O	-1797.4

 

Using data from the table, we calculate G⁰ as follows:

G⁰ = Gf⁰(AgCl(s)) - Gf⁰(Ag⁺(aq)) - Gf⁰(Cl^-(aq)) 

= -109.789 - 77.107 - (-131.228) kJ/mole

= -55.668 kJ/mole

This number is substantially negative. Formation of solid AgCl(s) from its ions under standard conditions (i.e., [Ag⁺] = [Cl^-] = 1 M) is spontaneous-it will occur of its own accord. If ion concentrations were 0.1 M rather than 1 M, would precipitation still occur? Use equation 4.

G=G⁰+RTInQ

G = G⁰ + RT In 1/[Ag⁺] [Cl^-]

= -55.668 + (8.314 x 10^-3)(298) ln [1/(0.1)(0.1)] 

= -44.250 kJ/mole

Precipitation is still spontaneous, but somewhat less so than at 1 M ion concentrations.

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