Solubility
Dissolving and Electrolytes
When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. Water and other polar molecules are attracted to ions, as shown in Figure 1. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water.
Substances that produce ions in solutions when they dissolve constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called non-electrolytes. If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte. If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte.
Substances may be identified as strong, weak, or non-electrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit.
Demonstration: Electrolytes and electrical conductivity
Set up. The movie below demonstrates the measurement of conductivity using two graphite electrodes. If a solution is conductive, the light bulb will light up. The electrodes are initially tested using a conducting piece of metal, and then first placed in water and then in NaCl(aq).
Prediction. Before watching the video, make a prediction about whether the bulb will light up in water or a solution of NaCl(aq).
Explanation. There is no significant concentration of ions (charge carriers) in water as it consists of mainly neutral H2O molecules. The lamp does not light. However, the presence of freely mobile ions Na+(aq) and Cl–(aq) in NaCl(aq), a strong electrolyte, results in a conducting solution.
Set up. The next movie demonstrates the conductivity of a sucrose solution.
Prediction. Before watching the video, make a prediction about whether the bulb will light up in a sucrose solution.
Explanation. The electrode set up is once again tested to make sure that it is functioning correctly. Sucrose is a non-electrolyte and although the molecules of sucrose separate from each other when the solid dissolves in water, the result is mobile neutral molecules and no ions. The solution does not conduct electricity.
Set up. The next movie demonstrates the conductivity of liquid ethanol.
Prediction. Before watching the video, make a prediction about whether the bulb will light up in ethanol.
Explanation. The electrode set up is once again tested to make sure that it is functioning correctly. Ethanol is a non-electrolyte and the neutral molecules do not conduct electricity.
Set up. The fourth movie in this series demonstrates the conductivity of weak electrolytes.
Prediction. Before watching the video, make a prediction about whether the bulb will light up/how brightly the bulb will light up in the weak electrolyte solutions.
Explanation. This movie shows the conductivity of weak versus strong acids which is relevant to this discussion of electrolytes. The first solution tested is HCl(aq). HCl(g) is a molecule but when dissolved in water clearly produces many ions in solution as evidenced by the brightness of the bulb and is classified as a strong electrolyte. The second solution tested is acetic acid which also produces some ions in solution, although not many as evidenced by the dim glow of the bulb filament. Acetic acid is classified as a weak electrolyte. The third solution is the strong electrolyte NaOH(aq) and the fourth solution is another example of a weak electrolyte, NH3(g) dissolved in water to produce “aqueous ammonia”.
Set up. Another way to produce mobile ions that conduct electricity is to use a molten salt, as shown in the following video. A molten salt is a salt that has been heated to its melting point.
Explanation. The setup is once again tested to make sure that it is functioning properly. When the salt is a solid, the light bulb does not light because the ions are stationary. Once the salt has been heated so that it is now in its liquid state, the mobile ions allow the current to flow and the bulb to light.
The Formation of Solutions
The amount of a particular ionic compound that will dissolve in a given quantity of a solvent must be determined experimentally. One can dissolve increasing amounts of a compound in a fixed volume of solvent at a specific temperature until no more will dissolve and a saturated solution is formed. However, even a qualitative prediction of the relative solubility of ionic compounds is challenging as there are many factors involved.
Dissolving an ionic compound is a more complex process than melting the compound, although the application of Coulomb’s Law is important for understanding both of these physical processes. Melting can be considered as breaking apart the ionic lattice to produce freely mobile ions. This lattice energy is described by Coulomb’s law, U = constant × 
. Breaking bonds always require an input of energy and is therefore an endothermic process (energy absorbed). Hence, melting of any substance is always an endothermic process. Ionic compounds with stronger bonds (compounds that contain small, highly charged ions) require a greater amount of energy to separate the anions and cations and therefore have higher melting points.
Like melting, dissolving involves breaking apart the ionic lattice. However, it also involves an energy-releasing process as solvent molecules surround and interact with the anions and cations, a process known as solvation, or hydration when water is the solvent, as shown in Figure 1. Whenever energy is released during a chemical or physical change, the process is described as an exothermic process (energy released). The energy of attraction between cations (or anions) and water/sovlent is called the solvation energy.
For an ionic substance to dissolve, the anion-cation attractions must be disrupted while new water-ion attractions form. The lattice energy of the ionic compound is often comparable to the solvation energy. Therefore, dissolving is usually either a slightly endothermic or slightly exothermic process overall as the solvation (exothermic) and lattice bond breaking (endothermic) steps almost cancel each other out.
Why do some ionic compounds dissolve in water while others do not? The solubility partly depends on a comparison of the lattice energy and the solvation energy. Ionic compounds that dissolve exothermically are generally quite soluble. On the other hand, for many insoluble salts, lattice energy in the solid salt is greater than the solvation energy in the solution.
However, ionic compounds that dissolve endothermically are not always insoluble. This is because when a relatively ordered crystalline ionic structure becomes part of a much more mobile aqueous solution, there is generally an increase in entropy. This change in entropy is a very powerful driving force for physical (and chemical) processes. Whenever a substance dissolves, either exothermically or endothermically, there will be an accompanied change in entropy.
Let us consider what happens at the microscopic level when we add solid KCl to water, as illustrated in Figure 1. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) end of the polar water molecule to the positive potassium ions. The water molecules penetrate between K+ and Cl− ions and surround them, reducing the strong Coulombic forces that bind the ions together while moving them into solution as solvated ions. As the ions change from their fixed and ordered positions in the crystal to independently mobile hydrated ions in a dilute solution, there is an increase in the entropy of the system. It is this entropy increase that is responsible for the dissolution of many ionic compounds, including KCl, which dissolve with absorption of heat.
In other cases, the electrostatic attractions between the ions in a crystal are so large, or the ion-dipole attractive forces between the ions and water molecules are so weak, that the increase in entropy cannot compensate for the energy required to separate the ions, and the crystal is insoluble. Such is the case for compounds such as calcium carbonate (limestone), calcium phosphate (the inorganic component of bone), and iron oxide (rust).
Overall, Complete Ionic and Net Ionic Equations
Given the abundance of water on earth, it stands to reason that a great many chemical reactions take place in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of detail appropriate to their intended use. To illustrate this, consider a reaction between ionic compounds taking place in an aqueous solution. When aqueous solutions of CaCl2 and AgNO3 are mixed, a reaction takes place producing aqueous Ca(NO3)2 and solid AgCl:
CaCl2(aq) + 2 AgNO3(aq) ⟶ Ca(NO3)2(aq) + 2 AgCl(s)
This balanced equation, derived in the usual fashion, is called an overall equation because it doesn’t explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution. Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions. Unlike CaCl2, AgNO3 and Ca(NO3)2, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, s.
Explicitly representing all dissolved ions results in a complete ionic equation. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:
Ca2+(aq) + 2 Cl–(aq) + 2 Ag+(aq) + 2 NO3–(aq) ⟶ Ca2+(aq) + 2 NO3–(aq) + 2 AgCl(s)
Notice that only the compounds that were represented with “(aq)” dissociated in the complete ionic equation. AgCl(s) remained unchanged since it does not dissociate in solution. Also notice that CaCl2, for example, dissociates into 1 Ca2+ ion and 2 different Cl– ions. The subscript “2” represented that there were 2 Cl atoms for every 1 Ca atom, so when the compound dissociates, 2 Cl– ions are formed.
Examining the above equation shows that two chemical species are present in identical form on both sides of the arrow, Ca2+(aq) and NO3−(aq). These spectator ions—ions whose presence is required to maintain charge neutrality—are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a net ionic equation:
2 Cl–(aq) + 2 Ag+(aq) ⟶ 2 AgCl(s)
Following the convention of using the smallest possible integers as coefficients, this equation is then written:
Cl–(aq) + Ag+(aq) ⟶ AgCl(s)
This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These overall and complete ionic equations provide additional information, namely, the ionic compounds used as sources of Cl− and Ag+.
A vivid example of precipitation is observed when solutions of potassium iodide and lead(II) nitrate are mixed, resulting in the formation of solid lead(II) iodide:
KI(aq) + Pb(NO3)2(aq) ⟶ PbI2(s) + 2 KNO3(aq)
Lead(II) iodide is a bright yellow solid that was formerly used as an artist’s pigment known as iodine yellow. The properties of pure PbI2 crystals make them useful for fabrication of X-ray and gamma ray detectors.
Demonstration: Precipitation of lead(II) iodide
Set up. The following video shows the reaction of potassium iodide with lead(II) nitrate. The potassium iodide is in a beaker and the lead(II) nitrate is in the buret. As the lead(II) nitrate is added to the beaker, the reaction will proceed.
Explanation. This video shows the formation of lead(II) iodide as the reaction progresses. Lead(II) iodide is the bright yellow precipitate that forms upon the addition of lead(II) nitrate.
Example 1
When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write balanced overall, complete ionic, and net ionic equations for this process.
Solution
Begin by identifying formulas for the reactants and products and arranging them properly in chemical equation form:
CO2(aq) + NaOH(aq) ⟶ Na2CO3(aq) + H2O(ℓ) (unbalanced)
Balance is achieved easily in this case by changing the coefficient for NaOH to 2, resulting in the overall equation for this reaction:
CO2(aq) + 2 NaOH(aq) ⟶ Na2CO3(aq) + H2O(ℓ)
The two dissolved ionic compounds, NaOH and Na2CO3, can be represented as dissociated ions to yield the complete ionic equation:
CO2(aq) + 2 Na+(aq) + 2 OH–(aq) ⟶ 2 Na+(aq) + CO32-(aq) + H2O(ℓ)
Finally, identify the spectator ion(s), in this case Na+(aq), and remove it from each side of the equation to generate the net ionic equation:
CO2(aq) + 2 OH–(aq) ⟶ CO32-(aq) + H2O(ℓ)
Check Your Learning
Diatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:
NaCl(aq) + H2O(ℓ) ⟶ NaOH(aq) + H2(g) + Cl2(g)
Write balanced overall, complete ionic, and net ionic equations for this process.
Answer
balanced overall: 2 NaCl(aq) + 2 H2O(ℓ) ⟶ 2 NaOH(aq) + H2(g) + Cl2(g)
complete ionic: 2 Na+(aq) + 2 Cl–(aq) + 2 H2O(ℓ) ⟶ 2 Na+(aq) + 2 OH–(aq) + H2(g) + Cl2(g)
net ionic: 2 Cl–(aq) + 2 H2O(ℓ) ⟶ 2 OH–(aq) + H2(g) + Cl2(g)
