Precipitation Reactions 


You must learn the rules that describe the solubility of some common salts in water; understand the definition of a saturated solution and the process of recrystallization; and be able to write balanced equations for precipitation reactions. Imagine taking 10 g of a pure solid compound and stirring it into 100 g of water. One of three things can happen (assuming the compound doesn't react with water, of course): either all of the solid dissolves; or some of the solid dissolves; or none of the solid dissolves. In the first case we say the solid is soluble in water; if the solid dissolves quickly and easily, we say it is readily soluble; if we can go on dissolving another 10 g, and another, in the same 100 g of water we can say the compound is very soluble. In the second case, we can say the compound is moderately or slightly soluble, depending on how much of the original 10 g appears to dissolve. In the third case we say the compound is insoluble in water. Solubility is a physical property of a pure substance. Many observations over time have led to some rules (generalizations) about the solubility of certain salts. Using these rules, we can predict when a particular salt is likely to be soluble in water, and if we have an unidentified compound we can design experiments to find out what it is.
Remember to apply rule 1 first. Only consider the anion if the cation is not a group 1A metal or ammonium. The descriptions given above are qualitative. One way to describe solubility quantitatively is to measure the mass of salt that dissolves in 100 g of water at a particular temperature. For example, potassium nitrate has a solubility of 31.2 g per 100g of water at 20 ^{o}C, calcium hydroxide has a solubility of 0.17 g per 100g, and calcium carbonate has a solubility of 0.0013 g per 100g. Notice that even calcium carbonate, a salt that we would describe as insoluble, does have a very slight solubility in water. Solubility is dependent on temperature. The solubility of most, but not all, salts increases as the temperature increases. Imagine dissolving as much of a salt as possible in hot water, and then allowing the solution to cool. A solution that contains the maximum amount of a dissolved solute is said to be saturated. As it cools, the solution can no longer hold as much salt, so the excess amount separates as a solid. If the solution is cooled slowly enough the solid separates as crystals. The process of recrystallization is used to purify substances, because the ordered arrangement of ions in the crystal lattice disfavors the inclusion of impurities which might have been present in the original solid. The impurities stay in solution when the crystals are collected by filtration. If solutions of two different salts are mixed, then there may be two different cations and two different anions in the solution. If one combination of cation and anion corresponds to an insoluble salt, then this salt will precipitate (separate as a solid) as soon as the original solutions are mixed. We can represent such a chemical reaction using ionic equations just as we did for neutralization reactions. Consider for example what happens when a solution of lithium bromide is mixed with silver nitrate.
Equations 1 and 2 represent the formation of the original solutions of lithium bromide and silver nitrate. When these solutions are mixed silver ions and bromide ions come together to form the insoluble salt, silver bromide, which precipitates from the solution. The remaining ions, lithium cation and nitrate, stay in solution. Equation 3 is the total ionic equation, and after canceling out the spectator ions we get equation 4, the net ionic equation . Equation 5 represents the overall chemical reaction. Precipitation reactions are classified as double replacement reactions (as are neutralization reactions). In effect, the reactants exchange anions to give the products. To write an overall equation for a precipitation reaction we must recognize when exchange of anions results in an insoluble salt being formed, then write an equation in terms of the formulas for the compounds involved. Finally, the equation is balanced by inspection, making sure that there are the same number of each type of ion on both sides of the equation. Let's write, for example, a balanced equation for the reaction of potassium phosphate with magnesium chloride. We recognize that the products of a double replacement reaction will be potassium chloride and magnesium phosphate (exchanging the anions!) and that magnesium phosphate is insoluble (applying solubility rules!). The formula for magnesium phosphate must be Mg_{3}(PO_{4})_{2} (three Mg^{2+} cations and two PO_{4}^{3} anions!), so we can write an equation in terms of formulas:
Starting with magnesium, we need a coefficient 3 before the formula for magnesium chloride so that we will have three magnesium atoms on each side of the equation. That gives us 6 chlorine atoms so we need a coefficient 6 before the formula for potassium chloride; that leads us to put a 2 before the formula for potassium phosphate, and checking phosphate and rechecking the other ions confirms that the equation is now balanced:
Depending on which ion you choose to start with you may have to cycle through the process, changing coefficients until the equation is balanced. If you do this, make sure your final coefficients don't have a common factor. If they do, divide through so that your coefficients are the smallest possible set of integers. 


