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Chemistry 113.1

Introduction to Chemical Techniques

Experiment 3. Precipitation reactions (May 2012)

I. INTRODUCTION

In Experiment 3, you applied heat from a Bunsen burner to decompose a hydrate into an anhydrous salt and gaseous water.  A decomposition reaction is one of four broader categories of chemical reactions.   The remaining categories are precipitation reactions, acid/base reactions, and  oxidation/reduction  reactions.     In  this  experiment,  you  will  investigate  precipitation reactions.  In a precipitation reaction, two aqueous solutions of soluble salts are mixed and yield an aqueous solution of a soluble salt and a solid compound.  The formation of the solid is called precipitation and the solid is called the precipitant.

When ionic compounds dissolve in water, the water interacts with the cation and anion to weaken the Coulombic interaction holding the two ions together as a solid.   Thus, as the ions break  apart  because  of water  surrounding  the  individual  ions  in  the  compound,  the  solid

dissolves.  For example, when CuSO4 dissolves in water, the chemical reaction equation is

CuSO4 (s)  →  Cu2+ (aq)  +  SO42- (aq)  .

Again, the (s) stands for solid, while the (aq) stands for aqueous.   Precipitation occurs when aqueous cations and anions form Coulombic interactions that are strong enough to overcome the interaction of the water molecules with the separate ions in solution.  In other words, the solid is more stable than an aqueous solution containing the two aqueous ions. Compounds that do not dissolve in water are called insoluble, while those that do are called soluble.   Table 5.1 gives some basic rules for the solubility of ionic salts.  These rules should be memorized, although in Chemistry 114, you will learn that these rules are not as black and white as they are presented here.

Table 5.1. Rules for determining the solubility of ionic compounds

1. Compounds of the alkali metal ions (i.e., the column to the far left on the periodic table) are soluble.

2. Compounds containing ammonium ion are soluble.

3. Nitrates, chlorates, perchlorates, and acetates are soluble.

4. Chlorides, bromides, iodides are soluble except when combined with lead(II), silver(I) and mercury(I) cations. Mercury(II) iodide is also insoluble.

5. All sulfates are soluble except when combined with strontium, barium, calcium, lead(II), mercury(I), and silver cations. Small amounts of calcium, silver, and mercury(I) sulfates will dissolve in solution (i.e., slightly soluble).

6. Carbonates, phosphates, oxalates, and chromates are insoluble unless they fall under the categories 1 and 2 (i.e., alkali metal or ammonium ion salts).

7. Sulfides are insoluble unless they fall under categories  1 and 2. Alkaline earth metals  (i.e., calcium, strontium, and barium) form slightly soluble sulfides.

8. Hydroxides and oxides are insoluble except for those that fall under categories 1 and 2. Alkaline earth metals form slightly soluble hydroxide and oxide salts.

In a precipitation reaction, two solutions containing soluble ionic salts are mixed.  However, some of the ions, when the solutions are combined, can interact to form insoluble salts.   When this occurs, a solid forms (i.e., the precipitant) and produces a cloudy solution and often falls to the bottom of the container.  For example,

Solution A:  Aqueous copper sulfate:  CuSO4 (aq) or  Cu2+ (aq) + SO42- (aq)

Solution B:  Aqueous sodium carbonate: Na2CO3 (aq) or 2 Na+ (aq) + CO32- (aq)

Notice that we can represent aqueous solutions (aq) in two ways, namely one with the ionic compound formula (aq) and one with the individual ions (aq).   Both ways are equally valid. Also notice that when sodium carbonate goes into solution, we obtain one mole of carbonate anion for every two moles of sodium cations.   Bulk solutions can have no net charge.   Thus, when you sum the charges for all ions in solution, you must obtain zero.  The integer number to the left of the sodium cation insures that the solution remains at zero net (or total) charge.  Then,

when the two solutions are mixed, we have the following chemical reaction equation:

Cu2+ (aq) + SO42- (aq) + 2 Na+ (aq) + CO32- (aq)  →  CuCO3 (s) + 2 Na+ (aq) + SO42- (aq)

The equation written above is called an ionic reaction equation.  Notice that the sodium cation and the sulfate anion appear on both sides of the equation.   This is because he combination Na2 SO4  is soluble. These ions are called spectator ions because they do not participate in the

chemical reaction that yields the solid copper(II) carbonate.  Thus, the net ionic reaction is

Cu2+ (aq) + CO32- (aq)  →  CuCO3 (s)

Notice how both of these reaction equations have the same number of each atom on the right hand and left hand side of the equation and that both of these equations also have the same net charge on both sides of the equation.   Insuring that the charge and mass on both sides of a chemical reaction equation is the same are both part of balancing a chemical reaction equation.

II. EXPERIMENT

1.   Given the following aqueous solutions, predict which mixtures of any two of these solutions would yield precipitants and determine the formula for the precipitant:

a. Copper(II) sulfate                     b. Barium nitrate

c. Sodium chloride                       d. Silver nitrate

e. Lead nitrate                               f. Sodium sulfate

2.   Show your predictions to the instructor.

3.   Test to see if your predictions where correct by placing a small amount of the appropriate solution into a clean test tube and adding the second solution.  You must test all of the reactions that you predict would form a precipitant.  When the precipitant forms, record   the color of the precipitant.  You must also test at least two reactions that, based on Table 5. 1, you predict would not form a precipitant.

III.  Post-laboratory discussion and questions

You must always balance chemical reactions before using these reactions to predict the amount of product one might expect from the reaction (i.e., the theoretical yield of the reaction). In this experiment, one has probed solubility and, therefore, must now learn how to write a chemical reaction equation and balance it.   When balancing a chemical reaction equation, the mole ratio of various compounds in the reaction can be adjusted to balance both mass and charge for the reaction.  However, the formula of a compound or a polyatomic ion does not change.  Therefore, do not change atomic symbols or subscripts in chemical formulas while attempting to balance reactions.  Some basic guidelines are

1.   Balance atoms other than H and O first

2.   Pure Elements [e.g.   Fe (s) or Cl2 (g) ] should be balanced last

3.   Balance as a unit any polyatomic ions that appear unchanged on both sides of the arrow.

As an example, let us balance the reaction of aqueous copper  sulfate with aqueous  sodium hydroxide.  The first step is to write the ionic reaction.  Thus,

CuSO4 (aq) + NaOH (aq) → products

To determine the products, we need to look at the solubility rules in Table 5.1.  Rule 8 states that hydroxides are insoluble unless that are formed with alkali metal cations or with ammonium. Since copper is not an alkali metal, this tells us that copper hydroxide will be insoluble in water. Thus, one of the products will be copper(II) hydroxide.  We also know from Rule 5 that almost all sulfate salts are soluble.  Thus, sodium sulfate will not form an insoluble salt.  Therefore, the sodium ion and the sulfate ion are spectator ions in this reaction and can be removed from both sides  of the  equation.   Using this  information, we  can now write the unbalanced net  ionic reaction

Cu2+ (aq) + OH- (aq) → Cu(OH)2 (s)

To balance the reaction, we can see that we have one mole of Cu2+   on the left side of the equation and one mole of Cu2+  on the right side of the equation.  However, we only have one mole of OH-  on the left side of the equation and two moles on the right side of the equation. Therefore, we need to place a 2 in front of the OH-  on the left side of the equation.  Once we have done so, we obtain the balanced reaction (check the net charge on both sides)

Cu2+ (aq) + 2 OH- (aq) → Cu(OH)2 (s)

What happens if one mixes  aqueous  sodium  sulfate with  aqueous potassium nitrate.   Since potassium sulfate is soluble (Rule 5) and sodium nitrate is soluble (Rule 3), no reaction occurs! Thus, one would write  K2 SO4 (aq) + NaNO3 (aq) → no reaction.

POST-LABORATORY QUESTIONS

1. Write the balanced chemical reaction equations for all reactions that yielded a precipitate.

2. Did the color of the solution and the color of the precipitate match? Speculate why or why not.   (This is a very complicated question. We are not expecting four pages to answer this question after extensive research. We are asking that you think about what might cause a color difference.)

3. Did any of the solutions that you expected should not have yielded a precipitate actually yield a precipitate? If so, state why you think this may have happened ?