Experiment 8 | Slides Medicine

Experiment 8 - Double Displacement Reactions

A double displacement reaction involves two ionic compounds that are dissolved

in water. In a double displacement reaction, it appears as though the ions are “trading

places,” as in the following hypothetical reaction:

AB (aq) + CD (aq) → AD + CB

Where AB exists as A+ and B- ions in solution and CD exists as C+ and D- ions in

solution. As the ions come in contact with each other, there are six possible combinations

that might conceivably cause a chemical reaction. Two of these combinations are the

meeting of ions of like charge; that is, A++ C+ and B-+ D-. But since like charges repel, no

reaction will occur with these combinations. Two other possible combinations are those

of the original two compounds; that is, A++ B- and C++ D-. Since we originally had a

solution containing each of these pairs of ions, they can mutually exist in the same

solution; therefore they do not recombine. Thus the two possibilities for chemical

reaction are the combination of each of the positive ions with the negative ion of the other

compound; that is, A++ D- and C++ B-.

In summary, when the solutions are mixed, these ions can all come into contact

with each other, and new products could be formed. If new products are to be formed,

there is only one possible combination of products: since like charges repel each other,

we cannot have new compounds containing two negative ions or two positive ions. The

only other possible new combination comes from the positive and negative ions of the

two compounds switching places.

There are three types of equations that can be written for reactions that involve

ions in solution. The first type is just the regular, overall (often called “molecular”)

equation. An example of an overall equation for a double displacement reaction is:

Na2CO3 (aq) + 2 AgNO3 (aq) → 2 NaNO3 (aq) + Ag2CO3 (s)

In this type of equation, the complete formulas are shown along with the

appropriate state symbols. The formulas of the products are obtained as follows:

1. Determine what ions are present in the reactants, without worrying about how

many of each there are. Include their charges: Na+

(aq), CO3

(aq), Ag+

(aq), and NO3

(aq).

2. Switch the ions so that the cation of the first compound is paired with the anion of

the second compound, and vice versa. In this case, the Na+

(aq is paired with the

NO3

(aq), and the Ag+

(aq) is paired with the CO3

(aq).

3. Determine the correct formulas of the products, keeping in mind the charges on

each of the ions and remembering that the overall formula of an ionic compound

must have no net charge. In this case, since sodium ion has a charge of +1 and

nitrate ion has a charge of –1, the correct formula of the compound sodium nitrate

contains one of each ion: NaNO3. For the other compound, silver ion has a charge

of +1 and carbonate ion has a charge of –2. Therefore, two silver ions are needed

to balance out the charge on one carbonate ion, and the formula of the compound

is Ag2CO3.

4. Write the formulas of the products, and check the solubility rules to determine the

appropriate state symbol to include next to the products. In this case, according to

the solubility rules, sodium nitrate is soluble, but silver carbonate is insoluble.

5. Balance the equation.

Another type of chemical equation, the complete ionic equation, shows all

substances in the reaction in their predominant form in solution. Since soluble ionic

compounds consist of separated ions in solution, they are shown in the equation as

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Experiment 8 - Double Displacement Reactions

A double displacement reaction involves two ionic compounds that are dissolved in water. In a double displacement reaction, it appears as though the ions are “trading places,” as in the following hypothetical reaction: AB (^) (aq) + CD (^) (aq) → AD + CB Where AB exists as A+^ and B-^ ions in solution and CD exists as C+^ and D-^ ions in solution. As the ions come in contact with each other, there are six possible combinations that might conceivably cause a chemical reaction. Two of these combinations are the meeting of ions of like charge; that is, A++ C+^ and B-+ D-. But since like charges repel, no reaction will occur with these combinations. Two other possible combinations are those of the original two compounds; that is, A++ B-^ and C++ D-. Since we originally had a solution containing each of these pairs of ions, they can mutually exist in the same solution; therefore they do not recombine. Thus the two possibilities for chemical reaction are the combination of each of the positive ions with the negative ion of the other compound; that is, A++ D-^ and C++ B-. In summary, when the solutions are mixed, these ions can all come into contact with each other, and new products could be formed. If new products are to be formed, there is only one possible combination of products: since like charges repel each other, we cannot have new compounds containing two negative ions or two positive ions. The only other possible new combination comes from the positive and negative ions of the two compounds switching places. There are three types of equations that can be written for reactions that involve ions in solution. The first type is just the regular, overall (often called “molecular”) equation. An example of an overall equation for a double displacement reaction is: Na 2 CO 3 (aq) + 2 AgNO 3 (aq) → 2 NaNO 3 (aq) + Ag 2 CO 3 (s) In this type of equation, the complete formulas are shown along with the appropriate state symbols. The formulas of the products are obtained as follows:

Determine what ions are present in the reactants, without worrying about how many of each there are. Include their charges: Na+ (aq), CO 32 - (aq), Ag+ (aq), and NO 3 - (aq).
Switch the ions so that the cation of the first compound is paired with the anion of the second compound, and vice versa. In this case, the Na+ (aq is paired with the NO 3 - (aq), and the Ag+ (aq) is paired with the CO 32 - (aq).
Determine the correct formulas of the products, keeping in mind the charges on each of the ions and remembering that the overall formula of an ionic compound must have no net charge. In this case, since sodium ion has a charge of +1 and nitrate ion has a charge of – 1, the correct formula of the compound sodium nitrate contains one of each ion: NaNO 3. For the other compound, silver ion has a charge of +1 and carbonate ion has a charge of – 2. Therefore, two silver ions are needed to balance out the charge on one carbonate ion, and the formula of the compound is Ag 2 CO 3.
Write the formulas of the products, and check the solubility rules to determine the appropriate state symbol to include next to the products. In this case, according to the solubility rules, sodium nitrate is soluble, but silver carbonate is insoluble.
Balance the equation. Another type of chemical equation, the complete ionic equation , shows all substances in the reaction in their predominant form in solution. Since soluble ionic compounds consist of separated ions in solution, they are shown in the equation as

separated ions. Likewise, any substances which exist mostly as molecules are shown as molecules in the equation. Insoluble ionic compounds are not shown ionized, since their ions are not actually separated in the solution. The complete ionic equation for the above reaction would be: 2 Na+ (aq) + CO 32 - (aq) + 2 Ag+ (aq) + 2 NO 3 - (aq) → 2 Na+ (aq) + 2 NO 3 - (aq) + Ag 2 CO3(s) A third type of equation is called a net ionic equation. In a net ionic equation, only those species (ions or molecules) that actually change are shown. All of the spectator ions (those ions that are present in the solution but which are not reacting) are not written. In the above reaction, the spectator ions are sodium ion and nitrate ion: they appear on both sides of the reaction unchanged. The net ionic equation for the above reaction is therefore: CO 32 - (aq) + 2 Ag+ (aq) → Ag 2 CO3(s) A net ionic equation is not really a complete equation, since it does not give complete formulas; it is nevertheless quite useful, since it focuses attention on the main event. In the present case, the equation says the Ag+^ ion (from an unspecified source) combines with a CO 32 -^ ion (also from an unspecified source) to form a precipitate of Ag 2 CO 3. This will happen any time these two ions are put together in the same solution. In each part of this experiment two aqueous solutions, each containing positive and negative ions, will be mixed in a test tube. You will observe whether or not a reaction occurs in each case, and predict the products. You will also practice writing different types of equations for reactions. Let us look at some examples. Example 1. When solutions of sodium chloride and silver nitrate are mixed, the equation for the hypothetical double displacement reaction is: NaCl + AgNO 3 → NaNO 3 + AgCl (The formulas of the products are obtained by switching the ions.) A white precipitate is produced when these solutions are mixed. This precipitate is definite evidence of a chemical reaction. One of the two products, sodium nitrate (NaNO 3 ) or silver chloride (AgCl) is insoluble. Although the precipitate could be identified by further chemical testing, we can instead look at the solubility table to find that sodium nitrate is soluble but silver chloride is insoluble. We may then conclude that the precipitate is silver chloride and indicate this in the equation with the state symbol (s), which stands for “solid”. Thus the overall equation should read: NaCl (^) (aq) + AgNO 3 (aq) → NaNO 3 (aq) + AgCl (^) (s) The complete ionic equation for the above reaction would be: Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO 3 - (aq) → Na+ (aq) + NO 3 - (aq) + AgCl(s) The net ionic equation for the above reaction (canceling out the spectator ions sodium ion and nitrate ion) is: Cl- (aq) + Ag+ (aq) → AgCl(s) Example 2. When solutions of sodium chloride and potassium nitrate are mixed, the equation for the hypothetical double displacement reaction is NaCl + KNO 3 → KCl + NaNO 3 (The formulas of the products are obtained by switching the ions.) We get the hypothetical products by simply combining each positive ion with the other negative ion. But has there been a reaction? When we do the experiment we see no evidence of reaction. There is no precipitate formed, no gas evolved, and no obvious temperature

H+ (aq) + OH- (aq) → H 2 O(l) Water is the most common slightly ionized substance formed in double displacement reactions; other examples are acetic acid (HC 2 H 3 O 2 ) and phosphoric acid (H 3 PO 4 ). (Any weak acid is slightly ionized.) From the four examples cited we see that a double displacement reaction will occur if at least one of the following classes of substances is formed by the reaction:

A precipitate.
A gas.
A slightly ionized compound (a weak electrolyte).
A non-ionized compound (a nonelectrolyte). Useful information: there are very few inorganic strong acids. They are: HCl, HBr, HI, HNO 3 , HClO 3 , HClO 4 , and H 2 SO 4 in its first ionization. It is helpful to commit these to memory, and assume that any other acids that you encounter are weak. The following solubility rules are helpful in predicting precipitates that might form when solutions are mixed. Solubility Rules for Ionic Compounds at 25°C
A compound will be soluble if it contains at least one of the following ions: Li+, Na+, K+, NH 4 +, NO 3 - , or C 2 H 3 O 2 -.
A compound containing Cl-, Br-, or I-^ is soluble unless the cation is Ag+, Pb2+, or Hg 2 2+.
Compounds that contain SO 42 -^ are soluble except those sulfates that also contain Ba2+, Ag+, Pb2+, or Ca2+.
Most other ionic compounds are insoluble (in other words, the amount of solute that dissolves is extremely small and will be regarded as negligible). Compounds containing S^2 - , CO 32 - , PO 43 - , and OH-^ are insoluble unless the cation is one of those listed in #1 above. Safety Precautions:

Wear safety goggles.
Silver nitrate solutions will stain skin and clothing. If you notice any silver nitrate splashing your skin or clothing, rinse it off immediately. Silver nitrate is a colorless solution that is light-sensitive. The stains don’t show up immediately, but you will see them by the next day! Stains on your skin will eventually wear off as your skin wears off; stains on your clothing will be permanent. It is therefore best to wear old clothing for this lab.
Part 3 involves testing the electrical conductivity of several solutions. The electrical conductivity meter operates on 120-volt “house” current. To prevent electrical shock, the exposed wires of the conductivity tester are covered with a non-conducting material, but it is still possible to touch the wires accidentally. Therefore, to avoid possible severe electrical shock, make sure that you are instructed in the use of the conductivity tester in advance, and exercise caution when you use it. Slowly bring the beaker containing the solution to be tested up to the wires from below. Hold the

beaker near the bottom so that your hand stays as far from the wires as possible. To be on the safe side, do not touch any part of the tester. Waste Disposal:

While you are doing the experiment, pour your waste into a beaker. When you are finished with the experiment, pour the contents of the waste beaker into the INORGANIC WASTE container (with a blue label) in the fume hood.

Procedure

Part 1 – Observing Double Displacement Reactions Each reaction in this part of the experiment (except number 12) consists of mixing equal volumes of two solutions in a test tube. Make sure the test tubes are clean and have been rinsed with deionized water and shaken dry. Use about one milliliter of each solution. It is not necessary to measure each volume accurately (why?). Record your observations at the time of mixing. Where there is no visible evidence of reaction, feel each tube, or check with a thermometer, to determine if heat is evolved. In each case where a reaction has occurred, complete and balance the overall equation, properly indicating precipitates and gases, and write the net ionic equation for the reaction. When there is no evidence of reaction, write the words “no reaction” as the right-hand side of the equation.

Mix 0.1 M sodium chloride (NaCl) and 0.1 M potassium nitrate (KNO 3 ) solutions.
Mix 0.1 M sodium chloride (NaCl) and 0.1 M silver nitrate (AgNO 3 ) solutions.
Mix 0.1 M sodium carbonate (Na 2 CO 3 ) and 6 M hydrochloric acid (HCl) solutions.
Mix 10% sodium hydroxide (NaOH) and 6 M hydrochloric acid (HCl) solutions.
Mix 0.1 M barium chloride (BaCl 2 ) and 3 M sulfuric acid (H 2 SO 4 ) solutions.
Mix 6 M ammonia (NH 3 ) and 3 M sulfuric acid (H 2 SO 4 ) solutions in the hood.
Mix 0.1 M copper (II) sulfate (CuSO 4 ) and 0.1 M zinc nitrate (Zn(NO 3 ) 2 ) solutions.
Mix 0.1 M sodium carbonate (Na 2 CO 3 ) and 0.1 M calcium chloride (CaCl 2 ) solutions.
Mix 0.1 M copper (II) sulfate (CuSO 4 ) and 0.1 M ammonium chloride (NH 4 Cl) solutions.
Mix 10% sodium hydroxide (NaOH) and 6 M nitric acid (HNO 3 ) solutions.
Mix 0.1 M iron (III) chloride (FeCl 3 ) and 10% sodium hydroxide (NaOH) solutions.
Add 1 g of solid sodium bicarbonate (NaHCO 3 ) to 3 mL of water and shake to dissolve. Add about 1 mL of 1 M hydrochloric acid (HCl) solution, dropwise, using a medicine dropper. Part 2 – Making Predictions of Double Displacement Reactions In the hood you will find dilute solutions of NH 4 Cl, CuSO 4 , NaOH, and HCl.
Predict which two of these solutions you could mix to produce a precipitate. Test your prediction by combining about 1 mL of each of the two solutions in a test tube or small beaker. Record the results of the test. Write the net ionic equation for the reaction.
Predict which two of these solutions you could mix to produce a gas. Before you test the reaction, wet a piece of red litmus paper and attach it to the bottom of a

Using the solubility rules, predict the products and write the balanced overall equations (including phase symbols) for the following reactions. Write the net ionic equations. a. Mix aqueous cobalt (III) nitrate with aqueous sodium sulfide. b. Mix aqueous phosphoric acid with aqueous potassium hydroxide. c. Mix aqueous ammonium phosphate with aqueous calcium chloride. d. Mix aqueous potassium bicarbonate with aqueous hydrochloric acid.
In part 3, steps 5-7 of this experiment, you tested the conductivity of aqueous barium hydroxide, aqueous sulfuric acid, and a equimolar mixture of the two. Why is it that barium hydroxide and sulfuric acid each conducted electricity, but the mixture of the two did not? Explain thoroughly.

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