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Objectives

In this laboratory you will become familiar with:

• writing equations for metathesis reactions, including net ionic equations
• the concept of solubility and the effect of temperature and crystallization

 Grading

You will be assessed on

• the correctness and thoroughness of your observations.
• answers to the pre-lab questions.
• the graph of the solubilities as a function of temperature.
• the equations for the metathesis reactions, including net ionic equations.
• the answers to the questions on the report form.

Click here for the Pre-Lab Questions which has to be handed in at the beginning of the lab. Don't forget that you need to plot your graph beforehand too.

The Report Form is to be filled out and turned in at the beginning of the next lab period.

Introduction

In molecular equations for many aqueous reactions, cations and anions appear to exchange partners. These reactions conform to the following general equation:

Equation 1: AX + BY ---> AY + BX

These reactions are known as metathesis reactions. For a metathesis reaction to lead to a net change in solution, ions must be removed from the solution. In general, three chemical processes can lead to the removal of ions from solution, comcomitantly serving as a driving force for metathesis to occur:

1. The formation of a precipitate
2. The formation of a weak electrolyte or nonelectrolyte
3. The formation of a gas that escapes from solution

The reaction of barium chloride with silver nitrate is a typical example:

Equation 2: BaCl₂(aq) + 2AgNO₃(aq) ---> Ba(NO₃)₂(aq) + 2AgCl(s)

This form of the equation for this reaction is referred to as the molecular equations. Since we know that the salts BaCl₂, AgNO₃, and Ba(NO₃)₂ are strong electrolytes and are completely dissociated in solution, we can more realistically write the equation as follows:

Equation 3: Ba²⁺(aq) + 2Cl^-(aq) + 2Ag⁺(aq) + 2NO₃^-(aq) ---> Ba²⁺(aq) + 2NO_3^-aq) + 2AgCl(s)

This form, in which all ions are shown, is known as the complete ionic equation. Reaction occurs because the insoluble substance AgCl precipitates out of solution. The other product, barium nitrate, is soluble in water and remains in solution. We see that Ba²⁺ and NO₃^- ions appear on both sides of the equation and thus do not enter into the reaction. Such ions are called spectator ions. If we eliminate or omit them from both sides, we obtain the net ionic equation:

Equation 4: Ag⁺(aq) + Cl^-(aq) ---> AgCl(s)

This equation focuses our attention on the salient feature of the reaction: the formation of the precipitate AgCl. It tells us that solutions of any soluble Ag⁺salt and any soluble Cl^- salt, when mixed, will form insoluble AgCl. When writing net ionic equations, remember that only strong electrolytes are written in the ionic form. Solids, gases, nonelectrolytes, and weak electrolytes are written in the molecular form. Frequently the symbol (aq) is omitted from ionic equations. The symbols (g) for gas and (s) for solid should not be omitted. Thus, Equation 4 can be written as

Equation 5: Ag⁺ + Cl^-----> AgCl(s)

Consider mixing solutions of KCl and NaNO₃. The ionic equation for the reaction is

Equation 6: K⁺(aq) + Cl^-(aq) + Na⁺(aq) + NO₃^-(aq) ---> K⁺(aq) + NO₃^-(aq) + Na⁺(aq) + Cl^-(aq)

Because all the compounds are water-soluble and are strong electrolytes, they have been written in the ionic form. They completely dissolve in water. If we eliminate spectator ions from the equation, nothing remains. Hence, there is no reaction:

Equation 7: K⁺(aq) + Cl^-(aq) + Na⁺(aq) +NO₃^-(aq) ---> no reaction

Metathesis reactions occur when a precipitate, a gas, a weak electrolyte, or a nonelectrolyte is formed. The following equations are further illustrations of such processes.

FORMATION OF A GAS

Molecular equation:

Equation 8: 2HCl(aq) + Na₂S(aq) ---> 2NaCl(aq) +H₂S(g)

Complete ionic equation:

2H⁺(aq) + 2Cl^-(aq) + 2Na⁺ (aq) + S^2-(aq) ---> 2Na⁺(aq) + 2Cl^-(aq) + H₂S(g)

Net ionic equation:

2H⁺(aq) + S^2- (aq) ---> H₂S(g)

or

2H⁺ + S^2- ---> H₂S(g)

FORMATION OF A WEAK ELECTROLYTE

Molecular equation:

HNO₃(aq) + NaOH(aq) ---> H₂O(l) + NaNO₃(aq)

Complete ionic equation:

H⁺(aq) + NO₃^-(aq) + Na⁺(aq) + OH^-(aq) ---> H₂O(l) + Na⁺(aq) NO₃^-(aq)

Net ionic equation:

H⁺(aq) + OH^-(aq) ---> H₂O(l)

In order to decide if a reaction occurs, we need to be able to determine whether or not a precipitate, a gas, a nonelectrolyte, or a weak electrolyte will be formed. The following brief discussion is intended to aid you in this regard. Table 1 summarizes solubility rules and should be consulted while performing this experiment.
The common gases are CO₂, SO₂, H₂S, and NH₃. Carbon dioxide and sulfur dioxide may be regarded as resulting form the decomposition of their corresponding weak acids, which are initially formed when carbonate and sulfite salts are treated with acid:

H₂CO₃(aq) ---> H₂O(l) + CO₂(g)

and

H₂SO₃(aq) ---> H₂O(l) +SO₂(g)

Ammonium salts form NH₃ when they are treated with strong bases:

NH₄⁺(aq) + OH^----> NH₃(g) + H₂O(l)

TABLE 1 Solubility Rules

*Slightly soluble.

Table 2 Strong Electrolytes

Which are the weak electrolytes? The easiest way of answering this question is to identify all of the strong electrolytes, and if the substance does not fall in that category then it is a weak electrolyte. Note, water is a nonelectrolyte. Strong electrolytes are summarized in Table.2.

In the first part of this experiment, you will study some metathesis reactions. In some instances it will be very evident that a reaction has occurred, whereas in others it will not be so apparent. In the doubtful case, use the guidelines above to decide whether or not a reaction has taken place. You will be given the names of the compounds to use but not their formulas. This is being done deliberately to give practice in writing formulas from names.

In the second part of this experiment, you will study the effect of temperature on solubility. The effect that temperature has on solubility varies from salt to salt. We conclude that mixing solutions of KCl and NaNO₃ resulted in no reaction (see Equations 6 and 7). What would happen if we cooled such a mixture? The solution would eventually become saturated with respect to one of the salts, and crystals of that salt would begin to appear as its solubility was exceeded. Examination of Equation 6 reveals that crystals of any of the following salts could appear initially: KNO₃, KCl, NaNO₃, or NaCl.
Consequently, if a solution containing Na⁺, K⁺, Cl-, and NO₃^- ions is evaporated at a given temperature, the solution becomes more and more concentrated and will eventually become saturated with respect to one of the four compounds. If a evaporation is continued, that compound will crystallize out, removing its' ions from solution. The other ions will remain in solution and increase in concentration. Before beginning this laboratory exercise you are to plot a graph of the solubilities of the four salts given in Table 3 on your report sheet.

Experimental Procedure

PART A Metathesis Reactions

CAUTION WEAR EYE PROTECTION

1. The report sheet lists 16 pairs of chemicals that are to be mixed. Use about 1 mL of the reagents to be combined as indicated on the report sheet.
2. Mix the solutions in small test tubes and record your observations on the report sheet. If there is no reaction, write N.R. (The reactions need not be carried out in the order listed. In order to reduce congestion at the reagent shelf, half the class will start in reverse order). Dispose of the contents of your test tubes in the designated receptacles.

PART B Solubility, Temperature and Crystallization

1. Place 8.5 g of sodium nitrate and 7.5 g of potassium chloride in a 100-mL beaker and add 25mL of water. Warm the mixture, stirring, until the solids completely dissolve.
2. Assuming a volume of 25mL for the solution, calculate the molarity of the solution with respect to NaNO₃, KCl, NaCl, and KNO₃, and record these molarities on your report form.
3. Cool the solution to about 10°C by placing the beaker in ice water in a 600-mL beaker and stir the solution carefully with a thermometer, being careful not to break it.
4. When no more crystals form, at approximately 10°C, filter the cold solution quickly and allow the filtrate to drain thoroughly into an evaporating dish. Dry the crystals between two dry pieces of filter paper or paper towels.
5. Examine the crystals with a magnifying glass (or fill a Florence flask with water and look at the crystals through it). Describe the shape of the crystals—that is, needles, cubes, plates, rhombs, and so forth on your report form.
6. Based upon your solubility graph, which compound crystallized out of solution and write that in the appropriate place on your report form
7. Evaporate the filtrate to about half of its volume using a Bunsen burner and ring stand. A second crop of crystals should form. Record the temperature and rapidly filter the hot solution, collecting the filtrate in a clean 100-mL beaker.
8. Dry the second batch of crystals between two pieces of filter paper and examine their shape. Compare their shape with the first batch of crystals.
9. Based upon your solubility graph, what is this substance?
10. Finally, cool the filtrate to 10°C while stirring carefully with a thermometer to obtain a third crop of crystals. Carefully observe their shapes and compare them with those of the first and second batches.
11. What compound is the third batch of crystals? Dispose of the chemicals in the designated receptacles.