Experiment

14

 

Overview 

In this experiment you ultimately want to try to identify two "unknown" solutions of ionic substances, by comparing the reactions of the "unknown" substances with the reactions of a set of "known" substances.

 

When a chemical reaction takes place, there is usually some sort of observable evidence for the reaction: a gas may be produced (causing bubbling); the color of the sample may change; an odor may be noticed; heat may be evolved. Perhaps the most common bit of evidence that a reaction has taken place, however, is the formation of a solid precipitate when the reactants are mixed. For example, if clear, colorless solutions of barium nitrate [Ba(NO₃)₂] and sodium sulfate (Na₂SO₄) are mixed, a precipitate of barium sulfate, BaSO₄ immediately forms:    

 

This reaction is an example of a general class of reactions called double displacement (or metathesis) reactions. Such reactions have the general format    

 

In a sense, each positive ion in the reactants has "displaced" the other positive ion. One of the new combinations of ions is insoluble in water and forms a precipitate. The precipitate could be filtered, and the water could be evaporated from the ions remaining in solution, to give two new substances. In this experiment you will use the appearance (or nonappearance) of a precipitate when two reagents are mixed as a means of identifying your two unknown solutions.

 

The most important part of this experiment is learning to relate the Table of Solubilities on Page 106 with your experimental observations. For example, suppose you mixed together CaCl₂ solution with AgNO₃ solution: a precipitate forms. But what is the identity of the precipitate?? The two possible products of the reaction are AgCl and Ca(NO₃)₂. Find the intersection of Ca²⁺ and NO₃^- in the Table of Solubilities: there is an "S" there, which means that Ca(NO₃)₂ is soluble and must not be the substance that precipitated. However, if you look at the intersection of Ag⁺ and Cl^-, there is an "I" there, which means that AgCl is insoluble and therefore must be the precipitate that formed.

 

Data 

Your data in Parts I and II on Page 109 should reflect your observations of the solutions you have available for the experiment and their mutual reactions when systematically combined. In particular, make certain that you note the color of the individual solutions (e.g., NiSO₄ is bright green, whereas CuSO₄ is bright blue). Be sure you understand the difference between saying a solution is "clear" (meaning there are no cloudy solids present) and "colorless" (which means that the solution shows no apparent inherent color). In Part II, record carefully your observations of what happens when you mix two solutions together. If nothing happens, write "no reaction". If a precipitate or cloudiness forms, describe the color and texture of the precipitate. For example, when NiSO₄ (itself a green solution) reacts to form a precipitate, is the resulting precipitate green or is the supernatant liquid above the precipitate green? Since the green color is due to the presence of the Ni²⁺ ion, where the green color ends up can help you in identifying things.

 

Chemical Equations 

Page 110, Part III

In this part of the experiment, you have to write the balanced overall and net ionic equations for the reaction that resulted in precipitate formation. To do this, you need to apply the Table of Solubilities on Page 106 as described in the Overview above. When you look at the Table of Solubilities to determine the formula of the precipitate, realize that this table essentially gives you the net ionic equation for the reaction. For example, for the reaction given earlier    

 

I knew that BaSO₄ was the precipitate by looking up the intersection of Ba²⁺ and SO₄^2- in the Table of Solubilities. The net ionic reaction for this overall reaction is just what I do in my head when looking in the Table of Solubilities: "barium ion plus sulfate ion gives an insoluble precipitate of barium sulfate", which gives the net ionic equation    

 

Page 111, Part IV

Identification of your two unknown solutions requires your looking back at your data for the known and "extra" solutions, and also checking frequently the Table of Solubilities. You are told on Page 107 the possible positive and negative ions in the unknowns-only these ions have been used. You can determine the identity of your unknowns by a "process of elimination". First of all, there are many substances which can be excluded as possibilities for what the unknowns are. Looking at the Table of Solubilities, you will see that there are many combinations which are insoluble in water. If a substance is insoluble in water, then we couldn't have given you a solution of the substance as an unknown!! This may seem obvious, but you'd be surprised how many students report insoluble substances as their unknown's identity.

 

The positive ions listed on Page 107 as being candidates for the unknowns are Ba²⁺, Pb²⁺, Ag⁺, and Ni²⁺. As mentioned earlier, Ni²⁺ solutions are bright green, so if your unknown is not green, it cannot contain Ni²⁺. So let's say we have a colorless unknown, and perform reactions of the unknown with each of the known and "extra" solutions. Since there are only four possible positive ions, just go through the Table of Solubilities and your reaction observations trying to make a match.

 

Suppose you thought the positive ion in your unknown might be either Ba²⁺ or Pb²⁺. For example, Ba²⁺ and Pb²⁺ both form precipitates with CrO₄^-. However, Pb²⁺ forms a precipitate with Cl^-, whereas Ba²⁺ does not. If your unknown formed a precipitate with Cl^-, then, of the two positive ions considered, it probably is Pb²⁺.

 

Take notes as you go through the data, and treat it like a puzzle. It may not be even possible to completely identify both ions in the unknowns, but as long as your observations, reasoning, and conclusions are correct, you should be able to narrow things down to a few possibilities at the least. You will receive credit not for just the right answer, but for how you arrived at the answer. Guessing does not help! Reasoning things out is the way to go.

 

Questions 

After identifying your unknowns, writing a few more equations should be a piece of cake. Just use the Table of Solubilities to identify the precipitate in each case.  

 

General Solubility Rules

In addition to the Table of Solubilities in your lab manual, there are some "general solubility rules" that you might find helpful in identifying your unknown.  There is a list of these in your textbook, or you can refer to the list below taken from http://www.csudh.edu/oliver/chemdata/solrules.htm

 

1. Salts containing Group I elements are soluble (Li⁺, Na⁺, K⁺, Cs⁺, Rb⁺). Exceptions to this rule are rare. Salts containing the ammonium ion (NH₄⁺) are also soluble.

2. Salts containing nitrate ion (NO₃^-) are generally soluble.

3. Salts containing Cl ^-, Br ^-, I ^- are generally soluble. Important exceptions to this rule are halide salts of Ag⁺, Pb²⁺, and (Hg₂)²⁺. Thus, AgCl, PbBr₂, and Hg₂Cl₂ are all insoluble.

4. Most silver salts are insoluble. AgNO₃ and Ag(C₂H₃O₂) are common soluble salts of silver; virtually anything else is insoluble.

5. Most sulfate salts are soluble. Important exceptions to this rule include BaSO₄, PbSO₄, Ag₂SO₄, and CaSO₄.

6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al³⁺ are insoluble. Thus, Fe(OH)₃, Al(OH)₃, Co(OH)₂ are not soluble.

7. Most sulfides of transition metals are highly insoluble. Thus, CdS, FeS, ZnS, Ag₂S are all insoluble. Arsenic, antimony, bismuth, and lead sulfides are also insoluble.

8. Carbonates are frequently insoluble. Group II carbonates (Ca, Sr, and Ba) are insoluble. Some other insoluble carbonates include FeCO₃, PbCO₃. Carbonates become soluble in acid solution.

9. Chromates are frequently insoluble. Examples: PbCrO₄, BaCrO₄

10. Phosphates are frequently insoluble. Examples: Ca₃(PO₄)₂, Ag₃PO₄

11. Fluorides are frequently insoluble. Examples: BaF₂, MgF₂ PbF₂.