Full Lab Manual
Introduction & Goals
Chemistry & Background
In Your Write-up
Chemistry & Background
It is important to understand the structure of the five anions, as you observe their reactions and balance the corresponding chemical equations. Chloride and iodide ions are formed when a chlorine or iodine atom gains an extra electron, to form a negatively charged ion. These anions form ionic compounds, or salts, with positively charged cations, often metal ions, such as sodium, Na+, as in sodium chloride, NaCl. The formula of an ionic compound is the ratio of cation and anion that gives a balanced, neutral charge. Carbonate, hydrogen carbonate, and sulfate ions are polyatomic ions. They consist of groups of atoms covalently bonded together, with an overall charge. They also form ionic compounds by the attraction of an oppositely charged ion, like sodium or calcium, Ca2+, as in calcium carbonate, CaCO3. Polyatomic atoms are discussed in your textbook in section 2.9.
The first step in writing chemical equations for aqueous ionic reactions is to recognize the anions and cations and write them as dissolved species. In aqueous solution, soluble ionic compounds separate into cations and anions, which are surrounded by water molecules. For example, sodium chloride dissolves in water to form sodium cations, Na+, and chloride anions, Cl-. Ionic compounds containing polyatomic ions also dissolve in water to form separate anions and cations, but the polyatomic ions remain intact. For example, sodium hydrogen carbonate dissolves in water to form sodium cations, Na+, and hydrogen carbonate anions, HCO3-. When writing chemical equations for the reactions observed, we generally write the net ionic equation, which shows only those ions that react, not those which are spectator ions. For more information on balancing chemical equations, see the following section.
A Note on Balanced Chemical Equations
All chemical equations in your lab notebook and write-up should be properly balanced. This means that each side of the equation should have the same number of moles of each element, so that matter is not created or destroyed. In addition, the electrical charges must be the same on both sides of the equation. When writing chemical equations for ionic species, one generally writes the net ionic equation which shows only those species which react. Ions which do not react, but serve to balance the electrical charge of reacting ions, are called spectator ions.
For example, when a solution of sodium chloride reacts with silver nitrate, the balanced equation can be written in complete form as
NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)
We can break each dissolved ionic compound into cations and anions to give
Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq)
AgCl(s) + Na+ (aq) + NO3- (aq)
Note that the insoluble precipitate, AgCl, is not broken down into separate ions, since it is not disolved. The sodium and nitrate ions are spectators to the precipitation reaction between silver and chloride ions. If we remove the spectator ions, the net ionic equation can be written
Cl- (aq) + Ag+ (aq) AgCl(s)
For more on balancing chemical equations, refer to the sections of your textbook listed on the Introduction and Goals page.
The following paragraphs present the chemical reactions that you will observe in the lab this week. Compounds containing the carbonate ion (or the related ion bicarbonate or hydrogen carbonate, HCO3-) react with acids to give carbon dioxide (CO2), a colorless, odorless, and slightly acidic gas. For example, the reaction of baking soda (NaHCO3) with acids such as those in lemon juice or buttermilk releases carbon dioxide which causes baked goods to rise:
HCO3- + H+ CO2 (g) + H2O
In this equation the sodium ion, Na+, does not participate in the reaction and is a spectator ion. It is not included in the net ionic equation, given above.
Many other gases are colorless and odorless, like CO2. To confirm that the gas formed is carbon dioxide, it can be allowed to react with Ba(OH)2, barium hydroxide, to form a white insoluble precipitate of barium carbonate, BaCO3:
CO2 + Ba2+ (aq) + 2 OH- (aq) BaCO3 (s) + H2O
Eggshells, seashells, chalk, and limestone are some of the common substances containing calcium carbonate.
Chloride salts react with sulfuric acid to make hydrogen chloride gas, which is also an acid:
Cl- + H+ (aq) HCl (g).
In this reaction, the sulfate ion, SO4-2, from the sulfuric acid, is a spectator ion and is not included in the net ionic equation.
Another useful reaction for identification of the chloride ion is its reaction with silver nitrate to form a white precipitate of silver chloride:
Cl- + Ag+ (aq) AgCl (s)
As discussed above, nitrate and sodium are spectator ions and do not appear in the net ionic equation.
To test the properties of the sulfate ion, you will examine the reactions of Epsom salts. This salt, MgSO4.7H2O, is used as a purgative and, as an aqueous solution, to soak tired feet. Since it dissolves in water to form Mg2+ and SO42- ions, it will not react with a solution containing SO42- and HSO4- ions, such as H2SO4. It will, however, form insoluble precipitate of BaSO4 when in the presence of Ba2+ ion:
Ba2+ + SO42- BaSO4 (s)
Iodide salts can be identified by their reaction with chlorine (Cl2). Chlorine is a toxic, pale, yellow-green gas with an irritating odor and low solubility in water. Its structure consists of two chlorine atoms covalently bonded together, with a single bond of two shared electrons. A convenient laboratory source of chlorine is commercial bleach, which is usually a 5 percent aqueous solution of sodium hypochlorite, NaOCl. The solution behaves as if it contains dissolved chlorine. This is because hypochlorite and chloride ions are in equilibrium with chlorine and hydroxide ions:
OCl- (aq) + Cl- (aq) + H2O Cl2 (aq) + 2 OH- (aq)
Note that sodium is again a spectator ion and does not appear in the net ionic equation. In this experiment, you will use a solution of bleach to supply Cl2 (aq) for reaction with iodide ion. The products of this reaction are molecular iodine and chloride ion.
Cl2 + 2 I- I2 + 2 Cl-
Iodine (I2) exists as a purple solid and dissolves sparingly in water. When iodine (I2) is in solution with iodide ion (I-), the brown triiodide ion forms:
I2 (aq) + I- (aq) I3- (aq)
Purple colorless brown
Thus, reaction of iodide ion with a solution of bleach produces iodine, which then reacts with remaining iodide ion to form the brown triiodide ion. This brown color will help you to recognize the presence of iodide and iodine in your reaction solutions.
A second test for iodide ion is formation of a silver iodide precipitate. Like the chloride ion, iodide reacts with silver nitrate solution to give a precipitate, although AgI is yellow, while AgCl is white:
Ag+ + I- AgI (s)
In this reaction, sodium and nitrate are spectator ions and are not included in the net ionic equation.
Yet another test for iodide is the reaction of an iodide salt with sulfuric acid to form a brown solid and cause gas evolution. In this series of reactions, iodide ion is first oxidized to I2. The gas formed is hydrogen sulfide (H2S), from the reduction of sulfuric acid. Since both iodide ion (I-) and molecular iodine (I2) are present, the triiodide ion forms, giving the characteristic brown color.
8 I- (aq) + 10 H+ (aq) + SO42-
H2S (g) + 4 I2(aq) + 4 H2O
I2 (aq) + I- (aq) I3- (aq)
A final test for the presence of iodide ion can be performed using starch to detect I2, produced from I- in one of the above reactions. Starch contains amylose, a polymer of a-D glucose that reacts to form a characteristic blue-black complex with I2 in the presence of I-. The iodine and iodide form an I5- chain which fits inside the helix formed by the sugar chains. This complex is a dark blue-black that is easily observed. Starch can be used to confirm the results of the reaction between iodide and chlorine or iodide and H2SO4, since both reactions produce with I2 in the presence of I-.
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