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EXPERIMENT

Ionic and Covalent Substances

and Aqueous Reactions

PURPOSE

The goals of this experiment are multiple:

1. To observe the electrical conductivity of various liquids and solutions of ionic and covalent

compounds using an LED electrical conductivity indicator;

2. To classify these substances as strong electrolytes, weak electrolytes or nonelectrolytes;

3. To observe the changes in electrical conductivity during double displacement reactions;

4. To learn to write general molecular, complete ionic and net ionic equations.

MATERIALS AND EQUIPMENT

250 mL beaker, LED conductivity indicator, glass evaporating dish, 12-well spot plate, grease

pencil, dropper, 175 mL test tubes, #2 stopper, 0.1 M NH3, 0.1 M NaCl, NaCl (s), 6M, 0.1 M and

glacial HC2H3O2, 0.1 M C12H22O11, 95 % C2H5OH, 0.1 M NaC2H3O2, 0.1 M NaOH, 6M and 0.1 M

HCl, mossy zinc, CaCO3 (s), phenolphthalein, 0.1 M CuSO4, 0.1 M Na3PO4, 0.05 M Ba(C2H3O2)2,

0.05 M H2SO4, 0.05 M Ba(OH)2, 10 mL graduated cylinder.

DISCUSSION

Ionic Compounds

When a metallic element reacts with a non–metallic element, the latter becomes negatively charged

because one or more electrons are transferred from the metal to the non-metal during the redox

reaction. The metal, in turn, becomes positively charged because of the loss of electrons. Such

electrically charged atoms are called ions. The positive ion (cation) is attracted to the negative ion

(anion) by electrostatic forces forming an ionic bond. Compounds containing ionic bonds are called

ionic compounds. In their solid state, the ions are arranged in a crystal lattice, and as such are not

free to move. Upon heating, the solids will become fluid. On an atomic level the ions are freed

from their positions in the crystal lattice and become mobile. These mobile ions can carry an

electrical current, so that molten compounds become good conductors of electricity.

Covalent Compounds

When two non-metallic elements combine, they do so by mutually sharing a pair (or pairs) of

electrons between the atomic nuclei. Such a bond is called a covalent bond; the compound is called

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a covalent compound. If the electrons comprising the bond are shared equally, the bond is non-

polar. If the electrons comprising the bond are shared unequally by the nuclei, the bond is polar.

Covalent compounds do not conduct electricity even when molten because the molten liquid is

comprised of neutral particles whose movement cannot carry an electrical current. Thus, these

compounds cannot conduct electricity.

Electrical Conductivity of Compounds in Aqueous Solution

Water is a good solvent for many ionic and covalent compounds. Substances that dissolve to form

solutions that conduct electricity are called electrolytes. All soluble ionic compounds are

electrolytes. The water molecules that comprise the solvent dissolve the solid by separating the ions

in the solid crystal lattice. The water molecules align themselves around each separated ion in the

solution (become hydrated) through an ion-dipole interaction. The number of water molecules that

align with each ion is unique (for example Zn(H2O)4+2

and Al(H2O)6+3

). We shall use the hydrated

formula whenever doing so contributes to better understanding the reaction, but in general, for the

sake of simplicity, the non-hydrated formula (Zn+2

and Al+3

) is used instead. Dissolving sodium

chloride in water can thus be written as (the solvent is omitted for simplicity):

NaCl (s) Na+ (aq) + Cl

− (aq).

Since the dissolved ionic solid consists entirely of ions in aqueous solution, this solution is a good

conductor of electricity and will be a strong electrolyte. Ionic compounds that are insoluble in

water will not form electrolytes because the solid crystal lattice comprising the ionic solid stays

intact when mixed with water and no ions will form. Solubility of ionic compounds in water is

determined by using solubility rules.

Substances with covalently bonded atoms generally have definite neutral molecules as units in any

state of matter, and as such will not conduct electricity in aqueous solution. Substances that dissolve

to form nonconducting solutions or do not conduct electricity in pure form are called

nonelectrolytes. Solubility in water will be determined by a molecule’s polarity. If the electrons

comprising the covalent bonds are equally shared between the two nuclei or if the vector sum of the

bond dipoles within a molecule is zero, the molecule is non-polar (for example PCl5 and I2) and will

not dissolve in water. If the electrons comprising the covalent bond are not evenly shared between

the nuclei or if the vector sum of the bond dipoles is not equal to zero, the molecule is polar (for

example C6H12O6 and CH3OH) and will dissolve in water.

However, some polar covalent compounds (notably acids and bases) will react with water to form

ions and so the aqueous solution will conduct electricity. So, these solutions are electrolytes. The

process of forming ions in this way is called ionization. If all of the polar covalent molecules react

in this way, the substance is 100 % ionized (all ions) and as such will behave as a strong electrolyte,

just like a soluble salt. The reaction with water can be shown as:

HCl (g) + H2O () H3O+ (aq) + Cl

− (aq) or HCl (g) H

+ (aq) + Cl

− (aq).

Water can be included as part of the equation or omitted. Acids and bases that ionize completely

(100 % ionized) are called strong acids and strong bases. If only a fraction (usually < 10 %) of the

dissolved molecules react with water (ionize), the solution becomes weakly conducting or a weak

electrolyte. Acids and bases that behave in this way are called weak acids and weak bases. This can

be shown as follows:

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The two arrows pointing in opposite directions show that both reactants and products exist at the

same time. Characteristics of strong electrolytes, weak electrolytes and nonelectrolytes are shown

the Table 1.

Table 1: Characteristics of Electrolyte Solutions

* 8.4 % of an aqueous sample of HF will exist as ions. The remainder is HF molecules. Most weak

electrolytes ionize to a far lesser extent.

Double Displacement Reactions These reactions begin with two compounds. The cation from the first compound is exchanged with

the cation from the second compound. This double exchange can be summarized as:

AB + CD CB + AD. For example,

AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)

Type of

Solution

Physical

Representation

Compounds/Ions in

Solution

Other Examples

Strong

Electrolyte

Cl-

Cl–

Cl-

H+(aq) and Cl

– (aq)

NaCl (aq)

AgNO3 (aq)

K2SO4 (aq)

NaOH (aq)

HClO4 (aq)

Weak

Electrolyte

HF HF

HF HF

HF HF

HF F–

HF

HF HF

HF

Mostly HF (aq),

A small amount of

H+(aq) and F

– (aq)

*

HC2H3O2 (aq)

NH3 (aq) - can

also be written as

NH4OH (aq)

Nonelectrolyte

CH3OH

CH3OH

CH3OH

CH3OH

CH3OH

CH3OH (aq)

C6H12O6 (aq)

CH3CH2OH (aq)

H+ H

+

H+

Cl–

HF

HF

CH3OH

H+

H+

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So AB in water can be thought of A+ and B

− ions, and CD be thought of C

+ and D

− ions. When the

reaction occurs, C+ exchanges partners, combining with B

−, and A

+ combines with D

−.

If after the reaction is completed, we form or consume either a precipitate (an ionic compound that is

insoluble in water), or a weakly ionized molecule (such as water, acetic acid, ammonium hydroxide,

etc.), or if we produce a gas, we can say that a reaction has occurred. This can be summarized as a

reaction occurs in ALL cases whenever there is a change in type of electrolyte for substances from

reactants to products. The reaction will be accompanied by one or more physical changes as

evidence of a chemical reaction. This is summarized in Table 2.

Table 2: Driving Forces for Double Displacement Reactions

If none of the driving forces in Table 2 has occurred, we would then have no reaction. If this

situation occurs, when writing a chemical equation, we simply write the reactants, an arrow as we

usually would and then write “NR” for no reaction. For example,

NaCl (aq) + KNO3 (aq) NR.

Example 1: Gas-Forming Reaction

Substances mixed: hydrochloric acid and sodium bicarbonate

Observation: Bubbles form

The products formed are carbonic acid (H2CO3) and sodium chloride (NaCl). Using the Solubility

Table at the end of the procedure, we see that NaCl and NaHCO3 is soluble in water and using the

Electrolyte Table (at the end of the experiment), H2CO3 is a weak electrolyte and HCl is a strong

electrolyte.

Balanced equation: HCl (aq) + NaHCO3 (aq) H2CO3 (aq) + NaCl (aq) (strong elect.) (strong elect.) (weak elect.) (strong elect.)

Driving Force Physical Evidence that a

Reaction has Occurred

Example

A gas is formed (Typically

CO2, SO2, H2S and NH3)

Bubbles Form (Many of the

gases listed also have a

distinctive odor.)

2 HCl + Na2S H2S + 2 NaCl (gas)

A weakly ionized molecule

such as water is formed or

consumed

A temperature change

occurs.

HCl + KOH KCl + H2O (weakly- ionized)

A solid (precipitate) is formed

or consumed

A solid is formed or

disappears. There may also

be a change in color.

AgNO3 + NaI AgI + NaNO3 (solid ppt)

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However, H2CO3, a weak unstable acid, decomposes to give H2O and CO2, so the balanced equation

is:

HCl (aq) + NaHCO3 (aq) NaCl (aq) + H2O () + CO2 (g)

(strong elect.) (strong elect.) (strong elect.) (very weak elect.) (nonelect.)

Example 2: Forming a weakly ionized substance (water)

Substances mixed: potassium hydroxide and sulfuric acid

Observation: Temperature of reaction mixture increases

The products formed are water (H2O) (a very weakly ionized substance) and potassium sulfate

(K2SO4). Using the Electrolyte Table at the end of the procedure, we see that KOH and H2SO4 are

both strong electrolytes. Using the Solubility Table at the end of the procedure, we see that K2SO4 is

soluble in water. H2O is the other product.

Balanced equation: 2 KOH (aq) + H2SO4 (aq) K2SO4 (aq) + 2 H2O () (strong elect.) (strong elect.) (strong elect.) (very weak elect.)

Example 3: Precipitation Reaction

Substances mixed: iron (II) sulfate and barium chloride

Observation: A white precipitate forms

The products formed are iron (II) chloride (FeCl2) and barium sulfate (BaSO4). Using the Solubility

Table at the end of the procedure, we see that FeCl2, FeSO4, and BaCl2 are soluble in water. BaSO4

is insoluble in water (this is the white precipitate).

Balanced equation: FeSO4 (aq) + BaCl2 (aq) FeCl2 (aq) + BaSO4 (s) (strong elect.) (strong elect.) (strong elect.) (nonelectrolyte)

Solution Equations

When we first talked about double displacement reactions, we mentioned that the cation and anion

that make up each compound exchange with each other to form products. We further said that a

reaction occurs whenever any of the following three things happen: a precipitate forms or is

consumed, a gas forms, or a weakly ionized substance forms or is consumed. Now, let us see why

this is so. In the process, we will learn how to write solution equations.

Let us examine the reaction between the ionic compounds Pb(NO3)2 (aq) and K2CrO4 (aq). In this

case, when an ionic solid dissolves in water, the solution will conduct electricity. The solids that are

used to make these solutions are called strong electrolytes. The aqueous solids in these solutions do

not exist as compounds at all, but rather separate, or ionize, to form freely floating ions that move

independently of each other. So Pb(NO3)2 actually exists in solution as one Pb+2

ion and two NO3−

ions, and K2CrO4 as two K+ ions and one CrO4

−2 ion. So our reactants, which previously were

written as Pb(NO3)2 (aq) + K2CrO4 (aq) when writing an equation may also be shown as: Pb+2

+

2 NO3− + 2 K

+ + CrO4

−2

Now, let’s actually consider what would happen if we mixed these reactants in the laboratory.

Adding Pb(NO3)2 (aq), a colorless solution, to K2CrO4 (aq), a yellow solution, will produce a yellow

solid in a colorless solution. Now let us see if we can explain these results. We have four ions

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freely floating in the solution after the reactants are mixed. Since we know a chemical change has

occurred, we may conclude that the Pb+2

ion could combine with either K+

or CrO4−2

. The NO3−

ion could combine with either K+

or CrO4−2

. Since compounds form because of the attraction

between ions with opposite charge, we may further say that the products are PbCrO4 and KNO3.

We now consult our Solubility Table to decide which of the products is the yellow solid. We see that

PbCrO4 is insoluble in water and KNO3 is soluble. This suggests that the yellow solid is PbCrO4

and KNO3 is dissolved in the colorless solution. As before, potassium nitrate may be written either

as KNO3 (aq) or K+ + NO3

−. This reaction is illustrated as follows:

+Pb+2

NO3–

K+

NO3– CrO4

–2

K+

NO3– K+

NO3–K+

PbCrO4 (s)

Now let us see how the illustration above can be written as a chemical equation. We will start out

with a balanced equation exactly as we would have written it before without taking into account the

compounds ionizing in solution. This called the molecular equation.

Pb(NO3)2 (aq) + K2CrO4 (aq) 2 KNO3 (aq) + PbCrO4 (s) (strong elect.) (strong elect.) (strong elect.) (nonelectrolyte)

Now, let us remember what really occurs in solution. In the complete ionic equation, we will show

aqueous ionic compounds as ions.

Pb+2

(aq) + 2 NO3− (aq) + 2 K

+ (aq)

+ CrO4

−2 (aq) 2 K

+ (aq) + 2 NO3

− (aq) + PbCrO4 (s)

Finally, you will notice that K+ and NO3

− remain unchanged during the reaction. They are present

both as reactants and as products. Substances that don’t undergo change during the reaction are

called spectator ions. If we want to show in our equation only substances that undergo a chemical

change, we simply cross out the spectator ions and write what remains. This last equation, which

summarizes the essence of the chemical change that happens, is called the net ionic equation.

Pb+2

(aq) + CrO4−2

(aq) PbCrO4 (s) (strong elect.) (nonelectrolyte)

Let us look at writing solution equations for other reactions.

Example 1: Forming a slightly ionized substance (water)

Molecular equation: HCl (aq) + NaOH (aq) NaCl (aq) + H2O () (strong elect.) (strong elect.) (strong elect.) (v. weak elect.)

Referring to the Electrolyte Table, NaOH and HCl are both strong electrolytes. From the Solubility

Table, we learn that NaCl is soluble in water. These substances ionize completely in water. Thus,

we will write them as ions in solution.

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+H+

Cl–Na+

Cl–

H2O

Na+

OH–

Complete ionic equation: H+

(aq) + Cl−

(aq) + Na+ (aq) + OH

− (aq) Na

+ (aq) + Cl

− (aq) + H2O ()

Net ionic equation: H+

(aq) + OH−

(aq) H2O () (strong elect.) (v. weak elect.)

Example 2: Gas-Forming Reaction

Molecular equation: HBr (aq) + KHCO3 (aq) KBr (aq) + H2O () + CO2 (g)

(strong elect.) (strong elect.) (strong elect.) (v. weak elect.) (nonelect.)

Complete ionic equation: H+ (aq) + Br

− (aq) + K

+(aq) + HCO3

− (aq)

K+ (aq) + Br

− (aq) + H2O () + CO2 (g)

Net ionic equation: H+ (aq) + HCO3

− (aq) H2O () + CO2 (g)

(strong elect.) (strong elect.) (v. weak elect.) (nonelect.)

Example 3: No Reaction

Molecular Equation: 2 NaCl (aq) + K2CO3 (aq) 2 KCl (aq) + Na2CO3 (aq) (strong elect.) (strong elect.) (strong elect.) (strong elect.)

Complete ionic equation: 2 Na+ (aq) + 2 Cl

− (aq) + 2 K

+(aq) + CO3

−2 (aq)

2 K+ (aq) + 2 Cl

− (aq) + 2 Na

+ (aq) + CO3

−2 (aq)

Note that all ions cancel when looking at the spectator ions or that there is no change in type of

electrolyte from reactants to products. So there is no net ionic equation. This is the same as saying

there is no chemical change, and therefore, no reaction.

Net ionic equation: None

So, when writing the molecular equation, we would simply write:

NaCl (aq) + K2CO3 (aq) NR.

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The Experiment

First, you will be asked to measure the electrical

conductivity of a series of substances and solutions and

from these measurements classify them as to type of

electrolyte. Then the role of the solvent in determining

conductivity will be

determined. Finally you

will perform different

chemical reactions

noting the change in

conductivity as the

reaction occurs. Either a

light bulb conductivity

apparatus (Figure 1) or a

LED conductivity

indicator (Figure 2) can

be used. The light bulb

conductivity apparatus

consists of an electric

lamp in series with open

electrodes. The

electrodes fit inside a beaker containing the liquid to be tested.

The LED conductivity detector works in much the same fashion

except the electrodes fit inside the well of a spot plate in which

the liquid to be tested has been placed. We will use the LED

conductivity indicator exclusively to make measurements. (Why?)

The LED conductivity meter has both a green and a red light emitting diode on it. By noting which

is glowing and the intensity of the light and by using Table 3, one can determine with a high degree

of accuracy the conductivity of the solution and classify the test solution (or substance) as to the type

of electrolyte.

Table 3: Conductivity Versus Type of Electrolyte for LED Conductivity Indicator

Scale Red LED Green LED Conductivity Type of

Electrolyte

0 Off Off Low/None Nonelectrolyte

1 Dim Off Low

Weak Electrolyte

2 Medium Off Medium

3 Bright Dim High

Strong Electrolyte

4 Very Bright Medium Very High

Figure 1: A light bulb

conductivity apparatus.

Figure 2: An LED

aaiiconductivity indicator.

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PROCEDURE

Part A: Electrolytes

1. Add 200 mL of deionized water to an empty 250 mL beaker. This will be used to rinse the

electrodes of the LED conductivity indicator. Obtain an LED conductivity indicator and a clean dry

12-well spot plate. With a grease pencil, number the wells on the spot plate 1 through 12. This will

correspond to the numbered substances and solutions in step 2.

2. Add five drops of substances 1 – 10 listed below in turn to each well in the spot plate. If testing a

solid, fill the well half full. Test the conductivity of each substance and record the result in your lab

report. The electrodes should just break the surface of each substance being tested. Make sure you

rinse and blot dry with a paper towel the electrodes after each substance is tested. No other part of

the LED conductivity indicator should get wet except for the electrodes!

1. deionized water 7. 0.1 M C12H22O11 (sugar)

2. tap water 8. C2H5OH (aq) S

3. 0.1 M NH3 (aq) (NH4OH) 9. 0.1 M NaC2H3O2 (aq)

4. 0.1 M NaCl 10. 0.1 M NaOH (aq)

5. NaCl (s) 11. HC2H3O2 () S

6. 0.1 M HC2H3O2 (aq) 12. C6H12 () S

3. Test liquids 11 and 12 under the fume hood. Wear gloves and goggles! Enter the results in the

lab report as before. Dispose of samples 11 and 12 down the sink in the fume hood. (Use a dropper

to remove them from the well.) Clean and dry the spot plate.

Part B: Effect of Solvent

1. Now test under the fume hood as before 5 drops of the HCl (g) dissolved in cyclohexane.

2. Add 2 mL of the HCl (g) in cyclohexane to a 175 mL test tube. Now add 2 mL of deionized

water, stopper and shake. The solution should separate into two discrete layers, the aqueous layer

SAFETY Glacial and 6 M acids are corrosive. If you spill these reagents on

yourself, rinse the affected area with water. Glacial acetic acid and cyclohexane

solutions must be used in the fume hood because their vapors are toxic! Wear gloves

and goggles!

WASTE DISPOSAL Glacial acetic acid and cyclohexane solutions should be

disposed of in the fume hood sink. Left over solid CaCO3 and Zn should be placed

in the trash. Barium solutions need to be placed in a toxic waste container.

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being the bottom layer. Use a dropper to remove approximately 0.5 mL of the aqueous layer and

place it in the well of the spot plate. Use the LED conductivity indicator to test as before.

Part C: Difference between Strong and Weak Acids

1. Add 5 mL of 6 M HC2H3O2 in a 175 mL test tube. Add 5 mL of 6 M HCl to a second test tube.

Place one marble chip (solid CaCO3) in each test tube. Note the results in the lab report.

2. Repeat step 1 using clean test tubes and fresh acetic acid, using a piece of mossy zinc in place of

the marble chip.

Part D: Double Displacement Reactions

1. The following four double replacement reactions are to be performed on a clean dry spot plate.

Add five drops of the first reactant to a well and 5 drops of the second reactant to a separate well.

Test the conductivity of both solutions.

1. 0.1 M HCl + 0.1 M NaOH

2. 0.1 M M HC2H3O2 + 0.1 M NH3 (NH4OH)

3. 0.1 M CuSO4 + 0.1 M Na3PO4

4. 0.05 M Ba(C2H3O2)2 + 0.05 M H2SO4 (dispose of products in toxic waste container)

2. Now add 5 drops of the second reactant to the first well (contains the first reactant). The amount

added is critical. You want to add just enough of the reagent to react with no excess. Wait about

30 seconds and then check the conductivity of the mixture.

Part E: Titration of Barium Hydroxide with Sulfuric Acid

1. Place 5 mL of 0.05 M H2SO4 directly into a clean dry glass evaporating dish. Add one drop of

phenolphthalein. (This will not affect the conductivity of the solution.) Now add 8.00 mL of 0.05 M

Ba(OH)2 to a clean dry 10 mL graduated cylinder.

2. Use a clean dry dropper to remove 1.00 mL of the 0.05 M Ba(OH)2 from the graduated cylinder

and add it to the glass evaporating dish containing the 0.05 M H2SO4. Swirl or stir the solution to

mix and then test the solution’s conductivity. Repeat step 2 three more times. You should now

have 4 mL of 0.05 M Ba(OH)2 left in the graduated cylinder.

3. Remove one more milliliter of the 0.05 M Ba (OH) 2. Add this drop by drop to the solution in the

dish until the solution turns a very faint pink lasting for at least 30 seconds. (You may want to place

a piece of white paper under the dish so the color change is more apparent.) Swirl to mix and test

the solution’s conductivity.

4. Now add the remaining contents of the graduated cylinder to the evaporating dish, mix the

solution and again test the conductivity. Dispose of the contents of this evaporating dish into the

toxic waster container.

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Electrolyte Table

Solubility Table (Common Salts and Bases)

Ion Solubility Behavior of Salts of Ion

NO3– All nitrates are soluble

C2H3O2– All acetates are soluble (AgC2H3O2 is moderately soluble.)

Cl–, Br

–, I

–

All chlorides, bromides, and iodides are soluble, except those of Ag+,

Hg2+2

and Pb+2

SO4–2

All sulfates are soluble except PbSO4 and BaSO4. (CaSO4, Hg2SO4 and

Ag2SO4 are slightly soluble. The corresponding hydrogensulfates are

more soluble.)

CO3–2

, PO4–3

, C2O4–2

All carbonates, phosphates and oxalates are insoluble except those of Li

+,

Na+, K

+, and NH4

+. (Many hydrogenphosphates are soluble.)

OH–, O

–2

All hydroxides and oxides are insoluble, except NaOH, KOH, NH4OH,

Ba(OH)2, and Ca(OH)2. [Ca(OH)2 is moderately soluble.]

S–2

All sulfides are insoluble except those of Li

+, Na

+, K

+, NH4

+, Mg

+2, Ca

+2,

Sr+2

, and Ba+2

Na+, K

+, NH4

+

All salts and bases of Na+, K

+, and other ions of group 1A and NH4

+ are

soluble.

Ag+

All silver salts are insoluble, except AgNO3 and AgClO4. (AgC2H3O2 and

Ag2SO4 are only moderately soluble.)

Classification Acids Bases

Salts

Strong

Electolyte

Strong Acids: Strong Bases:

Almost all soluble salts. HCl O4

HBr O4

HI O4

HClO4

H2SO4

HNO3

NaOH )2*

KOH)2*

Ca(OH)2*

Ba(OH)2

*Saturated Ca(OH)2 = 0.02 M

Weak

Electrolyte

All other acids.NH3

NH3 (aq) - can also be

written as NH4OH

A few salts such as l2

HgCl2, CdCl2, and

Pb(C2H3O2)2, are only

slightly ionized or form

complex ions.

Nonelectrolytes

All covalent compounds

Page 12

Part A: Electrolytes

Complete the table below. Use Table 3 to find conductivity from experiment data. Label each

substance as to type of electrolyte (S = strong electrolyte, W = weak electrolyte, N = nonelectrolyte).

Indicate the formulas of all species present circling those in low concentration.

Part B: Effect of Solvent

How does the solvent (cyclohexane versus water) affect the ability of HCl to act as an electrolyte?

Explain.

Substance Conductivity Type of Elect. Species Present

1. deionized water

2. tap water

3. NH3 (aq) (NH4OH)

4. NaCl (aq)

5. NaCl (s)

6. HC2H3O2 (aq)

7. C12H22O11 (aq)

8. C2H5OH (aq)

9. NaC2H3O2 (aq)

10. NaOH (aq)

11. HC2H3O2 ()

12. C6H12 ()

Ionic and Covalent

Substances and Aqueous Reactions

Name ____________________________

Section _________ Date _____________

Instructor _________________________

LAB

REPORT

Page 13

Part C: Difference between Strong and Weak Acids

Write net ionic equations for the reaction of HCl with Zn and HCl with CaCO3.

_______________________________________________________________________

_______________________________________________________________________

Compare and explain the chemical behavior of 6 M HCl versus 6M HC2H3O2 when reacted with

either Zn or CaCO3. Use concentration of H+ as part of the explanation.

Part D: Double Displacement Reactions

Write complete and net ionic equations for the double displacement reactions you performed. Also

include electrolyte data (S = Strong, W = Weak, N = Non) where indicated.

Reaction

1 Type of Elect. HCl __________ NaOH __________ Mixture __________

Complete

Ionic

Net

Ionic

Reaction

2 Type of Elect. HC2H3O2 ________ NH4OH __________ Mixture __________

Complete

Ionic

Net

Ionic

Reaction

3 Type of Elect. CuSO4 __________ Na3PO4 __________ Mixture __________

Complete

Ionic

Net

Ionic

Reaction

4 Type of Elect.

Ba(C2H3O2)2

______ H2SO4 __________ Mixture __________

Complete

Ionic

Net

Ionic

Page 14

Part E: Titration of Barium Hydroxide with Sulfuric Acid Compare the conductivities of the solution tested by filling in the table below. To the right of the

table, explain the conductivity behavior of the solution (a) before an equal volume of Ba(OH)2 is

added, (b) at the equivalence point, and (c) when an excess volume of Ba(OH)2 is added. Finally,

write a general molecular, complete ionic and net ionic equation for the reaction that occurred:

(a) Conductivity behavior before Ba(OH)2 is added

(b) Conductivity behavior at equivalence point

(c) Conductivity behavior when excess Ba(OH)2 is added

mL Ba(OH)2added toH2SO4

Conductivity

0

1

2

3

4

5

8

(molecular)_______________________________________________________________________

(complete)_______________________________________________________________________

(net ionic)_______________________________________________________________________

APPLICATION OF PRINCIPLES

1. Using the Electrolyte Table and the Solubility Table, write the formulas of the principle species

present when the following are mixed with water. Indicate solids by placing an (s) after the formula.

If the substance is a weak, soluble electrolyte, put (aq) after the formula.

CaC2O4 ____________________ Al2(SO4)3 ____________________

CaS ____________________ H3PO4 ____________________

PbI2 ____________________ KOH ____________________

ZnO ____________________ HClO4 ____________________

SnCl4 ____________________ Ni2S3 ____________________

Ba(C2H3O2)2 ____________________ (NH4)2SO4 ____________________

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2. Using the Electrolyte Table and the Solubility Table, write (a) general molecular, (b) complete

ionic and (c) net ionic equations for the mixing of the indicated reactants. Assume water is the

solvent. Indicate solids by writing an (s) and weak, soluble electrolytes by (aq) after the formula.

aluminum sulfide and nitric acid

(a) ___________________________________________________________________________

(b) ___________________________________________________________________________

(c) ___________________________________________________________________________

ammonia (ammonium hydroxide) and hydrochloric acid

(a) ___________________________________________________________________________

(b) ___________________________________________________________________________

(c) ___________________________________________________________________________

ammonium chloride and zinc nitrate

(a) ___________________________________________________________________________

(b) ___________________________________________________________________________

(c) ___________________________________________________________________________

aluminum and sulfuric acid

(a) ___________________________________________________________________________

(b) ___________________________________________________________________________

(c) ___________________________________________________________________________

barium sulfide and cobalt (III) acetate

(a) ___________________________________________________________________________

(b) ___________________________________________________________________________

(c) ___________________________________________________________________________

oxalic acid and sodium hydroxide

(a) ___________________________________________________________________________

(b) ___________________________________________________________________________

(c) ___________________________________________________________________________