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
Page 1
a covalent compound. If the electrons comprising the bond are shared equally, the bond is nonpolar. 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
+2
+3
align with each ion is unique (for example Zn(H2O)4 and Al(H2O)6 ). We shall use the hydrated
formula whenever doing so contributes to better understanding the reaction, but in general, for the
+2
+3
sake of simplicity, the non-hydrated formula (Zn and Al ) 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:
Page 2
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
Type of
Solution
Physical
Representation
Cl
Compounds/Ions in
Solution
+
H
+
+
Strong
Electrolyte
–
+
H
Cl
NaCl (aq)
AgNO3 (aq)
K2SO4 (aq)
NaOH (aq)
HClO4 (aq)
-
Cl
HF
HF
HF
HF
HF
+
H
–
HF
Weak
Electrolyte
–
H (aq) and Cl (aq)
H
Cl
Other Examples
HF
–
HF
F
+
H
HF
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)
CH3OH
CH3OH
CH3OH
Nonelectrolyte
CH3OH
CH3OH (aq)
C6H12O6 (aq)
CH3CH2OH (aq)
CH3OH
CH3OH
*
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)
Page 3
+
−
+
−
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
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
2 HCl + Na2S H2S + 2 NaCl
distinctive odor.)
(gas)
A weakly ionized molecule
such as water is formed or
consumed
A temperature change
occurs.
A solid (precipitate) is formed
or consumed
A solid is formed or
disappears. There may also
be a change in color.
HCl + KOH  KCl + H2O
(weaklyionized)
AgNO3 + NaI  AgI + NaNO3
(solid ppt)
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) +
(strong elect.)
NaHCO3 (aq)

(strong elect.)
Page 4
H2CO3 (aq) + NaCl (aq)
(weak elect.)
(strong elect.)
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) +
(strong elect.)
BaCl2 (aq)
(strong elect.)

FeCl2 (aq)
(strong elect.)
+
BaSO4 (s)
(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
+2
−
independently of each other. So Pb(NO3)2 actually exists in solution as one Pb ion and two NO3
+
−2
ions, and K2CrO4 as two K ions and one CrO4 ion. So our reactants, which previously were
+2
written as Pb(NO3)2 (aq) + K2CrO4 (aq) when writing an equation may also be shown as: Pb +
−
+
−2
2 NO3 + 2 K + CrO4
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
Page 5
freely floating in the solution after the reactants are mixed. Since we know a chemical change has
+2
+
−2
−
occurred, we may conclude that the Pb ion could combine with either K or CrO4 . The NO3
+
−2
ion could combine with either K or CrO4 . 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–
+
NO3–
NO3–
K+
CrO4–2
K+
K+
K+
NO3–
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.
+2
Pb
−
+
(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.
+2
Pb
−2
(aq)  PbCrO4 (s)
(strong elect.) (nonelectrolyte)
(aq) + CrO4
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.
Page 6
H+
+
Cl–
Na+
+
−
Na+
OH–
Cl–
H2O
+
−
+
−
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
2 K (aq) + 2 Cl (aq) + 2 Na (aq) + CO3
(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.
Page 7
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
Figure 1: A light bulb
consists of an electric
conductivity apparatus.
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
Figure 2: An LED
except the electrodes fit inside the well of a spot plate in which
aaiiconductivity indicator.
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
4
Very Bright
Medium
Very High
Strong Electrolyte
Page 8
PROCEDURE
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.
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)
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 ()
6. 0.1 M HC2H3O2 (aq)
12. C6H12 ()
S
S
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
Page 9
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.
Page 10
Electrolyte Table
Classification
Strong
Electolyte
Acids
Bases
Strong Acids:
Strong Bases:
HCl O4
HBr O4
HI O4
NaOH )2
*
KOH)2
HClO4
H2SO4
HNO3
Nonelectrolytes
Ca(OH)2
Ba(OH)2
*
Almost all soluble salts.
*
Saturated Ca(OH)2 = 0.02 M
All other acids.NH3
Weak
Electrolyte
*
Salts
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.
All covalent compounds
Solubility Table (Common Salts and Bases)
Ion
NO3
Solubility Behavior of Salts of Ion
–
All nitrates are soluble
–
C2H3O2
–
–
All acetates are soluble (AgC2H3O2 is moderately soluble.)
+
All chlorides, bromides, and iodides are soluble, except those of Ag ,
+2
+2
Hg2 and Pb
–
Cl , Br , I
SO4
All sulfates are soluble except PbSO4 and BaSO4. (CaSO4, Hg2SO4 and
Ag2SO4 are slightly soluble. The corresponding hydrogensulfates are
more soluble.)
–2
–2
CO3 ,
–
–3
PO4 ,
–2
C2O4
All hydroxides and oxides are insoluble, except NaOH, KOH, NH4OH,
Ba(OH)2, and Ca(OH)2. [Ca(OH)2 is moderately soluble.]
–2
OH , O
S
+
+
Na , K , NH4
+
Ag
+
+
+
+2
+2
All sulfides are insoluble except those of Li , Na , K , NH4 , Mg , Ca ,
+2
+2
Sr , and Ba
–2
+
+
All carbonates, phosphates and oxalates are insoluble except those of Li ,
+
+
+
Na , K , and NH4 . (Many hydrogenphosphates are soluble.)
+
+
+
+
All salts and bases of Na , K , and other ions of group 1A and NH4 are
soluble.
All silver salts are insoluble, except AgNO3 and AgClO4. (AgC2H3O2 and
Ag2SO4 are only moderately soluble.)
Page 11
Name ____________________________
LAB
REPORT
Ionic and Covalent
Substances and
Aqueous Reactions
Section _________ Date _____________
Instructor _________________________
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.
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 ()
Part B: Effect of Solvent
How does the solvent (cyclohexane versus water) affect the ability of HCl to act as an electrolyte?
Explain.
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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
Type of Elect.
1
Complete
Ionic
Net
Ionic
Reaction
Type of Elect.
2
Complete
Ionic
Net
Ionic
Reaction
Type of Elect.
3
Complete
Ionic
Net
Ionic
Reaction
Type of Elect.
4
Complete
Ionic
Net
Ionic
HCl __________
NaOH __________
HC2H3O2 ________
NH4OH __________ Mixture __________
CuSO4 __________
Na3PO4 __________
Mixture __________
Ba(C2H3O2)2
______
H2SO4 __________
Mixture __________
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Mixture __________
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:
mL Ba(OH)2
added to
H2SO4
Conductivity
(a) Conductivity behavior before Ba(OH)2 is added
0
1
(b) Conductivity behavior at equivalence point
2
3
4
(c) Conductivity behavior when excess Ba(OH)2 is added
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
____________________
(NH4)2SO4
____________________
Ba(C2H3O2)2 ____________________
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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) ___________________________________________________________________________
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