E
I
S
B
-B
I
By now you should be pretty good at writing chemical equations, determining limiting reactants and yields, and converting moles to grams or liters or any other unit. Today we’ll take a slightly different approach towards understanding chemical reactions. We’ll put down our pencils and calculators and just consider what the molecules are doing. We’ll look for qualitative results that don’t have a number or unit attached to them. You have seen qualitative chemistry in action in the discussion of Le Châtelier’s principle in lecture. Qualitative approaches like
Le Châtelier’s principle allow us to predict what happens when chemicals are mixed or reaction conditions change.
Primarily, we will examine reactions at equilibrium which form the basis for our qualitative understanding of chemistry. In a few weeks, we’ll see that the equilibrium position of a reaction is related to the energy exchange between products and reactants. Energy changes can be measured and assigned a number, so when we explore energy changes with thermochemistry and thermodynamics, we will be returning to a more quantitative description of chemical reactions and equilibria.
A qualitative understanding of how chemical reactions occur can help us design chemical processes which maximize the type of products we want while minimizing the type of products that we don’t want. For example you can use solubility rules to figure out how to remove contaminants from water. This week’s lab will help demonstrate how reactions can be controlled by the addition of reagents with known solubility. Additionally, we will see how equilibrium concepts can apply to the synthesis of a bio-based ink.
When you write a chemical equation, you are really describing two reactions; the forward reaction, which converts reactants into products , and the corresponding reverse reaction, which converts products into reactants . The back reaction prevents the reactants from completely converting to products, and in time, the forward and reverse reactions strike a balance and the reaction appears to stop. In reality, the forward and reverse reactions are still going. What does stop changing is the ratio of the amount of products to reactants because, at some point, for any reactants that complete the forward reaction some products complete the reverse reaction to keep the balance. This condition, where molecules are switching equally between products and reactants while maintaining a constant overall ratio, is called dynamic equilibrium.
Because the ratio of products to reactants doesn’t change at equilibrium, it is a constant and it can be measured. We give the ratio the symbol K and call it the equilibrium constant. In lecture, you’ll learn rules for writing equilibrium constant expressions. In this lab, we’re not interested in calculating numbers so the following description of the equilibrium constant will work fine. Let’s start with a hypothetical reaction that converts reactants to products. We can write:
Reactants

Products
K

Amount of
Amount of products at equilibriu reactants at equilibriu m m



Products
Reactants

 .
The square brackets, [ ], refer to the concentration of whatever is inside. Concentration is a nice unit to use for reactions in solution like those we’ll be looking at today. Different reactions yield different ratios of products to reactants, so K can be almost any number. For qualitative purposes however, the relative size of K is used to compare different reactions to determine the relative amounts of products and reactants present at equilibrium. For instance, a reaction that yields plenty of products always uses up most of its reactants. That means the numerator of
K (top) is large and the denominator (bottom) small making K itself a large number. Conversely, inefficient reactions that don’t produce much product have a small K because a large amount of reactants are left over in the denominator, while only a small amount of products appear in the numerator. Many reactions are not so clear cut.
49

The amounts of products and reactants at equilibrium are more similar so the ratio is closer to unity. Nevertheless, it’s the relative size of K, and not the actual number, that lets us compare and predict the extent of chemical reactions. Reactions that go nearly to completion have large K values whereas reactions that barely go at all have small K values. In between are reactions that reach equilibrium with similar amounts of products and reactants and have ‘medium sized’ Ks.
Here are a few reactions where K is very large (>10
10
). They are dissolution reactions, which mean the reaction shows a solid dissolving into ions in solution.
Pb(NO
3
)
2
Na
2
SO
4
NaNO
(s)
3

Pb
 Na
+
2+
(s)

2 Na
+
(aq) + 2 NO
(aq) + SO
4
2-
3
-
(aq)
(aq)
(aq) + NO
3
-
(aq)
The K for these reactions is so large that it is essentially unnecessary to use a double arrow; no significant back reaction occurs because the forward reaction totally dominates. In other words if you put solid sodium sulfate or lead nitrate in water, they dissolve completely to form the ion products shown. These are soluble salts. For other solids the forward reaction is extremely unfavorable and the reverse reaction dominates. For these solids, virtually nothing dissolves (no products are formed) and K is very small (<10
-10
). Lead sulfate and copper (II) hydroxide are examples of insoluble salts.
PbSO
4
(s)

Pb
2+
(aq) + SO
4
2-
Cu(OH)
2
(s)  Cu
2+
(aq)
(aq) + 2 OH
-
(aq)
We have made a simple qualitative classification of K here, but it can be used to predict the outcome of a chemical reaction between these compounds. For instance, if we mix a beaker of Pb(NO
3
)
2
solution with one of Na
2
SO
4 solution we know, because of the large Ks, the following ions are the mixture:
Pb
2+
, NO
3
-
, Na
+
, SO
4
2-
Notice that Pb
+2
and SO
4
-2
are now present in the same solution (they were in separate beakers before the mixing).
If these two ions combine, the solid PbSO
4
formed has a very small K for dissolution, and will not redissolve. In fact, we should predict based on the size of K that whenever Pb
+2
and SO
4
-2
meet in the solution, they combine to form solid PbSO
4
. Thus, the net result of mixing Pb(NO
3
)
2
and Na
2
SO
4
is complete precipitation of PbSO
4
. By using chemistry skills and some data, we have predicted the result of mixing two random chemicals, and discovered a way to remove lead from water!
Le Châtelier’s principle is the quintessential tool for qualitative analysis of chemical reactions. It states that reactions in equilibrium are dynamic and respond to stress or perturbations in a logical way. For instance, add more reactants to a reaction at equilibrium, and the reaction shifts to make more products and return to a balanced equilibrium. Remove some reactants and the reaction shifts into reverse to replace them. Heat can be a reactant or product too. A reaction that produces heat as a product shifts back to making reactants if you add more heat (by warming it). It will shift toward products if you remove heat (by cooling it). This makes perfect sense if you remember that chemical equilibrium is dynamic. The forward and back reactions are in a constant battle to maintain
K; if you disturb the equilibrium the forward and back reactions bring it back into balance.
We will put this principle to work in this investigation by coupling it with our understanding of the meaning of the size of K. Consider the following process. A beaker containing NaOH solution is mixed with a beaker containing
Cu(NO
3
)
2
. From Table 1, we can see that this is the same situation as the lead nitrate and sodium sulfate discussed before. There are ions in solution that can form an insoluble salt (a compound with a very small K for dissolution), namely Cu(OH)
2
(s). You will work out the details in the prelab. For now, the more important issue is whether we can influence the reaction and cause some more copper(II) hydroxide to go back into solution. Le Châtelier would say yes — here’s why. The dissolution for Cu(OH)
2
is
Cu(OH)
2
(s)

Cu
2+
+ 2OH
-
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Even though the K for dissolution is small, some Cu
+2
and OH
-
ions are in equilibrium with the solid. Our goal is to remove one of these ion products, say the OH-, from the Cu(OH)
2
equilibrium. That will shift the equilibrium to the right, dissolving more solid to replace the OH- and restore the balance. To remove OH-, we need to either physically remove it, or tie it up in some other equilibrium (i.e. react it to form something else). We can find a suitable reaction in two steps. First locate a reaction in Table 1 that has two properties: (i) it involves OH-, and (ii) it has a small K. The reaction
H
2
O  H
+
+ OH
is a good candidate. It indicates that if we can introduce some H
+
ions into our Cu(OH)
2
solution, they will react with OH
-
to form water. The OH
-
ions that become water can no longer participate in the Cu(OH)
2
equilibrium, so some solid copper(II) hydroxide dissolves to provide more and keep the balance. The second step in the process is to find a compound that dissolves into the H
+
we need in solution. This compound should have a large K for dissolution so it provides plenty of H
+
(you can see that water is not a good choice here because it provides H
+ only in small amounts). A quick glance at Table 1 provides several candidates in the ‘Strong Acid’ category.
, but
Addition of any of these will increase the solubility of copper(II) hydroxide by Le Châtelier’s principle.
This series of experiments is designed to help you think about chemical reactions. This is the most important skill that you can develop on your way to becoming a green chemist. So while this lab only directly exemplifies one principle of green chemistry it actually prepares you to apply many of the principles.
Renewable Feedstocks: The synthesis of ink from plant extract is a great of example of using a renewable feedstock. In fact for many years from the middle ages, through the beginning of the 20 th
century this type of ink was used extensively. Maestros including both Leonardo DaVinci and J.S. Bach wrote their masterpieces with ink produced from these materials.
Learning Le Châtelier’s principle, along with the basics of solubility and equilibrium, will help you design safer and more efficient processes. By avoiding heavy metals like Hg, Pb, Cr, and Cd you will be designing less harmful materials. For example, previous versions of this lab used chromium compounds instead of copper compounds to evaluate equilibrium. By understanding how acid-base indicators work you are developing the tools which will be used for future monitoring of reactions and pollutants.
51

Table 1: Selected Chemical Reactions: Dissociation and Dissolution
Reaction
HCl (aq)

H
+
+ Cl
-
H
2
SO
4
(aq)  H
+
+ HSO
4
-
NH
4
+
(aq)

H
+
+ NH
3
HInd (aq)

H
+
+ Ind
-
CH
3
COOH(aq)

H
+
+ CH
3
COO
-
C
4
H
6
O
6
(aq)

2 H
+
+ C
4
H
4
O
6
2-
H
2
O

H
+
+ OH
-
NaOH  Na
+
+ OH
-
K
2
CO
3

2 K
+
+ CO
3
2-
NaHCO
3
 Na
+
+ HCO
3
-
KCl

K
+
+ Cl
-
NaHSO
4

Na
+
+ HSO
4
-
KHSO
4
 K
+
+ HSO
4
-
Hg(CO
3
)  Hg
2+
+ CO
3
2-
Al
2
(CrO
4
)
3

2 Al
3+
+ 3 CrO
4
2-
Pb(NO
3
)
2

Pb
2+
+ 2 NO
3
-
Cu(SO
4
)

Cu
2+
+ SO
4
2-
AlCl
3
 Al
3+
+ 3 Cl
-
PbSO
4

Pb
2+
+ SO
4
2-
K
3
PO
4

3K
+
+ PO
4
3-
Cu(CO
3
)

Cu
2+
+ CO
3
2-
CdSO
4
 Cd
2+
+ SO
4
2-
CdSO
3

Cd
2+
+ SO
3
2-
Na
2
C
4
H
4
O
6

2 Na
+
+ C
4
H
4
O
6
2-
Al(OH)
3

Al
3+
+ OH
-
Hg(C
4
H
4
O
6
)  Hg
2+
+ C
4
H
4
O
6
2-
Cd(C
4
H
4
O
6
)

Cd
2+
+ C
4
H
4
O
6
2-
K
2
CrO
4

2 K
+
+ CrO
4
2-
Hg(NO
3
)
2

Hg
2+
+ 2 NO
3
-
K
Huge
Medium
Medium
Tiny
Huge
Tiny
Huge
Medium
Tiny
Small
Huge
Huge
Huge
Huge
Tiny
Tiny
Huge
Classified As
Strong acid
Strong acid
Weak acid
General form for acidbase indicators
Weak acid
Weak acid
Strong base
Moderate base, soluble salt
Weak base
Very soluble salt
Very soluble salt
Insoluble salt
Very soluble salt
Very soluble salt
Insoluble salt
Soluble salt
Slightly soluble salt
Very soluble salt
Insoluble salt
Slightly soluble salt
Very soluble salt
Name of Compound
Methlyorange and thymophthanein
Acetic acid (vinegar)
Tartaric acid
Water
Potassium carbonate
Sodium bicarbonate
Sodium bisulfate
Aluminum chromate
Potassium phosphate
Cadmium Sulfite
Mercury tartarate
52

Reaction
K
2
SO
3

K
+
+ SO
3
2-
Cu(OH)
2

Cu
2+
+ 2 OH
-
K
Tiny
Classified As
Soluble
Name of Compound
Copper(II) Hydroxide
53

The Problem
Observe, predict and record the response of reactions in equilibrium.
The Approach
Work in groups of two.
In this qualitative lab, your observations and predictions should be recorded in your lab book. With a little thought and help from Table 1, you should be able explain everything you observe in terms of equilibria and Le Châtelier’s principle.
Because this is a qualitative investigation, the exact amounts of the solutions you work with are not critical. You’ll be transferring around 1 mL of solution. You can help in your estimates by putting 1 mL of water in a test tube to get an idea of what the volume looks like. Don’t use too much of any solution. After all waste prevention is the cornerstone of green chemistry.
Equipment Needed:
Test tubes
10 mL graduated cylinder
Pipettes
Chemicals Needed:
CuSO
4
Na
2
CO
3
1) Dissolve 100 mg of CuSO
4
in 10 mL of water. Write down the chemical reactions and any observations you have in the boxes provided below.
2) Dissolve 100 mg of Na
2
CO
3
in 5 mL of water. Write down the chemical reactions and any observations you have in the boxes provided below.
3) Place 5-10 drops of the CuSO
4
into a medium test-tube. Fill the test-tube ¾ full with deionized water.
4) After all of the Na
2
CO
3
has dissolved, start adding a few drops of the Na
2
CO
3
solution to the diluted CuSO
4 solution. Write down the chemical reactions and any observations you have in the boxes provided below.
54

An acid is a compound that ionizes to form H
+
ions in water. A base is a compound that ionizes to form OH
-
ions in water. Several acid dissociation reactions are shown in Table 1. Weak acids have a small or medium K. Some weak acids also have a different color depending on whether they are ionized (products) or neutral (reactants). This is handy for equilibrium problems because you can see when the equilibrium shifts by the color change in the solution. For this reason we call acids that change color “indicators”. The general reaction for an indicator in solution is given in Table 1 where HInd represents the neutral species (we don’t care exactly what it is) and H
+
and
Ind
-
are the ions that form in solution. A weak base either reacts with H
+
, or produces OH
-
with a small K for the reaction.
The Problem
Determine the color of the HInd and Ind
- species in solution for two indicators and investigate acid/base equilibria.
The Approach
Chemicals Needed
Methyl Orange
Thymolphthalein
0.1
M HCl
0.1 M Acetic Acid (vinegar)
0.1 M NaOH
0.1 M Ammonium Hydroxide
1 M Sodium Acetate
1 M Ammonium Chloride
1.
Place several drops of methyl orange indicator in each of three small test tubes.
2.
To the first tube add about 1 mL of 0.1 M hydrochloric acid (HCl). Record what happens.
3.
To the second tube add about 1 mL of 0.1 M acetic acid (CH3COOH). Record what happens.
4.
To the third tube add about 1 mL of 0.1 M sodium hydroxide (NaOH). Record what happens.
5.
Using the information in Table 1 and Le Châtelier's Principle determine the color of the HInd complex and the
Ind
-
species.
6.
Predict what will happen when you add 1 M sodium acetate (CH
3
COONa) to the contents of the second tube.
Hint: write the reaction of the acetate ion (CH
3
COO-) and water.
7.
Add 1 M sodium acetate drop wise to the second tube. Record your observations.
55

1.
Place several drops of thymolphthalein indicator into each of three new test tubes. Record your observations.
2.
To the first tube add about 1 mL of 0.1 M hydrochloric acid (HCl). Record what happens.
3.
To the second tube add about 2 mL of 0.1 M ammonium hydroxide (NH
4
OH). Record what happens.
4.
To the third tube add about 1 mL of 0.1 M sodium hydroxide (NaOH). Record what happens.
5.
Using Le Châtelier’s Principle and the information in Table 1. Determine the color of each indicator species.
Hint: It might help to write the ammonia reaction as the reaction between NH3 and water, rather than as the dissociation of NH
4
OH.
6.
Divide the contents of the second tube into two parts by pouring about half of its contents into a new test tube.
Call these tubes "2a" and "2b."
7.
Predict what will happen to these tubes when hydrochloric acid (HCl) is added to tube 2a, and ammonium chloride (NH
4
Cl) is added to tube 2b.
8.
Add a few drops of 1M hydrochloric acid to tube 2a, and a few drops of 1M ammonium chloride to tube 2b.
Record your observations.
56

Chem Connections
This reaction involves another type of reaction equilibrium, a mixture of iron and tannins to make ink. In this investigation, you will be able to manipulate the equilibrium of oxygen reacting with a chemical to produce a colored compound. Many plants contain tannins, large molecules that are often brown in color. Tea is a good example of a brown solution of tannins. In this experiment you will use a very concentrated source of tannins, the oak gall, which is a knotty growth on the plant.
For this investigation it is helpful to know that in the presence of oxygen the following can occur:
FeSO
4
(s)

SO
4
2-
(aq) + Fe
2+
(aq) and then 2 Fe
2+
(aq) + 3/2 O
2
(g)

FeOOH (aq),
Oxygen in the air will slowly turn the ink darker but the process can be sped up by that addition of hydrogen peroxide which decomposes readily to form oxygen and water.
2H
2
O
2
(aq)  2H
2
O (l) + O
2
(g)
FeOOH and other Fe
3+
compounds that you will be making in the ink solution have different colors and are less soluble when compared to the FeSO
4
(Fe
2+
) that you start with. Without tannins from the oak galls the Fe solution is a reddish brown color, but in the presence of oak tannins the Fe
3+
3+
in
compounds turn a deep purplish black color and begin to precipitate.
The overall process can be simply represented as [tannin-Fe
2+
] (brown)  [tannin-Fe
3+
] (black)
The Problem
Synthesize and use a biobased ink. Understand the relationships between solubility, equilibrium, and oxygen, as they relate to the synthesis of a biobased ink.
The Approach
Chemicals Needed:
FeSO
4
Tea/Oak gall extract solution
3% H
2
O
2
solution
1.
In a 50 mL beaker add 10 mL of DI water.
2.
Dissolve 0.2 g of FeSO
4
in the water.
3.
Obtain 1 mL of a solution prepared by steeping Tea/Aleppo Galls in water for 12 hours.
4.
Mix the two solutions and record your observations.
5.
Using a pipette or spatula, try and use this ink to write on a sheet of paper. Set this paper aside and observe changes over the course of the next few minutes.
6.
Now add a few drops of your ink to a medium test tube. Fill the test tube ¾ full with water.
7.
Try adding a few drops of dilute hydrogen peroxide (H
2
O
2
) which acts as a good source of soluble oxygen.
In solution 2H
2
O
2

2H
2
O + O
2
. Record your observations as you add the hydrogen peroxide to the ink solution.
57

Report Sheet Name___________________________
For the reactions you performed in lab, write the relevant chemical equations and describe what you observed.
You do not have to write net ionic equations; just explain what happens based on Le Chatelier’s Principle.
Write the color of each species in solution under that term in the chemical reaction.
Investigation 1: Solubility of Copper Salts
(A) Prepare a solution of copper sulfate
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
(B) Prepare a solution of sodium carbonate
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
(C) Mix the solutions of copper sulfate and sodium carbonate
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
58

Investigation 2: Weak Acids and Bases
Indicator 1: Methyl Orange
(A) Starting with a solution of an acid-base indicator, use your observations of the next few experiments to determine the color of HInd and Ind
-
HInd

Ind
-
+ H
+
(B) Hydrochloric acid was added to methyl orange
Description of observations List the relevant reactions Chemically, what caused the results that you observed?
(C) Acetic acid was added to methyl orange
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
(D) Sodium hydroxide was added to methyl orange
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
59

(E) Sodium acetate was added to the contents of the 2 nd
tube
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
Indicator 2: Thymolphthalein
(A) Starting with a solution of an acid-base indicator, use you observations of the next few experiments to determine the color of HInd and Ind
-
HInd  Ind
-
+ H
+
(B) Hydrochloric acid was added to thymolphthalein
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
(C) Ammonium hydroxide was added to thymolphthalein
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
60

(D) Sodium hydroxide was added to thymolphthalein
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
(E) Ammonium chloride was added to the mixture above (C)
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
(F) Hydrochloric acid was added to the mixture above (C)
Description of observations Chemically, what caused the results that you observed?
List the relevant reactions
Investigation 3: A biobased ink
(A) A FeSO
4
solution
Description of observations Identify the all of the ions in solution.
List the relevant reactions
61

(B) Add the solution of Galls
Description of observations Chemically, what caused the results that you observed? (no reactions necessary)
(C) Writing with the ink.
Description of observations What process is occurring over time with added oxygen? Explain.
Attribution
"Experiment was borrowed in large part from a University of California-Berkley Laboratory Manual and used with permission of the author; Dr. Michelle Douskey."
62

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