1. Chapter 4
Reactions in Aqueous Solution

2. Solutions
A solution is a homogeneous mixture composed of a solute and a solvent. The solute is dissolved in the solvent.
A solute is the substance present in smaller amount in a solution. It can also be thought of as the dissolved material.
A solvent is the substance present in larger amount in a solution. It can also be thought of as the dissolving material.
A solution can be gaseous (air), solid (alloy), or liquid (seawater). In this chapter, we will only discuss aqueous solutions, in which the solute initially is a liquid or a solid and the solvent is water.
First, if it is not clear what the term dissolve means, then I should define it. Dissolve means to melt or liquefy. Basically, it changes the physical state of the matter. A solution is a homogeneous mixture. Therefore, the solute must dissolve in the solvent in order to have a constant composition of solute and solvent throughout the mixture.
An example of a solution would be salt water. The solute would be the salt and the solvent would be the water.
The way in which I memorize the terms solute and solvent is that I think of the fact that solute has a u in it and solvent has a v in it. Because u comes before v in the alphabet, it makes sense that solute would be what you would add to a solvent in order to dissolve the solute.
You must memorize the terms solute, solvent and solution. And you must know that the term molarity is used in order to describe the amount of solute inside a solution. The higher the amount of solute, the higher the molarity and the higher the concentration of the solution will be.
Also, you should all know that the terms concentrated and dilute are relevant to the amount of solute dissolved in a solvent. Concentrated solutions have a high amount of solute inside of a solvent. Dilute solutions have a low amount of solute inside of a solvent.

3. Molarity and Dilution
Molarity (M) is the number of moles of the solute per liters of solution.
Molarity (M) = moles of solute = mol
Liters of Solution L
Molarity (M) is the number of moles of the solute per liters of solution (mol/L) only in these units!!!!
Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. This means that it would have the same number of moles, but less volume, therefore it’s Molarity would be stronger
In the picture of 2 different solutions of Potassium Permanganate, you can see that the solution on the left has more moles per liter, more concentrated = higher M, because it has a stronger color and the darker the color, the more moles it has per the same size of Liters.
Click on the box to see an example of how dilution only changes the volume of the solution, not the moles of the solution.

4. When Diluting use: M1V1 = M2V2
Click on the blue box to see an example of creating a known Molarity concentration of solution.
Also memorize the equation M1V1 = M2V2 in order to dilute a known molarity of higher concentration (this would be considered M1 and V1) to a known volume of lower molarity (this would be considered M2 and V2).
When Diluting use: M1V1 = M2V2

5. Hydration
Water is a very effective solvent for ionic compounds. Although water is an electrically neutral molecule, it has a positive (H atoms) and negative (O atoms) region, or “positive and negative poles”. This is why water is called a polar solvent.
When an ionic compound, such as NaCl, dissolves in water, the three dimensional network of ions in the solid is destroyed. The Na+ and the Cl- ions are separated from one another and they undergo hydration.
Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. Let’s see this in action.

6. NaCl(s) H2O Na+(aq) + Cl-(aq)
The term Dissociation means that the compound breaks up into cations and anions like in the above equation for salt.
Solid NaCl, salt, is an ionic compound and breaks up into Na+ and Cl-, cations and anions when dissolved in water. The Na+ ions are attracted to the negative electrode and the Cl- anions are attracted to the positive electrode. This movement sets up an electric current that is equivalent to the flow of electrons along a metal wire.
Because NaCl conducts electricity, we say that NaCl is an electrolyte.
Pure Water contains very few ions and therefore we call it a nonelectrolyte.
The above equation also shows that all of the salt has dissociated into ions and there is no undissociated NaCl left over in the solution. This would be the same as saying that salt is a very strong electrolyte.

7. Ionic vs. Molecular Compounds Dissolving in Water
As we have already talked about in earlier chapters, Molecular Compounds dissolve in water but they break apart into molecules floating around in the water and therefore they do not have a charge. This is why they are called Molecular Compounds!
Ionic Compounds break apart into ions when they are dissolved in water. The ions have a charge in the solution and are good electrolytes! Remember all bases are ionic except ammonia!
Exception: Acids and the weak base ammonia (NH3) are considered molecular compounds but they do break apart into ions in water:
NH3 + H2O D NH OH-
HCl + H2O D H3O+ + Cl-
Also NOTE: Dissolving in water does not make something a strong electrolyte (think of sugar = molecular and dissolves in water, it is not a strong electrolyte!)

8. A nonelectrolyte does not conduct electricity when dissolved in water.
All solutes that dissolve in water fit into one of 2 categories: electrolytes and nonelectrolytes.
An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.
A nonelectrolyte does not conduct electricity when dissolved in water.
A pair of inert electrodes (copper or platinum) is immersed in a beaker of water. To light the bulb, electric current must flow from one electrode to the other, thus completing the circuit. Pure water is a very poor conductor of electricity, such as shown in (a), also remember they are talking about pure D.I. water or deionized water. However, if we add a small amount of sodium chloride, the bulb will glow as soon as the salt dissolves in water, such as shown in (b).
Solid NaCl, salt, is an ionic compound and breaks up into Na+ and Cl-, cations and anions when dissolved in water. The Na+ ions are attracted to the negative electrode and the Cl- anions are attracted to the positive electrode. This movement sets up an electric current that is equivalent to the flow of electrons along a metal wire.
Because NaCl conducts electricity, we say that NaCl is an electrolyte.
Comparing the lightbulb’s brightness for the same molar amounts of dissolved substances helps us distinguish between strong and weak electrolytes. A characteristic of strong electrolytes is that the solute is assumed to be 100 percent dissociated into ions in solution. (By dissociation we mean the breaking up of the compound into cations and anions).
You can see that in (c) a stronger electrolyte is used than in (b)

9. Strong Electrolyte Weak Electrolyte Nonelectrolyte CH3COOH HF HNO2 NH3
Always USE g arrow
Always USE D arrow
Strong Electrolyte
Weak Electrolyte
Nonelectrolyte
CH3COOH HF
HNO2
NH3
H2O*
(all of the above loose an H+ ion when dissociated)
*Pure water is an extremely weak electrolyte
(NH2)2CO (urea)
CH3OH (methanol)
C2H5OH (ethanol)
C6H12O6 (glucose)
C12H22O11 (sucrose)
HCl
HBr HI
HNO3
HClO4
H2SO4*
All 1A: (LiOH, NaOH,
KOH, RbOH, CsOH)
2A: Ba(OH)2 & Sr(OH)2
Ca(OH)2
Ionic Compounds
H2SO4 has 2 ionizable H+ ions, the second form, HSO4- is a weak electrolyte.
The strong/weak parts of this chart should be memorized because it will help you to memorize your strong/weak acids and bases.
This slide gives you some examples of strong and weak electrolytes and nonelectrolytes. Ionic compounds, such as potassium iodide, KI and calcium nitrate, Ca(NO3)2 are strong electrolytes because of hydration and the water being able to pull apart and stabilize the anions from clumping together.

10. Ionization of Acids and Bases
We use the term Ionization in order to describe the separation of acids and bases into ions.
In order to determine whether or not you have a strong acid or base, you see whether the acid or base dissociates completely in water. If it does completely ionize in water, then it is considered to be a strong acid or base. This would also make it a Strong Electrolyte. To symbolize a strong electrolyte you use a single arrow g. Such as: HCl(aq) g H+(aq) + Cl-(aq)
When an acid or base does not completely ionize in water, it is a weak acid/base. These are also called Weak Electrolytes.
Acetic acid is a weak acid and we represent the ionization of acetic acid with a double arrow D to show that it is a reversible reaction, or that the reaction can occur in both directions.
CH3COOH(aq) D CH3COO-(aq) + H+(aq)
The ionization of acetic acid is written with a double arrow to show that it is a reversible reaction. This means that the reaction can occur in both directions.
Initially, a number of the CH3COOH molecules break up into CH3COO- and H+ ions. As time goes on, some of the CH3COO- and H+ ions recombine into CH3COOH molecules. Eventually, a state is reached in which the acid molecules ionize as fast as the ions recombine.
Such a state is called chemical equilibrium in which no net change can be observed (even though activity is still occurring at the molecular level).

11. Solubility and Precipitates
Solubility is a term that is used in order to describe the amount of solute that will dissolve in a solvent.
Most mixtures have a certain amount of solute that will dissolve inside of a solvent.
Once the maximum amount of solute particles have been added to a solvent, the solute particles will no longer dissolve in the solvent and they will turn into the solid form of the particle.
This change from liquid to solid is called a precipitate.
If something is considered to be insoluble in water, then it means it has a solubility of less than .01 M in water at 25 degrees Celsius.
When understanding solubility, I always think of the analogy of a human lifting free weights. For example, lets say that I can physically only pick up 50 lbs of weights with both of my hands. I can suspend the weights in the air and not allow them to touch the ground as long as I have equal to or less than 50 lbs of weight. Once I have more than 50 lbs of weight, I can no longer hold up the excess weights and I will have to let anything above 50 lbs drop. Therefore, the excess of 50 lbs will fall to the floor.
It works the same way with solubility. A homogeneous mixture will contain a certain amount of solute and solvent and they will both be in the same state, liquid. Think of the solute as being the weights and think of the solvent as being the human that can lift them into the air. The solvent can only lift so many of the solute particles and maintain a homogeneous mixture. Once the maximum amount of solute particles has been reached, by adding any more solute particles, or weights, the solvent particles must “drop” the extra solute particles and then the extra solute particles will fall to the floor. It will then no longer be a homogeneous mixture and you will no longer have two of the same states of liquid. The excess solute particles will turn into solid and will be dropped to the bottom of the container. This makes the mixture now heterogeneous. The particles of solute are now a solid are called a precipitate.
Please keep in mind that even if a precipitate does form. Their will still be the maximum amount of solute particles floating around inside of the solvent. This is the same as thinking that the person will still be holding up the 50 lbs that she can hold up! The excess of 50 lbs is the part that will become a precipitate and will drop to the ground. The excess solute particles above the maximum solubility will be the solute particles that will change into a solid precipitate.
Let’s take a look at the solubility of salt by clicking on the purple circle.
Lets click on the red box to take a microscopic look into what dissolving NaCl in water looks like.
Now let’s click on the green circle to see an example of what a different precipitate looks like when it forms in a solution, besides NaCl.

12. Solubility Rules for Common Ionic Compounds in Water at 25 oC
Soluble Compounds Exceptions
Insoluble Compounds Exceptions
Compounds containing alkali metal ions (Li+, Na+, K+, Rb+, Cs+) and the ammonium ion (NH4+)
Nitrates (NO3-), bicarbonates (HCO3-), and chlorates (ClO3-)
Halides (Cl-, Br-, I-)
Sulfates (SO42-)
Memorize these Rules!!
Halides of Ag+, Hg22+, and Pb2+
Sulfates of Ag+, Ca2+, Sr2+, Ba2+, Hg2+, and Pb2+
MEMORIZE THESE RULES!!!!
We use this table to verify whether some common ionic compounds will be soluble or insoluble in water.
Remember though, that even insoluble compounds dissolve to a certain extent.
Let’s do some examples:
Ag2SO4 is insoluble according to the table because Ag+ has the exceptions of sulfates being insoluble.
CaCO3 is an insoluble compound because Ca is not an alkali metal ion and the carbonate anion is considered insoluble.
Na3PO4 is soluble because although PO4 is a phosphate and is considered insoluble, Na+ is an alkali metal ion and therefore it is an exception to the rule.
Lets try and figure some other examples out by clicking on the green box.
Carbonates (CO32-)
Phosphates (PO43-)
Chromates (CrO42-)
Sulfides (S2-)
Hydroxides (OH-)
Compounds containing alkali metal ions and the ammonium ion
Compounds containing alkali metal ions, ammonium ion and the Ba2+ ion

13. Molecular Equations and Ionic Equations
Pb(NO3)2(aq) + 2NaI(aq) g PbI2(s) + 2NaNO3(aq)
The above equation is considered a molecular equation because the formulas of the compounds are written as though all species existed as molecules or whole units. A molecular equation is useful because it identifies the reagents, if we wanted to bring about this reaction in the lab.
However, a molecular equation does not accurately describe what actually is happening at the microscopic level.
An ionic equation is used to represent what is occurring on the microscopic level. The ionic equation shows dissolved species as free ions. The above molecular equation would have an ionic equation such as:
Pb2+(aq) + 2NO3- (aq) + 2Na+(aq) + 2I-(aq)g PbI2(s) + 2Na+(aq) + 2NO3- (aq)

14. Spectator Ions and Net Ionic Equations
An ionic equation includes spectator ions.
Spectator ions are ions that are not included involved in the overall reaction.
Spectator ions appear on both sides of the equation and are unchanged in the chemical reaction, therefore they can be canceled. In the previous ionic equation:
Pb2+(aq) + 2NO3- (aq) + 2Na+(aq) + 2I-(aq) g PbI2(s) + 2Na+(aq) + 2NO3- (aq)
You would cancel the 2NO3- and the 2Na+ spectator ions. The net ionic equation, which shows only the species that actually take part in the reaction, would be:
Pb2+(aq) + 2I-(aq) g PbI2(s)
Note: If everything in a net ionic equation is a spectator ion (and cancels on both sides) then you say: No Reaction Occurred!!
Use the following steps to write out ionic and net ionic equations:
Write a balanced molecular equation for the reaction.
Rewrite the equation to show the dissociated ions that form in solution. We assume that all strong electrolytes, when dissolved in solution, are completely dissociated into cations and anions. This procedure gives us the ionic equation.
Identify and cancel spectator ions on both sides of the equation to arrive at the net ionic equation.
Try it for this example: Predict the products of the following reactions and write a net ionic equation for the reaction of 2K3PO4(aq) + 3Ca(NO3)2(aq) -> ?
Balanced molecular equation is:
2K3PO4(aq) + 3Ca(NO3)2(aq) -> 6KNO3(aq) + Ca3(PO4)2(s)
Ionic equation is:
6K+(aq) + 2PO4^3- (aq) + 3Ca2+(aq) + 6NO3^ - (aq) -> 6K+(aq) + 6NO3^ - (aq) + Ca3(PO4)2(s)
Canceling the spectator ions K+ and NO3^ - , we obtain the net ionic equation:
2PO4^3- (aq) + 3Ca2+(aq) -> Ca3(PO4)2(s)
Note that because we balanced the molecular equation first, the net ionic equation is balanced.
Lets take a look at a Net Ionic Equation situation in a Lab, click on the blue box.
Here we see the eqn:
K2CrO4(aq) + Pb(NO3)2(aq) -> PbCrO4(s) + 2KNO3(aq)
The net ionic equation is:
CrO4^2- (aq) + Pb^2+ (aq) -> PbCrO4(s)

15. Acids and Bases
Arrhenius defined an acid as producing a H+ ion when dissolved in water. He also defined a base as producing an OH- ion when dissolved in water.
This definition was limiting because it only applied to aqueous solutions.
The chemists Johannes Bronsted and Thomas Lowry proposed a more broad definition. Their definition does not require that an acid be in an aqueous solution and includes more than just protons and hydroxide ions for acids and bases.
A Bronsted-Lowry acid is a proton (H+) donor.
A Bronsted-Lowry base is a proton (H+) acceptor.

16. Hydrochloric Acid Water Hydronium Ion Chloride Ion
Ammonia Water Ammonium Ion Hydroxide Ion
This slide shows a great example of why we would want to use the Bronsted Lowry definition of an acid and a base. Notice that the ionization of HCl in water only shows the hydronium ion as the product, not the proton. Which is the acid and which is the base? Hydrochloric Acid is the acid because it is the proton donor, water is the base because it is the proton acceptor.
For the ionization of Ammonia in water we see that water is the acid (proton donor) and ammonia is the base (proton acceptor).
The proton H+ does not exist in nature.
It has such a strong attraction to the negative pole of the oxygen within a water molecule that it will attach itself right away to the water molecule. Therefore, the H+ proton ion alone does not exist in nature. It is instead represented by a hydrated proton, H30+, which is called a hydronium ion. Lets look at the movie click on the red box.
An example is that the equation that follows does not actually exist in nature:
HCl(aq) -> H+(aq) + Cl-(aq)
Instead you what exists is the following:
HCl(aq) + H20(l) -> H3O+(aq) + Cl-(aq)
But, for shorthand, we use the proton H+. Remember though that it does not actually exist in nature this way! When we write the shorthand H+, we are actually talking about a hydronium ion H3O+.

17. Monoprotic, Diprotic and Triprotic Acids
When each unit of acid only produces one hydrogen ion upon ionization (or hydronium ion), that acid is said to be a monoprotic acid. Examples are: hydrochloric (HCl), nitric (HNO3) and acetic acid (CH3COOH). These are common acids you should memorize.
HNO3(aq) g H+ + NO3-(aq) (Strong Acid/Electrolyte)
When each unit of acid produces 2 hydrogen ions upon ionization (or hydronium ions), that acid is said to be a diprotic acid. Example: Sulfuric Acid (H2SO4). This is also a common acid that you should memorize.
H2SO4(aq) g H+(aq) + HSO4-(aq) (Strong Acid/Electrolyte)
HSO4-(aq) D H+(aq) + SO42-(aq) (Weak Acid/Electrolyte)
When each unit of acid produces 3 hydrogen ions upon ionization (or hydronium ions), that acid is said to be a triprotic acid. Example: Phosphoric Acid (H3PO4). This is also a common acid that you should memorize.
Let’s try and represent the triprotic acid of phosphoric acid on our own.
Make sure you know which of the common acids are strong and which are weak acids/electrolytes. Also, you should know that any diprotic or triprotic acid will have weak acid/electrolytes for all of the second and third reactions that occur.

18. Acid-Base Neutralization
A neutralization reaction is a reaction between an acid and a base. Generally, aqueous acid-base reactions produce water and a salt, which is an ionic compound made up of a cation other that H+ and an anion other that OH- or O2-:
acid + base g salt + water
HCl(aq) + NaOH(aq) g NaCl(aq) + H2O(l)
All salts are strong electrolytes. The substance we know as table salt, NaCl, is a familiar example.
However, since both the acid and the base are strong electrolytes, they are completely ionized in solution. The ionic equation is:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) g Na+(aq) + Cl-(aq) + H2O(l)
Therefore, the net ionic equation is:
H+(aq) + OH-(aq) g H2O(l)
Both Na+ and Cl- ions are spectator ions.
If we had started the reaction with equal molar amounts of the acid and the base, at the end of the reaction we would have only a salt and no leftover acid or base. This is a characteristic of acid-base neutralization reactions.
Lets try some other examples (need to add aq to everything and l to water):
HF + KOH -> KF + H20
H2SO4 + 2NaOH -> Na2SO4 + 2H20
HNO3 + NH3 -> NH4NO3
The last reaction does not look like it should be an acid-base neutralization because you do not see water as the product. However, remember the ionization of ammonia in water?
The NH3 in the reactant side is aqueous, therefore it would react with water to form NH4 + and OH-.
The equation then looks like:
HNO3 + NH4^+ + OH- -> NH4NO3 + H2O
Let’s look at some acid-base neutralizations in the lab.
Click on the red box:
This shows the reaction of : HCl(aq) + NH3(aq) -> NH4+ (aq) + Cl-(aq) The indicator shows the change of pH when the neutralization occurs.
Lets look at how they would naturally appear, without indicators to show their neutralization. Click on the green and blue boxes.

19. Acid-Base Titration
In titration, a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete.
The point at which the acid has completely reacted with or has been neutralized by the base is called the equivalence point.
The endpoint is the point at which the solution should change in color due to the indicator, which changes color at or near the equivalence point.
Many times, solid sodium hydroxide is not pure because of its absorbance of water from the air. Therefore, when we make up solutions of sodium hydroxide in the lab, we are unsure of it’s exact molarity and we must standardize the solution. Standardizing the solution is when we use the acid-base titration technique in order to verify what the molarity of the unknown solution is.
Quantitative studies of Acid-Base neutralization reactions are most conveniently carried out using a technique known as titration.
One indicator that is commonly used is phenolphthalein. This is a colorless solution in acidic and neutral solutions but it is pink in basic solutions. If you add the acid and base together till it is neutral, then you add one more drop of base to the solution, the indicator will turn pink because you have made the solution basic.

20. Acid-Base neutralization reactions are most conveniently carried out using a technique known as titration.
In titration, a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete.
Typically, you place the acid in the beaker with an indicator, such as phenolphthalein, which will indicate when the solution has become basic by turning pink. The base is slowly added until a faint pink color is seen. At this point, you stop the reaction and calculate the volume of base and acid that was used. Then if you know the molarity of either the acid or the base, you can calculate the molarity of the unknown base or acid using stoichiometry.
The point at which the moles of acid have neutralized the moles of base and the pH is equal to 7 (neutral) is called the equivalence point.
Work out the following titration reaction question:
In a titration experiment, a student finds that g of KHP (KHC8H404) is needed to completely neutralize mL of a NaOH solution. What is the concentration (in Molarity) of the NaOH solution?
Given: KHC8H404(aq) + NaOH(aq) -> KNaC8H4O4(aq) + H20(l)
Moles of KHP = g KHP x (1 mol KHP/ g KHP) = x 10^ -3 mol KHP
1 mol KHP = 1 mol NaOH, therefore there must be x 10^ -3 mol NaOH in mL of NaOH solution.
The molarity of the NaOH must be:
Molarity of NaOH = (2.678 x 10^ -3 mol NaOH/23.48 mL) x (1000 ml / 1 L) = M NaOH

21. 2NaOH(aq) + H2SO4(aq) g Na2SO4(aq) + 2H2O(l)
Diprotic Titrations
How many milliliters of a M NaOH solution are needed to neutralize 20.0 mL of a M H2SO4 solution? The equation for the reaction is:
2NaOH(aq) + H2SO4(aq) g Na2SO4(aq) + 2H2O(l)
Notice that we need twice the amount of sodium hydroxide in order to neutralize the diprotic sulfuric acid. This is because:
H2SO4(aq) g H+(aq) + HSO4-(aq)
In the above equation, one mole of OH- ions will neutralize one mole of the H+ ions.
HSO4-(aq) D H+(aq) + SO42-(aq)
Therefore, we need a total of 2 moles of NaOH in order to neutralize one mole of H2SO4
Let’s Finish this problem on our own….
Answer to the problem:
Moles of H2SO4 = (0.245 mol H2SO4/1 L soln) x (1 L soln/1000 mL soln) x (20.0 mL of soln) = 4.90 x 10 ^ -3 mol H2SO4
Moles of NaOH reacted = (4.90 x 10 ^ -3 mol H2SO4) x (2 mol NaOH/1mol H2SO4) = 9.80 x 10 ^ -3 mol NaOH
Volume of M NaOH necessary to neutralize the H2SO4 = (9.80 x 10 ^ -3 mol NaOH) x (1 L NaOH / mol NaOH) x (1000 mL / 1 L) = 16.1 mL of M NaOH

22. Gas Forming Reactions
Many reactions release a gaseous product. Although a wide variety of these gas-forming reactions occur, some of the most important gases produced in reactions are the following:
Acid/Base Rxns that form Gases: S2- (Sulfides) and CO32-(Carbonate), HCO3-(Bicarbonate) ions react with acids to form gases:
CO2 Gas formation: A 2 Step Reaction:
1 step – Any CO32- or HCO3- ion reacting with an acid gives Carbonic Acid (H2CO3):
Example: HCl(aq) + NaHCO3(aq) g NaCl(aq) + H2CO3(aq)
2 step – Carbonic Acid is unstable and will decompose immediately into CO2 gas and liquid H2O:
Example: H2CO3(aq)g CO2(g) + H2O(l)
H2S Gas (smells like rotten eggs) formation: forms when an acid (like HCl) reacts with a Metal Sulfide (like Na2S):
Example: 2HCl(aq) + Na2S(aq) g H2S(g) + 2NaCl(aq)
H2 Gas forms in a single replacement (REDOX) reaction when an acid reacts with a metal:
Example: Ca(s) + 2HCl(aq) g CaCl2(aq) + H2(g)
Other Rxns that form Gases:
O2 formed in many ways, one example: Electrolysis of water
NO2 formed when in air when lightning hits to supply energy:2NO(g) + O2(g) g 2NO2(g)
Click on the Carbon Dioxide Box first The first reaction shows the production of carbon dioxide with an antacid. This reaction is: The dissolved components of the antacid is : C6H8O7(aq) + HCO3-(aq) -> C6H7O7-(aq) The acid neutralizing product + H20(l) + CO2(g)
Next, click on the red box, this shows the example of another carbon dioxide forming gas reaction:
CaCO3(s) limestone + 2HCl(aq) -> CaCl2(aq) + H2CO3(aq) then the further rxn: H2CO3(aq) -> CO2(g) + H2O(l)
Click on the Hydrogen gas producing box, this rxn occurs: 2Na(s) + 2H2O(l) -> H2(g) + 2NaOH(aq)
Click on the Nitrogen Dioxide producing reaction: Zn(s) + 2NO3-(aq) + 4H+(aq) -> 2NO2(g) + Zn^2+(aq) + 2H20(l)
Notice that the Zinc is oxidized because it looses to electrons.
Next, click on the Oxygen box, this is the gas producing oxygen reaction of photosynthesis.
CO2(g) + H2O(l) + sunlight -> O2(g) + [CH2O](s)

23. Rules for Assigning Oxidation States
The oxidation state of an atom in an element is 0. Atoms in their elemental form are 0, example H2 is the elemental form of Hydrogen therefore in the H2 molecule, each H = 0 (all diatomic atoms are the same) or in P4 each P = 0 or in S8 each S = 0.
In a neutral molecule, the sum of the oxidation numbers of all the atoms must equal 0.
The oxidation state of a monatomic ion is the same as its charge.
In its compound, fluorine is always assigned an oxidation sate of -1.
Oxygen is usually assigned an oxidation of -2 in its covalent compounds, such as CO, CO2, SO2. Exceptions to this rule includes peroxides (compounds containing the O22- group), where each oxygen is assigned an oxidation state of -1, as in hydrogen peroxide (H2O2), and OF2 in which oxygen is assigned a +2 oxidation state.
In its covalent compounds with nonmetals, hydrogen is assigned an oxidation state of +1. For example HCl. When hydrogen is bonded to a nonmetal in a binary compound, it is assigned an oxidation state of -1. For example LiH.
For an ion, the sum of the oxidation states must equal the charge of the ion. For example, the sum of the oxidation states must equal -2 in CO32-.
When Halogens combine with Oxygen, then halogens have a + charge.

24. This is a picture of all of the common oxidation numbers of the elements when they form compounds with other elements. Remember that oxidation numbers represent the charges on the elements within a compound and will show the way that the elements will combine with one another in order to get the overall zero charge on the compound.
When working with oxidation reactions, it is helpful to assign oxidation numbers to each element in a reaction, to verify right away whether a redox reaction is occuring.

25. Oxidation-Reduction Reactions Redox
Oxidation-Reduction Reactions are considered electron-transfer reactions.
Also known as Redox Reactions.
Remember: LEO goes GER
Loss of Electrons means Oxidized & Gain of Electrons means Reduced
If you say an element is Oxidized, then it is called a Reducing agent because it donates electrons to another element.
If you say an element is Reduced, then it is called an Oxidizing agent because it accepts electrons from another element.

26. 2Ca(s) + O2(g) -> 2CaO(s)
Let’s take a look at the formation of calcium oxide (CaO) from calcium and oxygen:
2Ca(s) + O2(g) -> 2CaO(s)
You should recognize this is a REDOX reaction by checking the individual oxidation charges of each atom/ion and verifying that Loss of Electrons & Gain of Electrons is occurring:
Loss of electrons: 2Ca -> 2Ca2+ + 4e-
Calcium is therefore being Oxidized
Therefore Calcium is a good Reducing Agent!
Gain of electrons: O2 + 4e- -> 2O2-
While Oxygen is being Reduced!
Therefore Oxygen is a good Oxidizing Agent!
Remember:
LEO goes GER

27. Single Replacement Reactions are one type of REDOX Reactions!Li Rb K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H2 Sb Bi Cu Hg Ag Pt Au
MOST ACTIVE
Single Replacement Reactions are one type of REDOX reactions (We will cover other types of REDOX reactions later – but you should be able to recognize a REDOX reaction by checking the oxidation charges of atoms/ions).
You must use the ACTIVITY SERIES to determine if a single replacement reaction will actually occur. The Activity Series for Metals is on the right. The Activity Series for nonmetals is on the left.
An element can only replace another element that is less active than itself. Sort of like the top boss can kick out anyone below him and use their office. Examples:
3Mg FeCl3 g 2 Fe + 3 MgCl2
Cl KI g I2 + 2KCl
Cl2 + KF g No Reaction! F2
Cl2
Br2 I2
LEAST ACTIVE

28. Redox Reactions In Net Ionic Equation Form
Cu(s) + 2NO3-(aq) + 4H+ -> 2NO2(g) + Cu2+(aq) + 2H2O(l)
2Ag+(aq) + Cu(s) -> 2Ag(s) + Cu2+ (aq)
2NO(g) + O2(g) -> 2NO2(g)
Click the green box to view the first redox rxn. The Copper (0) looses 2 electrons and changes to Cu2+, therefore it oxidizes. The nitrogen in the nitric acid is a +5 charge and then gains an electron to change into the nitric dioxide gas with the charge of +4, gaining electrons means reduced. Therefore you can say that for every 1 mole that copper looses 2 electrons, 2 moles of the nitrogen in the nitric acid will gain 1 electron (giving a total of 2 lost and 2 gained).
Click on the blue circle to view the next Redox Rxn: In this rxn, Ag+ gains 1 e- to form the neutral solid Ag, this is called a reduction because electrons are gained. Also, Cu (0) loses 2 e- to form the Cu2+ cation, this is called an oxidation because electrons are lost. You would say overall that for every 2 moles that Ag+ gains one electron, 1 mole of Cu (0) will lose 2 electrons, overall total (2 lost, 2 gained).
Click on the red box for the next Redox rxn: In this rxn, Nitrogen in the nitrogen monoxide has a +2 charge on it, the product of nitrogen dioxide has a nitrogen of +4 charge. This means that nitrogen loses 2 electrons in the reaction, it is oxidized. At the same time, the oxygen in the reactant side has a 0 charge (neutral) and it gains 2 electrons in the product side in nitrogen dioxide to form O with a -2 charge. Oxygen gains 2 electrons and is said to be reduced. Click on the purple circle to see the next redox rxn: The iron starts out in the neutral charge and so does the chlorine. Then, Iron goes from 0 to +3 charge, losing 3 electrons, therefore it is oxidized. The chlorine starts with a 0 charge and goes to a -1 charge, losing 1 electron, therefore it is reduced. The overall rxn can be stated that for every 2 moles of Iron, losing 3 electrons each (6 total), 6 moles of Chlorine will gain 1 electron each (for a total of 6).
2Fe(s) + 3Cl2(g) -> 2FeCl3(s)

29. These Reactions show the following:
Metallic zinc is added to a solution containing copper (II) sulfate. Zinc reduces Cu2+ by donating two electrons to it:
Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)
In the process, the solution will loose the blue color that is characteristic of the presence of hydrated Cu2+ cations.
Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)
The oxidation and reduction half-reactions are:
Loss of electrons: Zn -> Zn e-
Gain of electrons: Cu e- -> Cu
On the right side, we see a different oxidation reaction:
Metallic copper reduces silver ions in a solution of silver nitrate (AgNO3):
Cu(s) + 2AgNO3(aq) -> Cu(NO3)2(aq) + 2Ag(s) Or
Cu(s) + 2Ag+ (aq) -> Cu2+(aq) + 2Ag(s)