Basic Chemistry Review

Basic Chemistry Review Part A: Acid-Base Review Part B: Chemical Equilibrium Review Part C: Oxidation-Reduction Review Part D: Solutions 1

A. Acid-Base Review Part A Outline Arrhenius Acids and Bases Brønsted-Lowry Acids and Bases Strong and Weak Acids Amphoteric Substances Calculating ph The ph Values of Common Substances Practice Problems Acids were initially characterized as substances that tasted sour, and bases were characterized as substances that felt slippery to the touch. For example, vinegar has a sour taste and we now know that it is an aqueous solution of acetic acid. Household bleach has a slippery feel, and is now known to be an aqueous solution of sodium hypochlorite, a very strong base. When testing the acidity or alkalinity of a substance, an acidic solution changes blue litmus paper to red, and a basic solution changes red litmus paper to blue. Arrhenius Acids and Bases There are several variations on what defines a material as an acid or base. Svante Arrhenius, a Swedish physicist and chemist, proposed the first modern definition. He stated that an acid is a compound that produces hydrogen cations (H + ) in an aqueous solution, while a base is a compound that produces hydroxide anions (OH - ) when in an 2

aqueous solution. An example of a simple Arrhenius acid is hydrogen bromide, or HBr. When dissolved in water, the hydrogen and bromine disassociate according to the following reaction: HBr (g) H + (aq) + Cl - (aq) Once hydrogen bromide is in solution, it can be called hydrobromic acid. An example of an Arrhenius base is sodium hydroxide, or NaOH. When dissolved in water, the sodium and hydroxide ions dissociate according to the following reaction: NaOH (s) Na + (aq) + OH - (aq) This is an example of a strongly basic solution. Brønsted-Lowry Acids and Bases Another definition of acids and bases is named for Danish chemist Johannes Brønsted and English chemist Thomas Lowry. According to the Brønsted-Lowry definition, an acid is a molecule that loses or donates a proton (H + ), creating a conjugate base. A base is a molecule that gains or accepts a proton, creating a conjugate acid. In general, a Brønsted-Lowry acid-base reaction can be defined by the following: Acid + Base conjugate base + conjugate acid or 3

HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) When an acid (HA) dissolves in water, it donates its proton to the water to form a new acid, H 3 O +, and a new base, A -. These new acid and base species may also undergo an acid-base reaction to reform the original products: H 3 O + (aq) + A - (aq) HA (aq) + H 2 O (l) Both the A - and the H 2 O are competing to accept protons. Whichever species has the higher affinity for H + will be considered the stronger base, and as such, that reaction will predominate. Going back to our example, H 2 O is a much stronger base than A -, so the forward reaction HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) outweighs the reverse. Strong and Weak Acids In the preceding example, all of the acid HA is completely dissociated into H + and A -. Complete dissociation, also called complete ionization, categorizes the acid as a strong acid. If there is a situation where the acid only partially ionizes, the acid is considered to be a weak acid. Organic acids (acids that contain carbon double-bonded to one oxygen and single-bonded to another oxygen) are typically categorized as weak acids. Acetic acid (HC 2 H 3 O 2 ) is an example of a weak acid wherein only about 1% of its molecules dissociate into H + and C 2 H 3 O - 2 upon being dissolved in water. In the reaction of weak acids with water, the reverse reaction predominates: 4

HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O - (aq) Amphoteric Substances An amphoteric substance is a compound that can act as both an acid and a base. The most common amphoteric material is water. If two molecules of water interact, one molecule may donate a proton to the other, creating two ions: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) One of the water molecules acts as an acid by giving up a proton, and the other water molecule acts as a base by accepting the proton to form H 3 O +. Because water is neither a strong acid nor base, the forward reaction does not occur frequently. In other words, there is not very much H 3 O + (aq) + OH - (aq) in a sample of pure water. At standard temperature (25 C), the concentrations are as follows: [H 3 O + ] and [OH - ] = 1.0 x 10-7 If we multiply both concentrations together, we get what is known as the ion-product constant for water, or k w. We can simplify the term H 3 O + to just H +. k w = [H + ] x [OH - ] = (1.0 x 10-7 ) x (1.0 x 10-7 ) = 1.0 x 10-14 5

This constant indicates to us that regardless of what is in any aqueous solution, the product of the concentration of OH - and the concentration of H + is always equal to 1.0x10-14. In other words, if the concentration of H + increases, then the concentration of OH - must then decrease, creating an acidic solution. Conversely, if we increase the concentration of OH - in a solution, the concentration of H + must decrease, resulting in a basic solution. If both concentrations are equal, then the solution is said to be neutral. Calculating ph Whether an aqueous solution is basic or acidic depends on the concentration of H +. It is possible to describe the acidity of a solution by giving the value of the H + concentration, but you will notice that these numbers are very small. It is often more convenient to show these values in terms of ph, which is simply the negative logarithm of the molar H + ion concentration: ph = -log [H + ] For example, if a solution has a molar H + ion concentration of 1.0 x 10-5, the ph would be: -log [1.0 x 10-5 ] = 5.0 When the ph value is lower than 7.0, the substance is acidic, and when it is greater than 7.0, the substance is basic. A ph value of 7.0 is neutral. 6

The ph of strong acids can be calculated directly from the concentration of the acid, since there is 100% ion dissociation. For example, if an aqueous solution contains 1.0 moles of HCl per liter, we know that the HCl actually exists as H + and Cl - ions in solution. Therefore 1.0 moles of HCl contain 1.0 moles of H + ions and 1.0 moles of Cl - ions. The ph can be calculated based on the concentration of 1.0 H + : ph = -log [H + ] = -log (1.0) = 0 The ph Values of Common Substances Material ph 1.0 M NaOH 14 Basic Ammonia 12 Milk of Magnesia 10.5 Baking Soda 8.5 Blood 7.4 Milk 6.4 Neutral Beer 4.5 Vinegar 2.4-3.4 Lemon Juice 2.2 Stomach Acid 1.0-3.0 Acidic 1.0 M HCl 0 7

Part A Practice Problems 1. In the following reaction, identify the acid, base, conjugate acid and conjugate base: HBrO 4 + H 2 O BrO 4 - + H 3 O + 2. Write a chemical equation showing how each of the following behaves as an acid when dissolved in water: a. HF + b. NH 4 3. Write a chemical equation showing how each of the following behaves as a base when dissolved in water: a. I - b. NH 3 4. Calculate the concentration of H + in the following solution: [OH] = 6.00 X 10-8 Will the solution be basic or acidic? 5. Calculate the concentration of OH - in the following solution: [H + ] = 1.45 x 10-2 Will the solution be basic or acidic? 6. As the hydrogen ion concentration of a solution increases, does the ph of the solution increase or decrease? 8

7. Calculate the ph corresponding to each of the hydrogen ion concentrations given. Will the solution be basic or acidic? a. [H + ] = 4.79 x 10-9 b. [H + ] = 7.24 x 10-2 8. For the H+ ion concentration listed, calculate the ph of the solution as well as the concentration of OH- ion. a. [H + ] = 2.25 x 10-3 9. Calculate the hydrogen ion concentration (in moles per Liter), for solutions with the following ph values. a. ph = 7.98 b. ph = 1.02 10. Calculate the ph of the following strong acid solution: a. 1.12 x 10-3 HNO 3 9

B. Chemical Equilibrium Review Part B Outline Chemical Reactions Chemical Equilibrium The Law of Chemical Equilibrium Calculating the Equilibrium Constant Heterogeneous Equilibrium o Le Chatlier s Principle Practice Problems The general term equilibrium indicates a state of balance. In chemistry, equilibrium has a more specific definition wherein it is meant to define a balance of rates of a chemical reaction in both the forward and reverse directions. In order to understand chemical equilibrium, we must first review how chemical reactions occur. Chemical Reactions A chemical reaction is a process that involves one set of chemical substances transforming into another set of chemical substances. It is believed that this transformation of chemical species occurs via molecular collisions. Sometimes when molecules collide, enough energy is transferred in order to cause bonds to break. Once a molecule s bonds are broken, the atoms therein can rearrange to form new molecules. 10

When writing chemical equations for a reaction, you should always place the reactants (starting materials) on the left side and the products on the right side, with an arrow in between. Consider the following reaction between hydrogen and oxygen gas: 2H 2 (g) + O 2 (g) 2H 2 O (l) The bonds between the two hydrogen atoms and the two oxygen atoms must be broken by strong collisions in order for the reactants to rearrange and form water. In reactions in which molecules collide to form new compounds, the speed of the reaction depends on how many collisions can occur. For example, the reaction rate will be faster if the concentration (number) of molecules available for collisions increases, or if the speed at which the incoming molecules collide (temperature) increases. The minimum amount of energy colliding molecules must possess in order for bonds to break in a reaction is called the activation energy, or E a. If the colliding molecules have an amount of energy lower than E a, a reaction will not occur. Since temperature is simply a measure of energy, the higher the temperature of the reaction, the more energy the molecules possess, and the more likely it will be that they can collide and break bonds. The rate of a reaction can also be increased by adding a catalyst. A catalyst is a substance that provides a pathway for a reaction to proceed with a lower activation 11

energy, which allows more collisions to occur that have enough energy for breaking bonds. Chemical Equilibrium Chemical reactions are reversible, which means that they can occur both in the forward direction where reactants become products, and in the backward direction where products go back starting materials. We generally use a double arrow to indicate a reversible reaction. For example, water in the vapor form can react with gaseous carbon monoxide to form hydrogen gas and carbon dioxide, and the carbon dioxide and hydrogen gas may also react to form water and carbon monoxide: H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) In this example, chemical equilibrium is reached when water and carbon monoxide are reacting at the same rate as hydrogen and carbon dioxide. Keep in mind that just because the concentrations of reactants and products at equilibrium are the same, it does not mean the reactions have stopped. We use the term dynamic equilibrium to indicate that molecules are still highly active despite concentrations of reactants and products remaining the same. The Law of Chemical Equilibrium The law of chemical equilibrium is a general term we use to describe the condition of a reaction being at equilibrium. If we use the generic terms A, B, C, and D to represent 12

the chemical compounds in a reaction and the terms a, b, c, and d to represent their stoichiometric coefficient (number of moles of each) in a balanced equation as follows: aa + bb cc + dd then the law of chemical equilibrium can be represented by the following equilibrium expression, wherein K is the equilibrium constant: K = c d [(C) (D) ] a b [(A) (B) ] In other words, the equilibrium constant K is equal to the product of the concentration of products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants divided by their stoichiometric coefficients. The concentrations are represented in moles per Liter, or simply M. Calculating the Equilibrium Constant To construct a generic equilibrium constant, let s use the following chemical reaction between chlorine gas and water to generate hydrogen chloride gas and oxygen gas: 2Cl 2 (g) + 2H 2 O (g) 4HCl (g) + O 2 (g) 13

The stoichiometric coefficients of chlorine and water are both 2 moles, and the coefficients of hydrogen chloride and oxygen are 4 and 1, respectively. It is not necessary to write in coefficients of 1. The equilibrium constant would therefore be represented as: K = 4 [(HCl) (O 2 )] 2 2 2 2 [(CL ) (H O) ] The equilibrium constant tells us that for a specific reaction at a given temperature, the ratio of the concentration of products to reactants will always be the same number K. Take note that K for a reaction is temperature-dependent, and will be different at various temperatures even if concentrations remain the same. Consider the following reaction wherein ammonia gas is synthesized from nitrogen and hydrogen: N 2 (g) + 3H 2 (g) 2NH 3 (g) If one mole of each nitrogen and hydrogen gas were reacted in a 1-L vessel at 500 C, an experiment showed that the equilibrium concentration of nitrogen gas was 0.831M, the concentration of hydrogen gas was 0.874M, and the concentration of ammonia was 0.268M. We can calculate the equilibrium constant using the equilibrium expression shown previously: 14

K = [(NH )] [(0.268) ] 2 2 3 3 3 [(N 2)(H 2) ] = [(0.831)(0.874) ] = 0.129 Now let s take a look at a reaction that involves the decomposition of hydrogen iodide into hydrogen gas and iodine gas: 2 HI (g) H 2 (g) + I 2 (g) The amount of iodine gas that is in the reaction mixture can qualitatively be determined by the purple color in the reaction mixture. The darker the purple, the more iodine that is present. When 2.00 moles of hydrogen iodide are placed in a 2.50 L reaction vessel at 458 C, the equilibrium mixture was found to contain 0.221 moles of iodine gas. What would be the value of K for the decomposition of iodine at 458 C? In order to solve the problem, we first need to calculate the concentration of the substances. This is done by dividing the molar amounts by the volume of the reaction vessel. Initial concentration of HI = 2.00 moles 2.50 L = 0.800 M Equilibrium concentration of I 2 = 0.221 moles 2.50 L = 0.0884 M 15

Using these values, we can set up a table: Concentration (M) 2 HI (g) H 2 (g) + I 2 (g) Initial 0.800 0 0 Change -2x x x Equilibrium 0.800-2x x 0.0884 We can calculate the equilibrium concentration of substances from the last line in the table. We know that at equilibrium, the iodine concentration is 0.0884 M and that this is also equal to the unknown variable x. To find the concentration of HI: [HI] = (0.800 2x) M = (0.800 2 x 0.0884) = 0.623 M The concentration of hydrogen is equal to the iodine concentration, therefore: [H 2 ] = 0.0884 M We now can write the equilibrium constant expression from this equation as previously outlined: K = [(H )(I )] 2 2 2 [(HI) ] 16

K = [(0.0884)(0.0884)] 2 (0.623 ) = 0.0201 Heterogeneous Equilibrium So far, all of the example equations have only had substances in the gaseous phase. Sometimes, an equilibrium reaction contains reactants and products that are in different phases, which is known as heterogeneous equilibrium. For example, solid phosphorus pentachloride decomposes into liquid phosphorus trichloride and gaseous chlorine: PCl 5 (s) PCl 3 (l) + Cl 2 (g) Experimental results have shown that heterogeneous equilibrium positions do not depend on the amounts of pure solids or pure liquids. Therefore, the equilibrium expression for the decomposition of solid phosphorus pentachloride would simply be: K = [Cl 2 ] Le Chatlier s Principle There are three factors that can control the position of chemical equilibrium: concentration, volume, and temperature. Le Chatlier s principle can help us to predict what effect changing any of these factors have on equilibrium position. According to Le Chatlier s principle, when a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce the effect of that change. 17

Using the equation for ammonia synthesis, we will consider what effect an increase in concentration of nitrogen gas will have on the system. N 2 (g) + 3H 2 (g) 2NH 3 (g) Recall that a chemical reaction occurs when molecules collide together. If we were to increase the concentration of nitrogen gas in the system, there will be more collisions between hydrogen and nitrogen molecules. This will cause more ammonia to be formed, thus there will be a higher concentration of product. As the product concentration increases, the reverse reaction also increases, but since there is more ammonia and the formation of products is favored, we say that the equilibrium has shifted to the right. Note that the equilibrium constant will remain the same. Le Chatlier s principle predicts that as more nitrogen gas is added, the system will shift in a direction that consumes nitrogen to help offset the change of adding more nitrogen. In other words, when more reactant is added to an equilibrium reaction, the system will shift away from the added component. Part B Practice Problems 1. Write the chemical equilibrium expression for the following reaction: 2 Cl 2 (g) + 2 H 2 O (g) 4 HCl (g) + O 2 (g) 2. Write the chemical equilibrium expression for the following reaction: P 4 (g) + 6 Br 2 (g) 4 PBr 3 (g) 18

3. The equilibrium constant expression for a gaseous reaction is: K = [(H O) (SO ) ] 2 2 2 2 2 3 2 2 [(H S) (O ) ] Write a balanced chemical equation corresponding to this expression. 4. Calculate the value of K for the following reaction at a specific temperature: N 2 (g) + O 2 (g) 2NO (g) Assume the equilibrium concentrations were found to be [N 2 ] = 0.082M, [O 2 ] = 0.156M, and [NO] = 9.4 x 10-4 M. 5. Given the following reaction, wherein methanol decomposes to formaldehyde and hydrogen, calculate the value of K at a specific temperature if the equilibrium concentrations were found to be [CH 3 OH] = 0.00430M, [H 2 ] = 0.0662 M, and [CH 2 O] = 0.882 M: CH 3 OH (g) CH 2 O (g) + H 2 (g) 6. An 8.00 L reaction vessel at 491 C contained 0.650 mol H 2, 0.275 mol I 2 and 3.00 mol HI. Find the value of K for the following reaction at 491 C, assuming that the substances are in equilibrium: H 2 (g) + I 2 (g) 2HI (g) 7. Write the equilibrium expression for the following heterogeneous reaction: ZrI 4 (s) Zr (s) + 2I 2 (g) 8. Write the equilibrium expression for the following heterogeneous reaction: MgO (s) + CO 2 (g) MgCO 3 (s) 19

9. Consider the following exothermic reaction: C 2 H 2 (g) + 2Br 2 (g) C 2 H 2 Br 4 (g) What effect will an increase in temperature have on the position of the equilibrium? Will it shift toward the reactant or products? 10. Using Le Chatlier s principle, what three changes could be made to the reaction below to maximize the production of ammonia? N 2 (g) + 3H 2 (g) 2NH 3 (g) 20

C. Oxidation-Reduction Review Part C Outline Oxidation Numbers o Rules for Assigning Oxidation Numbers Half Reactions o Common Types of Oxidation-Reduction Reactions Balancing Oxidation-Reduction Reactions o Balancing an Oxidation-Reduction Reaction in an Acidic Solution o Balancing an Oxidation-Reduction Reaction in a Basic Solution Practice Problems There are many different classes of chemical reaction that take place. For example, a synthesis reaction occurs when two or more simple compounds combine to make a more complicated one. Decomposition reactions are the opposite of synthesis, wherein a complex molecule breaks down into simpler ones. In an acid-base reaction, a molecule that contains H+ ions in solution (acid) combines with a molecule that forms OH- ions in solution (base) to form water and a salt. An oxidation-reduction reaction is a special type of reaction wherein one or more electrons are transferred between chemical species. Oxidation Numbers To help keep track of where electrons are moving in oxidation-reduction, we can use the concept of oxidation numbers, otherwise known as the charge, on the various atoms 21

within a compound. In oxidation-reduction reactions, one or more atoms will change in oxidation number, meaning that there was some transfer of electrons that took place. Rules for Assigning Oxidation Numbers RULE # APPLIES TO STATEMENT EXAMPLE 1 Elements The oxidation number of an atom in an element is zero. This rule also applies to elements that exist as diatomic molecules. Fe (s) Oxidation number = 0 O 2 (g) Oxidation number = 0 2 Monatomic Ions The oxidation number of an atom in a monatomic ion is the same as the charge Fe 2+ Oxidation number = 2 on the ion. 3 Oxygen The oxidation number of oxygen is -2 in most compounds, except when in peroxides where the oxidation number is -1 SO 2 Oxidation number = -2 H 2 O 2 Oxidation number = -1 4 Hydrogen The oxidation number of hydrogen is +1 in most compounds, except in binary compounds with metals where it is -1 HClO 4 Oxidation number = +1 CaH 2 5 Group 7A (Halogens) 6 Compounds and Ions The oxidation number of the halogens is -1 in their compounds, except when in a diatomic molecule with themselves (rule 1) The sum of the oxidation numbers of the atoms in a compound is zero. The sum of the oxidation numbers in an ion must equal Oxidation number = -1 BF 3 Oxidation number = -1 F 2 Oxidation number = 0 22

the charge of the ion. Let s look at an example of an oxidation-reduction reaction wherein electrons are transferred between species. Fe (s) + Cu 2+ (aq) Fe 2+ (aq) + Cu (s) The oxidation number of Fe (s) = 0, Cu 2+ = +2, Fe 2+ = +2, and Cu (s) = 0. In other words, iron went from 0 to +2 (lost 2 electrons) and copper went from +2 to 0 (gained 2 electrons). Half Reactions We can rewrite the reaction in terms of two half reactions. A half reaction is simply one of two parts of an oxidation-reduction reaction. One part involves a loss of electrons (increase in oxidation number) and the other part involves a gain of electrons (decrease in oxidation number). The two half reactions for our equation are: Fe (s) Fe 2+ (aq) + 2 electrons Cu 2+ (aq) + 2 electrons Cu (s) Oxidation is the half reaction in which a chemical species lost one or more electrons. Reduction is the half reaction in which a chemical species gained one or more electrons. In the example above, iron was oxidized and copper was reduced. An 23

oxidizing agent is a chemical species that oxidizes another species. In other words, the oxidizing agent is the species that gets reduced. Similarly, a reducing agent is a chemical species that reduces another species. The reducing agent is the species that was oxidized. In our example, iron was oxidized and copper was reduced, so iron is the reducing agent and copper is the oxidizing agent. Common Types of Oxidation-Reduction Reactions Category of Reaction Combination Decomposition Displacement Combustion Definition Two substances combine to form a third A single compounds reacts to form two or more substances An element reacts with a compound and displaces another element from it A substance reacts with oxygen with the rapid release of heat, producing a flame Example 2Na (s) + Cl 2 (g) 2NaCl (s) CaCO 3 (s) CaO (s) + CO 2 (g) Zn (s) + 2 HCl (aq) ZnCl 2 (aq) + H 2 (g) 4Fe (s) + 3O 2 (g) 2Fe 2 O 3 (s) Balancing Oxidation-Reduction Reactions Oxidation-reduction reactions can often be very difficult to balance. Let s first look at a simple oxidation-reduction reaction, wherein we can use what is known as the half- 24

reaction method for balancing. Consider the following reaction between solid zinc and an aqueous silver solution: Zn (s) + Ag + (aq) Zn 2+ (aq) + Ag (s) The first step is to assign oxidation numbers to each species. In this case, the oxidation numbers are as follows: Zn (s) = 0, Ag + (aq) = +1, Zn 2+ (aq) = +2, and Ag (s) = 0. Zinc lost two electrons (was oxidized) while silver gained one electron (was reduced). We can now write each half reaction, including electrons, as follows: Zn Zn 2+ + 2 e - (oxidation half reaction) Ag + + e - Ag (reduction half reaction) It is important to note that the number of electrons zinc lost in the process does not equal the number of electrons silver gained. Since silver only gained one electron, we need to double the amount of Ag + in order to accept the two electrons lost by zinc. To accomplish this, we can multiply the half reaction by an integer to ensure the number of electrons are equal. We can then add both half reactions together, and the number of electrons should cancel out. 1 x (Zn Zn 2+ + 2 e - ) + 2 x (Ag + + e - Ag) Zn + 2Ag + + 2e - Zn 2+ + 2Ag + 2e - 25

The electrons cancel out, which yields the following balanced oxidation-reduction equation: Zn (s) + 2Ag + (ag) Zn 2+ (aq) + 2Ag (s) Balancing an Oxidation-Reduction Reaction in an Acidic Solution We can now extend this half-reaction to help balance more complex oxidation-reduction reactions that occur in acidic or basic solutions. The steps for balancing an oxidationreduction reaction in an acidic solution are as follows: 1. Assign oxidation numbers to each atom to determine which species was oxidized and which was reduced. 2. Split the equation into its two half reactions, the oxidation half reaction and the reduction half reaction. 3. Balance each half reaction. a. Begin by balancing all atoms besides oxygen and hydrogen. b. Balance oxygen atoms by adding H 2 O (water). c. Balance hydrogen atoms by adding H + (hydrogen ions). 4. Combine the two half reactions to obtain a balanced oxidation-reduction equation. a. Multiply each half reaction, as with the simple example, by an integer to ensure that the same number of electrons exists on either side of the equation. 26

b. Cancel out any other chemical species that exist on either side of the equation. Consider the following reaction wherein Cr 2 O 7 2- reacts with Fe 2+ in an acidic solution to give Cr 3+ and Fe 2+ : Cr 2 O 7 2- + Fe 2+ Cr 3+ + Fe 3+ We can begin by assigning oxidation numbers and dividing the reaction into its two half reactions. Chromium in Cr 2 O 7 2- has a charge of +6 and gains three electrons to form Cr 3+ (reduction). Iron loses one electron to go from Fe 2+ to Fe 3+ (oxidation). Cr 2 O 7 2- Cr 3+ Fe 2+ Fe 3+ The half reactions now need to be balanced. First balance any atoms that aren t hydrogen or oxygen. Cr 2 O 7 2-2Cr 3+ Fe 2+ Fe 3+ 27

Next, balance oxygen by adding water to the side that needs oxygen. There are seven oxygen atoms on the left side of the first half reaction, therefore seven water molecules should be added to the right side. Cr 2 O 7 2-2Cr 3+ + 7H 2 O Fe 2+ Fe 3+ While the number of oxygen atoms is balanced, we have created an imbalance in the number of hydrogen atoms. We balance hydrogen atoms by adding hydrogen ions to the side that lacks hydrogen. 14 H + + Cr 2 O 7 2-2Cr 3+ + 7H 2 O Fe 2+ Fe 3+ Finally, we need to balance electrical charge using electrons. First calculate the net charge on each side. For the first half reaction, we have (14+) + (2-) = 12+ on the left, and 2x(3+) = 6+ on the right. The algebraic difference between the net charges is the number of electrons that should be added to the more positive side. So, in this case, the left side has a difference of 6+, so six electrons should be added to the left. 6e - 14 H + + Cr 2 O 7 2-2Cr 3+ + 7H 2 O To balance the other half reaction, we simply need to add one electron to the right. 28

Fe 2+ Fe 3+ + e - Since 6 electrons were gained in the first half reaction and only one was lost in the second, we need to multiply the second half reaction by six so that the electrical charge is balanced. 6 x (Fe 2+ Fe 3+ + e - ) We can now combine the two half reactions 6e - 14 H + + Cr 2 O 7 2-2Cr 3+ + 7H 2 O + 6 x (Fe 2+ Fe 3+ + e - ) 6e - 14 H + + Cr 2 O 7 2- + 6 Fe 2+ 2Cr 3+ + 7H 2 O + 6Fe 3+ + 6e - This gives the final balanced equation: 14 H + + Cr 2 O 7 2- + 6 Fe 2+ 2Cr 3+ + 7H 2 O + 6Fe 3+ Balancing an Oxidation-Reduction Reaction in a Basic Solution If the reaction is in basic solution, we just need to include three more steps: 1. Add to BOTH sides of the equation the same number of OH - as there are H +. 2. Combine H + and OH - to form H 2 O 29

3. Cancel any H 2 O you can. Part C Practice Problems 1. Determine the oxidation number for the element noted in the following: a. Br in KBrO 4 b. Nb in NbO 2 2. Determine the oxidation number for all of the elements in each of the following: a. CoSeO 4 b. Cu 2 SO 3 3. Assign oxidation numbers to nitrogen in each of the following: a. NO b. N 2 O 5 c. N 2 d. NO 2 4. If atoms of a metallic element react with atoms of a nonmetallic element, which species will lose electrons and which species will gain electrons? 5. Label the oxidizing agent and the reducing agent in the following reaction: Co (s) + Cl 2 (g) CoCl 2 (s) 6. Label the oxidizing agent and the reducing agent in the following reaction: PbS (s) + 4H 2 O 2 (aq) PbSO 4 (s) + 4H 2 O (l) 7. Balance the following oxidation-reduction reaction using the half-reaction method: FeI 3 (aq) + Mg (s) Fe (s) + MgI 2 8. Balance the following oxidation-reduction reaction using the half-reaction method: 30

H 2 (g) + Ag + (aq) Ag (s) + H + (aq) 9. Balance the following reaction that occurs in an acidic solution: Sn + NO - 3 SnO 2 + NO 10. Balance the following reaction that occurs in a basic solution: CrO 2-4 + S 2- - S + CrO 2 31

D. Solutions Part A Solutions 1. i. The acid loses or donates a proton, so HBrO 4 is the acid. ii. The base gains or accepts a proton, so H 2 O is the base. iii. The conjugate acid is created when a proton is lost to the base, so BrO - 4 is the conjugate acid. iv. The conjugate base is created when a proton is accepted from the acid, so H 3 O + is the conjugate base. 2. i. Acids are molecules that donate protons. ii. Equation a: HF + H 2 O H 3 O + + F - iii. Equation b: NH + 4 + H 2 O H 3 O + + NH 3 3. i. Bases are molecules that accept protons. ii. Equation a: I - + H 2 O HI + OH - + iii. Equation b: NH 3 + H 2 O NH 4 + OH - 4. i. We know that Kw = [H + ] [OH - ] = 1.0 X 10-14 so we need to calculate [H + ], but [OH - ] is given as 6.00 X 10-8 M. ii. We can solve for [H + ] by dividing both sides by [OH - ]. [H + ] = 1.0 X 10 - [OH ] -14 = 1.0 X 10 6.00 X 10-14 -8 = 1.7 X 10-7 M 32

iii. Since the concentration of OH - is less than the concentration of H +, the solution is acidic. 5. i. In this case, [H + ] is given, so we can solve for [OH - ]. [OH - ] = 1.0 X 10 + [H ] -14 = 1.0 X 10 1.45 X 10 14 = 6.9 X 2 10-13 M ii. Since the [H + ] is greater than the [OH - ], the solution is acidic. 6. i. As [H + ] in a solution increases (becomes more acidic), ph decreases. ii. Lower ph = higher acidity = higher [H + ]. 7. i. To determine ph, use the formula ph = -log [H + ]. ii. Solution a: -log [4.79 x 10-9 ] = a ph of 8.32, which is basic. iii. Solution b: -log [7.24 x 10-2 ] = a ph of 1.14, which is acidic. 8. i. [OH - ] = 1.0 X 10 [H ] 14 = 1.0 X 10 2.25 X 10 14 = 4.4 X 3 10-12 M ii. ph = -log [2.25 x 10-3 ] = 2.65 9. i. Since ph = -log [H + ], [H + ] = inverse log (-ph) ii. Solution a: ph = 7.98 [H + ] = inverse log (-7.98) or [H + ] = 10 -ph [H + ] = 10-7.98 = 1.29 X 10-8 M 33

iii. Solution b: [H + ] = 10-1.02 = 9.54 X 10-2 M 10. i. Strong acids completely dissociate in solution. Therefore, molarity is the same as [H + ]. ii. 1.12 X 10-3 M HNO 3 = 1.12 X 10-3 M H + iii. ph = -log [H + ] = -log (1.12 X 10-3 ) = 2.95 Part B Solutions 1. i. K = ii. K = c d [(C )(D )] a b [(A )(B )] 4 (HCl) (O 2) 2 2 2 2 (Cl ) (H O) 2. i. K = ii. K = c d [(C )(D )] a b [(A )(B )] (PBr ) (Cl ) (H O) 4 3 2 2 2 2 3. i. 2H 2 S + 3O 2 2H 2 O + 2SO 2 4. i. K = 2 (NO) (N )(O ) 2 2 ii. K = -4 2 (9.4 x 10 ) (0.082)(0.156) 34

iii. K = 6.9 X 10-5 5. (CH2O)(H 2) i. K = (CH OH) ii. K = (0.882)(0.0662) (0.00430) iii. K = 13.6 3 6. i. First calculate the molar concentrations of the reactants and the product. (H 2 ) = (I 2 ) = 0.650 mol 8.00 L 0.275 mol 8.00 L = 0.0813 M = 0.0344 M (HI) = 3.00 mol 8.00 L = 0.375 M ii. Use these molar concentrations in the equilibrium expression. K = (HI) (0.375) (0.0813)(0.0344) = 50.3 2 2 (H 2)(I 2) = 7. i. The equilibrium expression for heterogeneous reactions does not rely on the amounts of pure solids or liquids. ii. K = (I 2 ) 2 8. i. K = 1 (CO ) 2 9. 35

i. For an exothermic reaction where energy is released from reactants to products, an increase in temperature will shift the equilibrium to the left (toward reactants). 10. i. Increasing the concentration of N 2 (g) and H 2 (g) would help to produce more NH 3 (g). ii. When the volume of a gas is decreased, the pressure increases. Therefore, when the volume of a gaseous reaction is reduced, leading to an increase in pressure, by Le Chatlier s principle, the system will shift in the direction that reduces pressure (the direction that gives the smaller number of gas molecules). So, for the reaction of N 2 (g) + 3H 2 (g) 2NH 3 (g), reducing the pressure would help produce more NH 3 (g). iii. Since the reaction is endothermic, decreasing temperature will help produce more NH 3 (g). Part C Solutions 1. i. All oxidation numbers should add up to zero. ii. Solution a: K = 1x + 1 = 1 O = 4x 2 = -8 (-8) + 1 = -7 Bromine must be +7. iii. Solution b: O = 2x 2 = -4 36

Niobium must be +4. 2. i. All of the oxidation numbers should add up to zero. ii. Solution a: SeO 2-4 is the selenate polyatomic ion. Oxygen is always -2 in a compound. -2 x 4 = -8 Se must be +6 to make a -2 ion. Cobalt must be +2 to balance with the SeO 2-4. iii. Solution b: SO 2-3 is the sulfite polyatomic ion. Oxygen is always -2 in a compound. -2 X 3 = -6 Sulfur must be +4 to make a -2 ion. The total charge from copper must be +2. 2 2 = +1 Each copper is +1. 3. i. Solution a: Oxygen is -2, so nitrogen is +2. ii. Solution b: Oxygen is -2 X 5 = -10. The total charge from nitrogen is +10. 37

10 2 = +5 Each nitrogen is +5. iii. Solution c: The oxidation number of a molecule is zero. iv. Solution d: Oxygen is -2 X 2 = -4. The nitrogen is +4. 4. i. In general, metals will lose electrons and non-metals will gain electrons. 5. i. Assign oxidation numbers. Co (s) + Cl 2 (g) CoCl 2 (s) 0 0 +2-1 ii. Cobalt went from 0 to +2, so it was oxidized and is the reducing agent. iii. Chlorine went from 0 to -1, so it was reduced and is the oxidizing agent. 6. i. Assign oxidation numbers. PbS (s) + 4H 2 O 2 (aq) PbSO 4 (s) + 4H 2 O (l) +2-2 +1-1 +2 +6-2 +1-2 ii. This reaction involves hydrogen peroxide, H 2 O 2, wherein the oxidation number of oxygen is -1. iii. The sulfur in PbS is -2, and in PbSO 4: SO 4 2- iv. Oxygen is -2 X 4 = -8, so sulfur must be +6. 38

v. Oxygen went from -1 to -2, so it was reduced and is the oxidizing agent. vi. Sulfur went from +6 to -2, so it was oxidized and is the reducing agent. vii. Lead and hydrogen s charges did not change. 7. i. Assign oxidation numbers. FeI 3 (aq) + Mg (s) Fe (s) + MgI 2 +6-2 0 0 +4-2 ii. Write the balanced half reactions. Fe 6+ + 6e - Fe (reduction) Mg Mg 4+ + 4e - (oxidation) iii. Multiply each half reaction by a factor that will cancel electrons. 2x (Fe 6+ + 6e - Fe) + 3x (Mg Mg 4+ + 4e - ) 2Fe 6+ + 12e - + 3Mg 2Fe + 3Mg 4+ + 12e - iv. Canceling out electrons, we get the balanced oxidation-reduction reaction. 8. 2Fe 6+ + 3Mg 2Fe + 3Mg 4+ i. Assign oxidation numbers. H 2 (g) + Ag + (aq) Ag (s) + H + (aq) 0 +1 0 +1 ii. Write the balanced half reactions. H 2 H + + 1e - (oxidation) + Ag + + 1e - Ag (s) (reduction) H 2 + Ag + + 1e - H + +1e - + Ag iii. Canceling out electrons, we get the balanced oxidation-reduction reaction. 39

H 2 + Ag + H + + Ag 9. i. Assign oxidation numbers. Sn + NO 3 - SnO 2 + NO 0 +5-2 +4-2 +2-2 ii. Split the reaction into half reactions. Sn SnO 2 NO 3 - NO iii. All atoms besides O are balanced, so balance O atoms by adding H 2 O. Sn + 2H 2 O SnO 2 NO - 3 NO + 2H 2 O iv. Balance H atoms by adding H +. Sn + 2H 2 O SnO 2 + 4H + NO - 3 + 4H + NO + 2H 2 O v. Balance electric charge by adding electrons to the more positive side. Sn + 2H 2 O SnO 2 + 4H + + 4e - NO - 3 + 4H + + 3e - NO + 2H 2 O vi. Each half reaction must be multiplied by an integer so electrons cancel out. 3x (Sn + 2H 2 O SnO 2 + 4H + + 4e - ) + 4x (NO - 3 + 4H + + 3e - NO + 2H 2 O) 3Sn + 6H 2 O + 4NO - 3 + 16H + + 12e - 3SnO 2 + 12H + + 12e - + 4NO + 8H 2 O vii. Canceling H+ and H2O on either side, we get: 3Sn + 4NO - 3 + 4H + 3SnO 2 + 4NO + 2H 2 O 40

10. i. CrO 2-4 + S 2- - S + CrO 2 CrO 2- - 4 CrO 2 S 2- S ii. Balance O by adding H 2 O. CrO 2-4 CrO - 2 + 2H 2 O S 2- S iii. Balance H by using H +. 4H + + CrO 2-4 CrO - 2 + 2H 2 O S 2- S iv. Balance electric charge by adding electrons to the more positive side. 4H + + CrO 2-4 + 3e - CrO - 2 + 2H 2 O S 2- S + 2e - v. Each half reaction must be multiplied by an integer so electrons cancel out. 2x (4H + + CrO 2-4 + 3e - CrO - 2 + 2H 2 O) + 3x (S 2- S + 2e - ) 8H + + 2CrO 2-4 + 6e - + 3S 2-2CrO - 2 + 4H 2 O +3S + 6e - 8H + + 2CrO 2-4 + 3S 2-2CrO - 2 + 4H 2 O +3S vi. Since the reaction is in a basic solution, we note the number of H+ and add this number of OH- to both sides. 8H + + 80H - + 2CrO 2-4 + 3S 2-2CrO - 2 + 4H 2 O +3S + 80H - vii. Simplify by combining H + and OH - to H 2 O. 8H 2 O + 2CrO 2-4 + 3S 2-2CrO - 2 + 4H 2 O +3S + 80H - 41

viii. Cancel any H 2 O that exists on both sides. 4H 2 O + 2CrO 4 2- + 3S 2-2CrO 2 - + 3S + 80H - 42