Acids and Bases and Oxidation-Reduction Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 8 Acids and Bases and Oxidation-Reduction Denniston Topping Caret 5th Edition

8.1 Acids and Bases Acids: Taste sour, dissolve some metals, cause plant dye to change color Bases: Taste bitter, are slippery, are corrosive Two theories that help us to understand the chemistry of acids and bases Arrhenius Theory Brønsted-Lowry Theory

Arrhenius Theory of Acids and Bases Acid - a substance, when dissolved in water, dissociates to produce hydrogen ions Hydrogen ion: H+ also called “protons” HCl is an acid: HCl(aq)  H+(aq) + Cl-(aq) 8.1 Acids and Bases

Arrhenius Theory of Acids and Bases Base - a substance, when dissolved in water, dissociates to produce hydroxide ions NaOH is a base NaOH(aq)  Na+(aq) + OH-(aq) 8.1 Acids and Bases

Arrhenius Theory of Acids and Bases Where does NH3 fit? When it dissolves in water it has basic properties but it does not have OH- ions in it The next acid-base theory gives us a broader view of acids and bases 8.1 Acids and Bases

Brønsted-Lowry Theory of Acids and Bases Acid - proton donor Base - proton acceptor Notice that acid and base are not defined using water When writing the reactions, both accepting and donation are evident 8.1 Acids and Bases

Brønsted-Lowry Theory of Acids and Bases HCl(aq) + H2O(l)  Cl-(aq) + H3O+(aq) What donated the proton? HCl Is it an acid or base? Acid What accepted the proton? H2O Is it an acid or base? Base acid base 8.1 Acids and Bases

Brønsted-Lowry Theory of Acids and Bases . NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) base acid Now, let us look at NH3 and see why it is a base. Did NH3 donate or accept a proton? Accept Is it an acid or base? Base What is water in this reaction? Acid 8.1 Acids and Bases

Acid-Base Properties of Water Water possesses both acid and base properties Amphiprotic - a substance possessing both acid and base properties Water is the most commonly used solvent for both acids and bases Solute-solvent interactions between water and both acids and bases promote solubility and dissociation 8.1 Acids and Bases

Acid and Base Strength 8.1 Acids and Bases Acid and base strength – degree of dissociation Not a measure of concentration Strong acids and bases – reaction with water is virtually 100% (Strong electrolytes) 8.1 Acids and Bases

Strong Acids and Bases 8.1 Acids and Bases Strong Acids: Strong Bases: HCl, HBr, HI Hydrochloric Acid, etc. HNO3 Nitric Acid H2SO4 Sulfuric Acid HClO4 Perchloric Acid Strong Bases: NaOH, KOH, Ba(OH)2 All metal hydroxides 8.1 Acids and Bases

Weak Acids 8.1 Acids and Bases Weak acids and bases – only a small percent dissociates (Weak electrolytes) Weak acid examples: Acetic acid: Carbonic Acid: 8.1 Acids and Bases CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+(aq)

Weak Bases 8.1 Acids and Bases Weak base examples: Ammonia: Pyridine: Aniline: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) 8.1 Acids and Bases C5H5NH2(aq) + H2O(l) C5H5NH3+(aq) + OH-(aq) C6H5NH2(aq) + H2O(l) C6H5NH3+(aq) + OH-(aq)

Conjugate Acids and Bases The acid base reaction can be written in the general form: Notice the reversible arrows The products are also an acid and base called the conjugate acid and base HA + B A– + HB+ 8.1 Acids and Bases acid base

8.1 Acids and Bases HA + B A- + HB+ acid base Conjugate acid - what the base becomes after it accepts a proton Conjugate base - what the acid becomes after it donates its proton Conjugate acid-base pair - the acid and base on the opposite sides of the equation HA + B A- + HB+ base acid 8.1 Acids and Bases

Acid-Base Dissociation HA + B A– + HB+ The reversible arrow isn’t always written Some acids or bases essentially dissociate 100% One way arrow is used HCl + H2O  Cl- + H3O+ All of the HCl is converted to Cl- HCl is called a strong acid – an acid that dissociates 100% Weak acid - one which does not dissociate 100% 8.1 Acids and Bases

Conjugate Acid-Base Pairs Which acid is stronger: HF or HCN? HF Which base is stronger: CN- or H2O? CN - 8.1 Acids and Bases

Acid-Base Practice 8.1 Acids and Bases Write the chemical reaction for the following acids or bases in water Identify the conjugate acid-base pairs HF (a weak acid) H2S (a weak acid) HNO3 (a strong acid) CH3NH2 (a weak base) Note: The degree of dissociation also defines weak and strong bases 8.1 Acids and Bases

The Dissociation of Water Pure water is virtually 100% molecular Very small number of molecules dissociate Dissociation of acids and bases is often called ionization Called autoionization Very weak electrolyte 8.1 Acids and Bases H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

Hydronium Ion 8.1 Acids and Bases H3O+ is called the hydronium ion In pure water at room temperature: [H3O+] = 1 x 10-7 M [OH-] = 1 x 10-7 M What is the equilibrium expression for: Remember, liquids are not included in equilibrium expressions 8.1 Acids and Bases H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

Ion Product of Water 8.1 Acids and Bases This constant is called the ion product for water and has the symbol Kw Since [H3O+] = [OH-] = 1.0 x 10-7 M, what is the value for Kw? 1.0 x 10-14 It is unitless 8.1 Acids and Bases

8.2 pH: A Measurement Scale for Acids and Bases pH scale - a scale that indicates the acidity or basicity of a solution Ranges from 0 (very acidic) to 14 (very basic) The pH scale is rather similar to the temperature scale assigning relative values of hot and cold The pH of a solution is defined as: pH = -log[H3O+]

8.2 pH: A Measurement Scale for Acids and Bases A Definition of pH Use these observations to develop a concept of pH if know one concentration, can calculate the other if add an acid, [H3O+]  and [OH-]  if add a base, [OH-]  and [H3O+]  [H3O+] = [OH-] when equal amounts of acid and base are present In each of these cases 1 x 10-14 = [H3O+][OH-] 8.2 pH: A Measurement Scale for Acids and Bases

8.2 pH: A Measurement Scale for Acids and Bases Measuring pH pH of a solution can be: Calculated if the concentration of either is known [H3O+] [OH-] Approximated using indicator / pH paper that develops a color related to the solution pH Measured using a pH meter whose sensor measures an electrical property of the solution that is proportional to pH 8.2 pH: A Measurement Scale for Acids and Bases

8.2 pH: A Measurement Scale for Acids and Bases Calculating pH How do we calculate the pH of a solution when either the hydronium or hydroxide ion concentration is known? How do we calculate the hydronium or hydroxide ion concentration when the pH is known? Use two facts: 8.2 pH: A Measurement Scale for Acids and Bases pH = -log[H3O+] 1 x 10-14 = [H3O+][OH-]

Calculating pH from Acid Molarity What is the pH of a 1.0 x 10-4 M HCl solution? HCl is a strong acid and dissociates in water If 1 mol HCl is placed in 1 L of aqueous solution it produces 1 mol [H3O+] 1.0 x 10-4 M HCl solution has [H3O+]=1.0x10-4M = -log [H3O+] = -log [1.0 x 10-4] = -[-4.00] = 4.00 8.2 pH: A Measurement Scale for Acids and Bases pH = -log[H3O+]

Calculating [H3O+] From pH What is the [H3O+] of a solution with pH = 6.00? 4.00 = -log [H3O+] Multiply both sides of equation by –1 -4.00 = log [H3O+] Take the antilog of both sides Antilog -4.00 = [H3O+] Antilog is the exponent of 10 1.0 x 10-4 M = [H3O+] pH = -log[H3O+] 8.2 pH: A Measurement Scale for Acids and Bases

Calculating the pH of a Base What is the pH of a 1.0 x 10-3 M KOH solution? KOH is a strong base (as are any metal hydroxides) 1 mol KOH dissolved and dissociated in aqueous solution produces 1 mol OH- 1.0 x 10-3 M KOH solution has [OH-] = 1.0 x 10-3 M Solve equation for [H3O+] = 1 x 10-14 / [OH-] [H3O+] = 1 x 10-14 / 1.0 x 10-3 = 1 x 10-11 pH = -log [1 x 10-11] = 11.00 8.2 pH: A Measurement Scale for Acids and Bases 1 x 10-14 = [H3O+][OH-] pH = -log[H3O+]

Calculating pH from Acid Molarity What is the pH of a 2.5 x 10-4 M HNO3 solution? We know that as a strong acid HNO3 dissociates to produce 2.5 x 10-4 M [H3O+] pH = -log [2.5 x 10-4] = 3.60 8.2 pH: A Measurement Scale for Acids and Bases pH = -log[H3O+]

Calculating [OH-] From pH What is the [OH-] of a solution with pH = 4.95? First find [H3O+] 4.95 = -log [H3O+] [H3O+] = 10-4.95 [H3O+] = 1.12 x 10-5 Now solve for [OH-] [OH-] = 1 x 10-14 / 1.12 x 10-5 = 1.0 x 10-9 pH = -log[H3O+] 8.2 pH: A Measurement Scale for Acids and Bases 1 x 10-14 = [H3O+][OH-]

8.2 pH: A Measurement Scale for Acids and Bases The pH Scale 8.2 pH: A Measurement Scale for Acids and Bases

8.2 pH: A Measurement Scale for Acids and Bases For a strong acid HCl molarity pH For a strong base NaOH molarity pH More Acidic 1.0 x 100 0.00 1.0 x 10-1 1.00 1.0 x 10-2 2.00 1.0 x 10-3 3.00 1.0 x 10-4 4.00 1.0 x 10-5 5.00 1.0 x 10-6 6.00 1.0 x 10-7 7.00 1.0 x 100 14.00 1.0 x 10-1 13.00 1.0 x 10-2 12.00 1.0 x 10-3 11.00 1.0 x 10-4 10.00 1.0 x 10-5 9.00 1.0 x 10-6 8.00 1.0 x 10-7 7.00 More basic 8.2 pH: A Measurement Scale for Acids and Bases Each 10 fold change in concentration changes the pH by one unit

The Importance of pH and pH Control Any change that takes place in aqueous solution generally has at least some pH dependence Agriculture - crops grow best in soil with proper pH Physiology - blood pH shift of 1 pH is fatal Acid Rain - lowers pH of water in aquatic systems causing problems for native fishes Municipal services - sewage treatment and water purification require optimal pH Industry - many processes require strict pH control for cost-effective production 8.2 pH: A Measurement Scale for Acids and Bases

8.3 Reactions Between Acids and Bases Neutralization reaction - the reaction of an acid with a base to produce a salt and water HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Acid Base Salt Water Break apart into ions: H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O Net ionic equation Show only the changed components Omit any ions appearing the same on both sides of equation = Spectator ions H+ + OH-  H2O

8.3 Reactions Between Acids and Bases Net Ionic Neutralization Reaction The net ionic neutralization reaction is more accurately written: H3O+(aq) + OH-(aq)  2H2O(l) This equation applies to any strong acid / strong base neutralization reaction An analytical technique to determine the concentration of an acid or base is titration Titration involves the addition of measured amount of a standard solution to neutralize the second, unknown solution Standard solution - solution of known concentration 8.3 Reactions Between Acids and Bases

8.3 Reactions Between Acids and Bases Acid – Base Titration Buret – long glass tube calibrated in mL which contains the standard solution Standard solution is slowly added until the color changes The equivalence point is when the moles of H3O+ and OH- are equal Indicator – a substance which changes color as pH changes 8.3 Reactions Between Acids and Bases Flask contains a solution of unknown concentration plus indicator

Relationship Between pH and Color in Acid-Base Indicators 8.3 Reactions Between Acids and Bases

Determine the Concentration of a Solution of Hydrochloric Acid Place a known volume of acid whose concentration is not known into a flask Add an indicator, experience guides selection, here phenol red is good Known concentration of NaOH is placed in a buret Drip NaOH into the flask until the indicator changes color 8.3 Reactions Between Acids and Bases

Determine the Concentration of a Solution of Hydrochloric Acid Indicator changes color Equivalence point is reached Mol OH- = Mol H3O+ present in the unknown acid Volume dispensed from buret is determined Calculate acid concentration: Volume of Hydrochloric Acid: 25.00 mL Volume of NaOH added: 35.00 mL Concentration of NaOH: 0.1000 M Balanced reaction shows that HCl and NaOH react 1:1 8.3 Reactions Between Acids and Bases

Determine the Concentration of a Solution of Hydrochloric Acid 35.00 mL NaOH x 1L NaOH x 0.1000 mol NaOH 103 mL NaOH 1L NaOH = 3.500 x 10-3 mol NaOH 3.500 x 10-3 mol NaOH x 1 mol HCl 1 mol NaOH = 3.500 x 10-3 mol HCl this amount of HCl is contained in 25.00 mL 3.500 x 10-3 mol HCl x 103 mL HCl 25.00 mL HCl 1 L HCl = 1.400 x 10-1 mol HCl / L HCl = 0.1400 M HCl 8.3 Reactions Between Acids and Bases

Determine the Concentration of a Solution of Hydrochloric Acid Alternative strategy to solve the acid concentration (Macid)(Vacid) = (Mbase)(Vbase) (Macid) = (Mbase)(Vbase) (Vacid) (Macid) = (0.1000 M) (35.00 mL) (25.00 mL) = 0.1400 M HCl 8.3 Reactions Between Acids and Bases

Calculating the Concentration of Sodium Hydroxide Calculate [NaOH] if 25.00 mL of this solution were required to neutralized 45.00 mL of 0.3000 M HCl Alternative strategy to solve the acid concentration (Macid)(Vacid) = (Mbase)(Vbase) (Mbase) = (Macid)(Vacid) (Vbase) (Mbase) = (0.3000 M) (45.00 mL) (25.00 mL) = 0.5400 M NaOH 8.3 Reactions Between Acids and Bases

Polyprotic Substances The previous examples have the acid and base at a 1:1 combining ratio Not all acid-base pairs do this Polyprotic substance - donates or accepts more than one proton per formula unit Hydrochloric acid is monoprotic, producing one H+ ion for each unit of HCl Sulfuric acid is diprotic, each unit of H2SO4 produces 2 H+ ions 8.3 Reactions Between Acids and Bases H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2 H2O(l)

Dissociation of Polyprotic Substances Step 1. H2SO4(aq) + H2O(l) HSO4-(aq) + H3O+(aq) Step 2. HSO4-(aq) + H2O(l) SO42-(aq) + H3O+(aq) In Step 1, H2SO4 behaves as a strong acid – dissociating completely In Step 2, HSO4-( behaves as a weak acid – reversibly dissociating, note the double arrow 8.3 Reactions Between Acids and Bases

8.4 Acid-Base Buffers Buffer solution - solution which resists large changes in pH when either acids or bases are added These solutions are frequently prepared in laboratories to maintain optimum conditions for chemical reactions Buffers are also used routinely in commercial products to maintain optimum conditions for product behavior

A buffer is Le Chatelier’s principle in action The Buffer Process Buffers act to establish an equilibrium between a conjugate acid – base pair Buffers consist of either a weak acid and its salt (conjugate base) a weak base and its salt (conjugate acid) Acetic acid (CH3COOH) with sodium acetate (CH3COONa) An equilibrium is established in solution between the acid and the salt anion A buffer is Le Chatelier’s principle in action 8.4 Acid-Base Buffers CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Addition of Base (OH-) to a Buffer Solution Adding a basic substance to a buffer causes changes The OH- will react with the H3O+ producing water Acid in the buffer system dissociates to replace the H3O+ consumed by the added base Net result is to maintain the pH close to the initial level The loss of H3O+ (the stress) is compensated by the dissociation of the acid to produce more H3O+ 8.4 Acid-Base Buffers CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Addition of Acid (H3O+) to a Buffer Solution Adding an acidic substance to a buffer causes changes The H3O+ from the acid will increase the overall H3O+ Conjugate base in the buffer system reacts with the H3O+ to form more acid Net result is to maintain the H3O+ concentration and the pH close to the initial level The gain of H3O+ (the stress) is compensated by the reaction of the conjugate base to produce more acid 8.4 Acid-Base Buffers CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Buffer Capacity 8.4 Acid-Base Buffers Buffer capacity - a measure of the ability of a solution to resist large changes in pH when a strong acid or strong base is added Also described as the amount of strong acid or strong base that a buffer can neutralize without significantly changing pH 8.4 Acid-Base Buffers

Preparation of a Buffer Solution Buffering process is an equilibrium reaction described by an equilibrium-constant expression In acids, this constant is Ka If you want to know the pH of the buffer, solve for [H3O+], then calculate pH CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) 8.4 Acid-Base Buffers

Calculating the pH of a Buffer Solution Calculate the pH of a buffer solution in which Both the acetic acid (acid) and sodium acetate (salt) concentrations are 2.0 x 10-2 M Sodium acetate, the salt, is also the conjugate base Ka = 1.75 x 10-5 [H3O+] = [acid]Ka [conjugate base] = {(2.0 x 10-2 M) x (1.75 x 10-5)} / 2.0 x 10-2 M = 1.75 x 10-5 pH = -log 1.75 x 10-5 = 4.76 8.4 Acid-Base Buffers CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Henderson-Hasselbalch Equation Solution of equilibrium-constant expression and pH can be combined into one operation Henderson-Hasselbalch equation is this combined expression Using these two equations: pKa = -log Ka just as pH = -log[H3O+] pKa = pH – log ( [CH3COO-] / [CH3COOH] ) Henderson-Hasselbalch – pH = pKa + log( [CH3COO-] / [CH3COOH] ) 8.4 Acid-Base Buffers CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Henderson-Hasselbalch Equation pH = pKa + log ([CH3COO-] / [CH3COOH]) can be rewritten pH = pKa + log ([conjugate base] / [weak acid]) 8.4 Acid-Base Buffers

8.5 Oxidation-Reduction Processes Oxidation-reduction processes are responsible for many types of chemical change Oxidation - defined by one of the following loss of electrons loss of hydrogen atoms gain of oxygen atoms Example: NaNa+ + e- Oxidation half reaction

8.5 Oxidation-Reduction Processes Reduction - defined by one of the following: gain of electrons gain of hydrogen loss of oxygen Example: Cl + e-  Cl- Reduction half reaction Cannot have oxidation without reduction 8.5 Oxidation-Reduction Processes

Oxidation and Reduction as Complementary Processes Na  Na+ + e- Cl + e-  Cl- 8.5 Oxidation-Reduction Processes Na + Cl Na+ + Cl- Oxidizing Agent Is reduced Gains electrons Causes oxidation Reducing Agent Is oxidized Loses electrons Causes reduction

Applications of Oxidation and Reduction Corrosion - the deterioration of metals caused by an oxidation-reduction process Example: rust (oxidation of iron) 4Fe(s) + 3O2(g)  2Fe2O3(s) Combustion of Fossil Fuels Example: natural gas furnaces CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) 8.5 Oxidation-Reduction Processes

Applications of Oxidation and Reduction Bleaching Most bleaching agents are oxidizing agents The oxidation of the stains produces compounds that do not have color Example: Chlorine bleach - sodium hypochlorite (NaOCl) 8.5 Oxidation-Reduction Processes

8.5 Oxidation-Reduction Processes Biological Processes Respiration Electron-transport chain of aerobic respiration uses reversible oxidation and reduction of iron atoms in cytochrome c Metabolism Break down of molecules into smaller pieces by enzymes 8.5 Oxidation-Reduction Processes

8.5 Oxidation-Reduction Processes Voltaic Cells Voltaic cell - electrochemical cell that converts stored chemical energy into electrical energy Let’s consider the following reaction: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) 8.5 Oxidation-Reduction Processes Is Zn oxidized or reduced? Oxidized Copper is reduced

8.5 Oxidation-Reduction Processes Voltaic Cells Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) If the two reactants are placed in the same flask they cannot produce electrical current A voltaic cell separates the two half reactions This makes the electrons flow through a wire to allow the oxidation and reduction to occur 8.5 Oxidation-Reduction Processes

Voltaic Cell Generating Electrical Current 8.5 Oxidation-Reduction Processes Zn  Zn2+ + 2e- Oxidation anode – electrode where oxidation occurs Cu2+ + 2e-  Cu Reduction cathode – electrode where reduction occurs

Voltaic Cell Generating Electrical Current 8.5 Oxidation-Reduction Processes

8.5 Oxidation-Reduction Processes Silver Battery Batteries use the concept of the voltaic cell Modern batteries are: Smaller Safer More dependable 8.5 Oxidation-Reduction Processes

8.5 Oxidation-Reduction Processes Electrolysis Electrolysis reactions - use electrical energy to cause nonspontaneous oxidation-reduction reactions to occur These reactions are the reverse of a voltaic cell Rechargeable battery When powering a device behaves as a voltaic cell With time, the chemical reaction nears completion Battery appears to “run down” Cell reaction is reversible when battery attached to charger 8.5 Oxidation-Reduction Processes

Comparison of Voltaic and Electrolytic Cells 8.5 Oxidation-Reduction Processes