Chemistry Grade 12 Textbook PDF

CHEMISTRY CHEMISTRY STUDENTS TEXTBOOK CHEMISTRY GRADE 12...

CHEMISTRY CHEMISTRY STUDENTS TEXTBOOK CHEMISTRY GRADE 12 Students Textbook GRADE 12 STUDENTS TEXTBOOK GRADE 12 FEDERAL DEMOCRATIC REPUBLIC OF ETHIOPIA MINSTRY OF EDUCATION FEDERAL DEMOCRATIC REPUBLIC OF ETHIOPIA MINSTRY OF EDUCATION  የኢትዮጵያ ፌደራላዊ ዴሞክራሲያዊ ሪፐብሊክ የትምህርት ሚኒስቴር FEDERAL DEMOCRATIC REPUBLIC OF ETHIOPIA MINISTRY OF EDUCATION CHEMISTRY STUDENT TEXTBOOK GRADE 12 Writers Hailu Shiferaw (PhD) Muluken Aklilu (PhD) Content Editor Chala Regasa(MSc) Instructional Design Editor Taye Hirpassa (Bsc, MA) Language Editor Meseret Getnet(PhD) Illustrator Asresahegn Kassaye (MSc) Book Designer Daniel Tesfay (MSc) Evaluators Tolessa Mergo Roro (MSc., MEd) Nega Gichile (BSc, MA) Sefiw Melesse (MSc.)  CHEMISTRY GRADE 12 CONTENTS UNIT 1: ACID-BASE EQUILIBRIA 1.1 ACID-BASE CONCEPTS...........................................................3 1.1.1 Arrhenius concept of acids and bases...................................................3 1.1.2 Brønsted-Lowry Concept of Acids and Bases.......................................5 1.1.3 Lewis concept of acids and bases.......................................................... 11 1.2 IONIC EQUILIBRIA OF WEAK ACIDS AND BASES.......... 14 1.2.1 Ionization of water................................................................................. 14 1.2.1 Measures of the strength of acids and bases in aqueous solution....20 1.3 COMMON ION EFFECT AND BUFFER SOLUTION..........28 1.3.1 The Common ion Effect........................................................................28 1.3.2 Buffer Solutions....................................................................................... 31 1.4 Hydrolsis of Salts......................................................................35 1.4.1 Hydrolysis of Salts of Strong Acids and Strong Bases........................35 1.4.2 Hydrolysis of Salts of weak acids and strong bases............................36 1.4.3 Hydrolysis of Salts of strong acids and weak bases............................37 1.4.4 Hydrolysis of Salts of weak acids and week bases.............................37 1.5 Acid–Base Indicators and Titrations........................................38 1.5.1 Acid–Base Indicators..............................................................................38 1.5.2 Equivalents of Acids and Bases.............................................................40 1.5.3 Acid–Base Titrations..............................................................................41 UNIT 2: ELECTRO CHEMISTRY 2.1 OXIDATION-REDUCTION REACTIONS............................50 2.1.1 Oxidation................................................................................................50 2.1.2 Reduction...............................................................................................51 2.1.3 Balancing Oxidation-Reduction (Redox) Reactions.........................52 2.2 ELECTROLYSIS OF AQUEOUS SOLUTIONS...................... 57 2.2.1 Electrolytic cells.....................................................................................60 2.2.2 Preferential Discharge...........................................................................62 2.2.3 Electrolysis of Some Selected Aqueous Solutions...............................66 2.3 QUANTITATIVE ASPECTS OF ELECTROLYSIS.................72 2.3.1 Faraday’s First Law of Electrolysis........................................................72 2.3.2 Faraday’s Second Law of Electrolysis...................................................75 2.4 INDUSTRIAL APPLICATION OF ELECTROLYSIS............. 76 2.5 VOLTAIC CELLS...................................................................... 81 II  UNIT 3: INDUSTRIAL CHEMISTRY 3.1 Introduction...........................................................................120 3.2 Natural Resources and Industry............................................ 122 3.2.1 Natural resources (Raw materials).....................................................122 3.2.2 Industry..................................................................................................123 3.3 Manufacturing of valuable products/ Chemicals.................. 125 3.3.1 Production of Ammonia (NH3).........................................................126 3.3.2 Nitric Acid Manufacturing Process...................................................133 3.3.3 Manufacturing of Sulphuric Acid.....................................................138 3.3.4 Nitrogen Based Fertilizers.................................................................. 141 3.3.5 Some Common Pesticide and Herbicides........................................144 3.3.6 3.3.7 Manufacturing of Sodium Carbonate.....................................148 3.3.7 3.3.8 Manufacturing of Sodium Hydroxide NaOH.........................150 3.4 Some Manufacturing Industries in Ethiopia........................ 152 3.4.1 3.4.1 Glass Manufacturing..................................................................153 3.4.2 Manufacturing of Ceramics...............................................................155 3.4.3 Cement...................................................................................................157 3.4.4 3.4.4 Sugar Manufacturing..................................................................159 3.4.5 Paper and Pulp...................................................................................... 161 3.4.6 Tannery..................................................................................................163 3.4.7 Food Processing and Preservation.....................................................164 3.4.8 Manufacturing of Alcoholic Beverages.............................................166 3.4.9 Soaps and Detergents..........................................................................172 UNIT 4: POLYMER 4.1 Introduction to polymers........................................................ 184 4.2 Classification of Polymers...................................................... 185 4.3 Polymerization Reactions....................................................... 193 III  CHEMISTRY GRADE 12 UNIT 5: INTRODUCTION TO ENVIRONMENTAL CHEMISTRY 5.1 Introduction...........................................................................206 5.1.1 Components of the Environment......................................................207 5.1.2 Natural Cycles in the Environment................................................... 210 5.1.3 Concepts Related to Environmental Chemistry...............................215 5.2 Environmental Pollution....................................................... 216 5.2.1 Air Pollution.......................................................................................... 217 5.2.2 Water Pollution.....................................................................................220 5.2.3 Land Pollution.......................................................................................221 5.2.4 Pollutants of the Environment............................................................223 5.3 Global Warming and Climate Change..................................224 5.3.1 Global Warming and Climate Change.............................................224 5.3.2 Chemistry of Greenhouses Gasses and their effects on Climate change...............................................................................................226 5.4 Green Chemistry and Cleaner Production...........................228 5.4.1 Principle of Green Chemistry.............................................................230 5.4.2 Cleaner production in chemistry......................................................234 IV  Acid-Base Equilibria UNIT 1 ACID-BASE EQUILIBRIA Unit outcomes At the end of this unit, you will be able to:  describe the draw backs of Arrhenius,  define Bronsted-Lowery &Lewis concepts of acids & bases  describe the dissociation of water, weak mono-protic & polyprotic acids, & weak bases;  solve equilibrium problems involving concentration of reactants & products, ka, kb, PH& POH  discuss the common ion effect, buffer solution, hydrolysis of salts, acid- base indicators and acid-base titrations  explain how buffering action affects our daily lives using examples  determine the equivalents of acid or base that are required to neutralize specific amount of acid or base  predict, in qualitative terms, whether a solution of a specific salt will be acidic, basic or neutral  explain how to solve problems involving concentration and pH of acid- base titration 1  CHEMISTRY GRADE 12 Startup Activity You have learnt the acid-base concepts and properties. Remember this and discuss the following questions in group. After discussion write a report and present to the class: 1. List the properties of acids and bases 2. What do you observe when the following indicators added in the solutions listed in the table? Color of indicator Red Blue Phenol- Methyl litmus litmus phthalein Orange SOLUTION paper paper acetic acid solution hydrochloric acid solution sodium chloride solution ammonia solution Lemon juice The concepts of acids and bases are probably among the most familiar chemistry concepts. The reason is that acids and bases have been used as laboratory chemicals for centuries, as well as in the home. Common household acids include acetic acid ( CHCOOH , vinegar), citric acid ( H 3C6 H 5O7 , in citrus fruits), and phosphoric acid ( H 3 PO4 , a flavoring in carbonated beverages). Sodium hydroxide ( NaOH , drain cleaner) and ammonia ( NH 3 , glass cleaner), are household bases. Weak acids and weak bases are important weak electrolytes. They are found in many chemical and biological processes of interest. Amino acids, for example, are both weak acids and weak bases. In this unit, we will learn some ways of expressing concentrations of hydronium ions and of hydroxide ions in solutions of weak acids and weak bases. Then you will examine equilibria involving these weak electrolytes. You will also see that the indicators, used in titration, such as phenolphthalein, are weak acids or weak bases. Finally, you will learn how to use these properties to select an appropriate indicator for a titration. 2  Acid-Base Equilibria 1.1 ACID-BASE CONCEPTS At the end of this subunit, you will be able to:  define acid by the Arrhenius concept  define base by the Arrhenius concept  Describe the drawbacks of Arrhenius concept  define acid by the Bronsted-Lowry concept;  give examples of Bronsted-Lowry acids  define base by the Bronsted-Lowry concept  give examples of Bronsted-Lowry bases.  explain what conjugate acids and conjugate bases are  identify the acid-base conjugate pairs from the given reaction;  write an equation for self-ionization of water and ammonia.  explain what is meant by amphiprotic species  give examples of reactions of amphiprotic species;  define acid by the Lewis concept  give examples of Lewis acids  define base by the Lewis concept  give examples of Lewis bases  calculate pH from  H +  and  H +  from pH calculate pOH from OH −  and OH −  from pOH 1.1.1 Arrhenius concept of acids and bases Activity 1.1 Form groups and discuss the following questions and write a report of your discussion. 1. Explain Arrhenius acid and base concepts using suitable examples? 2. Does hydrogen ion exist freely in water? 3. What are the drawbacks of the Arrhenius’ concepts of acids and bases? 3  CHEMISTRY GRADE 12 The Swedish chemist Svante Arrhenius framed the first successful concept of acids and bases. He defined acids and bases in terms of the effect these substances have on water. According to Arrhenius, acids are substances that increase the concentration of H + (proton ion) in aqueous solution, and bases increase the concentration of OH − (a hydroxide ion) in aqueous solution. In Arrhenius’s theory, a strong acid is a substance that completely ionizes in aqueous solution to give H 3O + (aq) and an anion. An example is perchloric acid, HClO4 HClO4 (aq ) + H 2O(l ) → H 3O + (aq ) + ClO4 − (aq ) Other examples of strong acids are H 2 SO4 , HI , HBr , HCl , and HNO3. A strong base completely ionizes in aqueous solution to give OH − and a cation. Sodium hydroxide is an example of a strong base. NaOH ( s )  H 2O → Na + (aq ) + OH − (aq ) The principal strong bases are the hydroxides of Group IA elements and Group IIA elements (except Be). Despite its early successes and continued usefulness, the Arrhenius theory does have limitations. Exercise 1.1 1. Based on their dissociations in water solution, classify each of the following compounds as Arrhenius acid, Arrhenius base, or as a compound that cannot be classified as an Arrhenius acid or Arrhenius base. a. HBr (aq ) + H 2O(l ) → H 3O + (aq ) + 2 Br − (aq ) b. NaCl ( s ) + H O(l ) → Na + (aq ) + Cl − (aq ) + − c. NH 3 (l ) + H 2O(l ) → NH 4 (aq ) + OH (aq ) + − 2. NaOH (aq ) + H 2O(l ) → Na (aq ) + 2OH (aq ) 4  Acid-Base Equilibria 1.1.2 Brønsted-Lowry Concept of Acids and Bases Activity 1.2 From what you have learnt in Grade 10 chemistry, discuss the following questions. 1. How does Brønsted-Lowry concept of acids and bases differ from Arrhenius definition? What are the similarities? 2. Give two Brønsted-Lowry bases that are not Arrhenius bases. 3. Are there any Brønsted acids that do not behave as Arrhenius acids? Consider the ionization of hydrochloric acid in water: O H + H Cl H H + Cl O H H Which one is a Brønsted-Lowry acid and which one is a Brønsted-Lowry basem? In 1923, J. N. Brønsted in Denmark and T. M. Lowry in Great Britain independently proposed a new acid base theory. They pointed out that acid–base reactions can be seen as proton-transfer reactions and those acids and bases can be defined in terms of this proton (H) transfer. According to their concept, an acid is a proton donor and a base is a proton acceptor. Let’s use the Brønsted–Lowry theory to describe the ionization of ammoniain aqueous solution In this reaction water acts as an acid. It gives up a proton (H+) to NH 3 , a base. As a result of this transfer the polyatomic ions NH 4 + and OH − are formed- the same ions produced by the ionization of the hypothetical NH 4OH of the Arrhenius theory. However, they cannot be called Arrhenius bases since in aqueous solution they do not dissociate to form OH −. The advantage of this definition is that it is not limited to aqueous solutions. Bronsted-Lowry acids and bases always occur in pairs called conjugate acid base pairs. 5  CHEMISTRY GRADE 12 1.1.2.1 Conjugate acid-base pairs Conjugate acid-base pairs can be defined as an acid and its conjugate base or a base and its conjugate acid. The conjugate base of a Brønsted-Lowry acid is the species that remains when one proton has been removed from the acid. Conversely, a conjugate acid results from the addition of a proton to a Brønsted-Lowry base. Every Brønsted-Lowry acid has a conjugate base, and every Brønsted- Lowry base has a conjugate acid. For example, the chloride ion (Cl − ) is the conjugate base formed from the acid HCl , and H 2O is the conjugate base of the acid H 3O +. Similarly, the ionization of acetic acid can be represented as Conjugate acid - base pair CH 3COOH (aq )+ H 2O(l )  CH 3COO − (aq ) + H 3O + (aq ) Acid Base Base Acid Conjugate acid - base pair In the above reaction, CH 3COOH acts as an acid. It gives up a proton, H + , which is taken up by H 2O. Thus, H 2O acts as a base. In the reverse reaction, the hydronium ion, H 3O + , acts as an acid and CH COO − acts as a base. When CH 3COOH loses a proton, it is converted into CH 3COO −. Notice that the formulas of these two species differ by a single proton, H +. Species that differ by a single proton ( H + ) constitute a conjugate acid–base pair. Within this pair, the species with the added H + is the acid, and the species without the H + is the base..For any conjugate acid-base pair,  The conjugate base has one fewer H and one more minus charge than the acid.  The conjugate acid has one more H and one fewer minus charge than the base. 6  Acid-Base Equilibria Table 1.1: The conjugate pairs in some Acid-Base Reactions. Conjugate pair Acid + Base ⇌ Base + Acid Conjugate pair HF + H 2O F- + H3O+ HCOOH + CN- HCOO- + HCN NH4+ + CO32- NH3 + HCO3- H2PO4- + OH- HPO4- + H 2O H2SO4 + N 2H 5+ HSO4- + N2H62+ HPO42- + SO32- PO43- + HSO3- Example 1.1 For each of the following reactions, which occur in aqueous solution, identify the Brønsted-Lowry acids and bases and their respective conjugates in each of the following reactions. a. NH 3 + H 2 PO4 −  NH 4 + + HPO4 2 − b. HCl + H 2 PO4 −  Cl − + H 3 PO4 Solution: To identify Brønsted-Lowry acids and bases, we look for the proton donors and proton-acceptors in each reaction. a. H 2 PO4 − is converted to HPO4 2 − by donating a proton. So, H 2 PO4 − is an acid, and HPO4 2 − is its conjugate base. NH 3 accepts the proton lost by the H 2 PO4 −.As a result, NH 3 is a base, and NH 4 + is its conjugate acid. NH 3 + H 2 PO4 −  NH 4 + + HPO4 2− Base Acid Acid Base b H 2 PO4 − accepts a proton from HCl. Therefore, H 2 PO4 − is a base and H 3 PO4 is its conjugate acid. HCl donates a proton to H 2 PO4 −. Thus, HCl is an acid, and Cl − is its conjugate base HCl + H 2 PO4 −  Cl − + H 3 PO4 Acid Base Base Acid 7  CHEMISTRY GRADE 12 Exercise 1.2 1. Identify the Brønsted-Lowry acids, bases, conjugate acids and conjugate bases in each of the following reaction − + b. HClO2 + H 2O  ClO2 + H 3O − − c. OCl + H 2O  HOCl + OH 2− − − d. H 2O + SO3  OH + HSO3 Strengths of Conjugate Acid-Base Pairs How do you know the strengths of conjugate acid- base pairs? The net direction of an acid-base reaction depends on relative acid and base strengths: A reaction proceeds to the greater extent in the direction in which a stronger acid and stronger base form a weaker acid and weaker base. The stronger the acid, the weaker is its conjugate base. Similarly, the stronger the base, the weaker is its conjugate acid. For example, HCl is a strong acid, and its conjugate base Cl − , is a weak base. Acetic acid, CH 3COOH , is a weak acid, and its conjugate base, CH 3COO − ,is a strong base. The following chart shows the strength of conjugate acid-base pairs. 8  Acid-Base Equilibria Chart 1.1: Strengths of Conjugate Acid-Base Pairs. 9  CHEMISTRY GRADE 12 1.1.2.2 Auto ionization of substances Name the ions present in water. How are they formed? Molecular auto ionization (or self-ionization) is a reaction between two identical neutral molecules, especially in a solution, to produce an anion and a cation. If a pure liquid partially dissociates into ions, it is said to be self-ionizing. Water, as we know, is a unique solvent. One of its special properties is its ability to act either as an acid or as a base. Water functions as a base in reactions with acids such as HCl and CH 3COOH , and it functions as an acid in reactions with bases such as NH 3. Water is a very weak electrolyte and therefore a poor conductor of electricity, but it does undergo ionization to a small extent: 2 H 2O(l )  H 3O + (aq ) + OH − (aq ) H H O H O H O H + O H H H Base Base Acid Acid This reaction is sometimes called the auto ionization of water. Note that, in this reaction, some water molecules behave as acids, donating protons, while the other water molecules behave as bases, accepting protons. 1.1.2.3 Amphiprotic Species Do you think that a molecule or an ion can be both a donor or acceptor of protons? Molecules or ions that can either donate or accept a proton, depending on the other reactant, are called amphiprotic species. For example, HCO3− acts as an acid in the presence of OH − but as a base in the presence of HF. The most important amphiprotic species is water itself. When an acid donates a proton to water, the water molecule is a proton acceptor, and hence a base. Conversely, when a base reacts with water, a water molecule donates a proton, and hence acts as an acid. Consider, for example, the reactions of water with the base NH 3 and with the acid CH 3COOH (acetic acid) 10  Acid-Base Equilibria - NH3(aq) + H 2O(l) NH4+(aq) + OH (aq) acid acid base base - H C2H3O2(aq) + H2O(aq) C2H3O2 (aq) + H3O+(aq) acid base base acid In the first case, water reacts as an acid with the base NH 3. In the second case, water reacts as a base with the acid CH 3COOH. Exercise 1.3 1. Define each of the following terms and give examples for each. a. autoionization b. amphiprotic species 2. Write equations to show the amphiprotic behavior of a H 2 PO4 − b. H 2O 3. Predict the relative strengths of each of the following groups: − − a. OH − , Cl − , NO , CH 3COO , and NH 3 b. HClO4 , CH 3COOH , HNO3 and HCl 3. What is the weakness of the Brønsted-Lowry acids and bases theory? 4. Write the self-ionization of ammonia. 1.1.3 Lewis concept of acids and bases Activity 1.3 Form groups and discuss the following questions and report the result of your discussion to your teacher. 1. What is the main difference between Lewis acid -base and Brønsted- Lowry acid - base concepts? 2. Are all Brønsted-Lowry acids and bases are also acids and bases according to Lewis concept? 3. Is there any limitation to the Brønsted-Lowry definition of acids and bases? Explain if any 11  CHEMISTRY GRADE 12 G. N. Lewis, who proposed the electron-pair theory of covalent bonding, realized that the concept of acids and bases could be generalized to include reactions of acidic and basic oxides and many other reactions, as well as proton-transfer reactions. According to this concept, the Lewis acid-base definition holds that  A base is any species that donates an electron pair to form a bond.  An acidis any species that accepts an electron pair to form a bond. Consider, for example, the reaction of boron trifluoride with ammonia H F H F + N H F B N H B F H F H F Lewis acid Lewis base The boron atom in boron trifluoride, BF3 , has only six electrons in its valance shell and needs two electrons to satisfy the octet rule. Consequently, BF3 (Lewis acid) accepts a pair of electrons from NH 3 (Lewis base). This example suggests that in a Lewis acid-base reaction, we should look for: 1. a species that has an available empty orbital to accommodate an electron pair such as the B atom in BF3 , and 2. a species that has lone-pair electrons such as N in NH 3 The Lewis definition allows us to consider typical Brønsted-Lowry bases, such as OH − , NH 3 , and H 2O , as Lewis bases. They all have electron pairs available to donate for electron-deficient species. Note that any molecule or negatively charged species having an excess of electrons can be considered as a Lewis base, and any electron-deficient molecule or positively charged species can be considered as a Lewis acid. 12  Acid-Base Equilibria Example 1.2 1. Identify the acid and the base in each Lewis acid–base reaction. BH 3 + (CH 3 ) 2 S → H 3 BS (CH 3 ) 2 CaO + CO2 → CaCO3 BeCl2 + 2Cl − → BeCl4 2− Solution: In BH 3 , boron has only six valence electrons. It is therefore electron deficient and can accept a lone pair of electrons. Like oxygen, the sulfur atom in (CH 3 ) 2 S has two lone pairs. Thus (CH 3 ) 2 S donates an electron pair on sulfur to the boron atom of BH 3. The Lewis base is (CH 3 ) 2 S , and the Lewis acid is BH 3. CO2 accepts a pair of electrons from the O 2− ion in CaO to form the carbonate ion. The oxygen in CaO is an electron-pair donor, so CaO is the Lewis base. Carbon accepts a pair of electrons, so CO2 is the Lewis acid. The chloride ion contains four lone pairs. In this reaction, each chloride ion donates one lone pair to BeCl2 , which has only four electrons around Be. Thus, the chloride ions are Lewis bases, and BeCl2 is the Lewis acid. 2. Identify the acid and the base in each Lewis acid–base reaction. (CH 3 ) 2 O + BF3 → (CH 3 ) 2 O : BF3 H 2O + SO3 → H 2 SO4 Solution Lewis’s base: (CH 3 ) 2 O ; Lewis’s acid: BF3 Lewis’s base: H 2O ; Lewis’s acid: SO3 Exercise 1.4 1. Identify Lewis acids and Lewis bases in each of the following reactions. a. H + + OH −  H 2O b. Cl − + BCl3  BCl4 − c. K + + 6 H 2O  K ( H 2O)6 + d. OH − + Al (OH )3  Al (OH ) 4 − e. CO2 + H 2O  H 2CO3 f. Ni + 4CO  Ni (CO) 4 13  CHEMISTRY GRADE 12 1.2 IONIC EQUILIBRIA OF WEAK ACIDS AND BASES At the end of this subunit, you will be able to:  describe the ionization of water;  derive the expression of ion product for water, K w  explain why water is a weak electrolyte;  use K w to calculate  H 3O +  or OH −  in aqueous solution;  write an expression for the percent ionization of weak acids or weak bases;  calculate the percent dissociation of weak acids and bases;  write the expression for the acid-dissociation constant, K a ;  calculate K a for an acid from the concentration of a given solution and it’s pH ;  calculate  H +  and pH of an acidic solution from given values of K a and the initial concentration of the solution;  write the expression for the base-dissociation constant, K b ;  calculate K b for a base from the concentration of a basic solution and its pOH ; and OH −   calculate the and pOH of a basic solution from a given value of K b and the initial concentration of the solution. 1.2.1 Ionization of water How do you calculate the concentration of H 3O + ions if the concentrations of OH − ions and K w at 25°C are given? Although pure water is often considered a non-electrolyte (nonconductor of electricity), precise measurements do show a very small conduction. This conduction results from self-ionization (or autoionization) of water, a reaction in which two like molecules react to give ions. The H 2O molecule can act as either an acid or a base; it is amphiprotic. It should come as no surprise that amongst themselves water molecules can produce H 3O + and OH − ions via the following self-ionization reaction or autoionization reaction: 14  Acid-Base Equilibria H 2O(l ) + H 2O(l )  H 3O + (aq ) + OH − (aq ) Like any equilibrium process, the auto-ionization of water is described quantitatively by an equilibrium constant:  H 3O +  OH −  Kc = [ H 2O ] 2 Because the concentration of ions formed is very small and the concentration of H 2O remains essentially constant, about 56 Mat 25oC, we multiply Kc by [ H 2O ] to obtain a 2 new equilibrium constant, the ion-product constant for water, K w : [ H 2= O] Kc =  H 3O +  OH −  2 constant We call the equilibrium value of the ion product  H 3O +  OH −  , the ion-product constant for water, K w. At 25oC, the value of K w is 1.0 x 10-14. Like any equilibrium constant, K w varies with temperature. At body temperature (37oC), K w equals 2.5 x 10-14. =K w = H 3O +  OH −  1.0 x 10−14 at 25 oC Because we often write H + (aq ) for H 3O + (aq ) , the ion-product constant for water can be written K w =  H 3O +  OH −  Using K w , you can calculate the concentrations of H 3O + and OH − ions in pure water. These ions are produced in equal numbers in pure water, so their concentrations are equal. Let x = = H 3O +  OH − . Then, substituting into the equation for the ion-product constant, K w =  H 3O +  OH −  you get at 25 oC, 1.0 x 10-14 = x2 hence x equals 1.0 x 10-7. Thus, the concentrations of H O + and OH − are both 1.0 x 10 -7 M in pure water. If you add an acid or a base to water, the concentrations of H 3O + and OH − will no longer be equal. The equilibrium-constant equation K w =  H 3O +  OH −  will still hold. 15  CHEMISTRY GRADE 12 In any aqueous solution at 25°C, no matter what it contains, the product of  H +  and OH −  must always equal 1.0 × 10–14. The equilibrium nature of auto-ionization allows us to define “acidic” and “basic” solutions in terms of relative magnitudes of  H 3O +  and OH −  : i. a neutral solution, where  H 3O +  = OH − . ii. an acidic solution, where  H 3O +  > OH − . iii. a basic solution, where OH −  >  H 3O +  Activity 1.4 Form a group and discuss the following. Write a report on the discussion and present to the class. Many substances undergo auto-ionization in analogous to water. For example, the auto-ionization of liquid ammonia is:– 2NH 3  NH 4 + + NH 2 − a. Write K c expression for auto-ionization of ammonia that is analogous to the K w expression for water. b. Name the strongest acids and strongest bases that can exist in liquid ammonia? c. For water, a solution with OH −  <  H 3O +  is acidic. What are the analogous relationships in liquid ammonia? Example 1.3 A research chemist adds a measured amount of HCl gas to pure water at 25 0Cand obtains a solution with  H O   + -4  = 3.0 x 10 M. Calculate OH − . Is the solution neutral, acidic, or basic? Solution: We use the known value of K w at 25 0C (1.0 x 10─14) and the given  H 3O +  (3.0x10-4M) to solve for OH − . Then we compare  H 3O +  with OH −  to determine whether the solution is acidic, basic, or neutral Kw 1.0 x 10−14 OH −  Calculate for the= = = −4 3.3 x 10−11 M  H 3O  + 3.0 x 10 Because  H 3O +  > OH −  , the solution is acidic 16  Acid-Base Equilibria Exercise 1.5 1. Calculate  H +  or OH −  , as required, for each of the following solutions at 25°C,and state whether the solution is neutral, acidic, or basic. a.  H +  = 1.0 x 10−7 M b. OH −  = 1.0 x 10−4 M c. OH −  = 1.0 x 10−8 M 2. Calculate the concentration of OH − in a solution in which a.  H 3O +  = 2.0 x 10−5 M b.  H 3O +  = OH −  c.  H 3O +  = 102 x OH −  3. Calculate  H 3O +  in a solution that is at 25°C and has OH −  = 6.7 x 10−2 M. Is the solution neutral, acidic, or basic? 4. At 40°C, the value of K w is 2.92 × 10–14. Calculate the  H +  and OH −  of pure water at 40°C. 5. Why water is a weak electrolyte? The pH scales It is a well-known fact that whether an aqueous solution is acidic, neutral, or basic depends on the hydronium-ion concentration. You can quantitatively describe the acidity by giving the hydronium-ion concentration. But because these concentration values may be very small, it is often more convenient to give the acidity in terms of pH , which is defined as the negative of the logarithm of the molar hydronium-ion concentration. pH is a measure of the hydronium ion content of a solution. It is also restated in terms of  H 3O + . − log  H 3O +  or pH = pH = − log  H +  Thus, in a solution that has  H 3O +  = 2.5 x 10−3 M − log(2.5 x 10−3 ) = pH = 2.60 Note that the negative logarithm gives us positive numbers for pH. Like the equilibrium constant, the pH of a solution is a dimensionless quantity. To determine the  H 3O +  , that corresponds to a particular pH value, we do an inverse calculation. In a solution with pH = 4.5, log  H 3O +  = − 4.50 and  H 3O +  =10−4.50 = 3.2 x 10−5 M The pH of a solution is measured by a pH - meter. 17  CHEMISTRY GRADE 12 Figure 1.1: pH -meter. A pH meter is commonly used in the laboratory to determine the pH of a solution. Although many pH meters have scales marked with values from 1 to 14, pH values can, in fact, be less than1 and greater than 14. pH value decreases as the concentration of H + ions increase; in other words, the more acidic the solution, the lower its pH ; the more basic the solution, the higher its pH. For a neutral solution, pH = 7 Acidic solutions have pH < 7 Basic solutions have pH > 7 Activity 1.5 In your group, measure the pH of the following substances, using a pH-meter. Copy and fill in the following table. Compare your results with those of other groups. Find the pH values of the substances in reference books and other sources and compare your results with the values you observed. Substance Acidic, Basic or Neutral pH Beer Milk of magnesia Tomato juice Lemon juice Drinking water 18  Acid-Base Equilibria The pH notation has been extended to other exponential quantities. A pOH scale analogous to the pH scale can be devised using the negative logarithm of the hydroxide ion concentration of a solution. Thus, we define pOH as: pOH = − log OH −  If we are given the pOH value of a solution and asked to calculate the OH − ion concentration, we can take the antilog of the above equation as follows OH −  = 10− pOH Another useful expression can be derived by taking the negative logarithmof the expression (written for 25 °C) and introducing the symbol pK w = − log K w. Now = consider again the ion product for water at 25°C. K w = H 3O +  OH −  1.0 x 10−14 Taking the negative logarithm of both sides, we obtain - (log  H 3O +  + log OH -  ) = - log(1.0 x 10-14 ) − log  H 3O +  = − log OH −  = 14 From the definition of pH and pOH we obtain: pH + pOH = 14 Thus, in general, the sum of the pH and pOH values must equal pK w. This equation provides us with another way to express the relationship between the H + ion concentration and the OH − - ion concentration. Example 1.4 Calculate: a. the pH and pOH of a juice solution in which  H 3O +  is 5.0 ×10–3 M b. the  H 3O +  and OH −  of human blood at pH = 7.40 Solution:  H 3O +  = 5.0 x 10−3 M a. = pH ?= and pOH ? pH = − log  H 3O +  = − log(5.0 x 10−3 ) = 3 − log 5.0 =2.3 pH + pOH = 14 pOH= 14 − pH pOH= 14 − 2.3 19  CHEMISTRY GRADE 12 = 11.7 b. pH =7.40, pOH = 14 − 7.40 =6.6  H 3O +  ?= = OH −  ? pH = − log  H 3O +  and pOH = − log[OH − ] log  H 3O +  = −7.40 and log[OH − ] = −6.6  H 3O= +  10−= 7.40 4.0 x 10−8 M and [OH = − = ] 10 −6.6 2.51 x 10−7 M Exercise 1.6 1. A solution formed by dissolving an antacid tablet has a pH of 9.18 at 25°C. Calculate  H +  , OH −  and pOH. A solution is prepared by diluting concentrated HNO3 to 2.0 M, 0.30 M and 0.0063 M HNO3 at 25°C. Calculate  H +  , OH −  , pH and pOH of the three solutions. 1.2.1 Measures of the strength of acids and bases in aqueous solution The strength of acids and bases depends on a number of factors. Some of the ways to compare the strengths of acids and bases are the concentration of hydrogen and hydroxide ions, pH and pOH , percent dissociation, K a and K b for the reaction describing its ionization in water. 1.2.1.4 Concentration of hydrogen and hydroxide ions How do the concentration of hydrogen and hydroxide ions affect acid and base strength? By definition a strong acid is one that completely dissociates in water to release proton. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions ( H 3O + ) than do weaker acids. Examples of the strong acids are: hydrochloric acid ( HCl ) , nitric acid ( HNO3 ) , perchloric acid ( HClO4 ) , and sulfuric acid ( H 2 SO4 ). Each of these acids ionize essentially 100% in solution. By contrast, however, a weak acid, being less willing to donate its proton, will only partially 20  Acid-Base Equilibria dissociate in solution. Base strength refers to the ability of a base to accept protons. A strong base accepts more protons readily than a weak base. A solution of a stronger base will contain a larger concentration of hydroxide ions than a solution of a weaker base if both solutions are of equal concentration. The net direction of an acid-base reaction depends on relative acid and base strengths: a reaction proceeds to the greater extent in the direction in which a stronger acid and stronger base form a weaker acid and weaker base. Activity 1.6 By referring to this text book and other chemistry books, list strong acids, strong bases, weak acids and weak bases. Then discuss what you have written with the rest of the class. What is the reason of your classification of acids and bases as strong and weak? 1.2.1.5 pH and pOH If the pH of a solution at 25°C is 2, is it acidic, neutral or basic? One way to determine the strength of acids is using the pH values. The smaller the pH value, the stronger the acid. The concentration of hydroxide ions in a solution can be expressed in terms of the pOH of the solution. Hence, the strength of bases can also be determined from their pOH values. The smaller the pOH value, the stronger the base. 1.2.1.6 Percent Ionization How is the percent ionization and acid strength related? Another measure of the strength of an acid is its percent ionization, which is defined asthe proportion of ionized molecules on a percentage basis. Mathematically: Ionized acid concentration at equilibrium Percent ionization = x 100 % Initial concentration of acid The strength of an acid depends on the percentage of the acid molecules that dissociate inwater solution. The stronger the acid, the greater the percent ionization.Strong acids and strong bases ionize nearly completely in water. But weak acids and weak bases dissociate partially in water, and their percent of ionization is small. 21  CHEMISTRY GRADE 12 1.2.1.7 Dissociation (Ionization) Constants Acid ionization Constant, K a What is the relationship between strength of acids with their acid-dissociation constant values? The acid Ionization constant or dissociation constant is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for the reaction of dissociation of acid into its conjugate base and hydrogen ion. Acid dissociation constant of weak acid HA like acetic acid, formic acid can be written as: HA(aq ) + H 2O(l )  H 3O + (aq ) + A− (aq ) the dissociation-constant expression can be written as:  H 3O +   A−  K= [ H 2O ][ HA] Since the concentration of water is nearly constant, we can write;  H 3O +   A−  K [ H 2O ] = [ HA] The product of the two constants, K and [ H 2O ] , is itself a constant. It is designated as Ka, which is the acid-dissociation constant or the acid-ionization constant. Hence for a weak acid, HA :  H 3O +   A−  Ka = [ HA] Acids are classified as either strong or weak, based on their ionization in water. Therefore, the numerical value of K a is a reflection of the strength of the acid. Acids with relatively higher K a values are stronger than acids with relatively lower K a values. The ionization- constants of some weak monoprotic acids are tabulated in table 1.2 22  Acid-Base Equilibria Table 1.2: Ionization constant of some weak monoprotic acids at 25°C. Name of the Acid Formula Ka Acetic acid CH 3COOH 1.8 × 10–5 Ascorbic acid C6 H 8O6 8.0 × 10–5 Benzoic Acid C6 H 5COOH 6.5 × 10–5 Formic acid HCOOH 1.7 × 10–4 Hydrocyanic acid HCN 4.9 × 10–10 Hydrofluoric acid HF 6.8 × 10–4 Hypobromous acid HOBr 2.5 × 10–9 Hypochlorous acid HOCl 3.0 × 10–8 Nitrous acid HNO2 4.5 × 10–4 Example 1.5 1. A 0.250 M aqueous solution of butyric acid is found to have a pH of 2.72. Determine K a for butyric acid Solution: CH 3 (CH 2 ) 2 COOH (aq ) + H 2O(l )  H 3O + (aq ) + CH 3 (CH 2 ) 2 COO − (aq ) Initial conc. 0.250 M ------- ------- Changes −x +x +x equilib. conc. 0.250 M − x x x x is a known quantity, It is the  H 3O  in solution, which we can determine + from the pH. log  H 3O +  = − pH = −2.72  H= + 3O   10 = −2.72 1.9 x 10−3 M Now we can solve the following expression for K w by substituting in the value x = 1.9 x 10 -3 M (1.9 x 10−3 )(1.9 x 10−3 ) =Ka = −3 1.5 x 10−5 0.250 − 1.9 x 10 2. Calculate the pH of a 0.50 M HF solution at 25°C. The ionization of HF is given by HF (aq ) + H 2O(l )  H 3O + (aq ) + F − (aq ) Solution: The species that can affect the pH of the solution are HF , and the conjugate base F − , Let x be the equilibrium concentration of H 3O + and F − ions in molarity (M). Thus, 23  CHEMISTRY GRADE 12 HF (aq ) + H 2O(l )  H 3O + (aq ) + F − (aq ) Initial conc. 0.50 ----- ----- Changes −x +x +x Equil. conc. 0.50 − x x x  H 3O +   F −  Ka =  [ HF ] Substituting the concentration of HF , H+ and F − , in terms of x, gives: ( x)( x) =Ka = 6.8 x 10−4 0.50 − x Rearranging this expression provides: x2 + (6.8 × 10–4 ) x – 3.4 × 10–4 = 0 This is a quadratic equation that can be solved, using the quadratic formula, or youcan use the approximation method for x. Because HF is a weak acid, and weak acids ionize only to a slight extent, x must be small compared to 0.50. Therefore, you can make this approximation: 0.50 − x ≈ 0.50 Now, the ionization constant expression becomes x2 x2 ≈ = 6.8 x 10−4 0.50 − x 0.50 Rearranging this equation gives: = = x 2 (0.50)(6.8 x 10−4 ) 3.4 x 10−4 =x 3.4 x 10−4 1.8 x 10−2 M = Thus, we have solved for x without using the quadratic equation. At equilibrium, we have [ HF ] = (0.50 − 0.018) M = 0.48 M =  F −  = H 3O +  0.018 M and the pH of the solution is pH = − log(0.018) = 1.74 How good is this approximation? Because K a values for weak acids are generally known to an accuracy of only ± 5%, it is reasonable to require x to be less than 5 % of 0.50, the number from which it is subtracted. In other words, the approximation is valid if the percent ionization is equal to or less than 5%. 0.018 x 100 % = 3.6 % Is the approximation valid? 0.50 24  Acid-Base Equilibria Example: 1.6 1. Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100M solution of acetic acid, CH 3COOH ? Solution: CH 3COOH (aq ) + H 2O(l )  H 3O + (aq ) + CH 3COO − (aq ), K a = 1.8 x 10−5 Initial conc. 0.100 M ----- ----- Changes -x +x +x Equilib conc. 0.100 M - x x x x is a known quantity, it is  H 3O +  in solution, which we can determine from the K a  H 3O +  CH 3COO −  Ka = [CH 3COOH ] Substituting the concentration of CH 3COOH , H 3O + and CH 3COO − , in terms of x, gives: ( x)( x) Ka = (0.100 − x) −5 x2 1.8 x 10 = 0.100 − x Since CH 3COOH is a weak acid, and weak acids ionize only to a slight extent, x must be small compared to 0.10. Therefore, you can make the approximation: 0.10 − x ≈ 0.10. Now, the equation becomes (1.8 x 10−5 )(0.100) = x 2 =x 1.8 x 10−6 1.342 x 10−3 = 1.342 x 10−3 = Percent ionization = x 100 % 1.342 % 0.100 How do you calculate the pH of weak acids? Generally, we can calculate the hydrogen-ion concentration or pH of an acid solution at equilibrium, given the initial concentration of the acid and its K a value. 25  CHEMISTRY GRADE 12 Exercise 1.7 Calculate the percent ionization of a 0.10M solution of acetic acid with a pH of 2.89. Calculate the pH of a 0.10 M solution of acetic acid. Ka = 1.8 x 10-5 For a 0.036 MHNO 2 solution. a. Write a chemical equation that shows the ionization of nitrous acid in water. b. Calculate the equilibrium concentration of hydrogen ions and nitrous acid at 25°C, using the approximation method. Then check whether the approximation is valid or not. c. If the approximation is invalid, use the quadratic formula to calculate the concentration of hydrogenions. d. Calculate the pH of the solution. Base dissociation constant, K b Equilibria involving weak bases are treated similarly to those for weak acids. Ammonia, for example, ionizes in water as follows: NH 3 (aq ) + H 2O(l )  NH 4 + (aq ) + OH − (aq ) The corresponding equilibrium constant is:  NH 4 +  OH −  Kc = [ NH 3 ][ H 2O ] Because the concentration of H 2O is nearly constant, you can rearrange this equation as you did for acid ionization.  NH 4 +  OH −  = c[ K b K= H 2 ] O [ NH 3 ] where K b is the base dissociation constant. In general, a weak base B with the base ionization B (aq ) + H 2O(l )  HB + (aq ) + OH − (aq ) has a base-ionization constant, K b (the equilibrium constant for the ionization of a weak base), equal to:  HB +  OH −  Kb = [ B] Note that K b values for strong bases are large, while K b values for weak bases are small. 26  Acid-Base Equilibria Table 1.3: shows the K b values of some common weak bases at 25°C Base Formula Kb Ammonia 1.8 × 10–5 NH 3 Aniline 4.0 × 10–10 C6 H 5 NH 2 Ethylamine 4.7 × 10–4 C2 H 5 NH 2 Hydrazine 1.7 × 10–6 C2 H 4 Hydroxylamine 1.1 × 10–6 NH 2OH Methylamine 4.4 × 10–4 CH 3 NH 2 Pyridine 1.7 × 10–9 C5 H 5 N In solving problems involving weak bases, you should follow the same guidelines as you followed for weak acids. The main difference is that we calculate OH −  first, instead of  H +  Example 1.7 What is the hydronium-ion concentration of a 0.20 M solution of ammonia in water? Kb = 1.8 × 10–5 Solution: NH 3 (aq ) + H 2O(l )  NH 4 + (aq ) + OH − (aq ) initial conc. 0.20 M ----- ----- changes −x +x +x equlib.conc. 0.20M-x x x  NH 4 +  OH −  The equilibrium equation for the reaction is given by: Kb = NH 3 Substituting the concentration of NH 3 , NH 4 + and OH − in terms of x, gives: ( x)( x) Kb = (0.20 − x) −5x2 1.8 x 10 = 0.20 − x Since NH 3 is a weak base, and weak bases ionize only to a slight extent, x must be small compared to 0.20. Therefore, you can make the approximation: 0.20 − x ≈ 0.20 x = 3.6 x 10−6 = 1.897 x 10−3 Now, the equation becomes (1.8 x 10−5 )(0.20) = x 2 27  CHEMISTRY GRADE 12 Exercise 1.8 1. For a 0.040 M ammonia solution: a. Write a chemical equation that shows the ionization of ammonia in water. b. Calculate the equilibrium concentration of ammonia, ammonium ions and hydroxide ions, using the approximation method. Check whether the approximation is valid or not. c. If the approximation is invalid, use the quadratic formula to calculate the concentration of ammonia, ammonium ions and hydroxide ions. d. Calculate the pOH and pH of the solution. 1.3 COMMON ION EFFECT AND BUFFER SOLUTION At the end of this subunit, you will be able to:  define the common-ion effect;  explain the importance of the common-ion effect;  define buffer solution;  give some common examples of buffer systems;  explain the action of buffer solutions and its importance in chemical processes;  calculate the pH of a given buffer solution; and  demonstrate the buffer action of CH 3COOH / CH 3COONa 1.3.1 The Common ion Effect Activity 1.7 In Grade 11 Chemistry, you learned Le Chatelier’s principle. Make a group and discuss the following and present your report to the class. Industrially, ammonia is produced by the Haber process. 1. Write a chemical equation for the production of ammonia in the process. 2. Assume that the reaction is at equilibrium. What is the effect of a. adding more ammonia to the equilibrium system? b. removing ammonia from the equilibrium system? c. adding more hydrogen gas to the equilibrium system? d. decreasing the concentration of both hydrogen and nitrogen gases from the equilibrium system? 28  Acid-Base Equilibria e. increasing temperature? f. decreasing pressure? g. adding finely divided iron as a catalyst? The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion that takes part in the equilibrium. The common-ion effect occurs when a given ion is added to an equilibrium mixture that already contains that ion. It is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion that takes part in the equilibrium. Consider a solution of acetic acid, CH 3COOH , in which you have the following acid- ionization equilibrium: CH 3COOH (aq ) + H 2O(l )  CH 3COO − (aq ) + H 3O + (aq ) Suppose you add HCl (aq) to this solution. What is the effect on the acid-ionization equilibrium? Because HCl (aq) is a strong acid, it provides H 3O + ion, which is present on the right side of the equation for acetic acid ionization. According to LeChâtelier’s principle, the equilibrium composition should shift to the left CH 3COOH (aq ) + H 2O(l ) ← CH 3COO − (aq ) + H 3O + (aq ) added The degree of ionization of acetic acid is decreased by the addition of a strong acid. This repression of the ionization of acetic acid by HCl (aq) is an example of the common-ion effect. Another example is, if sodium acetate and acetic acid are dissolved in the same solution, they both dissociate and ionize to produce CH 3COO − ions, we can represent the effect of acetate salts on the acetic acid equilibrium as: 29  CHEMISTRY GRADE 12 Example 1.8 Determine the  H 3O +  and CH 3COO −  in a solution that is 0.10 M in both CH 3COOH and HCl. Solution: 0.10 M HCl ionizes completely to form 0.10 M H 3O + and 0.10 M Cl − ions. The Cl − ion is a spectator ion, and it has no influence on the concentrations of CH 3COO − and H 3O + CH 3COOH (aq ) + H 2O(l )  CH 3COO − (aq ) + H 3O + (aq ) initial conc. 0.10 M ----- 0.10 M changes −x +x +x equlib.conc. 0.10 M − x x 0.10 M + x  H 3O +  CH 3COO −  (0.1 + x)( x) =Ka = [CH 3COOH ] (0.1 − x) (0.10 + x)( x) 1.8 x 10−5 = (0.10 − x) If x is very small, you can approximate (1.00 –x) and (1.00 +x) to 1.00 (0.10)( x) 1.8 x 10−5 = (0.10) =x = CH 3COO −  1.8 x 10−5 M  H 3O +  =+ 0.10 1.8 x 10−5 = 1.9 x 10−5 M 30  Acid-Base Equilibria Exercise 1.9 a. Calculate the pH of a solution containing 0.20 M CH 3COOH and 0.30 M CH 3COONa b. What would be the pH of a 0.20 M CH 3COOH solution if no salt were present? 1.3.2 Buffer Solutions How does a buffer solution resist a pH change? A buffer is commonly defined as a solution that resists changes in pH when a small amount of acid or base is added or when the solution is diluted with pure solvent. This property is extremely useful in maintaining the pH of a chemical system at an optimum value to appropriately influence the reaction kinetics or equilibrium processes. A buffer solution actually is a mixture of a weak acid and its conjugate base or a mixture of a weak base and its conjugate acid. The conjugate forms are commonly referred to as “salts”. Table1.4: Comparison of buffered and un buffered solutions. Initial pH of pH after addition pH after addition 1.0 L sample of 0.010 mol NaOH of 0.010 mol HCl Un buffered solution:1.28 × 4.8 12.0 2.0 10–5 M HCl Buffered solution:0.099M 4.8 4.8 4.7 CH 3COOH 0.097M CH 3COONa A buffer solution must contain a relatively large concentration of acid to react with any OH − ions that may be added to it. Similarly, it must contain a large concentration of base to react with any added H + ions. Furthermore, the acid and the base components of the buffer must not consume each other in a neutralization reaction. These requirements are satisfied by an acid-base conjugate pair (a weak acid and its conjugate base or a weak base and its conjugate acid). A simple buffer solution can be prepared by adding comparable amounts of acetic acid ( CH 3COOH ) and sodium acetate ( CH 3COONa ) to water. The equilibrium concentrations 31  CHEMISTRY GRADE 12 of both the acid and the conjugate base (from CH 3COONa ) are assumed to be the same as the starting concentrations. This is so because − 1. CH 3COOH is a weak acid and the extent of hydrolysis of the CH 3COO ion is very small and 2. the presence of CH 3COO − ions suppress the ionization of CH 3COOH , and the presence of CH 3COOH suppresses the hydrolysis of the CH 3COO − ions A solution containing these two substances has the ability to neutralize either added acid or added base. Sodium acetate, a strong electrolyte, dissociates completely in water: H O 2 → CH COO − ( aq ) + Na + ( aq ) CH 3COONa ( s )  3 If an acid is added, the H + ions will be consumed by the conjugate base in the buffer, CH 3COO − ,according to the equation CH 3COO − (aq ) + H + (aq ) → CH 3COOH (aq ) If a base is added to the buffer system, the OH − ions will be neutralized by the acid in the buffer: CH 3COOH (aq ) +OH − (aq ) → CH 3COO − (aq ) + H 2O(l ) Thus, a buffer solution resists changes in pH through its ability to combine with the H + and OH − ions. Buffers are very important to chemical and biological systems. The pH in the human body varies greatly from one fluid to another; for example, the pH of blood is about 7.4, whereas the gastric juice in our stomachs has a pH of about 1.5. These pH values, which are crucial for the proper functioning of enzymes and the balance of osmotic pressure, are maintained by buffers in most cases. The pH of a buffer solution can be estimated with the help of Henderson–Hasselbalch equation when the concentration of the acid and its conjugate base, or the base and the corresponding conjugate acid, are known. The Henderson-Hasselbach equation is derived from the definition of the acid dissociation constant as follows. 32  Acid-Base Equilibria Consider the hypothetical compound HA in water. The dissociation equation and K a ex- pression are HA + H 2O  H 3O + + A−  H 3O +   A−  Ka = [ HA] The key variable that determines [ H 3O + ] is the concentration ratio of acid species to base species, so, rearranging to isolate [ H 3O + ] gives  H 3O +  = K a x [ HA]  A−  Taking the negative logarithm of both sides gives  [ HA]  − log K a − log  −  − log  H 3O +  =  A      [ HA]  pH pK a − log  −  from which definitions gives =  A   Rearranging the equation gives     A−   pH pK a + log     =  [ HA]    Generalizing the previous equation for any conjugate acid-base pair gives the Henderson- Hasselbalch equation:  [Conjugate base]  = pH pK a + log   [ weak acid ]    Similarly, for a weak base dissociation:  [Conjugate acid ]  = pK b + log   [ weak base]  pOH   33  CHEMISTRY GRADE 12 Example 1.9 1. What is the pH of a buffer solution consisting of 0.035 M NH 3 and 0.050 M NH 4 + (Ka for NH4+ is 5.6 x 10-10)? The equation for the reaction is: NH 4 +  H + + NH 3 Assuming that the change in concentrations is negligible in order for the system to reach equilibrium, the Henderson-Hasselbalch equation will be:  [ NH ]  pH pK a + log  = 3    NH 4 +      0.035  = 9.23 + log  pH   0.050  pH = 9.095 2. A buffer is made by mixing 0.060 M NH 3 with 0.040 M NH4Cl. What is the pH of this buffer? K b = 1.8 x 10−5 Solution The buffer contains a base and its conjugate acid in equilibrium. The equation is NH 3 (aq ) + H 2O(l )  NH 4 + (aq ) + OH − (aq ) Assuming that the change in concentrations is negligible in order for the system to reach equilibrium, the Henderson-Hasselbalch equation will be:  [Conjugate acid ]  = pK b + log   [ weak base]  pOH    NH 4 +  = pK b + log  pOH   NH 3   0.04  = 4.745 + log  pOH   0.06  = 4.745 − 0.1761 pOH = 4.5689 pH= 14 − pOH = 14 − 4.5689 = 9.4311 34  Acid-Base Equilibria Exercise 1.10 1. Calculate the pH : a. of a buffer solution containing 0.1 M CH 3COOH and a 0.1 M solution of CH 3COONa. b. when 1.0 mL of 0.10 M HCl is added to 100 mL of the buffer in (a); c. when 1.0 mL of 0.10 M NaOH is added to 100 mL of the buffer in (a); d. of an unbuffered solution containing 1.8 × 10 –5 HCl ; e. change of an unbuffered solution in (d) after adding 2. i. 1.0 mL of 0.1 M NaOH to 100 mL of the solution, ii.1.0 mL of 0.10 M HCl to 100 mL of the solution. 1.4 Hydrolsis of Salts At the end of this subunit, you will be able to:  define hydrolysis;  explain why a salt of weak acid and strong base gives a basic solution;  explain why a salt of strong acid and weak base gives an acidic solution; and  explain why salts of weak acids and weak bases give acidic, basic or neutral solutions. What does salt hydrolysis mean? Hydrolysis is a common form of a chemical reaction where water is mostly used to break down the chemical bonds that exist between particular substances. Hydrolysis is derived from a Greek word hydro meaning water and lysis meaning break or to unbind. Usually in hydrolysis the water molecules get attached to two parts of a molecule. One molecule of a substance will get H + ion and the other molecule receives the OH − group. The term salt hydrolysis describes the “interaction of anion and cation of a salt, or both, with water. Depending on the strengths of the parent acids and bases, the cation of a salt can serves as an acid, base or neutral. 1.4.1 Hydrolysis of Salts of Strong Acids and Strong Bases The reaction between a strong acid (say, HCl ) and a strong base (say, NaOH ) can be represented by NaOH (aq ) + HCl (l ) → NaCl (aq ) + H 2O(l ) or in terms of the net ionic equation 35  CHEMISTRY GRADE 12 H + (aq ) + OH − (aq ) → H 2O(l ) The anions derived from strong acids are weak conjugate bases and do not undergo hydrolysis. The cations derived from strong bases are weak conjugate acids and also do not hydrolysis. The strong bases are the ionic hydroxides of Group IA and IIA metals. The cations of these metals also do not hydrolyze. It involves only ionization of water and no hydrolysis..For the above reaction, chloride ions, Cl− , and sodium ions, Na + , do not hydrolyze, the solution of the salt will be neutral ( pH =7). Can you give more examples? 1.4.2 Hydrolysis of Salts of weak acids and strong bases Consider the neutralization reaction between acetic acid (a weak acid) and sodium hydroxide (a strong base): CH 3COOH (aq ) + NaOH (aq ) → CH 3COONa(aq)+H 2O(aq ) This equation can be simplified to CH 3COOH (aq ) + OH − (aq ) → CH 3COO − (aq)+H 2O(aq ) The acetate ion undergoes hydrolysis as follows: CH 3COO − (aq)+H 2O(aq ) → CH 3COOH (aq ) + OH − (aq ) Therefore, at the equivalence point, when we only have sodium acetate present, the pH will be greater than7 as a result of the excess OH − ions formed. Note that this situation is analogous to the hydrolysis of sodium acetate ( CH 3COONa )Therefore, solutions of these salts are basic because the anion of the weak acid is a moderately strong base and can be hydrolyzed. Activity 1.8 Consider NaNO2 and discuss the following: a. What are the ‘parents’ (acid and base) of this salt? b. Which ions of the salt can be hydrolyzed? c. What will be the nature of NaNO2 solution? Will it be acidic, basic or neutral? 36  Acid-Base Equilibria 1.4.3 Hydrolysis of Salts of strong acids and weak bases When we neutralize a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. This conjugate acid is a strong acid. For example, ammo- nium chloride, NH 4 Cl , is a salt formed by the reaction of the weak base ammonia with the strong acid HCl : NH 3 (aq)+HCl (aq ) → NH 4Cl (aq ) A solution of this salt contains ammonium ions and chloride ions. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acid of ammonia, reacts with water and increases the hydronium ion concentration: NH 4 + (aq ) +H 2 O(l) → H 3O + (aq ) + NH 3 (aq) 1.4.4 Hydrolysis of Salts of weak acids and week bases Solutions of a salt formed by the reaction of a weak acid and a weak base involve both cationic and anionic hydrolysis. To predict whether the solution is acidic or basic you need to compare the K a of the weak acid and the K b of the weak base. If the K a is larger than the K b , the solution is acidic, and if the K a is less than the K b , the solution is basic. If they are equal the solution is neutral. Consider solutions of ammonium formate, NH 4CHO2. These solutions are slightly acidic, because the K b for NH 4 + ( 5.6x 10-10) is somewhat larger than the K b for formate ion, CHO2 − ( 5.9 x 10-11). 37  CHEMISTRY GRADE 12 Activity 1.9 In the following table you are given K a and K b values of some cations and anions, respectively. Anion Kb Cation Ka F− 1.4 × 10–11 NH 4 + 5.6 × 10–10 CNS − 2.0 × 10–5 5.6 × 10–10 CH 3COO − Using the above table, determine whether the solutions of NH 4 F , NH 4 CNS and CH 3COONH 4 are acidic, basic or neutral. Discuss your results with your classmates. 1.5 Acid–Base Indicators and Titrations At the end of this subunit, you will be able to:  define acid-base indicators;  write some examples of acid-base indicators;  suggest a suitable indicator for a given acid-base titration;  explain the equivalents of acids and bases;  calculate the normality of a given acidic or basic solution;  define acid-base titration;  distinguish between end point and equivalent point; and  discuss the different types of titration curves 1.5.1 Acid–Base Indicators How do acid-base indicators change color? Acid-base indicators are weak organic acids (denoted here as HIn ) or weak organic bases ( In − ) that indicate whether a solution is acidic, basic or neutral. The color of the indicator depends on the pH of the solution to which it is added. When just a small amount of indicator is added to a solution, the indicator does not affect the pH of the solution. Instead, the ionization equilibrium of the indicator is affected by the prevailing  H 3O +  in the solution 38  Acid-Base Equilibria HIn + H 2O  H 3O + + In − Acid colour Ba se c olour If the indicator is in a sufficiently acidic medium [increasing  H 3O +  ], the equilibrium, according to Le Châtelier’s principle, shifts to the left and the predominant color of the indicator is that of the non-ionized form ( HIn ). On the other hand, in a basic medium [Decreasing  H 3O +  ] the equilibrium shifts to the right and the color of the solution will be due mainly to that of the conjugate base ( In − ). An acid–base indicator is usually prepared as a solution (in water, ethanol, or some other solvent). In acid–base titrations, a few drops of the indicator solution are added to the solution being titrated. In other applications, porous paper is impregnated with an indicator solution and dried. When this paper is moistened with the solution being tested, it acquires a color determined by the pH of the solution. This paper is usually called pH test paper. Table 1.5: Some common indicators. Indicator change Acid Color Base Color pH range of Color Methyl violet Yellow Violet 0.0 – 1.6 Methyl orange Red Yellow 3.2 – 4.4 Bromcresol green Yellow Blue 3.8 – 5.4 Methyl red Red Yellow 4.8 – 6.0 Litmus Red Blue 5.0 – 8.0 Phenolphthalein Colorless Pink 8.2 – 10.0 Example 1.10 What is the pH of a buffer prepared with 0.40 M CH 3COOH and 0.20 M CH 3COO − if the K a of CH 3COOH is 1.8 x 10 -5 ? Which type of indicator is used to check the acidity or basicity of this solution? =  H 3O +  K= [CH 3COOH ] 1.8 x 10−5 M x 0.40 a x CH 3COO −  0.20 Solution:  H 3O +  = 3.6 x 10−5 M − log(3.6 x 10−5 ) = pH = 4.44 The pH value indicates that the solution is acidic 39  CHEMISTRY GRADE 12 1.5.2 Equivalents of Acids and Bases The reactive capacity of a chemical species, the ions or electrons, depends on what is being transferred in a chemical reaction. In acid-base reactions, an equivalent is the amount of a substance that will react with one mole of hydrogen ions. For acids, an equivalent is the number of hydrogen ions a molecule transfer. For example, for Sulfuric acid ( H 2 SO4 ) equivalents is equal to 2. For bases, it is the number of hydroxide ions ( OH − )ions provided for a reaction, for example for Barium hydroxide ( Ba (OH ) 2 )equivalents is equal to 2. The normality of a solution refers to the number of equivalents of solute per Liter of solu- tion. Number of equivalents of solutes Normality = Liters of solution The definition of chemical equivalent depends on the substance or type of chemical re- action under consideration. Because the concept of equivalents is based on the reacting power of an element or compound, it follows that a specific number of equivalents of one substance will react with the same number of equivalents of another substance. When the concept of equivalents is taken into consideration, it is less likely that chemicals will be wasted as excess amounts. Keeping in mind that normality is a measure of the reacting power of a solution, we use the following equation to determine normality: Thus, according to the definition of normality, the number of equivalents is the normality multiplied by the volume of solution, in litters. If we add enough acid to neutralize a given volume of base, the following equation holds: N1V1 = N 2V2 Where N1and V1refer to the normality and volume of the acid solution, respectively, and N2 and V2 refer to the normality and volume of the base solution, respectively. 40  Acid-Base Equilibria Example 1.11 What volume of 2.0 N NaOH is required to neutralize 25.0 mL of 2.70 N H 2 SO4 ? Solution: N1V1 = N 2V2 N1V1 V2 = N2 (2.7 N H 2 SO4 )(25.0 mL H 2 SO4 ) V2 NaOH = 2.0 N NaOH = 33.8 mL 1.5.3 Acid–Base Titrations An acid–base titration is a procedure for determining the amount of acid (or base) in a solution by determining the volume of base (or acid) of known concentration that will completely react with it. In a titration, one of the solutions to be neutralized say, the acid is placed in a flask or beaker, together with a few drops of an acid base indicator (Figure 1.2). The other solution (the base) used in a titration is added from a burette and is called the titrant. The titrant is added to the acid (Titrand or analyte), first rapidly and then drop by drop, up to the equivalence point. The equivalence point of the titration is the point at which the amount of titrant added is just enough to completely neutralize the analyte solution. At the equivalence point in an acid-base titration, moles of base are equal to moles of acid and the solution contains only salt and water. The equivalence point is located by noting the color change of the acid base indicator. The point in a titration at which the indicator changes color is called the end point of the indicator. The end point must match the equivalence point of the neutralization. That is, if the indicators end point is near the equivalence point of the neutralization, the color change marked by that end point will signal the attainment of the equivalence point. This match can be achieved by use of an indicator whose color change occurs over a pH range that includes the pH of the equivalence point. Knowing the volume of titrant added allows the determination of the concentration of the unknown analyte using the following relation volume of base x concentration of base = volume of acid x unknown concentration of acid volume of base x concentration of base unknown concentration of acid = volume of acid 41  CHEMISTRY GRADE 12 Figure 1.2: The Techniques of Titration. An acid–base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). Such curves are used to gain insight into the titration process. Experiment 1.1 Acid-base Titration Objective: To find the normality of a given hydrochloric acid solution by titrating against 0.1 N standard sodium hydroxide solution. Apparatus: 10 mL pipette, burette, 150 mL Erlenmeyer flask, beaker, funnel, bu- rette clamp and metal stand. Procedure: 1. Clean the burette with distilled water and rinse it with the 0.1 N sodium hydrox- ide solution; and fix the burette on the burette clamp in vertical position (Figure 1.3). 2. Using a funnel, introduce 0.1 N sodium hydroxide solutions into the burette. Al- low some of the solution to flow out and make sure that there are no air bubbles in the solution (why?). 3. Record level of the solution, corresponding to the bottom of the meniscus, to the nearest 0.1 mL. 42  Acid-Base Equilibria 4. Measure exactly 10 mL of hydrochloric acid solution (given) with the help of a10 mL pipette and add it into a clean 150 mL Erlenmeyer flask and add two or three drops of phenolphthalein indicator. 5. Caution: When you suck hydrochloric acid or any reagent solution, into a pi- pette, have the maximum caution not to suck it into your mouth. 6. Titration: First hold the neck of the Erlenmeyer flask with one hand and the stopcock with the other. As you add the sodium hydroxide solution from the burette, swirl the content of the flask gently and continuously. 7. Add sodium hydroxide solution until the first faint pink color comes which dis- appears on swirling. 8. Add more sodium hydroxide drop wise until the pink color persists for a few seconds. 9. Find the difference between the initial level and the end point level of the bu- rette. Observations and analysis: 1. Color change at the end point is from __________ to _________. 2. What is the volume of sodium hydroxide added at the end point? 3. What is the normality of hydrochloric acid at the end point? 4. What is the similarity and difference between equivalence point and end point level after reaching the end point? Figure 1.3: Titration Setup. 43  CHEMISTRY GRADE 12 Unit Summary ~ By the Arrhenius definition, an Arrhenius acid produces H + and an Arrhenius base produces OH − in aqueous solutions. And an acid-base reaction (neutral- ization)is the reaction of H + and OH − to form H 2O.The Arrhenius definition of acids and bases has many limitations but still we cannot ignore it. ~ The Brønsted-Lowry acid-base definition does not require that bases contain OH − in their formula or that acid-base reactions occur in aqueous solution. An acid is a species that donates a proton and a base is one that accepts it, so an acid-base reaction is a proton-transfer process. When an acid donates a pro- ton, it becomes the conjugate base; when a base accepts a proton, it becomes the conjugate acid. In an acid-base reaction, acids and bases form their conju- gates. A stronger acid has a weaker conjugate base, and vice versa. ~ Brønsted-Lowry bases include NH 3 and amines and the anions of weak acids. All produce basic solutions by accepting H + from water, which yields OH − , thus making  H 3O +  < OH −  ~ The Lewis acid-base definition focuses on the donation or acceptance of an electron pair to form a new covalent bond in an addict, the product of an acid-base reaction. Lewis bases donate the electron pair, and Lewis’s acids ac- cept it. Thus, many species that do not contain H + act as Lewis’s acids; exam- ples are molecules with electron-deficient atoms and those with polar double bonds. Metal ions act as Lewis’s acids when they dissolve in water, which acts as a Lewis base, to form the adduct, a hydrated cation ~ Pure water auto ionizes to a small extent in a process whose equilibrium con- stant is the ion product constant for water, K w (1.0 x 10-14 at 25 0C).  H 3O +  and OH −  are inversely related: in acidic solution,  H 3O +  is greater than OH −  ; the reverse is true in basic solution; and the two are equal in neu- tral solution. To express small values of  H 3O +  , we use the pH scale ( pH = − log  H 3O +  ). Similarly, pOH = − log OH −  and. pK = − log K ~ A high pH represents a low  H 3O + . In acidic solutions, pH < 7; in basic solutions, pH > 7; and in neutral solutions pH = 7. The sum of pH and pOH equals pK w (14.00 at 25 0C). ~ Acid strength depends on  H 3O +  relative to [ HA] in aqueous solution. Strong acids dissociate completely and weak acids slightly. ~ The extent of dissociation is expressed by the acid-dissociation constant, K a. Most weak acids have K a values ranging from about 10-2 to 10-10. ~ The pH of a buffered solution changes much less than the pH of an unbuff- ered solution when  H 3O +  or OH − is added. 44  Acid-Base Equilibria ~ An acid-base reaction proceeds to the greater extentin the direction in which a stronger acid and base form a weaker base and acid. ~ Two common types of weak-acid equilibrium problems involve finding K a from a given concentration and finding a concentration from a given K a. ~ The extent to which a weak base accepts a proton from water to form OH − is expressed by a base-dissociation constant, K b. ~ By multiplying the expressions for K a of HA and K b of A− , we obtain K w. This relationship allows us to calculate either K a of BH + or K b of A−. ~ A buffer consists of a weak acid and its conjugate base (or a weak base and its conjugate acid).To be effective, the amounts of the components must be much greater than the amount of  H 3O +  or OH − added. ~ The buffer-component concentration ratio determines the pH ;the ratio and the pH are related by the Henderson-Hasselbalch equation. ~ When  H 3O +  or OH − is added to abuffer, one component reacts to form the other; thus  H 3O +  (and pH ) changes only slightly. ~ A concentrated (higher capacity) buffer undergoes smaller changes in pH than a dilute buffer. When the buffer pH equals the pK a of the acid component, the buffer has its highest capacity. Check List Key terms of the unit  Arrhenius acid-base concept  Bronsted-Lowry concept of acid and bases  Conjugate acid and Conjugate base  Lewis’s concept of acids and bases  Autoionization  Amphiprotic species  pH scale  percent ionization  Hydrolysis of salts  Buffer solution  Common ion effect  Equivalents of acids and bases  Acid-base titration 45  CHEMISTRY GRADE 12 REVIEW EXERCISE FOR UNIT 1 Part I Multiple Choice 1. Which of the following is a Brønsted-Lowry base but an Arrhenius base? a. NH3 b. NaOH c. Ca (OH)2 d. KOH 2. Use the following acid ionization constants to identify the correct decreasing order of base strengths. HF Ka = 7.2 x 10–4HNO2 Ka = 4.5 x 10–4 HCN Ka = 6.2 x 10–10 a. CN–>NO2–> F– b. F–>NO2–>CN– c. NO2–> F–> CN– d. NO2–> CN–>F– 3. In the Brønsted–Lowry definition of acids and bases, an acid __________ a. is a proton donor. b. breaks stable hydrogen bonds. c. is a proton acceptor. d. corrodes metals. 4. Which of the following is the conjugate acid of the hydrogen phosphate ion, HPO42–? a. H3PO4 b. PO43– c. H2PO4– d. H3O+ 5. Which one of the following is not a conjugate acid–base pair? a. NH3 and NH4+ b. HS– and H2S c. NH3 and NH2– d. H3O+ and OH– 6. Which one of the following is not a conjugate acid–base pair? a. NH3 and NH2– b. H2PO4– and HPO42– c. HNO3 and HNO2 d. H2O and OH– 7. Which one of the following is not a strong acid? a. nitric acid, HNO3 b. carbonic acid, H2CO3 c. sulfuric acid, H2SO4 d. perchloric acid, HClO4 8. Each of the following pairs contains one strong acid and one weak acid EXCEPT: a. H2SO4 and H2CO3 b. HNO3 and HNO2 c. HBr and H3PO2 d. HSO4─ and HCN 9. When [H+] = 4.0 × 10–9 M in water at 25°C, then __________ a. pH = 9.40. b. pH = 8.40. c. pH = 7.00. d. pH = –9.40 10. A solution with an [OH–] concentration of 1.20 × 10–7 M has a pOH and pH of __________ a. 6.92 and 7.08 b. 7.08 and 6.92 c. 1.00 and 13.00 d. 5.94 and 8.06 46  Acid-Base Equilibria 11. The acidic ingredient in vinegar is acetic acid. The pH of vinegar is around 2.4, and the molar concentration of acetic acid in vinegar is around 0.85 M. Based on this information, determine the value of the acid ionization constant, Ka, for acetic acid. a. 2.5 × 10–5 b. 1.9 × 10–5 c. 5.0 × 10–5 d. 7.4 × 10–3 12. A cup of coffee has a hydroxide ion concentration of 1.0 × 10–10 M. What is the pH of this coffee? a. 6 b. 7 c. 4 d. 2 13. What is the concentration of [OH–] in a 0.20 M solution of ammonia? The Kb value for ammonia is 1.8 × 10–5. a. 3.6 × 10–6 M b. 1.8 × 10–5 M c. 1.9 × 10–3 M d. 4.2 × 10–4 M 14. Whcih of the following molecules is not a lewis base? a. HCl b NH3 c. BH3 d. NaOH 15. Which one of the following salts forms aqueous solutions with pH = 7? a. Na2S b. NaNO2 c. NaBr d. Na2CO3 Part II: Short Answer Questions 1. Define acids and bases based on the concept of a. Arrhenius b. Brønsted-Lowry c. Lewis 2. In each of the following equations, identify the reactant that is a Brønsted–Lowry acid and the reactant that is a Brønsted–Lowry base: HBr (aq ) + H 2O(l ) → H 3O + (aq ) + Br − (aq ) a. − − b. CN (aq) + H 2O(l )  HCN(aq) + OH (aq) 3. Identify the Bronsted -Lowery acid - base pairs in each of the following equations. − + a. H 3 PO (aq) + H 2O(l )  H 2 PO4 (aq ) + H 3O (aq ) b. CO32− (aq ) + H 2O(l )  HCO3− (aq ) + OH − (aq ) c. H 3 PO (aq) + NH 3 (aq)  H 2 PO4 − (aq) + NH 4 + (aq) 4 4. According to the Lewis theory, each of the following is an acid base reaction. Which species is the acid and which is the base? a. BF3 + F − → BF4 − b. OH − (aq ) + CO2 (aq ) → HCO3− (aq ) 47  CHEMISTRY GRADE 12 5. Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base: a. KOH b. (CH3)2CHCOOH c.H2SeO4 d. (CH3)2CHNH2 6. Which of the following are amphiprotic? a. OH─ b. NH3 c. H2O d. H2S e. NO3– f. HCO3– g.CH3COO– h. HNO3 7. A research chemist adds a measured amount of HCl gas to pure water at 25 0C and obtains a solution with [H3O+] = 5 3.0 x 10-4M. Calculate [OH─]. Is the solution neutral, acidic, or basic? 8. Calculate the percent ionization of a 0.125 M solution of nitrous acid (a weak acid), with a pH of 2.09 9. A vinegar solution has a [OH-] = 5.0 x 10-12 M at 25 °C. What is the [H3O+] of the vinegar solution? Is the solution acidic, basic, or neutral? 10. The Ka for acetic acid, HC2H3O2, is 1.8 x 10-5. What is the pH of a buffer prepared with 1.0 M HC2H3O2 and 1.0 M C2H3O2- ? HC2 H 3 O2 (aq) + H 2O(aq )  H 3O + (aq) + C2 H 3O 2 − (aq) 48  Electro chemistry UNIT ELECTRO CHEMISTRY 2 Unit Outcomes After completing this unit, students will be able to:  describe the fundamental concepts related to oxidation- reduction reaction;  explain the application of redox reactions in production of new substances and energy;  demonstr