Acids and Bases

Introductory Chemistry, 3rd EditionNivaldo Tro : Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Introductory Chemistry, 3rd EditionNivaldo Tro Chapter 14 Acids and Bases 2009, Prentice Hall

Types of Electrolytes : Tro's Introductory Chemistry, Chapter 14 2 Types of Electrolytes Salts are water-soluble ionic compounds. All strong electrolytes. Acids form H+1 ions in water solution. Bases combine with H+1 ions in water solution. Increases the OH-1 concentration. May either directly release OH-1 or pull H+1 off H2O.

Properties of Acids : Tro's Introductory Chemistry, Chapter 14 3 Properties of Acids Sour taste. React with “active” metals. I.e., Al, Zn, Fe, but not Cu, Ag or Au. 2 Al + 6 HCl ® 2 AlCl3 + 3 H2 Corrosive. React with carbonates, producing CO2. Marble, baking soda, chalk, limestone. CaCO3 + 2 HCl ® CaCl2 + CO2 + H2O Change color of vegetable dyes. Blue litmus turns red. React with bases to form ionic salts.

Common Acids : Tro's Introductory Chemistry, Chapter 14 4 Common Acids

Structures of Acids : Tro's Introductory Chemistry, Chapter 14 5 Structures of Acids Binary acids have acid hydrogens attached to a nonmetal atom. HCl, HF

Structure of Acids : Tro's Introductory Chemistry, Chapter 14 6 Structure of Acids Oxyacids have acid hydrogens attached to an oxygen atom. H2SO4, HNO3

Structure of Acids : Tro's Introductory Chemistry, Chapter 14 7 Structure of Acids Carboxylic acids have COOH group. HC2H3O2, H3C6H5O3 Only the first H in the formula is acidic. The H is on the COOH.

Properties of Bases : Tro's Introductory Chemistry, Chapter 14 8 Properties of Bases Also known as alkalis. Taste bitter. Alkaloids = Plant product that is alkaline. Often poisonous. Solutions feel slippery. Change color of vegetable dyes. Different color than acid. Red litmus turns blue. React with acids to form ionic salts. Neutralization.

Common Bases : Tro's Introductory Chemistry, Chapter 14 9 Common Bases

Structure of Bases : Tro's Introductory Chemistry, Chapter 14 10 Structure of Bases Most ionic bases contain OH ions. NaOH, Ca(OH)2 Some contain CO32- ions. CaCO3 NaHCO3 Molecular bases contain structures that react with H+. Mostly amine groups.

Arrhenius Theory : Tro's Introductory Chemistry, Chapter 14 11 Arrhenius Theory Bases dissociate in water to produce OH- ions and cations. Ionic substances dissociate in water. NaOH(aq) → Na+(aq) + OH–(aq) Acids ionize in water to produce H+ ions and anions. Because molecular acids are not made of ions, they cannot dissociate. They must be pulled apart, or ionized, by the water. HCl(aq) → H+(aq) + Cl–(aq) In formula, ionizable H is written in front. HC2H3O2(aq) → H+(aq) + C2H3O2–(aq)

Arrow Conventions : Tro's Introductory Chemistry, Chapter 14 12 Arrow Conventions Chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions. A single arrow indicates that all the reactant molecules are converted to product molecules at the end. A double arrow indicates that the reaction stops when only some of the reactant molecules have been converted into products.  in these notes.

Arrhenius Theory, Continued : Tro's Introductory Chemistry, Chapter 14 13 Arrhenius Theory, Continued

Brønsted–Lowry Theory : Tro's Introductory Chemistry, Chapter 14 14 Brønsted–Lowry Theory A Brønsted-Lowry acid–base reaction is any reaction in which an H+ is transferred. Does not have to take place in aqueous solution. Broader definition than Arrhenius. Acid is H+ donor; base is H+ acceptor. Since H+ is a proton, acid is a proton donor and base is a proton acceptor. Base structure must contain an atom with an unshared pair of electrons to bond to H+. In the reaction, the acid molecule gives an H+ to the base molecule. H–A + :B  :A– + H–B+

Comparing Arrhenius Theory and Brønsted–Lowry Theory : Tro's Introductory Chemistry, Chapter 14 15 Comparing Arrhenius Theory and Brønsted–Lowry Theory Arrhenius theory HCl(aq)  H+(aq) + Cl−(aq) HF(aq)  H+(aq) + F−(aq) NaOH(aq)  Na+(aq) + OH−(aq) NH4OH(aq)  NH4+(aq) + OH−(aq) Brønsted–Lowry theory HCl(aq) + H2O(l)  Cl−(aq) + H3O+(aq) HF(aq) + H2O(l)  F−(aq) + H3O+(aq) NaOH(aq) + H2O(l)  Na+(aq) + OH−(aq) + H2O(l) NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)

Amphoteric Substances : Tro's Introductory Chemistry, Chapter 14 16 Amphoteric Substances Amphoteric substances can act as either an acid or a base. They have both transferable H and an atom with a lone pair. HCl(aq) is acidic because HCl transfers an H+ to H2O, forming H3O+ ions. Water acts as base, accepting H+. HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) NH3(aq) is basic because NH3 accepts an H+ from H2O, forming OH–(aq). Water acts as acid, donating H+. NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)

Conjugate Pairs : Tro's Introductory Chemistry, Chapter 14 17 Conjugate Pairs In a Brønsted-Lowry acid-base reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process. Each reactant and the product it becomes is called a conjugate pair. The original base becomes the conjugate acid; the original acid becomes the conjugate base.

Example—Identify the Brønsted–Lowry Acids and Bases and Their Conjugates in the Reaction. : Tro's Introductory Chemistry, Chapter 14 18 Example—Identify the Brønsted–Lowry Acids and Bases and Their Conjugates in the Reaction. H2SO4 + H2O  HSO4– + H3O+ Acid Base Conjugate Conjugate base acid H2SO4 + H2O  HSO4– + H3O+ When the H2SO4 becomes HSO4, it loses an H+, so H2SO4 must be the acid and HSO4 its conjugate base. When the H2O becomes H3O+, it accepts an H+, so H2O must be the base and H3O+ its conjugate acid.

Example—Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction, Continued. : Tro's Introductory Chemistry, Chapter 14 19 Example—Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction, Continued. HCO3– + H2O  H2CO3 + HO– Base Acid Conjugate Conjugate acid base HCO3– + H2O  H2CO3 + HO– When the HCO3 becomes H2CO3, it accepts an H+, so HCO3 must be the base and H2CO3 its conjugate acid. When the H2O becomes OH, it donates an H+, so H2O must be the acid and OH its conjugate base.

Practice—Write the Formula for the Conjugate Acid of the Following: : Tro's Introductory Chemistry, Chapter 14 20 Practice—Write the Formula for the Conjugate Acid of the Following: H2O NH3 CO32− H2PO41− H3O+ NH4+ HCO3− H3PO4

Practice—Write the Formula for the Conjugate Base of the Following: : Tro's Introductory Chemistry, Chapter 14 21 Practice—Write the Formula for the Conjugate Base of the Following: H2O NH3 CO32− H2PO41− HO− NH2− Since CO32− does not have an H, it cannot be an acid. HPO42−

Practice—Write the Equations for the Following Reacting with Water and Acting as a Monoprotic Acid. Label the Conjugate Acid and Base. : Tro's Introductory Chemistry, Chapter 14 22 Practice—Write the Equations for the Following Reacting with Water and Acting as a Monoprotic Acid. Label the Conjugate Acid and Base. HSO4-1 HSO4-1 + H2O ® SO4-2 + H3O+1 Acid Base Conjugate Conjugate base acid CO32− CO32− + H2O ® HCO3− + OH− Base Acid Conjugate Conjugate acid base

Practice—Write the Equations for the Following Reacting with Water and Acting as a Monoprotic Acid. Label the Conjugate Acid and Base, Continued. : Tro's Introductory Chemistry, Chapter 14 23 Practice—Write the Equations for the Following Reacting with Water and Acting as a Monoprotic Acid. Label the Conjugate Acid and Base, Continued. HBr HBr + H2O ® Br-1 + H3O+1 Acid Base Conjugate Conjugate base acid I− I− + H2O ® HI + OH− Base Acid Conjugate Conjugate acid base

Neutralization Reactions : Tro's Introductory Chemistry, Chapter 14 24 Neutralization Reactions H+ + OH- H2O Acid + base salt + water Double-displacement reactions. Salt = cation from base + anion from acid. Cation and anion charges stay constant. H2SO4 + Ca(OH)2 → CaSO4 + 2 H2O Some neutralization reactions are gas evolving, where H2CO3 decomposes into CO2 and H2O. H2SO4 + 2 NaHCO3 → Na2SO4 + 2 H2O + 2 CO2

Example 14.2—Write the Equation for the Reaction of Aqueous Nitric Acid with Aqueous Calcium Hydroxide. : Tro's Introductory Chemistry, Chapter 14 25 Example 14.2—Write the Equation for the Reaction of Aqueous Nitric Acid with Aqueous Calcium Hydroxide. 1. Write the formulas of the reactants. HNO3(aq) + Ca(OH)2(aq)  2. Determine the ions present when each reactant dissociates. (H+ + NO3−) + (Ca2+ + OH−)  3. Exchange the ions. H+1 combines with OH-1 to make H2O(l). (H+ + NO3−) + (Ca2+ + OH−)  (Ca2+ + NO3−) + H2O(l)

Example 14.2—Write the Equation for the Reaction of Aqueous Nitric Acid with Aqueous Calcium Hydroxide, Continued. : Tro's Introductory Chemistry, Chapter 14 26 4. Write the formulas of the products. Cross charges and reduce. HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2 + H2O(l) 5. Balance the equation. May be quickly balanced by matching the numbers of H and OH to make H2O. Coefficient of the salt is always 1. 2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2 + 2 H2O(l) Example 14.2—Write the Equation for the Reaction of Aqueous Nitric Acid with Aqueous Calcium Hydroxide, Continued.

Example 14.2—Write the Equation for the Reaction of Aqueous Nitric Acid with Aqueous Calcium Hydroxide, Continued. : Tro's Introductory Chemistry, Chapter 14 27 6. Determine the solubility of the salt. Ca(NO3)2 is soluble (all NO3- are soluble). 7. Write an (s) after the insoluble products and an (aq) after the soluble products. 2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l) Example 14.2—Write the Equation for the Reaction of Aqueous Nitric Acid with Aqueous Calcium Hydroxide, Continued.

Example—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Nitric Acid, a Gas Evolves. : Tro's Introductory Chemistry, Chapter 14 28 Example—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Nitric Acid, a Gas Evolves. 1. Write the formulas of the reactants. Na2CO3(aq) + HNO3(aq)  2. Determine the ions present when each reactant dissociates. (Na+ + CO32−) + (H1 + NO3−)  3. Exchange the ions. (Na+ + CO32−) + (H+ + NO3−)  (Na+ + NO3−) + (H+ + CO32−)

Example—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Nitric Acid, a Gas Evolves, Continued. : Tro's Introductory Chemistry, Chapter 14 29 4. Write the formulas of the products. Cross charges and reduce. Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3 5. Check to see of product decomposes – Yes H2CO3 decomposes into CO2(g) + H2O(l) Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l) Example—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Nitric Acid, a Gas Evolves, Continued.

Example—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Nitric Acid, a Gas Evolves, Continued. : Tro's Introductory Chemistry, Chapter 14 30 6. Balance the equation. Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3 + CO2(g) + H2O(l) 7. Determine the solubility of other product. NaNO3 is soluble (all Na+ are soluble). 8. Write an (s) after the insoluble products and an (aq) after the soluble products. Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l) Example—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Nitric Acid, a Gas Evolves, Continued.

Practice–Complete Each Reaction. : Tro's Introductory Chemistry, Chapter 14 31 Practice–Complete Each Reaction. Ca(OH)2(s) + H2SO3(aq)  HClO3(aq) + Pb(OH)4(s)  CaCO3(s) + HNO3(aq)  Mg(HCO3)2(aq) + HC2H3O2(aq) 

Practice–Complete Each Reaction, Continued. : Tro's Introductory Chemistry, Chapter 14 32 Practice–Complete Each Reaction, Continued. Ca(OH)2(s) + H2SO3(aq)  CaSO3(s) + 2 H2O(l) 4 HClO3(aq) + Pb(OH)4(s)  Pb(ClO3)4(s) + 4 H2O(l) CaCO3(s) + 2 HNO3(aq)  Ca(NO3)2(aq) + CO2(g) + 2 H2O(l) Mg(HCO3)2(aq) + 2 HC2H3O2(aq)  Mg(C2H3O2)2(aq) + 2 CO2(g) + 2 H2O(l)

Acid Reactions:Acids React with Metals : Tro's Introductory Chemistry, Chapter 14 33 Acid Reactions:Acids React with Metals Acids react with many metals. But not all!! When acids react with metals, they produce a salt and hydrogen gas. 3 H2SO4(aq) + 2 Al(s) → Al2(SO4)3(aq) + 3 H2(g)

Acid Reactions:Acids React with Metal Oxides : Tro's Introductory Chemistry, Chapter 14 34 Acid Reactions:Acids React with Metal Oxides When acids react with metal oxides, they produce a salt and water. 3 H2SO4 + Al2O3 → Al2(SO4)3 + 3 H2O

Example 14.3a—Write an Equation for the Reaction of Hydroiodic Acid with Potassium Metal. : Tro's Introductory Chemistry, Chapter 14 35 Example 14.3a—Write an Equation for the Reaction of Hydroiodic Acid with Potassium Metal. Write formulas of reactants. HI(aq) + K(s)  Identify the type of reaction and predict the pattern. acid + metal  salt + H2(g) Determine the charge on the cation. K  K+ Determine the formula of the salt. K+ + I−  KI

Example 14.3a—Write an Equation for the Reaction of Hydroiodic Acid with Potassium Metal, Continued. : Tro's Introductory Chemistry, Chapter 14 36 Example 14.3a—Write an Equation for the Reaction of Hydroiodic Acid with Potassium Metal, Continued. 5. Write the skeletal equation. HI(aq) + K(s)  KI + H2(g) 6. Check the solubility of the salt. KI is soluble. 7. Write (aq) if soluble; (s) if insoluble. HI(aq) + K(s)  KI(aq) + H2(g) 8. Balance the equation. 2 HI(aq) + 2 K(s)  2 KI(aq) + H2(g)

Example 14.3b—Write an Equation for the Reaction of Hydrobromic Acid with Sodium Oxide(s). : Tro's Introductory Chemistry, Chapter 14 37 Example 14.3b—Write an Equation for the Reaction of Hydrobromic Acid with Sodium Oxide(s). 1. Write formulas of reactants. HBr(aq) + Na2O(s)  2. Identify the type of reaction and predict the pattern. Acid + metal oxide  salt + H2O(l) 3. Determine the charge on the cation. Na+ 4. Determine the formula of the salt. Na+ + Br−  NaBr

Example 14.3b—Write an Equation for the Reaction of Hydrobromic Acid with Sodium Oxide(s), Continued. : Tro's Introductory Chemistry, Chapter 14 38 Example 14.3b—Write an Equation for the Reaction of Hydrobromic Acid with Sodium Oxide(s), Continued. 5. Write the skeletal equation. HBr(aq) + Na2O(s)  NaBr + H2O(l) 6. Check the solubility of the salt. NaBr is soluble. 7. Write (aq) if soluble; (s) if insoluble. HBr(aq) + Na2O(s)  NaBr(aq) + H2O(l) 8. Balance the equation. 2 HBr(aq) + Na2O(s)  2 NaBr(aq) + H2O(l)

Practice—Complete and Balance the Following Reactions: : Tro's Introductory Chemistry, Chapter 14 39 Practice—Complete and Balance the Following Reactions: HCl(aq) + CaO(s)  HCl(aq) + Ca(s) 

Practice—Complete and Balance the Following Reactions, Continued: : Tro's Introductory Chemistry, Chapter 14 40 Practice—Complete and Balance the Following Reactions, Continued: 2 HCl(aq) + CaO(s)  CaCl2(aq) + H2O(l) 2 HCl(aq) + Ca(s)  CaCl2(aq) + H2(g)

Base Reactions : Tro's Introductory Chemistry, Chapter 14 41 Base Reactions The reaction all bases have in common is neutralization of acids. Strong bases will react with Al metal to form sodium aluminate and hydrogen gas. 2 NaOH + 2 Al + 6 H2O → 2 NaAl(OH)4 + 3 H2

Titration : Tro's Introductory Chemistry, Chapter 14 42 Titration Titration is a technique that uses reaction stoichiometry to determine the concentration of an unknown solution. Titrant (unknown solution) is added from a buret. Indicators are chemicals that are added to help determine when a reaction is complete. The endpoint of the titration occurs when the reaction is complete.

Titration, Continued : Tro's Introductory Chemistry, Chapter 14 43 Titration, Continued

Acid–Base Titration : Tro's Introductory Chemistry, Chapter 14 44 Acid–Base Titration The base solution is the titrant in the buret. As the base is added to the acid, the H+ reacts with the OH– to form water. But there is still excess acid present, so the color does not change. At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point, the indicator changes color: [H+] = [OH–]

Example 14.4—What Is the Molarity of an HCl Solution if 10.00 mL Is required to Titrate 12.54 mL of 0.100 M NaOH?NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) : Tro's Introductory Chemistry, Chapter 14 45 Example 14.4—What Is the Molarity of an HCl Solution if 10.00 mL Is required to Titrate 12.54 mL of 0.100 M NaOH?NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) The unit is correct, the magnitude is reasonable. Check: Solve: M = mol/L, 1 mol NaOH = 1 mol HCl, 1 mL = 0.001L Solution Map: Relationships: 12.54 mL NaOH, 10.00 mL HCl M HCl Given: Find:

Slide 46 : Tro's Introductory Chemistry, Chapter 14 46 Example 14.4: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? : Tro's Introductory Chemistry, Chapter 14 47 Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Write down the given quantity and its units. Given: 10.00 mL HCl 12.54 mL NaOH

Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? : Tro's Introductory Chemistry, Chapter 14 48 Write down the quantity to find and/or its units. Find: concentration HCl, M Information: Given: 10.00 mL HCl 12.54 mL NaOH Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? : Tro's Introductory Chemistry, Chapter 14 49 Collect needed equations and conversion factors: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)  1 mole HCl = 1 mole NaOH. 0.100 M NaOH 0.100 mol NaOH  1 L solution. Information: Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? : Tro's Introductory Chemistry, Chapter 14 50 Write a solution map: mL NaOH L NaOH mol NaOH mol HCl Information: Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl Conversion Factors: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? mL HCl L HCl

Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? : Tro's Introductory Chemistry, Chapter 14 51 Apply the solution map: = 1.25 x 10-3 mol HCl Information: Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl Conversion Factors: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L Solution Map: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? : Tro's Introductory Chemistry, Chapter 14 52 Apply the solution map: Information: Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl Conversion Factors: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L Solution Map: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? : Check the solution: HCl solution = 0.125 M The units of the answer, M, are correct. The magnitude of the answer makes sense since the neutralization takes less HCl solution than NaOH solution, so the HCl should be more concentrated. Information: Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl Conversion Factors: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L Solution Map: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

Practice—What Is the Molarity of a Ba(OH)2 Solution if 37.6 mL Is Required to Titrate 43.8 mL of 0.107 M HCl?Ba(OH)2(aq) + 2 HCl(aq)  BaCl2(aq) + 2 H2O(l) : Tro's Introductory Chemistry, Chapter 14 54 Practice—What Is the Molarity of a Ba(OH)2 Solution if 37.6 mL Is Required to Titrate 43.8 mL of 0.107 M HCl?Ba(OH)2(aq) + 2 HCl(aq)  BaCl2(aq) + 2 H2O(l)

Practice—What Is the Molarity of a Ba(OH)2 Solution if 37.6 mL Is Required to Titrate 43.8 mL of 0.107 M HCl?Ba(OH)2(aq) + 2 HCl(aq)  BaCl2(aq) + 2 H2O(l), Continued : Tro's Introductory Chemistry, Chapter 14 55 Practice—What Is the Molarity of a Ba(OH)2 Solution if 37.6 mL Is Required to Titrate 43.8 mL of 0.107 M HCl?Ba(OH)2(aq) + 2 HCl(aq)  BaCl2(aq) + 2 H2O(l), Continued The unit is correct, the magnitude is reasonable. Check: Solve: M = mol/L, 1 mol Ba(OH)2= 2 mol HCl, 1 mL = 0.001L Solution Map: Relationships: 37.6 mL Ba(OH)2, 43.8 mL HCl M Ba(OH)2 Given: Find:

Strong or Weak : Tro's Introductory Chemistry, Chapter 14 56 Strong or Weak A strong acid is a strong electrolyte. Practically all the acid molecules ionize, →. A strong base is a strong electrolyte. Practically all the base molecules form OH– ions, either through dissociation or reaction with water, →. A weak acid is a weak electrolyte. Only a small percentage of the molecules ionize, . A weak base is a weak electrolyte. Only a small percentage of the base molecules form OH– ions, either through dissociation or reaction with water, .

Strong Acids : Tro's Introductory Chemistry, Chapter 14 57 Strong Acids The stronger the acid, the more willing it is to donate H. Use water as the standard base. Strong acids donate practically all their Hs. 100% ionized in water. Strong electrolyte. [H3O+] = [strong acid]. [ ] = molarity.

Strong Acids, Continued : Tro's Introductory Chemistry, Chapter 14 58 Strong Acids, Continued

Strong Acids, Continued : Tro's Introductory Chemistry, Chapter 14 59 Strong Acids, Continued Pure water HCl solution

Weak Acids : Tro's Introductory Chemistry, Chapter 14 60 Weak Acids Weak acids donate a small fraction of their Hs. Most of the weak acid molecules do not donate H to water. Much less than 1% ionized in water. [H3O+] << [weak acid].

Weak Acids, Continued : Tro's Introductory Chemistry, Chapter 14 61 Weak Acids, Continued

Weak Acids, Continued : Tro's Introductory Chemistry, Chapter 14 62 Weak Acids, Continued Pure water HF solution

Degree of Ionization : Tro's Introductory Chemistry, Chapter 14 63 Degree of Ionization The extent to which an acid ionizes in water depends in part on the strength of the bond between the acid H+ and anion compared to the strength of the bond between the acid H+ and the O of water. HA(aq) + H2O(l)  A−(aq) + H3O+(aq)

Relationship Between Strengths of Acids and Their Conjugate Bases : 64 Relationship Between Strengths of Acids and Their Conjugate Bases The stronger an acid is, the weaker the attraction of the ionizable H for the rest of the molecule is. The better the acid is at donating H, the worse its conjugate base will be at accepting an H. Strong acid HCl + H2O → Cl– + H3O+ Weak conjugate base Weak acid HF + H2O  F– + H3O+ Strong conjugate base

Example 14.5—Determine the [H3O+] in the Following Solutions: : Tro's Introductory Chemistry, Chapter 14 65 Example 14.5—Determine the [H3O+] in the Following Solutions: 1.5 M HCl Since HCl is a strong acid, [H3O+] = [HCl] = 1.5 M. 3.0 M HC2H3O2 Since HC2H3O2 is a weak acid, [H3O+] << [HC2H3O2]. Therefore, [H3O+] << 3.0 M.

Practice—Determine the [H3O+] in the Following Solutions: : Practice—Determine the [H3O+] in the Following Solutions: 0.5 M HI 0.1 M HCHO2 Strong Weak

Practice—Determine the [H3O+] in the Following Solutions, Continued: : Tro's Introductory Chemistry, Chapter 14 67 Practice—Determine the [H3O+] in the Following Solutions, Continued: 0.5 M HI Since HI is a strong acid, [H3O+] = [HI] = 0.5 M. 0.1 M HCHO2 Since HCHO2 is a weak acid, [H3O+] << [HCHO2]. Therefore, [H3O+] << 0.1 M.

Strong Bases : Tro's Introductory Chemistry, Chapter 14 68 Strong Bases The stronger the base, the more willing it is to accept H. Use water as the standard acid. Strong bases, practically all molecules are dissociated into OH– or accept Hs. Strong electrolyte. Multi-OH bases completely dissociated. [HO–] = [strong base] x (# OH).

Strong Bases, Continued : Tro's Introductory Chemistry, Chapter 14 69 Strong Bases, Continued

Weak Bases : Tro's Introductory Chemistry, Chapter 14 70 Weak Bases In weak bases, only a small fraction of molecules accept Hs. Weak electrolyte. Most of the weak base molecules do not take H from water. Much less than 1% ionization in water. [HO–] << [strong base].

Weak Bases, Continued : Tro's Introductory Chemistry, Chapter 14 71 Weak Bases, Continued

Example 14.6—Determine the [OH−] in the Following Solutions: : Tro's Introductory Chemistry, Chapter 14 72 Example 14.6—Determine the [OH−] in the Following Solutions: 2.25 M KOH Since KOH is a strong base, [OH−] = [KOH] x 1 = 2.25 M. 0.35 M CH3NH2 Since CH3NH2 is a weak base, [OH−] << [CH3NH2]. Therefore, [OH−] << 0.35 M. 0.025 M Sr(OH)2 Since Sr(OH)2 is a strong base, [OH−] = [Sr(OH)2] x 2 = 0.050 M.

Practice—Determine the [OH−] in the Following Solutions: : Practice—Determine the [OH−] in the Following Solutions: 0.05 M Ba(OH)2 0.01 M C5H5N Strong Weak

Practice—Determine the [OH−] in the Following Solutions: : Tro's Introductory Chemistry, Chapter 14 74 Practice—Determine the [OH−] in the Following Solutions: 0.05 M Ba(OH)2 Ba(OH)2 is a strong base. [OH−] = [Ba(OH)2] x 2 = 0.1 M. 0.01 M C5H5N C5H5N is a weak base, [OH−] << [C5H5N]. Therefore, [OH−] << 0.01 M.

Autoionization of Water : Tro's Introductory Chemistry, Chapter 14 75 Autoionization of Water Water is actually an extremely weak electrolyte. Therefore, there must be a few ions present. About 1 out of every 10 million water molecules form ions through a process called autoionization. H2O Û H+ + OH– H2O + H2O Û H3O+ + OH– All aqueous solutions contain both H3O+ and OH–. The concentration of H3O+ and OH– are equal in water. [H3O+] = [OH–] = 1 x 10-7M at 25 °C in pure water.

Ion Product of Water : Tro's Introductory Chemistry, Chapter 14 76 Ion Product of Water The product of the H3O+ and OH– concentrations is always the same number. The number is called the ion product of water and has the symbol Kw. [H3O+] x [OH–] = 1 x 10-14 = Kw. As [H3O+] increases, the [OH–] must decrease so the product stays constant. Inversely proportional.

Acidic and Basic Solutions : Tro's Introductory Chemistry, Chapter 14 77 Acidic and Basic Solutions Neutral solutions have equal [H3O+] and [OH–]. [H3O+] = [OH–] = 1 x 10-7 Acidic solutions have a larger [H3O+] than [OH–]. [H3O+] > 1 x 10-7; [OH–] < 1 x 10-7 Basic solutions have a larger [OH–] than [H3O+]. [H3O+] < 1 x 10-7; [OH–] > 1 x 10-7

Example—Determine the [H3O+] for a 0.00020 M Ba(OH)2 and Determine Whether the Solution Is Acidic, Basic, or Neutral. : Tro's Introductory Chemistry, Chapter 14 78 Example—Determine the [H3O+] for a 0.00020 M Ba(OH)2 and Determine Whether the Solution Is Acidic, Basic, or Neutral. Ba(OH)2 = Ba2+ + 2 OH– therefore: [OH–] = 2 x 0.00020 = 0.00040 = 4.0 x 10−4 M [H3O+] = 2.5 x 10-11 M. Since [H3O+] < 1 x 10−7, the solution is basic.

Practice—Determine the [H3O+] Concentration and Whether the Solution Is Acidic, Basic, or Neutral for the Following: : Tro's Introductory Chemistry, Chapter 14 79 Practice—Determine the [H3O+] Concentration and Whether the Solution Is Acidic, Basic, or Neutral for the Following: [OH–] = 0.000250 M [OH–] = 3.50 x 10-8 M Ca(OH)2 = 0.20 M

Practice—Determine the [H3O+] Concentration and Whether the Solution Is Acidic, Basic, or Neutral for the Following, Continued: : Tro's Introductory Chemistry, Chapter 14 80 Practice—Determine the [H3O+] Concentration and Whether the Solution Is Acidic, Basic, or Neutral for the Following, Continued: [OH–] = 0.000250 M [OH–] = 3.50 x 10-8 M Ca(OH)2 = 0.20 M [H3O+] < [OH-1], therefore, base. [H3O+] > [OH-1], therefore, acid. [H3O+] < [OH-1], therefore, base. [OH-1] = 2 x 0.20 = 0.40 M

Complete the Table[H+] vs. [OH-] : Tro's Introductory Chemistry, Chapter 14 81 Complete the Table[H+] vs. [OH-] [OH-] [H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14

Complete the Table[H+] vs. [OH-] : Tro's Introductory Chemistry, Chapter 14 82 Complete the Table[H+] vs. [OH-] [OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100 [H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 Even though it may look like it, neither H+ nor OH- will ever be 0. The sizes of the H+ and OH- are not to scale because the divisions are powers of 10 rather than units. Acid Base

pH : Tro's Introductory Chemistry, Chapter 14 83 pH The acidity/basicity of a solution is often expressed as pH. pH = ─log[H3O+], [H3O+] = 10−pH Exponent on 10 with a positive sign. pHwater = −log[10-7] = 7. Need to know the [H+] concentration to find pH. pH < 7 is acidic; pH > 7 is basic; pH = 7 is neutral.

pH, Continued : 84 pH, Continued The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution. 1 pH unit corresponds to a factor of 10 difference in acidity. Normal range is 0 to 14. pH 0 is [H+] = 1 M, pH 14 is [OH–] = 1 M. pH can be negative (very acidic) or larger than 14 (very alkaline).

pH of Common Substances : 85 pH of Common Substances

Example—Calculate the pH of a 0.0010 M Ba(OH)2 Solution and Determine if It Is Acidic, Basic, or Neutral. : Tro's Introductory Chemistry, Chapter 14 86 Example—Calculate the pH of a 0.0010 M Ba(OH)2 Solution and Determine if It Is Acidic, Basic, or Neutral. pH > 7 therefore, basic. Ba(OH)2 = Ba2+ + 2 OH− therefore, [OH-] = 2 x 0.0010 = 0.0020 = 2.0 x 10-3 M. pH = −log [H3O+] = −log (5.0 x 10-12) pH = 11.3

Practice—Calculate the pH of the Following Strong Acid or Base Solutions. : Tro's Introductory Chemistry, Chapter 14 87 Practice—Calculate the pH of the Following Strong Acid or Base Solutions. 0.0020 M HCl 0.0050 M Ca(OH)2 0.25 M HNO3

Practice—Calculate the pH of the Following Strong Acid or Base Solutions, Continued. : Tro's Introductory Chemistry, Chapter 14 88 Practice—Calculate the pH of the Following Strong Acid or Base Solutions, Continued. 0.0020 M HCl therefore, [H3O+] = 0.0020 M. 0.0050 M Ca(OH)2 therefore, [OH–] = 0.010 M. 0.25 M HNO3 therefore, [H3O+] = 0.25 M. pH = −log (2.0 x 10-3) = 2.70 pH = −log (1.0 x 10−12) = 12.00 pH = −log (2.5 x 10−1) = 0.60

Complete the Table:pH : Tro's Introductory Chemistry, Chapter 14 89 [OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100 [H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 pH Complete the Table:pH

Complete the Table:pH, Continued : Tro's Introductory Chemistry, Chapter 14 90 Complete the Table:pH, Continued [OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100 [H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 pH 0 1 3 5 7 9 11 13 14 Acid Base

Example—Calculate the Concentration of [H3O+] for a Solution with pH 3.7. : Tro's Introductory Chemistry, Chapter 14 91 Example—Calculate the Concentration of [H3O+] for a Solution with pH 3.7. [H3O+] = 10-pH [H3O+] = 10-3.7 means 0.0001 < [H+1] < 0.001. [H3O+] = 2 x 10-4 M = 0.0002 M.

Practice—Determine the [H3O+] for Each of the Following: : Tro's Introductory Chemistry, Chapter 14 92 Practice—Determine the [H3O+] for Each of the Following: pH = 2.7 pH = 12 pH = 0.60

Practice—Determine the [H3O+] for Each of the Following, Continued: : Tro's Introductory Chemistry, Chapter 14 93 Practice—Determine the [H3O+] for Each of the Following, Continued: pH = 2.7 pH = 12 pH = 0.60 [H3O+] = 10−0.60 = 0.25 M [H3O+] = 10−2.7 = 2 x 10−3 M = 0.002 M [H3O+] = 10−12 = 1 x 10-12 M

pOH : Tro's Introductory Chemistry, Chapter 14 94 pOH The acidity/basicity of a solution may also be expressed as pOH. pOH = ─log[OH−], [OH−] = 10−pOH Exponent on 10 with a positive sign. pOHwater = −log[10−7] = 7. Need to know the [OH−] concentration to find pOH. pOH < 7 is acidic; pOH > 7 is basic, pOH = 7 is neutral.

pOH, Continued : Tro's Introductory Chemistry, Chapter 14 95 pOH, Continued The lower the pOH, the more basic the solution; the higher the pOH, the more acidic the solution. 1 pOH unit corresponds to a factor of 10 difference in basicity. Normal range is 0 to 14. pOH 0 is [OH−] = 1 M; pOH 14 is [H3O+] = 1 M. pOH can be negative (very basic) or larger than 14 (very acidic). pH + pOH = 14.00.

Complete the Table:pOH : Tro's Introductory Chemistry, Chapter 14 96 [OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100 [H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 pOH Complete the Table:pOH

Complete the Table:pOH, Continued : Tro's Introductory Chemistry, Chapter 14 97 Complete the Table:pOH, Continued [OH-]10-14 10-13 10-11 10-9 10-7 10-5 10-3 10-1 100 [H+] 100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14 pOH 14 13 11 9 7 5 3 1 0 Acid Base

Example—Calculate the pH of a 0.0010 M Ba(OH)2 Solution and Determine if It Is Acidic, Basic, or Neutral. : Tro's Introductory Chemistry, Chapter 14 98 Example—Calculate the pH of a 0.0010 M Ba(OH)2 Solution and Determine if It Is Acidic, Basic, or Neutral. pH > 7 therefore, basic. Ba(OH)2 = Ba+2 + 2 OH− therefore, [OH−] = 2 x 0.0010 = 0.0020 = 2.0 x 10−3 M. pOH = -log [OH−] = -log (2.0 x 10−3) pOH = 2.70 pH = 14.00 - pOH = 14.00 - 2.70 pH = 11.30

Practice—Calculate the pOH and pH of the Following Strong Acid or Base Solutions. : Tro's Introductory Chemistry, Chapter 14 99 Practice—Calculate the pOH and pH of the Following Strong Acid or Base Solutions. 0.0020 M KOH 0.0050 M Ca(OH)2 0.25 M HNO3

Practice—Calculate the pOH and pH of the Following Strong Acid or Base Solutions, Continued. : Tro's Introductory Chemistry, Chapter 14 100 Practice—Calculate the pOH and pH of the Following Strong Acid or Base Solutions, Continued. 0.0020 M KOH therefore, [OH–] = 0.0020 M. 0.0050 M Ca(OH)2 therefore, [OH–] = 0.010 M. 0.25 M HNO3 therefore, [H3O+] = 0.25 M. pOH = −log (1.0 x 10-2) = 2.00 pH = 14.00 – 2.00 = 12.00 pH = −log (2.5 x 10-1) = 0.60 pOH = 14.00 – 0.60 = 13.40 pOH = −log (2.0 x 10-3) = 2.70 pH = 14.00 – 2.70 = 11.30

Buffers : Tro's Introductory Chemistry, Chapter 14 101 Buffers Buffers are solutions that resist changing pH when small amounts of acid or base are added. They resist changing pH by neutralizing added acid or base. Buffers are made by mixing together a weak acid and its conjugate base. Or weak base and its conjugate acid.

How Buffers Work : Tro's Introductory Chemistry, Chapter 14 102 How Buffers Work The weak acid present in the buffer mixture can neutralize added base. The conjugate base present in the buffer mixture can neutralize added acid. The net result is little to no change in the solution pH.

How Buffers Work, Continued : Tro's Introductory Chemistry, Chapter 14 103 How Buffers Work, Continued HA  + H3O+ A− A− Added H3O+ New HA HA

How Buffers Work, Continued : Tro's Introductory Chemistry, Chapter 14 104 HA How Buffers Work, Continued HA  + H3O+ A− Added HO− New A− A−

A Buffer Made from Acetic Acid and Sodium Acetate : Tro's Introductory Chemistry, Chapter 14 105 A Buffer Made from Acetic Acid and Sodium Acetate A buffer solution with a pH of 4.75 can be made by mixing equal volumes of 1 M HC2H3O2 and 1 M NaC2H3O2. Adding 10 mL of 0.1 M HCl to 1 L of this solution will give a solution with a pH of 4.75. Adding 10 mL of 0.1 M HCl to 1 L of distilled water will give a solution with pH of 3.0. Adding 10 mL of 0.1 M NaOH to 1 L of this solution will give a solution with a pH of 4.75. Adding 10 mL of 0.1 M NaOH to 1 L of distilled water will give a solution with pH of 11.0.

Acetic Acid/Acetate Buffer : Tro's Introductory Chemistry, Chapter 14 106 Acetic Acid/Acetate Buffer

Nonmetal Oxides Are Acidic : Tro's Introductory Chemistry, Chapter 14 107 Nonmetal Oxides Are Acidic Nonmetal oxides react with water to form acids. Causes acid rain. CO2 (g) + H2O(l) → H2CO3(aq) 2 SO2(g) + O2(g) + 2 H2O(l) → 2 H2SO4(aq) 4 NO2(g) + O2(g) + 2 H2O(l) → 4 HNO3(aq)

What Is Acid Rain? : Tro's Introductory Chemistry, Chapter 14 108 What Is Acid Rain? Natural rain water has a pH of 5.6. Naturally slightly acidic due mainly to CO2. Rain water with a pH lower than 5.6 is called acid rain. Acid rain is linked to damage in ecosystems and structures.

What Causes Acid Rain? : Tro's Introductory Chemistry, Chapter 14 109 What Causes Acid Rain? Many natural and pollutant gases dissolved in the air are nonmetal oxides. CO2, SO2, NO2. Nonmetal oxides are acidic. CO2 + H2O  H2CO3 2 SO2 + O2 + 2 H2O  2 H2SO4 Processes that produce nonmetal oxide gases as waste increase the acidity of the rain. Natural—volcanoes and some bacterial action. Man-made—combustion of fuel. Weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced.

pH of Rain in Different Regions : pH of Rain in Different Regions

Sources of SO2 from Utilities : Sources of SO2 from Utilities

Damage from Acid Rain : Tro's Introductory Chemistry, Chapter 14 112 Damage from Acid Rain Acids react with metals and materials that contain carbonates. Acid rain damages bridges, cars, and other metallic structures. Acid rain damages buildings and other structures made of limestone or cement. Acidifies lakes affecting aquatic life. Dissolves and leaches more minerals from soil. Making it difficult for trees.

Damage from Acid Rain : Tro's Introductory Chemistry, Chapter 14 113 Damage from Acid Rain circa 1935 circa 1995

pH of Rain in Different Regions : Tro's Introductory Chemistry, Chapter 14 114 pH of Rain in Different Regions

Sources of SO2 from Utilities : 115 Sources of SO2 from Utilities

Acid Rain Legislation : Tro's Introductory Chemistry, Chapter 14 116 Acid Rain Legislation 1990 Clean Air Act attacks acid rain. Forces utilities to reduce SO2. The result is acid rain in the northeast is stabilized and begins to be reduced.