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Introductory Chemistry , 3 rd Edition Nivaldo Tro

Introductory Chemistry , 3 rd Edition Nivaldo Tro. Chapter 14 Acids and Bases. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. 2009, Prentice Hall. Types of Electrolytes. Salts are water-soluble ionic compounds. All strong electrolytes.

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Introductory Chemistry , 3 rd Edition Nivaldo Tro

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

  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. 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. Tro's Introductory Chemistry, Chapter 14

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

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

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

  9. Common 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. 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) 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. Tro's Introductory Chemistry, Chapter 14

  13. HCl ionizes in water, producing H+ and Cl– ions. NaOH dissociates in water, producing Na+ and OH– ions. Arrhenius Theory, Continued 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+ Tro's Introductory Chemistry, Chapter 14

  15. Comparing Arrhenius Theory and Brønsted–Lowry Theory • 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) • 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) 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) 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. 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+ 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. H2SO4 + H2O  HSO4– + H3O+ Acid Base Conjugate Conjugate base acid 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– 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. HCO3– + H2O  H2CO3 + HO– Base Acid Conjugate Conjugate acid base 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 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− Tro's Introductory Chemistry, Chapter 14

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

  23. 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 Tro's Introductory Chemistry, Chapter 14

  24. 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) Tro's Introductory Chemistry, Chapter 14

  25. 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) Tro's Introductory Chemistry, Chapter 14

  26. 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) Tro's Introductory Chemistry, Chapter 14

  27. 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. Tro's Introductory Chemistry, Chapter 14

  28. Titration, Continued Tro's Introductory Chemistry, Chapter 14

  29. 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–] Tro's Introductory Chemistry, Chapter 14

  30. mL L NaOH mol NaOH mol HCl M mL HCl L HCl 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) Given: Find: 12.54 mL NaOH, 10.00 mL HCl M HCl Solution Map: Relationships: M = mol/L, 1 mol NaOH = 1 mol HCl, 1 mL = 0.001L Solve: Check: The unit is correct, the magnitude is reasonable. Tro's Introductory Chemistry, Chapter 14

  31. 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

  32. mL L HCl mol HCl mol Ba(OH)2 M mL Ba(OH)2 L Ba(OH)2 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 Given: Find: 37.6 mL Ba(OH)2, 43.8 mL HCl M Ba(OH)2 Solution Map: Relationships: M = mol/L, 1 mol Ba(OH)2= 2 mol HCl, 1 mL = 0.001L Solve: Check: The unit is correct, the magnitude is reasonable. Tro's Introductory Chemistry, Chapter 14

  33. HCl ® H+ + Cl- HCl + H2O® H3O+ + Cl- 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. Tro's Introductory Chemistry, Chapter 14

  34. Strong Acids, Continued Tro's Introductory Chemistry, Chapter 14

  35. Strong Acids, Continued Pure water HCl solution Tro's Introductory Chemistry, Chapter 14

  36. HF Û H+ + F- HF + H2O Û H3O+ + F- 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]. Tro's Introductory Chemistry, Chapter 14

  37. Weak Acids, Continued Tro's Introductory Chemistry, Chapter 14

  38. Weak Acids, Continued Pure water HF solution Tro's Introductory Chemistry, Chapter 14

  39. 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) Tro's Introductory Chemistry, Chapter 14

  40. 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 acidHCl + H2O → Cl– + H3O+Weak conjugate base Weak acid HF + H2O  F– + H3O+Strong conjugate base

  41. NaOH ® Na+ + OH- 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). Tro's Introductory Chemistry, Chapter 14

  42. Strong Bases, Continued Tro's Introductory Chemistry, Chapter 14

  43. NH3 + H2O Û NH4+ + OH- 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]. Tro's Introductory Chemistry, Chapter 14

  44. 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. Tro's Introductory Chemistry, Chapter 14

  45. 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. Tro's Introductory Chemistry, Chapter 14

  46. 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 Tro's Introductory Chemistry, Chapter 14

  47. 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. Tro's Introductory Chemistry, Chapter 14

  48. 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 Tro's Introductory Chemistry, Chapter 14

  49. 1 x 10-14 2.50 x 10-4 [H3O+] = = 4.00 x 10-11 1 x 10-14 3.50 x 10-8 [H3O+] = = 2.86 x 10-7 1 x 10-14 4.0 x 10-1 [H3O+] = = 2.5 x 10-14 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. [OH-1] = 2 x 0.20 = 0.40 M [H3O+] < [OH-1], therefore, base. Tro's Introductory Chemistry, Chapter 14

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

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