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Section 2.1: Acids and Bases

Section 2.1: Acids and Bases

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Section 2.1: Acids and Bases

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  1. Section 2.1: Acids and Bases Solutions

  2. Learning Goals: • Identify the physical and chemical properties of acids and bases. • Classify solutions as acidic, basic, or neutral. • Compare the Arrhenius and Brønsted-Lowry models of acids and bases.

  3. Properties of Acids and Bases

  4. Properties of Acids and Bases • All water solutions contain hydrogen ions (H+) and hydroxide ions (OH–). • An acidic solutioncontains more H+ions than OH-ions. • A basic solutioncontains more OH-ions than H+ions.

  5. Properties of Acids and Bases • The usual solvent for acids and bases is water. • Water produces equal numbers of H+and OH-ions in a process called self-ionization. H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

  6. Arrhenius Model • States that an acid is a substance that contains hydrogen and ionizes to produce hydrogen ions in aqueous solution • States that a base is a substance that contains a hydroxide group and dissociates to produce a hydroxide ion in solution.

  7. Arrhenius Model • HCl ionizes to produce H+ ions. • HCl(g) → H+(aq) + Cl–(aq) • NaOH dissociates to produce OH– ions. • NaOH(s) → Na+(aq) + OH–(aq) • Some solutions produce hydroxide ions even though they do not contain a hydroxide group

  8. Brønsted-Lowry Model • States that an acid is a hydrogen ion donor, and a base is a hydrogen ion acceptor. • The Brønsted-Lowry Model is a more inclusive model of acids and bases.

  9. Brønsted-Lowry Model • A conjugate acidis the species produced when a base accepts a hydrogen ion. • A conjugate baseis the species produced when an acid donates a hydrogen ion.

  10. Brønsted-Lowry Model • A conjugate acid-base pair consists of two substances related to each other by donating and accepting a single hydrogen ion.

  11. Brønsted-Lowry Model • HF(aq) + H2O(l) ↔ H3O+(aq) + F–(aq)

  12. Brønsted-Lowry Model • HF = acid • H2O = base • H3O+ = conjugate acid • F– = conjugate base

  13. Brønsted-Lowry Model • NH3(aq) + H2O(l) ↔ NH4+(aq) + OH–(aq) • NH3 = • H2O = • NH4+ = • OH–=

  14. Acid Conjugate Base

  15. Section 2.2: pH and pOH Solutions

  16. Learning Goals: • Explain pH and pOH. • Relate pH and pOH to the ion product constant for water. • Calculate the pH and pOH of aqueous solutions.

  17. Water • Pure water contains equal concentrations of H+ and OH– ions. H2O  H+ + OH-

  18. Water • The ion product constant of water when water self-ionizes is Kw Kw= [H+][OH–]

  19. Water • With pure water at 25°C, both [H+] and [OH–] are equal to 1.0 × 10–7M. Kw = 1.0 × 10–14 1.0 x 10-14= [H+][OH–]

  20. Water • In a neutral solution, [H+] = [OH-] • In an acidic solution, [H+] > [OH-] • In a basic solution, [H+] < [OH-] [H+][OH–]= 1.0 x 10-14

  21. Practice • Calculate the ion concentration for each of the following solutions and state whether they are neutral, acidic, or basic. • 1.0 x 10-5 M OH- calculate H+ • 1.0 x 10-7 M OH-  calculate H+ • 10.0 M H+  calculate OH-

  22. pH and pOH • pH is a measurement of the concentration of hydrogen ions. • Low pH = acid • High pH = base

  23. pH and pOH • Calculating pH: • pH = –log [H+]

  24. Calculating pH • Enter the [H+] • Press the “log” key • Press the “+/-” (change of sign) key

  25. Practice • Calculate the pH value of the following: • A solution in which [H+] = 1.0x10-9 • A solution in which [OH-] = 1.0x10-6

  26. pH and pOH • pOHis a measurement of concentration of hydroxide ions. • Low pOH = base • High pOH = acid

  27. pH and pOH

  28. pH and pOH • Calculating pOH • pOH= –log [OH–]

  29. Calculating pOH • Enter the [OH-] • Press the “log” key • Press the “+/-” (change of sign) key

  30. Practice • Calculate the pH and pOH values of the following: • 1.0 x 10-3 M OH- • 1.0 M H+

  31. pH and pOH pH + pOH = 14

  32. Calculating [H+] from pH • Enter the pH • Press the “+/-” (change of sign) key • Take the inverse log by pressing “10x” button (inv log)

  33. Practice • The pH of a human blood sample was measured to be 7.41. What is the [H+] in this blood?

  34. Practice • Calculate [H+] and [OH-] in each of the following solutions: • Milk, pH = 6.50 • Ammonia, pH = 11.90

  35. Section 2.3: Neutralization Solutions

  36. Learning Goals: • Write chemical equations for neutralization reactions. • Explain how neutralization reactions are used in acid-base titrations. • Compare the properties of buffered and unbuffered solutions.

  37. Acid-Base Reactions • A neutralization reactionis a reaction in which an acid and a base in an aqueous solution react to produce a salt and water. • A salt is an ionic compound made up of a cation from a base and an anion from an acid. • Neutralization is a double-replacement reaction.

  38. Acid-Base Reactions

  39. Practice • What volume of a 1.00M HCl solution is needed to neutralize 25.0 mL of a 0.350 M NaOH solution?

  40. What volume of a 1.00M HCl solution is needed to neutralize 25.0 mL of a 0.350 M NaOH solution? 1.) Balanced equation: HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) H+(aq) + OH-(aq)  H2O(l)

  41. What volume of a 1.00M HCl solution is needed to neutralize 25.0 mL of a 0.350 M NaOH solution? 2.) Calculate moles of reactants:

  42. What volume of a 1.00M HCl solution is needed to neutralize 25.0 mL of a 0.350 M NaOH solution? 3.) Determine limiting reactant: