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AP Chapter 14

AP Chapter 14. Acids & Bases. Taste sour React with metals  H 2 (g) React with CO 3 - 2  CO 2 (g) Change color with indicators Neutralize bases Conduct electricity in solution. (Alkali) Taste bitter Feel soapy – slippery Reverse color change of indicators Neutralize acids

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AP Chapter 14

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  1. AP Chapter 14 Acids & Bases

  2. Taste sour React with metals  H2 (g) React with CO3-2  CO2 (g) Change color with indicators Neutralize bases Conduct electricity in solution (Alkali) Taste bitter Feel soapy – slippery Reverse color change of indicators Neutralize acids Conduct electricity in solution Classic PropertiesAcids Bases

  3. Arrhenius Theory of Acid and Base: Acids release H+ Bases release OH- A neutralization reaction involves: H+ + OH- H2O The problem comes in with chemicals such as NH3. It acts as a base, but where is the OH-? NH3 + H2O  NH4OH  NH4+1 + OH-1 There is no compelling evidence for the existence of NH4OH

  4. Bronsted – Lowry theory: Acid is a proton (H+) donor Base is a proton acceptor NH3 + H2O  NH4+1 + OH-1 Base Acid HCl + H2O  Cl-1 + H3O+1 Acid Base

  5. Conjugate Acid – Base Pair: 2 substances related to each other by the donation and acceptance of a proton (H+).

  6. Generic equations HA(aq) + H2O (L) H3O+1(aq) + A-1(aq) Acid Disassociation constant Ka = [H3O+][A-] [H+][A-] [HA] [HA] = B(aq) + H2O (L) BH+1(aq) + OH-1(aq) Base Disassociation constant Kb = [OH-][HB+] [B]

  7. Strong Acids & Bases1. Strong Acid & Base have large Ka and Kb2. Concentration is high (0.1 M or higher)Weak Acids & Bases1. Have a small Ka or Kb (so reaction sits far left)2. The weaker the acid or base, the stronger it’s conjugate A Bronsted – Lowry Acid Base Reaction always favor the direction: Strong Acid – Base  weaker Acid – Base HCl + OH-1(NaOH)  H2O + Cl-1 stong acid stong base weak acid weak base

  8. Demonstration of strong and weak – acids and bases Strong acids & bases, having large Ka or Kb ionize 100% HCl + H2O  H3O+1 + Cl-1 NaOH  Na+1 + OH-1 Weak acids & bases, having small Ka or Kb only partially ionize. HC2H3O2 + H2O  H3O+1 + C2H3O2-1 Ka = 1.8 x 10-5 NH3 + H2O  NH4+1 + OH-1 Ka = 1.8 x 10-5

  9. Determine which chemicals are the acids, which are the bases in each of the following equations. HIO3 + H2O  H3O+ + IO3-1 HS-1 + H2O  H2S + OH-

  10. Determine which are acids & bases, the strength of each, and the direction the reaction will proceed. HSO4-1 + NO3-1 HNO3 + SO4-2

  11. Determine which are acids & bases, the strength of each, and the direction the reaction will proceed. HNO2 + ClO4-1  HClO4 + NO2-1 H2CO3 + CO3-2  HCO3-1 + HCO3-1 End day 1

  12. Self Ionization of H2O Water is amphoteric. An amphoteric compound has the ability to act as both the acid and the base. Autoionization H2O + H2O  H3O+ + OH- Kw = [H3O+][OH-] = 1.0 x 10-14 (or = [H+][OH-] ) Pure H2O [H3O+] = [OH-] = 1.0 x 10-7@ 25 oC Kw = 1.0 x 10-14 applies to all aqueous solutions

  13. [H3O+] > [OH-] Acidic [H3O+] = [OH-] Neutral [H3O+] < [OH-] Basic

  14. Example: A 0.1 M solution of Ammonium chloride has a [H3O+] = 2.36 x 10-5M. What is [OH-]? Is the solution acidic or basic?

  15. Recognition of some acids: • Polyprotic acids: having more than 1 H+ to loose as protons H2SO4 + H2O  H3O+1 + HSO4-1 Large K, 100% ionization HSO4-1 + H2O  H3O+1 + SO4-2 Ka = 1.2 x 10-2 • Oxyacids: acidic proton is attached to an oxygen atom in the molecule • Organic acids: carbon based, with –COOH on the molecule (carboxylic acid) + H2O + H3O+1 -

  16. The strength of oxyacids increases as the number of oxygen atoms increases.

  17. Soren Sorenson defined pH as the potential of the hydrogen ion pH = - log 10 [H+] pH = - log 10 [H+] pOH = - log 10 [OH-] * Remember: Since pH changes 1 for every power of 10 change in [H+], change from pH 5 to pH 4 is a difference of (x 10) change from pH 5 to pH 3 is a difference of (10 x 10 = 100) If [H+] = 1.0 x 10-9M, pH = 9 Remember that the number of decimal places in the log is equal to the significant figures in the original number.

  18. Calculate the [OH-] or [H3O+] of each of the following solutions. Identify if the solution is neutral, acidic, or basic with the pH. [OH-] = 1.0 x 10-7M pH = 7 • [H3O+] = 1.0 x 10-7 M • [H3O+] = 8.3 x 10-16M • [H3O+] = 5.4 x 10-5M • [OH-] = 1.5 M • [OH-] = 3.6 x 10-15M • [OH-] = 7.3 x 10-4M [OH-] = 12.05 M pH = 15.08 [OH-] = 1.8 x 10-10M pH = 4.27 [H3O+] = 6.67 x 10-15M pH = 14.18 [H3O+] = 2.78 M pH = -0.444 [H3O+] = 1.37 x 10-11M pH = 10.86

  19. [H3O+] > [OH-] Acidic [H3O+] = [OH-] Neutral [H3O+] < [OH-] Basic

  20. Kw = 1.0 x 10-14= Ka. Kb pH + pOH = pKw = - log 10 [1 x 10-14] = 14 pKw = pH + pOH = 14 Example: Students found the pH of rainwater to be 4.35. What is the [H3O+]? They found that household ammonia to be pH = 11.28. What is the [H3O+] and [OH-]?

  21. Example: Strong acids go to completion! Calculate the [H3O+], [Cl-], and [OH-] in 0.015 M HCl. HCl(aq) + H2O(L) H3O+1(aq) + Cl-1(aq)

  22. Example: Calcium hydroxide (slake lime) is the cheapest strong base available and is generally used for industrial operations in which a high [OH-] is not required. Ca(OH)2 is soluble in H2O only to the extent of 0.16 g / 100 mL solution at 25 oC. What is the pH of the Ca(OH)2 solution? End day 2

  23. Slake lime Ca(OH)2 is used in chimney scrubbers SO2 + H2O  H2SO3 Ca(OH)2 + H2SO3  CaSO3 + 2 H2OAlso used in H2O treatment to remove Ca+2 CaO + H2O  Ca(OH)2 Ca(OH)2 + Ca+2 + 2 HCO3-1  2 CaCO3 (s) + 2 H2O

  24. Calculate the major species present in 0.250 M solution of each of the following acids, as well as the pH. HClO4 HNO3

  25. Example: Calculate the pH of a weak acid solution, therefore, it is required to use the KaWhat is the pH of 0.100 M HC2H3O2? Ka = 1.8 x 10-5 HC2H3O2 (aq) + H2O(L) H3O+1(aq) + C2H3O2-1(aq) Initial M Change in M End M

  26. Example: Find the pH of a weak base.Piperidine is a base found in small amount in black pepper. What is the pH of 315 mL H2O solution containing 114 mg piperidine? Kb = 1.6 x 10-3 C5H11N(aq) + H2O(L) C5H11NH+1(aq) + OH-1(aq) I C E

  27. Determine Kb if the initial [M] and pH are known.Cocaine, C17H21O4N, is an alkaloid. It has a bitter taste. It is soluble in H2O to the extent of 0.17 g / 100 mL solution. A saturated solution has a pH of 10.08. What is the value of Kb? C17H21O4N(aq) + H2O(L) C17H21O4NH+1(aq) + OH-1(aq) I C E End day 3

  28. Salt – Ionic compoundsReactions between ions and H2O • Salts made with cations from strong base and anions from strong acids have no effect on [H+] in H2O pH = 7 example: NaCl, KNO3 • Salts made with cations from strong base (no effect on H2O) and anions from weak acid (will react with H2O  OH-) pH > 7 example: NaC2H3O2C2H3O2-1 + H2O  HC2H3O2 + OH-1 • Salts made with cations from weak bases (will react with H2O  H+) and anions from strong acids (no effect) pH < 7 example: NH4Cl NH4+1 + H2O  NH3 + H3O+1 • Salts made with cations from weak base and anions from weak acids, pH ??? Must look at the Ka and Kb to determine if acidic or basic.

  29. Ions as Acids and BasesSometimes one of the ions formed with a salt dissolves can act as an acid or base. Add an OH- to the positive ion formed when a salt dissolves. If this base is weak, the positive ion can act like an acid with water.Add a H+ to the negative ion formed when the salt dissolves. If the acid formed is weak, the negative ion can act like a base with water.NH4Cl  NH4+1 + Cl-1 NH4+1 + OH-1  NH4OH (weak base, NH4+1 acts like an acid) H+1 + Cl-1  HCl (stong acid, has no ability to react with H2O)NaC2H3O2  Na+1 + C2H3O2-1 Na+1 + OH-1  NaOH (strong base, has no ability to react with H2O) H+1 + C2H3O2-1  HC2H3O2(weak acid, C2H3O2-1 acts like a base)

  30. Remember:Ka. Kb = Kw = 1.0 x 10-14

  31. Sodium cyanide, NaCN, is a very dangerous substance because it forms HCN in acidic solution. Are NaCN solutions normally acidic, basic, or neutral? What is the pH of 0.50 M NaCN? NaCN  + OH-1  H+1 +  + H2O(L)  I C E You will need to look up the Ka or Kb, which ever you can find.

  32. Percent Dissociation (% Ionization) % dissociation = ( )100 Amount dissociated in M initial M Determine the % ionization as a function of a weak acidconcentration. What is the % ionization of acetic acid in 1.0 M , 0.1 M, and 0.01 M of HC2H3O2 Ka = 1.8 x 10-5 HC2H3O2(aq) + H2O(L) H3O+1(aq) + C2H3O2-1(aq) I C E The % dissociation increases as the acid becomes more dilute.

  33. Calculate the Ka from the % dissociation.Lactic acid is a waste product that accumulates in muscle tissue during exercise, leading to pain and a feeling of fatigue. In a 0.100 M lactic acid, 3.7 % dissociates. Calculate Ka HC3H5O3 (aq) + H2O(L) H3O+1(aq) + C3H5O3+1(aq) I C E

  34. Properties of Polyprotic Acids: They furnish more than one proton. They always dissociate in a stepwise manner. As H+ is removed, it becomes harder to remove the next H+ (because a (-) charge exists), therefore Ka1 > Ka2 > Ka3 . . . 1. Essentially all H3O+1 is made in the first dissociation. The pH of a weak polyprotic acid is identical to that of the 1st dissociation. 2. Very little dissociation occurs on the 2nd H+1 loss, so essentially [H2A-1] approximately = [H3O+1] A large [H3O+1] from the 1st ionization represses further ionization. 3. Ka2 approximately = [HA-2]

  35. Calculate the concentrations of [H3O+1], [H2PO4-1], [HPO4-2], and [PO4-3] for 3.0 M H3PO4 H3PO4 (aq) + H2O(L) I C E + H2O(L)  I C E + H2O(L) 

  36. Binary Acids: 2 atoms examples: HI, HBr, HCl, HF HI HBr HCl HF Longer   shorter Bond length: shorter bond length, H+1 held tight, doesn’t dissociate Lower   higher Bond dissociation energy lower, will ionize easier Ka > Ka > Ka > Ka { { Weak Acid Strong Acids

  37. Comparing Binary Acid Strength: When polar bonds exist, look for electronegativity difference (DEN) Loss of e- occurs with polar bonds. CH4 NH3 H2O HF DEN 0.4 < 0.9 < 1.4 < 1.9 { { { Limited acid properties in H2O Moderate acid properties in H2O No acid properties in H2O See page 334 text

  38. Anhydrides • Metal Oxides form bases when they react with H2O CaO + H2O  Ca(OH)2 • Nonmetal oxides form acids when they react with H2O SO2 + H2O  H2SO3

  39. Lewis Acid & Base Theory: Acids accept an e- pair Bases donate an e- pair Coordinate covalent bond: One in which one of the chemicals has provided both e-’s to the bond

  40. Assignments • worksheet • Ch 14 Review 44, 45, 46, 53 a, 78 and worksheet • Ch 14 Review 33, 34, 35, 36, 73, 74, 75, 76, 53 b, 55, 84, 85 • Ch 14 Review 19 and Ions as Acids & Bases • Ch 14 Review 51, 52, 81, 82 • Ch 14 Review 56, 61 b, 63 a b c, 64 a b c d 80, 89 • Ch 14 Review 91, 96 • Hydrolysis lab • Ch 14 Review 22

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