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Acids and Bases

This text provides an overview of acid and base definitions, including the Arrhenius and Bronsted-Lowry models. It also discusses acid dissociation, self-ionization of water, and pH and pOH calculations. Helpful for understanding acid-base equilibria.

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Acids and Bases

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  1. Acids and Bases

  2. Acid/Base Definitions • Arrhenius Model • Acids produce hydrogen ions in aqueous solutions • Bases produce hydroxide ions in aqueous solutions • Bronsted-Lowry Model • Acids are proton donors • Bases are proton acceptors

  3. Acid Dissociation HA  H+ + A- AcidProtonConjugate base Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)

  4. HNO2(aq) + H2O(l) NO2-(aq) H3O+(aq) Acid Base Conjugate base Conjugate acid NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Base Acid Conjugate base Conjugate acid

  5. Dissociation of Strong Acids Strong acids are assumed to dissociate completely in solution. Large Ka or small Ka? Reactant favored or product favored?

  6. Dissociation Constants: Strong Acids

  7. Dissociation of Weak Acids Weak acids are assumed to dissociate only slightly (less than 5%) in solution. Large Ka or small Ka? Reactant favored or product favored?

  8. Dissociation Constants: Weak Acids

  9. PROBLEM: Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids. Hydrochloric acid (HCl) Acetic acid (HC2H3O2) The ammonium ion (NH4+) The anilinium ion (C6H5NH3+) The hydrated aluminum(III) ion [Al(H2O)6]3+

  10. Self-Ionization of Water H2O + H2O  H3O+ + OH- At 25, [H3O+] = [OH-] = 1 x 10-7 Kw is a constant at 25 C: Kw = [H3O+][OH-] Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

  11. Calculating pH, pOH pH = -log10(H3O+) pOH = -log10(OH-) Relationship between pH and pOH pH + pOH = 14 Finding [H3O+], [OH-] from pH, pOH [H3O+] = 10-pH [OH-] = 10-pOH

  12. pH and pOH Calculations

  13. pH Scale

  14. Problem: Calculate [H+] or [OH-] as required for each of the following solutions @ 25oC, and state whether the solution is neutral, acidic, or basic. • 1.0 x 10-5M OH- • 1.0 x 10-7M OH- • 10.0M H+

  15. Do Now: Please review your notes on yesterday’s acid/base problems. Also, get out your homework and notebook in preparation for today’s exciting continuation of Chapter 14! Objective: Describe equilibria involving strong and weak bases. Explain the method for calculating pH for basic solutions. Describe the dissociation equilibria of acids with more than one acidic proton. Homework: Acid/Base problem set. 012016

  16. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #1:Write the dissociation equation HC2H3O2 C2H3O2- + H+

  17. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #2:ICE it! HC2H3O2 C2H3O2- + H+ 0.50 0 0 +x +x - x x x 0.50 - x

  18. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #3:Set up the law of mass action HC2H3O2 C2H3O2- + H+ E 0.50 - x x x

  19. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #4:Solve for x, which is also [H+] HC2H3O2 C2H3O2- + H+ E 0.50 - x x x [H+] = 3.0 x 10-3 M

  20. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #5:Convert [H+] to pH HC2H3O2 C2H3O2- + H+ E 0.50 - x x x pH = -log(3.0 x 10-3) = 2.52

  21. Percent Dissociation/Ionization Amount dissociated (mol/L) = X 100% Initial concentration (mol/L) So….Calculate the percent dissociation for the previous problem. 60%? JK…How about 0.6%? Better? Tip of the day! For a given weak acid….the percent dissociation INCREASES as the acid becomes more dilute. Larger Percent dissociation for which one? 1.0 M HC2H3O2 or 0.1 M HC2H3O2?

  22. Aspirin, is a weak organic acid whose molecular formula is HC9H7O4. An aqueous solution of aspirin has a total volume of 350.0 mL and contains 1.26 g of aspirin. The pH of the solution is found to be 2.60 Calculate the Ka for aspirin Step #1:Write the dissociation equation HC9H7O4 C9H7O4- + H+

  23. Step #2:Determine concentration of acid! 1.26 g x 1 mol .350L 180.15 g = 0.02M HC9H7O4 Step #3:ICE it! HC9H7O4 C9H7O4- + H+ .02 0 0 +x +x - x x x .02 - x

  24. Step #4:Determine the [H+] pH = 2.60 [H+] = 10-2.60 [H+] = 2.5 x 10-3 M Step #5:Set up the law of mass action HC9H7O4 C9H7O4- + H+ E .02 - x x x 2 x2 .0025 Ka = Ka = Ka = 3.6 x 10-4 .02 - x .02-.0025

  25. The calculation of concentration and pH for weak acids is more complex than for strong acids due to: • The incomplete ionization of weak acids. • The low Ka value for strong acids. • The more complex atomic structures of strong acids. D. The low percent ionization of strong acids. E. The inconsistent Kb value for strong acids. Answer: A Strong acids ionize 100%, therefore, the [H+] is equal to the initial concentration of the strong acid. This is not the case for weak acids. Since not all of the acid ionizes, it is necessary to determine both how much of the acid forms H+ and how much is left in molecular form.

  26. The general reaction of an acid dissolving in water may be shown as HA (aq) + HOH (l) H3O (aq) + A-(aq) A conjugate acid base pair for this reaction is: • HA and HOH • HA and A- • HOH and A- • H3O+ and A- • HA and H3O+ Answer: B The conjugate acid differs from its conjugate base by a proton, (H+). The acid form has the proton (HA in this example) and the base form has lost the proton (A- in this case).

  27. Strong acids are those which • Have an equilibrium lying far to the left. • Yield a weak conjugate base when reacting with water • Have a conjugate base which is a stronger base than water D. Readily remove the H+ ions from water. E. Are only slightly dissociated (ionized) at equilibrium. Answer: B Strong acids give up their protons (H+ ions) easily. If their conjugate base form were strong, then the acid formed would hold strongly to the H+, which would not be a characteristic of a strong acid.

  28. When calculating the pOH of a hydrofluoric acid solution (Ka = 7.2 x 10-4) from its concentration, the contribution of water ionizing (Kw = 1.0 x 10-14) is usually ignored because • Hydrofluoric acid is such a weak acid. • Hydrofluoric acid can dissolve glass • The ionization of water provides relatively few H+ ions. • The [OH-] for pure water is unknown. • The conjugate base of HF is such a strong base. Answer: C By comparing the Ka for HF with Kw (for water), you can see that even though hydrofluoric acid is a very weak acid it is a much stronger acid than is water, so much so that the H+ ion contribution of water may be ignored.

  29. The percent dissociation (percent ionization) for weak acids • Is always the same for a given acid, no matter what the concentration. B. Usually increases as the acid becomes more concentrated. C. Compares the amount of acid that has dissociated at equilibrium with the initial concentration of the acid. D. May only be used to express the dissociation of weak acids. E. Has no meaning for polyprotic acids. Answer: C the percent dissociation does change with the concentration of the acid.

  30. The [OH-] of a certain aqueous solution is 1.0 x 10-5M. The pH of this same solution must be • 1.0 x 10-14 • 5.00 • 7.00 • 9.00 • 12.00 Answer: D Referring to pH + pOH = 14.00 at 25oC, if [OH-] = 1 x 10-5M, pOH = 5, pH = 14-5.0 =9.0

  31. In many calculations for the pH of a weak acid from the concentration of the acid, an assumption is made that often takes the form [HA]o – x = [HA]o This • Is valid because x is very small compared to the initial concentration of the weak acid. B. Is valid because the concentration of the acid changes by such large amounts. C. Is valid because actual value of x cannot be known. D. Is valid because pH is not dependent upon the concentration of the weak acid. E. Approximation is always shown to be valid and so need not be checked. Answer: A. We can expect x to be very small (as indicated by the low value for Ka for weak acids), so that the concentration of the weak acid does not significantly change from its initial value. As the final step in such problems, you must determine that the change in the initial [HA] is less than 5% for the approximation to be considered valid.

  32. HA is a weak acid which is 4.0% dissociated at 0.100M. Determine the Ka for this acid • 0.0040 • 0.00016 • 0.040 • 1.6 • 16.5 Answer: B [HA]o = 0.100M [H+] = [A-] = 0.004M Ka = [H+][A-]/[HA] = (0.004 x 0.004) / 0.100 = (4 x 10-3)(4 x 10-3) / (10-1) = 1.6 x 10-4

  33. Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H2O  BH+ + OH- (Yes, all weak bases do this – DO NOT endeavor to make this complicated!)

  34. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #1:Write the equation for the reaction NH3 + H2O  NH4+ + OH-

  35. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #2:ICE it! NH3 + H2O  NH4+ + OH- 0.50 0 0 +x +x - x x x 0.50 - x

  36. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #3:Set up the law of mass action NH3 + H2O  NH4+ + OH- E 0.50 - x x x

  37. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #4:Solve for x, which is also [OH-] NH3 + H2O  NH4+ + OH- E 0.50 - x x x [OH-] = 3.0 x 10-3 M

  38. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #5:Convert [OH-] to pH NH3 + H2O  NH4+ + OH- E 0.50 - x x x

  39. ANOTHER TIP OF THE DAY Ka x Kb = Kw

  40. Dissociation of Strong Bases MOH(s)  M+(aq) + OH-(aq) • Strong bases are metallic hydroxides • Group I hydroxides (NaOH, KOH) are very soluble • Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble • pH of strong bases is calculated directly from the concentration of the base in solution

  41. Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH3NH2 + H2O  CH3NH3+ + OH- Kb = 4.38 x 10-4

  42. Kb for Some Common Weak Bases Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

  43. Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH3NH2 + H2O  CH3NH3+ + OH- Kb = 4.38 x 10-4

  44. Kb for Some Common Weak Bases Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

  45. SALTS

  46. Acid-Base Properties of Salts These salts simply dissociate in water: KCl(s)  K+(aq) + Cl-(aq)

  47. Acid-Base Properties of Salts The basic anion can accept a proton from water: C2H3O2- + H2O  HC2H3O2 + OH- base acid acid base

  48. Acid-Base Properties of Salts The acidic cation can act as a proton donor: NH4+(aq)  NH3(aq) + H+(aq) Acid Conjugate Proton base

  49. Acid-Base Properties of Salts • IF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidic • IF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basic • IF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral

  50. Acid-Base Properties of Salts Step #1: AlCl3(s) + 6H2O  Al(H2O)63+(aq) + Cl-(aq) Salt water Complex ion anion Step #2: Al(H2O)63+(aq)  Al(OH)(H2O)52+(aq) + H+(aq) Acid Conjugate base Proton

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