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Chapter 15

Chapter 15. Aqueous Equilibria: Acids and Bases. Everyday Acids and Bases. Acids: vinegar, lemon juice, sulfuric acid Bases: antacids, ammonia. A. Acid-Base Concepts: The Bronsted-Lowry Theory. So far, we’ve discussed the Arrhenius theory of acids and bases:

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Chapter 15

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  1. Chapter 15 Aqueous Equilibria: Acids and Bases

  2. Everyday Acids and Bases. • Acids: vinegar, lemon juice, sulfuric acid • Bases: antacids, ammonia

  3. A. Acid-Base Concepts: The Bronsted-Lowry Theory • So far, we’ve discussed the Arrhenius theory of acids and bases: • Acids dissociate to produce H+ (examples: HCl, H2SO4) • Bases dissociate to produce OH- (examples: NaOH, Ba(OH)2 ) But -- as some bases don’t contain OH, this is limiting.

  4. Bronsted-Lowry Theory of Acids and Bases An acid is a substance that can transfer H+ A base is a substance that can accept H+ HA + B BH+ + A- Note conjugate acid base acid base acid/base pairs.

  5. Conjugate Pairs… In the previous reaction, A- is the conjugate base of the acid HA B is the conjugate base of the acid HB+ HA + B BH+ + A- In an acid/base reaction, keep your eye on the proton!

  6. examples Write balanced equations for the dissociation of the following Bronsted-Lowry acids in water: a. H2SO4 b. H3O+

  7. Examples -- answer • H2SO4 + H2O HSO4- + H3O+ • H3O+ + H2O H2O + H3O+ BOOKKEEPING -- be able to label acid, base, conjugate acid, conjugate base! (by convention, CA/CB are on the product side)

  8. Examples What is the conjugate acid of… • HCO3- • CO32-

  9. Examples -- answers • H2CO3 • HCO3-

  10. B. Acid Strength and Base Strength Think of the acid-base reaction as a “tug of war” between the two bases for a proton: HA + H2O H3O+ + A- Which is the stronger base: H2O or A-? Which wants the proton more? This will determine whether the equilibrium lies more to the right or to the left.

  11. Acid/Base Strength cont. • If H2O is a stronger proton acceptor than A-, H2O will get the protons, and the solution will mostly contain H3O+ and A-. • Conversely, if A- is the stronger proton acceptor,, the solution will mostly contain HA and H2O. **The proton is always transferred to the stronger base.

  12. Acid/Base Strength: Equilibrium The strength of the acids and bases in an acid-base reaction will dictate the position of the equilibrium in an acid-base reaction. HA + H2O A- + H3O+ if this is the stronger the equilibrium will lie to the right acid Because a strong acid will dissociate more thoroughly/completely.

  13. What does it mean to be a strong acid? • Almost completely dissociated in water • Equilibrium almost entirely to the right • Solution consists almost entirely of H3O+ and A- ions -- almost no HA molecules Strong acids have very weak conjugate bases. If HA has a strong tendency to lose its proton, A- will not be a good proton acceptor.

  14. What does it mean to be a weak acid? • Only partially dissociated in water. • Solution contains mostly undissociated HA. • Not much H3O+ and A- ion present in solution. • Equilibrium lies toward left. Weak acids have strong conjugate bases. If HA does not have a strong tendency to lose its proton, A- will be a good proton acceptor.

  15. C. Hydrated Protons and Hydronium Ions H+ does not exist by itself in aqueous solution, despite the fact that you frequently will see H+ (aq) in acid-base equilibria. In aqueous solution, H+ binds to a water molecule to form H3O+ ==> hydronium ion.

  16. D. Dissociation of Water Water can act either as an acid or base, depending on the situation. If acid is present, water is a base: HA + H2O A- + H3O+ If base is present, water is an acid: B + H2O HB + OH-

  17. Dissociation of Water cont. Or, water can act as both an acid and a base! H2O + H2O H3O+ + OH- acid base acid base This process is referred to as the dissociation of water, and has a dissociation constant, Kw.

  18. Dissociation of Water cont. For the dissociation of water, Kw = [H3O+][OH-] Remember that pure liquids, such as H2O, are not included in equilibrium expressions. = equilibrium constant for the dissociation of water (also referred to as the ion product constant for water)

  19. Dissociation of Water cont. Very little of water is ionized; the equilibrium lies far to the left. At 25oC, [H3O+] = [OH-] = 1.0 * 10-7 M so Kw = [H3O+][OH-] = 1.0 * 10-14 M **This is true for any aqueous solution at 25oC.

  20. [H3O+] vs. [OH-] Defining acidic/basic/neutral solutions: In an acidic solution, [H3O+] > [OH-] In a basic solution, [H3O+] < [OH-] In a neutral solution, [H3O+] = [OH-]

  21. example The concentration of OH- in a sample of 25oC seawater is 5.0 * 10-6 M. Calculate the concentration of H3O+ ions, and classify the solution as acidic, neutral, or basic.

  22. Example -- answer Kw = [H3O+][OH-] At 25oC, Kw = 1.0 * 10-14 M Given [OH-] = 5.0 * 10-6 M Solve for [H3O+] [H3O+] = Kw/[OH-] = (1.0 * 10-14)/(5.0 * 10-6) = 2 * 10-9 M Since [OH-] > [H3O+] -- basic

  23. E. The pH Scale pH is a more convenient way to express the concentration of hydronium ion in a solution. pH = -log[H3O+] pH < 7 -- acidic pH > 7 -- basic pH = 7 -- neutral

  24. example Calculate the pH of a solution that has an [H3O+] of 6.0 * 10-5 M.

  25. Example -- answer If pH = -log (6.0 * 10-5) then pH = 4.22 This solution is acidic.

  26. F. Measuring pH Color indicators are often used for approximation Examples: phenolphthalein (changes from colorless --> pink when going from acidic --> basic), bromothymol blue However, pH meters give you a more precise number, measuring the electrical potential of the solution (chapter 18)

  27. G. Equilibria in Solutions of Weak Acids Beware: a weak acid is not the same thing as a dilute solution of a strong acid! Even when dilute, the equilibrium for the strong acid will lie to the right (not for the weak acid)

  28. Equilibrium Constant for a Weak Acid Dissociation For the reaction HA(aq) + H2O(l) H3O+(aq) + A-(aq) Ka = [H3O+][A-]/[HA] In dilute aqueous solution, [H2O] is essentially constant. pKa = -log Ka

  29. example The pH of 0.10 M HOCl is 4.23. Calculate Ka for hypochlorous acid. How close did you come to the true value?

  30. Example -- answer Ka = [H3O+][-OCl]/[HOCl] [H3O+] = 10-pH = 10-4.23 = 5.89 * 10-5 M = [-OCl] [HOCl] = 0.10 - 5.89 * 10-5 = 0.0999 M Ka = (5.89 * 10-5)2/0.0999 = 3.47 * 10-8

  31. H. Calculating Equilibrium Concentrations in Solutions of Weak Acids How are Ka values useful? Allows us to calculate the pH of an acidic solution, as well as equilibrium concentrations of all species present Example: Calculate concentrations of all species present, and the pH of, a 0.10 M HCN solution.

  32. Concentration of all species in 0.10 M HCN soln. Step 1 List species present before any dissociation happens, and identify them as either acid or base. HCN H2O acid acid or base

  33. Concentration of… in 0.10 M HCN cont. Step 2 What proton transfer reactions can occur, given the aforementioned molecules? HCN + H2O H3O+ + CN- Ka = 4.9 * 10-10 H2O + H2O H3O+ + OH- Kw = 1.0 * 10-14 **Ka values in Table 15.2

  34. Concentration of… in 0.10 M HCN cont. Step 3 Label the reaction that proceeds farther to the right (larger equilibrium constant) as the principal reaction; the other reaction(s) is the subsidiary reaction. HCN reaction: principal dissociation of water: subsidiary

  35. Concentration of… in 0.10 M HCN cont. Step 4 Create an ICE table, expressing changes in concentration in terms of x. Principal rxn HCN + H2O H3O+ + CN- ---------------------------------------------------------------------------------------------------------------------- Initial (M) 0.10 ~0 0 Change (M) -x +x +x Equil. (M) 0.10 - x x x H2O is not part of the equilibrium expression, and is present in excess.

  36. Concentration of… in 0.10 M HCN cont. Step 5 Place the equilibrium values into the Ka expression. Ka = 4.9 * 10-10 = [H3O+][CN-]/[HCN] = (x)(x)/(0.10 - x)

  37. Concentration of… in 0.10 M HCN cont. Step 5 continued In this particular case -- Ka is small. This means the reaction does not proceed very far to the right. If this is the case, x is small, and to simplify our math, we can say that (0.10 - x) ~ 0.10 and 4.9 * 10-10 ~ x2/0.10 Thus x2 = 4.9 * 10-11 and x = 7.0 * 10-6

  38. Concentration of… in 0.10 M HCN cont. Step 6 Now, you can use x to find all equilibrium concentrations. [H3O+] = [CN-] = x = 7.0 * 10-6 M [HCN] = 0.10 - x = 0.10 M **x was small/negligible here. This is not always the case!

  39. Concentration of… in 0.10 M HCN cont. Step 7 The only concentration left to calculate is [OH-] from the subsidiary reaction. [OH-] = Kw/[H3O+] = 1.0*10-14/7.0*10-6 = 1.4 * 10-9 M Since [H3O+] from the dissociation of water is also 1.4*10-9 M, our assumption in the ICE table that [H3O+] = 0 was valid.

  40. Concentration of… in 0.10 M HCN cont. Step 8 Finally… calculate pH! pH = -log [H3O+] = -log (7.0*10-6) =5.15

  41. J. Percent Dissociation in Solutions of Weak Acids • Another way to measure/express acid strength: percent dissociation. • The stronger the acid, the more dissociated it will be in aqueous solution. percent dissociation = ([HA] dissoc./[HA] initial) * 100%

  42. K. Polyprotic Acids • Are acids that contain more than one dissociable proton • Examples: H2SO4 H3PO4 • Dissociate in a stepwise manner, and each dissociation step has its own Ka • Note that each successive Ka value decreases. After the first proton is removed, the remaining conjugate base will have a negative charge, making the next proton harder to remove.

  43. L. Equilibria in Solutions of Weak Bases Consider the reaction NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) There is a base-dissociation constant similar to that for an acid: Kb = [BH+][OH-]/[B] = [NH4+][OH-]/[NH3]

  44. Weak Bases… cont. • Weak bases are frequently amines • Amines are derivatives of ammonia where one or more of the H has been replaced by a hydrocarbon group

  45. example Calculate the pH and the concentrations of all species present in 0.40 M NH3 (Kb = 1.8 * 10-5). Step 1 What species are present prior to dissociation? NH3 H2O base acid or base

  46. Example -- cont Step 2/3 Seeing that Kb for NH3is 1.8 * 10-5, this is considered the principal reaction (as opposed to Kw) Step 4 NH3 + H2O NH4 + OH- ---------------------------------------------------------------------------------------------------------- Initial(M) 0.40 0 ~0 Change(M) -x +x +x Equil.(M) 0.40-x x x

  47. Example -- cont. Step 5 Kb = [NH4+][OH-]/[NH3] = x2/0.40-x ~x2/0.40 =1.8 * 10-5 x = 2.7 * 10-3 M = [NH4+] = [OH-] [NH3] = 0.40 - x = 0.40 M

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