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Acids and Bases

Acids and Bases. 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 Lewis Acid Model Acids are electron pair acceptors

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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 • Lewis Acid Model • Acids are electron pair acceptors • Bases are electron pair donors

  3. Three definitions of acid

  4. Some Definitions • Arrhenius acids and bases • Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions (protons, H+). • Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions.

  5. Some Definitions • Brønsted–Lowry: must have both 1. an Acid: Proton donor and 2. a Base: Proton acceptor

  6. If it can be either… ...it is amphiprotic. HCO3– HSO4 – H2O

  7. What Happens When an Acid Dissolves in Water? • Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid. • As a result, the conjugate base of the acid and a hydronium ion are formed. Movies…

  8. Conjugate Acids and Bases: • From the Latin word conjugare, meaning “to join together.” • Reactions between acids and bases always yield their conjugate bases and acids.

  9. Acid and Base Strength • Strong acids are completely dissociated in water. • Their conjugate bases are quite weak. • Weak acids only dissociate partially in water. • Their conjugate bases are weak bases.

  10. Acid and Base Strength • Substances with negligible acidity do not dissociate in water. • Their conjugate bases are exceedingly strong.

  11. Acid Dissociation HA  H+ + A- AcidProtonConjugate base Ka > 1 = product favored Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)

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

  13. Dissociation Constants: Strong Acids

  14. 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?

  15. Dissociation Constants: Weak Acids

  16. HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2–(aq) Acid and Base Strength Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1). The stronger base “wins” the proton.

  17. Self-Ionization of Wateraka “autoionization” H2O + H2O  H3O+ + OH- At 25, [H3O+] = [OH-] = 1 x 10-7 The ion-product constant for water at 25 C: Kw = [H3O+][OH-] Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

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

  19. pH • In pure water, Kw = [H3O+] [OH–] = 1.0  10-14 • Because in pure water [H3O+] = [OH-], [H3O+] = (1.0  10-14)1/2 = 1.0  10-7

  20. pH • Therefore, in pure water, pH = –log [H3O+] = –log (1.0  10-7) = 7.00 • An acid has a higher [H3O+] than pure water, so its pH is <7 • A base has a lower [H3O+] than pure water, so its pH is >7.

  21. Other “p” Scales • The “p” in pH tells us to take the negative log of the quantity (in this case, hydronium ions). • Some similar examples are • pOH –log [OH-] • pKw–log Kw If you know one, you know them all: [H+] [OH-] pH pOH

  22. pH and pOH Calculations

  23. pH ScaleYou need to know the acidity of common substances!

  24. How Do We Measure pH? • Litmus paper • “Red” paper turns blue above ~pH = 8 • “Blue” paper turns red below ~pH = 5 • An indicator • Compound that changes color in solution.

  25. Strong Acids • You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. • These are strong electrolytesand exist totally as ions in aqueous solution. • For the monoprotic strong acids, [H3O+] = [acid].

  26. Strong Bases • Strong bases are the soluble hydroxides, which are the alkali metal (NaOH, KOH)and heavier alkaline earth metal hydroxides (Ca(OH)2, Sr(OH)2, and Ba(OH)2). • Again, these substances dissociate completely in aqueous solution. [OH-] = [hydroxide added].

  27. Dissociation Constants • For a generalized acid dissociation, the equilibrium expression is • This equilibrium constant is called the acid-dissociation constant, Ka.

  28. Calculating Ka from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. To calculate Ka, we need all equilibrium concentrations. We can find [H3O+], which is the same as [HCOO−], from the pH.

  29. Calculating Ka from the pH pH = –log [H3O+] –2.38 = log [H3O+] 10-2.38 = 10log [H3O+] = [H3O+] 4.2  10-3 = [H3O+] = [HCOO–]

  30. Calculating Ka from pH In table form:

  31. [4.2  10-3] [4.2  10-3] [0.10] Ka = Calculating Ka from pH = 1.8  10-4

  32. 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+

  33. 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

  34. 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

  35. 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

  36. 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

  37. Calculating Percent Ionization In the example: [A-]eq = [H3O+]eq = 4.2  10-3 M [A-]eq + [HCOOH]eq = [HCOOH]initial = 0.10 M

  38. Calculating pH from Ka Calculate the pH of a 0.30 M solution of acetic acid, C2H3O2H, at 25°C. Ka for acetic acid at 25°C is 1.8  10-5. Is acetic acid more or less ionized than formic acid(Ka=1.8 x 10-4)?

  39. Calculating pH from Ka The equilibrium constant expression is:

  40. Calculating pH from Ka Use the ICE table:

  41. Calculating pH from Ka Use the ICE table: Simplify: how big is x relative to 0.30?

  42. Calculating pH from Ka Use the ICE table: Simplify: how big is x relative to 0.30?

  43. Calculating pH from Ka Now, (1.8  10-5) (0.30) = x2 5.4  10-6 = x2 2.3  10-3 = x Check: is approximation ok?

  44. Calculating pH from Ka pH = –log [H3O+] pH = – log (2.3  10−3) pH = 2.64

  45. Polyprotic Acids Have more than one acidic proton. If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

  46. 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

  47. 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- (all weak bases do this)

  48. Weak Bases Bases react with water to produce hydroxide ion.

  49. Weak Bases The equilibrium constant expression for this reaction is where Kb is the base-dissociation constant.

  50. 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

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