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AP Chemistry – Chapter 16

AP Chemistry – Chapter 16. Acid and Base Equilibrium HW:3 6 17 19 25 37 47 53 57 59 73 81 89 93 101. 16.1 – Brief Review. Arrhenius Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions.

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AP Chemistry – Chapter 16

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  1. AP Chemistry – Chapter 16 Acid and Base Equilibrium HW:3 6 17 19 25 37 47 53 57 59 73 81 89 93 101

  2. 16.1 – Brief Review • Arrhenius • Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions. • Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions.

  3. 16.2 – Bronsted-Lowry • Brønsted–Lowry • Acid: Proton donor • Base: Proton acceptor

  4. A Brønsted–Lowry acid… …must have a removable (acidic) proton. A Brønsted–Lowry base… …must have a pair of nonbonding electrons.

  5. If it can be either… ...it is amphoteric. HCO3− HSO4− H2O Remove H+ to form: Add H+ to form:

  6. 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.

  7. Relative Strengths of Acids and Bases • Strong Acid = Ionizes “completely” • Hydrochloric • Hydrobromic • Hydroiodic • Nitric • Chloric • Perchloric • Sulfuric (the first proton only) • Reactions:

  8. Strengths of Acids and Bases • Weak Acid = Ionizes only a limited amount • Hydrofluoric • Acetic • Ammonium ion • Reaction: • At equilibrium:

  9. Strengths of Acids and Bases • Strong Base = Ionizes “completely” • Metal hydroxides • Reactions:

  10. Strengths of Acids and Bases • Weak Base = Ionizes only a limited amount • Ammonia • Reactions: • At equilibrium:

  11. 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.

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

  13. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl(aq) + H2O(l) H3O+(aq) + Cl−(aq) H2O is a much stronger base than Cl−, so the equilibrium lies so far to the right K is not measured (K>>1).

  14. C2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Acid and Base Strength Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1).

  15. General Rules • If an acid is strong, the conjugate base is very weak • Hydronium is the strongest acid that can exist in aqueous solution • Hydroxide is the strongest base that can exist in aqueous solution

  16. 16.3 – Acid and Base Properties of Water In the presence of an acid: • 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.

  17. 16.3 - Water In the presence of a base: Water can act as an acid and add a proton to a base NH3 + H2O -> NH4+1 +OH-

  18. 16.3 - Water • Water self-ionizes: 2 H2O (l)  H3O+ (aq) + OH− (aq) OR H2O (l)  H+ (aq) + OH− (aq) • Ion-Product of Water: Kc = [H+][OH-] [H2O] Kw = [H+][OH-] (because [water] does not change)

  19. 16.3 - Water • For any aqueous solution at 25oC, Kw = 1 x 10-14 = [H+][OH-] • When NEUTRAL WATER [H+] = [OH-] = 1 x 10-7 • When acid or base – when one varies the other will by the Kw equation If [H+] > [OH-] = Acidic If [OH-]>[H+] = Basic

  20. Example: [H+] = 1.0 x 10-6 M Calculate hydroxide concentration

  21. 16.4 – The pH Scale • More practical than concentration due to small values usually seen in concentrations. pH = −log [H3O+] (Typical range 1 – 14, but can be outside of this)

  22. pH • Pure water, pH = −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. • Neutral solution has pH  7.0

  23. pH These are the pH values for several common substances.

  24. Other “p” Scales • The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions). • Some similar examples are • pOH −log [OH−] • pKw−log Kw

  25. Because [H3O+] [OH−] = Kw = 1.0  10−14, −log [H3O+] + −log [OH−] = −log Kw = 14.00 or, in other words, pH + pOH = pKw = 14.00 Algebra

  26. Examples: • 16.6 • 16.7

  27. How Do We Measure pH? • For less accurate measurements, one can use • Litmus paper • “Red” paper turns blue above ~pH = 8 • “Blue” paper turns red below ~pH = 5 • An indicator

  28. How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

  29. 16.5 - Calculations using Strong Acid and Strong Base • Monoprotic STRONG ACIDS: • NOT Equilibrium work! • Break apart completely Example – 0.001 M HCl

  30. 16.5 - Calculations using Strong Acid and Strong Base • STRONG BASES: • NOT Equilibrium work! • Break apart completely Example – 0.02 M Ba(OH)2

  31. Kc = [H3O+] [A−] [HA] HA(aq) + H2O(l) A−(aq) + H3O+(aq) 16.6 – Weak Acids and Ionization Constants • For a generalized acid dissociation, the equilibrium expression would be • This equilibrium constant is called the acid-dissociation constant, Ka.

  32. Dissociation Constants The greater the value of Ka, the stronger the acid.

  33. ICE Example • Calculate the equilibrium concentrations and pH of a 0.5 M HF solution. Option instead of quadratic equation…

  34. Accuracy of the approximation: =([H+] / [ ] initial ) x 100 Must be 5% or less! If not, redo with the quadratic equation.

  35. [H3O+] [COOH−] [HCOOH] Ka = Calculating Ka from the pH Example: 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. • We know that

  36. 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 the equilibrium concentrations of all three things. • We can find [H3O+], which is the same as [HCOO−], from the pH.

  37. Calculating Ka from the pH pH = −log [H3O+] 2.38 = −log [H3O+] −2.38 = log [H3O+] 10−2.38 = 10log [H3O+] = [H3O+] 4.2  10−3 = [H3O+] = [HCOO−]

  38. Calculating Ka from pH Now we can set up a table…

  39. [4.2  10−3] [4.2  10−3] [0.10] Ka = Calculating Ka from pH = 1.8  10−4

  40. [H3O+]eq [HA]initial Calculating Percent Ionization • Percent Ionization =  100 • In this example [H3O+]eq = 4.2  10−3 M [HCOOH]initial = 0.10 M • A more dilute solution would have a higher % ionization, due to LeChatlier’s Principle and a shift to MORE ionization to counteract the low concentration.

  41. 4.2  10−3 0.10 Calculating Percent Ionization Percent Ionization =  100 = 4.2%

  42. Calculating pH from Ka Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25°C. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Ka for acetic acid at 25°C is 1.8  10−5.

  43. [H3O+] [C2H3O2−] [HC2H3O2] Ka = Calculating pH from Ka The equilibrium constant expression is

  44. Calculating pH from Ka We next set up a table… We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

  45. 1.8  10−5 = (x)2 (0.30) Calculating pH from Ka Now, (1.8  10−5) (0.30) = x2 5.4  10−6 = x2 2.3  10−3 = x

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

  47. PolyproticAcids • Have more than one acidic proton. • Ionize one proton at a time. • 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. • [CB] of a second step is numerically equal to Ka2.

  48. Example –Example 16.14

  49. Example – Practice Exercise 16.14(if needed)

  50. 16.7 - Weak Bases Bases react with water to produce hydroxide ion. Kb =

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