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Understanding Acid-Base Equilibria: Key Concepts and Calculations in AP Chemistry

This review covers essential concepts of acid-base equilibria as outlined in Chapter 16 of AP Chemistry. It delves into definitions, including Arrhenius and Brønsted-Lowry theories, and the properties of strong and weak acids and bases. Key highlights include ionization, the importance of conjugate acids and bases, and equilibrium principles. The text also discusses water's role in acid-base reactions, the pH scale, and practical measurement techniques, providing a comprehensive understanding necessary for tackling associated homework problems.

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Understanding Acid-Base Equilibria: Key Concepts and Calculations in AP Chemistry

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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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