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Understanding Acid-Base Equilibria: Brønsted-Lowry Concept in Chemistry

Explore the fundamentals of acid-base equilibria with the Brønsted-Lowry concept, including water and the pH scale, strengths of acids and bases, and equilibrium involving weak acids and bases. Learn about buffer solutions, acid-base titration curves, and Lewis acids and bases.

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Understanding Acid-Base Equilibria: Brønsted-Lowry Concept in Chemistry

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  1. Chapter 8Acid-Base Equilibria • 8-1 Brønsted-Lowry Acids and Bases • 8-2 Water and the pH Scale • 8-3 The Strengths of Acids and Bases • 8-4 Equilibria Involving Weak Acids and Bases • 8-5 Buffer Solutions • 8-6 Acid-Base Titration Curves • 8-7 Polyprotic Acids • 8-8 Lewis Acids and Bases OFB Chapter 8

  2. Acid and Base Definitions • Arrhenius (Section 4.3) • Acids are H+ donors • Bases are OH-donors • Broadened Definition (Section 4.3) • Acids are substances that increase [H+] • Bases are substances that increase [OH-] • Brønsted-Lowry (Section 8.1) • Lewis (Section 8.8) OFB Chapter 8

  3. Chapter 8Acid-Base Equilibria • Brønsted-Lowry • Acids • Bases • Conjugate Base - • Conjugate Acid OFB Chapter 8

  4. Acid-Base Equilibria Brønsted-Lowry Acids and Bases A Brønsted-Lowry acid is a substance that can donate a hydrogen ion. A Brønsted-Lowry base is a substance that can accept a hydrogen ion. In the Brønsted-Lowry Acid and Base concept, acids and bases occur as conjugate acid-base pairs. OFB Chapter 8

  5. Conjugate Base - subtract an H+ from the acid • Conjugate Acid add H+ to the base • Examples • is the conjugate base ofH2O • is the conjugated base of H3O+(called the hydronium ion) • is the conjugated acid of OH- • (or shown as H+) is the conjugate acid of H2O OFB Chapter 8

  6. H2O + H+ - H+ conjugate base of H2O Hydronium ion conjugate acid of H2O Hydroxide ion OH- H3O+ + H+ - H+ H2O Water is the conjugate acid of OH- Water is the conjugate base of H3O+ • Conjugate Base - subtract an H+ from the acid • Conjugate Acid add H+ to the base OFB Chapter 8

  7. Point of View #1 CH3CO2H + H2O ↔ H3O+ + CH3CO2- acid base Conjugate acid of H2O Conjugate base of CH3CO2H Point of View #2 CH3CO2H + H2O ↔ H3O+ + CH3CO2- Conjugate acid of CH3CO2- Conjugate base of H3O+ acid base CH3CO2H + H2O ↔ H3O+ + CH3CO2- Acetic Acid Acetate Ion Pairs 1 2 2 1 acid base acid base OFB Chapter 8

  8. Point of View #1 H2O + H2O ↔ H3O+ + OH- # 1 # 2 acid base Conjugate acid of H2O #2 Conjugate base of H2O #1 Point of View #2 H2O + H2O ↔ H3O+ + OH- Conjugate acid of OH- Conjugate base of H3O+ acid base Autoionization of H2O H2O + H2O ↔ H3O+ + OH- Pairs 1 2 2 1 acid base acid base OFB Chapter 8

  9. Exercise 8-1: Trimethylamine (C3H9N) is a soluble weak base with a foul odor (it contributes to the smell of rotten fish). Write the formula of its conjugate acid. “… the formula of a conjugate acid is obtained by adding H+ to the formula of the base.” OFB Chapter 8

  10. Nomenclature When H+ is hydrated it is H3O+ and called a hydronium ion. Often H3O+ is written in a simpler notation H+ + 111.7° OFB Chapter 8

  11. Amphoterism - an ion or molecule can act as an acid or base depending upon the reaction conditions 1.) Water in NH3 serves as an acid H2O + NH3↔ NH4+ + OH- 2.) Water in acetic acid serves as a base H2O + CH3CO2H ↔ H3O+ + CH3CO2- OFB Chapter 8

  12. 3.) Acetic Acid is also amphoteric, if in the presence of a strong acid serves as a base H2SO4 + CH3CO2H ↔ CH3CO2H2+ + HSO4- Amphoterism - an ion or molecule can act as an acid or base depending upon the reaction conditions OFB Chapter 8

  13. [H3O+][OH-] [H2O]2 Water Autoionization of water: 2 H2O(l)  H3O+(aq) + OH-(aq) KW = 1.0  10-14(at 25oC) OFB Chapter 8

  14. Strong Acids and Bases A strong acid is one that reacts essentially completely with water to produce H3O+(aq). Hydrochloric acid (HCl) is a strong acid: HCl (aq) + H2O(l)  H3O+(aq) + Cl-(aq) (reaction essentially complete) Dissolving 0.10 mol of HCl in enough water to make 1.0 L of solution gives a final concentration of 0.10 M for H3O+(aq). [H3O+][OH-] = KW OFB Chapter 8

  15. A strong base is one that reacts essentially completely with water to produce OH-(aq) ions. Strong Acids and Bases Sodium hydroxide (NaOH) is a strong base: Others are NH2- (amide ion) and H- (Hydride ion) NaOH(s)  Na+(aq) + OH-(aq) (reaction essentially complete) Dissolving 0.10 mol of NaOH in enough water to make 1.0 L of solution gives a final concentration of 0.10 M for OH-(aq). [H3O+][OH-] = KW OFB Chapter 8

  16. The pH Function pH = -log10[H3O+] pH < 7 acidic solution [H3O+] > [OH-] pH = 7 neutral solution [H3O+] = [OH-] pH > 7 basic solution [H3O+] < [OH-] OFB Chapter 8

  17. The pH Function EXAMPLE 8-3 Calculate the pH (at 25oC) of an aqueous solution that has an OH-(aq) concentration of 1.2  10-6M. OFB Chapter 8

  18. pH = -log10[H3O+] If converting from pH to [H+] [H+] =10 -pH = 10 -X OFB Chapter 8

  19. Recall [H3O+][OH-] = KW = 10 -14 pH + pOH = pKw = 14 OFB Chapter 8

  20. The pH Function Exercise page 8-4: Calculate [H3O+] and [OH-] in saliva that has a pH of 6.60 at 25oC. OFB Chapter 8

  21. Strengths of Acids and Bases • Strong acids (completely ionized in water, i.e., 100%) • HI • HBr • HClO4 (perchloric acid) • HCl • HClO3 (chloric acid) • H2SO4 • HNO3 HA + H2O ↔ H3O+ + A – HA = generic acid OFB Chapter 8

  22. Table 8-2 Scale of pKa -11 To +14 Known Scale of pKa -11 To +50 E.g., Acetone = 19, methane = 48, ethane = 50 OFB Chapter 8

  23. OFB Chapter 8

  24. Strengths of Acids and Bases B + H2O ↔ BH+ + OH- B = Base Kb is the Basicity constant OFB Chapter 8

  25. B + H2O ↔ BH+ + OH- B = Base Kb is the Basicity constant BH+ + H2O ↔ B + H3O+ B = Base Ka is the Acidity constant OFB Chapter 8

  26. B + H2O ↔ BH+ + OH- B = Base BH+ + H2O ↔ B + H3O+ B = Base OFB Chapter 8

  27. Chapter 8Acid-Base Equilibria • Base Strength • strong acids have weak conjugate bases • weak acids have strong conjugate bases The strength of a base is inversely related to the strength of its conjugate acid; the weaker the acid, the stronger its conjugate base, and vice versa pKa + pKb = pKw OFB Chapter 8

  28. [H3O+][In-] = Ka [HIn] Indicators (denoted by In) A soluble compound, generally an organic dye, that changes its color noticeably over a fairly short range of pH. Typically, a weak organic acid that has a different color than its conjugate base. HIn(aq) + H2O(l) H3O+(aq) + In-(aq) OFB Chapter 8

  29. OFB Chapter 8

  30. Methyl Red Bromothymol blue Phenolphtalein OFB Chapter 8

  31. OFB Chapter 8

  32. Equilibria Involving Weak Acids and Bases Weak acids Ka < 1 i.e., pKa > 0 H3O+ (hydronium ion) Ka = 1 pKa =0 HA + H2O↔ H3O+ + A- HA is a weak acid, Ka < 1 or pKa > 1 OFB Chapter 8

  33. OFB Chapter 8

  34. OFB Chapter 8

  35. Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium. Assume 1.0M CH3CO2H initially CH3CO2H + H2O ↔ H3O+ + CH3CO2- Init. conc. ∆ conc. Equil. conc. OFB Chapter 8

  36. A trick to solving: Assume that y is small (less than 5% of the initial conc.) OFB Chapter 8

  37. Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium. Assume 1.0M CH3CO2H initially CH3CO2H + H2O ↔ H3O+ + CH3CO2- (b) Fraction CH3CO2H ionized at equilibrium OFB Chapter 8

  38. Weak Bases Similarly NH3 acts as a weak base in H2O Kb = 1.8 x 10-5 H20 + NH3↔ NH4+ + OH- acid1base2acid2base1 1 OFB Chapter 8

  39. As before, you can calculate a pH at equilibrium. Assume 0.01M NH3 initially. Calculate the pH of the resulting solution H20 + NH3↔ NH4+ + OH- Init. conc. ∆ conc. Equil. conc. OFB Chapter 8

  40. As before, you can calculate a pH at equilibrium. Assume 0.01M NH3 initially. Calculate the pH of the resulting solution OFB Chapter 8

  41. Hydrolysis is a term applied to reactions of aquated ions that change the pH from 7 • When NaCl is placed in water, the resulting solution is observed to be neutral (pH = 7) • However when sodium acetate (NaC2H3O2) is dissolved in water the resulting solution is basic • Other salts behave similarly, NH4Cl and AlCl3 give acid solutions. • These interactions between salts and water are called hydrolysis OFB Chapter 8

  42. Despite the special term, hydrolysis, there is no reason to treat hydrolysis in a special manner. It is still a Brønsted-Lowry Acid and Base Reaction OFB Chapter 8

  43. Example problem: Suppose a 0.1 mole solution sodium acetate is dissolved in 1 liter of water. What is the pH of the solution? Init. conc. ∆ conc. Equil. conc. • Find Kb • Find [OH-] • Find [H+] • Find pH OFB Chapter 8

  44. Example problem: What is the pH of the solution? CH3CO2- + H2O ↔ CH3CO2H + OH- Init. conc. 0.1M 0 ~0 ∆ conc. - y + y + y Equil. conc. 0.1– y y y Ka x Kb = Kw • Find Kb • Find [OH-] • Find [H+] • Find pH OFB Chapter 8

  45. Non-Hyrolyzed Ions (a few) 7 Anions, not hydrolyzed Cl -, Br -, I -, HSO4-, NO3-, ClO3-, ClO4- 10 Cations, not hydrolyzed Li+, Na+, K+, Rb+, Sc+, Mg++, Ca++, Sr++, Ba++, Ag+ OFB Chapter 8

  46. Can predict pH of some salts (relative pH) Na3PO4 is basic (a non hydrolyzed cation and a hydrolyzed anion) FeCl3 is acidic (a hydrolyzed cation and a non hydrolyzed anion) OFB Chapter 8

  47. Chapter 8Acid-Base Equilibria • Buffer Solutions: important in biochemical and physiological processes • Organisms (and humans) have built-in buffers to protect them against large changes in pH. • Buffers any solutions that maintain an approximately constant pH despite small additions of acids or bases OFB Chapter 8

  48. Buffers any solutions that maintain an approximately constant pH despite small additions of acids or bases Human blood (pH=7.4) is maintained by a combination of CO3-2, PO4-3 and protein buffers, which accept H+ Death = 7.0 < pH > 7.8 = Death How Do Buffers Work? OFB Chapter 8

  49. How Do Buffers Work? HA + H2O ↔ H3O+ + A – HA = generic acid • [H+] depends on Ka and the ratio of acid to salt. • Thus if both conc. HA and A- are large then small addictions of acid or base don’t change the ratio much OFB Chapter 8

  50. HA + H2O ↔ H3O+ + A – • H+ depends on Ka and the ratio of acid to salt. • Thus if both conc. HA and A- are large then small addictions of acid or base don’t change the ratio much Try Example 8-11 OFB Chapter 8

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