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Chapter 12 Acids and Bases

Chapter 12 Acids and Bases. 12.1 The Nature of Acids and Bases 12.2 Acid Strength 12.3 The pH Scale 12.4 Calculating the pH of Strong Acid Solutions 12.5 Calculating the pH of Weak Acid Solutions 12.6 Bases 12.7 Polyprotic Acids 12.8 Acid-Base Properties of Salts

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Chapter 12 Acids and Bases

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  1. Chapter 12Acids and Bases 12.1 The Nature of Acids and Bases 12.2 Acid Strength 12.3 The pH Scale 12.4 Calculating the pH of Strong Acid Solutions 12.5 Calculating the pH of Weak Acid Solutions 12.6 Bases 12.7 Polyprotic Acids 12.8 Acid-Base Properties of Salts 12.9 Acid Solutions in Which Water Contributes to the H+ Concentration (skip) 12.10 Strong Acid Solutions in Which Water Contributes to the H+ Concentration (skip) 12.11 Strategy for Solving Acid-Base Problems: A Summary

  2. Acids and Bases and Their Reactions Definitions • Arrhenius Acids and Bases Acids are • H+ donors • Bases are OH- donors • Arrhenius Broadened Definition • Acids increase H+ concentration or [H+] increases • Bases increase OH- concentration or [OH-] increases • Brønsted-Lowry Acids and Bases (1923) • Acids donate H+ • Bases accept H+ Arrhenius 1903 Nobel Prize

  3. Acids-Bases Brønsted-Lowry Acids and Bases A Brønsted-Lowry acid is a substance that can donate a hydrogen ion (aka H+, proton). A Brønsted-Lowry base is a substance that can accept a hydrogen ion. Acids and bases occur as conjugate acid - base pairs.

  4. Conjugate Base - subtract an H+ from the acid • Conjugate Acid add H+ to the base • Examples • OH– is the conjugate base of • H2O is the conjugated base of • H2O is the conjugated acid of • H3O+(or often shown as H+) is the conjugate acid of

  5. Point of View #1 CH3CO2H + H2O H3O+ + CH3CO2- Conjugate base of CH3CO2H acid base Conjugate acid of H2O Point of View #2 CH3CO2H + H2O H3O+ + CH3CO2- Conjugate acid of CH3CO2- acid base Conjugate base of H3O+ CH3CO2H + H2O H3O+ + CH3CO2- Acetic Acid Acetate Ion Pairs 1 2 2 1 acid base acid base

  6. Nomenclature When H+ is hydrated, it is H3O+ and is called a hydronium ion. A hydronium ion has the same molecular geometry as NH3. + 111.7°

  7. HA + H2O H3O+ + A – HA = generic acid • There is Competition for the proton between two bases H2O and A– • If H2O is a much stronger base than A– the equilibrium lies far to the right. • If A– is a much stronger base than H2O the equilibrium lies far to the left.

  8. Acid Strength: graphical representation of the behavior of acids of different strengths in aqueous solution. A strong acid: equilibrium lies far to the right HA + H2O H3O+ + A – A weak acid: equilibrium lies far to the left HA + H2O H3O+ + A – A weak acid yields a relatively strong conjugate base

  9. Relationship of acid strength and conjugate base strength HA + H2O H3O+ + A-

  10. HA + H2O H3O+ + A– SAME AS HA H+ + A–

  11. Ka is the Acidity constant

  12. Point of View #1 H2O + H2O H3O+ + OH- # 1 # 2 Conjugate base of H2O #1 Conjugate acid of H2O #2 acid base Point of View #2 H2O + H2O H3O+ + OH- Conjugate acid of OH- base Conjugate base of H3O+ acid Autoionization of H2O H2O + H2O H3O+ + OH- Pairs 1 221 base acid acid base Amphoterism - an ion or molecule can act as an acid or base depending upon the reaction conditions

  13. 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- acid base acid base 2.) Water in acetic acid serves as a base H2O + CH3CO2H H3O+ + CH3CO2- base acid acid base Conjugate base of acetic acid Conjugate acid of H2O

  14. [H3O+][OH-] [H2O(l)]2 Water as an Acid and a Base Autoionization of water: 2 H2O (l) H3O+ (aq) + OH- (aq) Kw = [H3O+][OH-] = [H+][OH-] Kw = 1.0 x 10-14(at 25oC) In pure water [H+] = [OH-] Kw = [H3O+][OH-]

  15. NH4+H+ + NH3 Ka/Kb/Kw NH3 + H2O OH- + NH4+

  16. NH4+H+ + NH3 Ka/Kb/Kw NH3 + H2O OH- + NH4+ H2O OH- + H+

  17. The pH Scale pH = -log10[H3O+] SAME AS pH = -log10[H+] pH < 7 acidic solution [H3O+] > [OH-] pH = 7 neutral solution [H3O+] = [OH-] pH > 7 basic solution [H3O+] < [OH-] Sig figs: for logs: the number of decimal places in the log is equal to the number of sig figs in the original number

  18. Calculate the pH of Strong Acid-Base Solutions pH = -log10[H3O+] EXAMPLE Calculate the pH (at 25oC) of an aqueous solution that has an OH-(aq) concentration of 1.2 x 10-6M (i.e., mol/liter) Solution The concentration of H+ (aq) is [H+][OH-] = KW [H+] = KW/[OH-] = 10-14/1.2x10-6 = 8.3 x 10-9 pH = -log[8.3 x 10-9] = 8.1

  19. Strong Acids A strong acid is one that dissociates completely in water to produce H+(aq). E.g., Hydrochloric acid (HCl) is a strong acid: HCl (aq) → H+ (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 H+(aq). H2O (l) H+ (aq) + OH- (aq) [H+][OH-] = Kw [OH-] = Kw / [H+] = 10-14/10-1 = 10-13

  20. Weak Acids Write the major species in solution and Ka values. Which is dominant? #1: HA H+ + A – #2: H2O H+ + OH – Compare the value for Ka and Kw. Which is larger? This is the dominate source of H+

  21. Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium. The concentration of CH3CO2H is 1 M (initial, or total). The Ka for acetic acid is 1.8 x 10-5 Estimate major species in solution CH3CO2H (a weak acid) andH2O. CH3CO2H H+ + CH3CO2─ Ka = 1.8 x 10-5 H2O H+ + OH – Kw = 1.0 x 10-14

  22. Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium. CH3CO2H H+ + CH3CO2─ Initial 1.0M ~ 0 0 Change - y + y + y Equilibrium 1.0 – y y y

  23. Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium. CH3CO2H H+ + CH3CO2─ Initial 1.0M ~ 0 0 Change - y + y + y Equilibrium 1.0 – y y y

  24. Problem: (a) Calculate pH and (b) the fraction of CH3CO2H ionized at equilibrium. CH3CO2H H+ + CH3CO2─ Initial 1.0M ~ 0 0 Change - y + y + y Equilibrium 1.0 – y y y

  25. Effect of dilution on the % dissociation and [H+] and pH HA H+ + A – Le Chatelier’s Priniciple: If a chemical system at equilibrium experiences a change in concentration (or temperature, volume, or partial pressure) then the system shifts to counteract the imposed change. The more dilute the weak acid solution, the greater the percent dissociation

  26. Strong BasesA strong base reacts completely with water to produce OH-(aq) ions. Strong Acids and Bases Sodium hydroxide (NaOH) is a strong base: Others are KOH–, 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). From this you can calculate pH and pOH. [H+][OH-] = Kw [H+] = Kw/[OH-] =10-14/10-1 = 10-13 [OH-] = 10-1 [H+] = 10-13pOH = 1 pH = 13

  27. Weak Bases B (aq) + H2O (l) BH+(aq) + OH–(aq) B = Base Kb is the basicity constant Kb = [BH+][OH–]/[B]

  28. Calculations for solutions of weak bases are similar to those for weak acids. Bases (B) compete with OH–, a very strong base, for H+ ions. B (aq) + H2O (l) BH+(aq) + OH–(aq)

  29. Acid-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 • This equation applies to an acid and its conjugate base.

  30. Weak Bases NH3 is a base Kb = 1.8 x 10-5 NH3 + H20 NH4+ + OH-acid1 base2acid2base1 NH4+is an acid Ka = 5.6 x 10-10 NH4+ NH3 + H+acid1base1

  31. Weak Bases NH3 is a base Kb = 1.8 x 10-5 NH3 + H20 NH4+ + OH- NH4+is an acid Ka = 5.6 x 10-10 NH4+ NH3 + H+ Ka Kb = Kw = (5.6 x 10-10) (1.8 x 10-5) = 10-14 pKa + pKb = pKw = (-9.25) + (-4.75) = -14

  32. Weak Bases with Weak Acids NH4+is an acid Ka(am) = 5.6 x 10-10 NH4+ NH3 + H+ CH3COOH is an acid Ka(aa) = 1.8 x 10-5 CH3COOH CH3COO- + H+ change the direction of the aa reaction and add them together NH4+ + CH3COO- NH3 + CH3COOH K = Ka(am)/Ka(aa) = 5.6 x 10-10/1.8 x 10-5 =3.1 x 10-5

  33. Assuming 0.1M NH3 initial, calculate the pH of the resulting solution H20 + NH3 NH4+ + OH-

  34. Assuming 0.1M NH3 initial, calculate the pH of the resulting solution H20 + NH3 NH4+ + OH- Init. conc. 0.1M 0 ~0 Change - y + y + y Equil. conc. 0.1– y y y

  35. Assuming 0.1M NH3 initial, calculate the pH of the resulting solution H20 + NH3 NH4+ + OH- Init. conc. 0.1M 0 ~0 Change - y + y + y Equil. conc. 0.1– y y y [H+] = Kw/[OH-] = 10-14/y pH=-log[H+]

  36. It is always easier to remove the first proton in a     polyproticacid than the second.That is, Ka1 > Ka2 > Ka3 Polyprotic Acids Note: Sulfuric acid is a strong acid in its first dissociation step and a weak acid in its second step. H2SO4 (aq) →H+ (aq) + HSO4– (aq) Ka1 ≈ 1.0x102 HSO4– (aq) H+ (aq) + SO42– (aq) Ka2 ≈ 1.2x10-2

  37. Polyprotic Acids • Polyprotic acids have more than one ionizable proton. • The protons are removed in steps, not all at once. • The extent of the steps can be calculated sequentially. • From the successive acidity constants, the equilibrium concentrations of all species present can be calculated at any value of the pH. H2SO3 (aq) + H2O (l) → H3O+ (aq) + HSO3- (aq) Ka1 = 103 HSO3- (aq) + H2O (l) H3O+ (aq) + SO32- (aq) Ka2 = 1.2 x 10-2

  38. Polyprotic Acids • pH calculations for solutions of Polyprotic acids appear complicated, most common cases (with weak acids) are surprising straightforward. • Setup equations, K values, set up ICE type table, assume dissociation is small (check assumption) and solve. • For typical weak Polyprotic acids. Ka1 >> Ka2 >> Ka3 • The first dissociation step dominates the H+ concentration and thus the pH

  39. A plot of the fractions of H2CO3, HCO-3 and CO32- At pH = 9.00H2CO3 ≈ 0%,HCO-3 = 95% and CO32- = 5% At pH = 10.00H2CO3 ≈ 0%,HCO-3 = 68% and CO32- = 32%

  40. Acid-Base Properties of Salts “Hydrolysis”

  41. Hydrolysis is a Brønsted-Lowry Acid and Base Reaction Anion: A─ + H2O ↔ HA + OH─ Cation: M+2 + H2O ↔ H3O+ + M(OH)+ E.g., Cations Ni+2 and Fe+2

  42. 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

  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? CH3CO2– + H2O ↔ CH3CO2H + OH– base acid acid base Init. conc. 0.1M 0 ~0 ∆ conc. - y + y + y Equil. conc. 0.1– y y y • Find Kb • Find [OH-] • Find [H+] • Find pH

  44. Find Kb • Find [OH-] • Find [H+] • Find pH 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

  45. 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(Problem: found at 0.1M NaAc, pH =8.89) • a non-hydrolyzed cation (Na+) • a hydrolyzed anion (acetate ion) • Other salts behave similarly, NH4Cl and AlCl3 give acid solutions. • a non-hydrolyzed anion (Cl-) • a hydrolyzed cation (NH4+ or Al+3) • These interactions between salts and water are called hydrolysis • Hydrolysis is a term applied to reactions of aquated ions that change the pH from 7

  46. 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+

  47. Predict pH of salts in water (relative pH) Na3PO4 is basic (raised pH) (a non hydrolyzed cation and a hydrolyzed anion) Na3PO4? FeCl3 is acidic (lowers pH) (a hydrolyzed cation and a non hydrolyzed anion) FeCl3?

  48. Summary Acid-Base Properties of Salts “Hydrolysis”

  49. Summary • Acid-Base Equilibrium Problems • Which major species are present • Does a reaction occur that can be assumed to go to completion? • Which equilibrium dominates the solution? • Set up ICE type table • Solve for equilibrium concentrations using know K values • Check any simplifying assumptions • Typically solve for pH or % species in solution

  50. Chapter 7Acids and Bases 7.1 The Nature of Acids and Bases 7.2 Acid Strength 7.3 The pH Scale 7.4 Calculating the pH of Strong Acid Solutions 7.5 Calculating the pH of Weak Acid Solutions 7.6 Bases 7.7 Polyprotic Acids 7.8 Acid-Base Properties of Salts 7.9 Acid Solutions in Which Water Contributes to the H+ Concentration (skip) 7.10 Strong Acid Solutions in Which Water Contributes to the H+ Concentration (skip) 7.11 Strategy for Solving Acid-Base Problems: A Summary

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