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Unit 4-Systems and Equilibrium Acid-Base Equilibrium

Unit 4-Systems and Equilibrium Acid-Base Equilibrium. Properties of Acids and Bases. Which one will destroy the coke can faster, the acid or the base? http://youtu.be/WnPrtYUKke8 Acids and Bases have many similar properties, including the fact that they can both be dangerous.

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Unit 4-Systems and Equilibrium Acid-Base Equilibrium

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  1. Unit 4-Systems and EquilibriumAcid-Base Equilibrium

  2. Properties of Acids and Bases • Which one will destroy the coke can faster, the acid or the base? • http://youtu.be/WnPrtYUKke8 • Acids and Bases have many similar properties, including the fact that they can both be dangerous

  3. Properties of Acids and Bases Produce H3O+ in solution Produce OH- in solution React with bases to form a salt and water React with acids to form a salt and a water Have a pH less than 7 Have a pH greater than 7 Are Electrolytes Are Electrolytes O-----H Hydroxide Ion OH-

  4. Acid-Base Definitions Definition 1- Arrhenius Theory of Acids and Bases “An acid is a substance that produces H3O+ in water, a base is a substance that produces OH- in water” Problem: there are compounds that are neither oxides nor hydroxides but are still bases e.g. Na2CO3(aq) and NH3(aq)

  5. Acid-Base Definitions Definition 2- Bronsted-Lowry Theory of Acids and Bases “An acid is a proton donor, a base is a proton acceptor” “Proton” = hydrogen without its electron

  6. Acid-Base Definitions Definition 2- Bronsted-Lowry (B-L) Theory of Acids and Bases “An acid is a proton donor, a base is a proton acceptor” • B-L Theory defines acids and bases in terms of the effect the substance has on the chemical reaction • NOT defined in terms of general properties of the substance itself • This means that a compound can be either an acid or a base depending on how it affects the chemical system it is in • Compound that can act as either acids or bases are called amphoteric (amphiprotic) compounds

  7. Acid-Base Definitions • Amphoteric Example: H2O Acid BaseConjugateConjugate Acid Base Base Acid Conjugate Conjugate Acid Base

  8. Conjugate Acids and Bases Conjugate Pairs differ by one hydrogen!

  9. Conjugate Acids and Bases Label each conjugate acid/base pair • HClO4(aq) + H2O(l)⇄ H3O+(aq) + ClO4–(aq) Acid BaseConjugateConjugate Acid Base • HF(aq) + HSO3–(aq) ⇄ F–(aq) + H2SO3(aq) Acid Base Conjugate Conjugate Base Acid

  10. Strength of Conjugate Acids and Bases • A strong acid has a weak conjugate base • A strong base has a weak conjugate acid

  11. Acid-Base Definitions • B-L Theory does not account for all substances, including those that form complexes in water • Examples: Al+3 and BF3 Definition 3- Lewis Theory of Acids and Bases “An acid is an electron-pair acceptor, a base is an electron-pair donor” • Note: In order for a substance to be a Lewis base, it must have a lone electron pair to donate!

  12. Lewis Theory of Acids and Bases • Example 1: Lewis Acid Lewis Base Adduct • A bond is formed between Nitrogen and Boron, but it is the Nitrogen atom originating on the Lewis base that contributes (donates) to the bond

  13. Lewis Theory of Acids and Bases • Example 2: Autoionization of water Lewis Lewis Hydroxide Hydronium Acid Base Ion Ion • A bond is formed between Oxygen and Hydrogen, but it is the Oxygen atom originating on the Lewis base that contributes (donates) to the bond • H2O can act as both and acid and a base!

  14. [C]c[D]d [A]a[B]b More About Water • In a solution of pure water, the forward reaction, autoionization of water, is favoured. ⇄ H20(l) H20(l) OH-(aq) H30+ (aq) Memorize this! Kw= 1 x 10-14 @ 250C [OH-][H3O+] Keq= Kw= Equilibrium Constant for water

  15. Water and pH Memorize this! Taking the –log of both sides gives us 14 = -log[OH-] + -log[H3O+] 14 = pOH + pH Kw= 1 x 10-14 @ 250C ∴ @ 250C 1 x 10-14= [OH-][H3O+] Memorize this! [OH-][H3O+] Kw=

  16. pH and the Environment

  17. Acids and Bases-pH • The strength of acids and bases is measured by how much they ionize to produce H+ or OH- • That means the pH is just a way of reporting the strength of an acid or base • pH = 7 neutral • pH < 7 Acidic • pH > 7 Basic ….why?

  18. Acids and Bases-pH • pH = 7 neutral pH =-log [H30+] • pH < 7 Acidic • pH > 7 Basic If pH = 7 then 14 = pH + pOH 14 = 7 + pOH 7 = pOH pH = pOH = 7, which means that [H30+] = [OH-]  neutral

  19. Acids and Bases-pH Example: If pH = 5.5 then what is the concentration of H30+? OH-? pH =-log [H30+] -pH = log[H30+] 10-pH = [H30+] (antilog = 10x) 10-5.5 = [H30+] [H30+] = 3.16 * 10-6 mol/L pOH = 14 - 5.5 = 8.5 8.5 = -log[OH-] -8.5 = log[OH-] 10-8.5 = [OH-] [OH-] = 3.16 * 10-9 mol/L ……. <[H3O+] ∴ acidic

  20. pH =-log [H30+] Acids and Bases-pH Your turn: If pH = 9.3 then what is the concentration of H30+? OH-? pH =-log [H30+] -pH = log[H30+] (antilog = 10x) 10-pH = [H30+] 10-9.3 = [H30+] [H30+] = 5.01 * 10-10 mol/L pOH = 14 – 9.3 = 4.7 4.7 = -log[OH-] -4.7 = log[OH-] 10-4.7 = [OH-] [OH-] = 2.00 * 10-5 mol/L ……. > [H3O+] ∴ basic

  21. Strong Acids and Bases • The strength of an acid or base is measured by how many ions it forms in solution, NOT BY ITS CONCENTRATION! • A strong acid is defined as “an acid that undergoes complete dissociation to form ions in solution” • It has nothing to do with the concentration of the acid, only how much of the acid reacts to form H3O+ in solution • A strong base is defined as “a base that undergoes complete dissociation to form ions in solution” • It has nothing to do with the concentration of the base, only how much of the base reacts to form OH- in solution

  22. Strong Acids and Bases • Strong acids and bases are 100% ionized HA(aq) +H2O(l)  H3O+(aq)+ A-(aq) B(aq) + H2O(l)  BH+(aq) + OH-(aq) • There is no equilibrium setup because all strong acids and bases becomes ions • The initial [HA] = [H3O+] (for a 1:1 ratio) • The initial [B] = [OH-] • The pH (and pOH) of strong acids and bases is easy to determine 100% 0%

  23. Strong Acids and BasesFinding pH pH =-log [H30+] • Determine the pH for a 0.15 mol/L solution of HCl pH =-log [H30+] pH =-log [0.15M] pH =-log [-0.82] pH =-(-0.82) pH = 0.82

  24. Strong Acids and BasesFinding pH pH =-log [H30+] • Determine the pH for a 0.0010 mol/L solution of NaOH pOH =-log [OH-] Kw= [OH-][H3O+] [0.0010M][H3O+] 1 x 10-11= 1 x 10-14 = 1 x 10-14= [OH-][H3O+] [H3O+] pOH =-log [0.0010M] OR pOH = 3 14 = pOH + pH pH = 14 - pOH pH = -log [1 x 10-11] pH = 11 pH = 14 - 3 pH = 11

  25. Strong Acids and BasesFinding pH pH =-log [H30+] • What is the pH of a 2.0 x 10-3 M solution of HBr? • HBr is a strong acid100% dissociation pH =-log [H+] pH =-log [0.0020M] pH = 2.7 pH = 14 - pOH

  26. Strong Acids and Bases Versus Weak Acids and Bases • Strong Acids and Bases • Are 100% ionized • There is no equilibrium, only ions • HNO3, HCl, HBr, HI, HClO4, H2SO4strong acids • Hydroxides with groups 1 and 2strong bases • E.g. Ca(OH)2, KOH • Weak Acids and Bases • Do not dissociate completely • There is an equilibrium formed between the ions in solution and molecules of acid/base • The equilibrium constant for acids and bases tells us the ratio of ions to molecules (acid/base)

  27. [A-][H3O+] [HA] Weak Acids and Bases HA(aq) + H2O (l) ⇄ A-(aq)+ H3O+(aq) • A weak acid means that the [H3O+] and [A-] is small compared to the [HA] • Ka < 1 acid Ka= NO H2O(l)

  28. [C2H3O2 -][H3O+] [HC2H3O2] Weak Acids and Bases Example 1: Acetic Acid in Water HC2H3O2(aq) + H2O (l) ⇄C2H3O2-(aq)+ H3O+(aq) Ka= NO H2O(l)

  29. [BH+][OH-] [B] Weak Acids and Bases B(aq) + H2O (l) ⇄ BH+(aq)+ OH-(aq) • A weak base means that the [OH-] and [BH+] is small compared to the [B] • Kb < 1 base Kb= NO H2O(l)

  30. [NH4+][OH-] [NH3] Weak Acids and Bases Example 1: Ammonia in Water NH3(aq) + H2O (l) ⇄ NH4+(aq)+ OH-(aq) Kb= NO H2O(l)

  31. [CH3COO- ][H3O+] [CH3COOH ] Setting Up the Equilibrium for Weak Acids and Bases-Weak Acid Example 1: Acetic Acid in Water What is the pH of a solution of 0.100 mol/L of acetic acid in water? What are the equilibrium concentration of all species in equilibrium in this system? Ka = 1.8 *10-5 Step 1: Balanced Equation CH3COOH(aq) + H2O(l)⇄ H3O+(aq) + CH3COO- (aq) Step 2: Equilibrium Expression Ka= NO H2O(l)

  32. x*x [0.100-x ] Setting Up the Equilibrium for Weak Acids and Bases-Weak Acid Example 1: Acetic Acid in Water CH3COOH(aq) + H2O(l)⇄ H3O+(aq) + CH3COO- (aq) Step 3: I.C.E. Table Step 4: Equilibrium Expression Ka=

  33. x2 [0.100 ] Setting Up the Equilibrium for Weak Acids and Bases-Weak Acid Example 1: Acetic Acid in Water CH3COOH(aq) + H2O(l)⇄ H3O+(aq) + CH3COO- (aq) Step 5: Can we approximate? Since Ka is so small, we can assume that x will also be very small and that 0.100-x ≈ 0.100 Step 6: Solve with approximation and then check to make sure the approximation is valid! 1.8 * 10-5=

  34. x2 [0.100 ] Setting Up the Equilibrium for Weak Acids and Bases-Weak Acid Example 1: Acetic Acid in Water CH3COOH(aq) + H2O(l)⇄ H3O+(aq) + CH3COO- (aq) X = 1.3 * 10-3 mol/L 0.100 – 0.0013 ≈ 0.100 The approximation is valid! [CH3COOH] = 9.9 * 10-2 mol/L [H3O+] = 1.3 *10-3 mol/L [CH3COO-] = 1.3 * 10-3 mol/L 1.8 * 10-5=

  35. Setting Up the Equilibrium for Weak Acids and Bases-Weak Acid Example 1: Acetic Acid in Water CH3COOH(aq) + H2O(l)⇄ H3O+(aq) + CH3COO- (aq) [CH3COOH] = 9.9 * 10-2 mol/L [H3O+] = 1.3 *10-3 mol/L [CH3COO-] = 1.3 * 10-3 mol/L pH = -log[H3O+] = -log[1.3 *10-3 ] = 2.88

  36. [NH4+ ][OH-] [NH3] Setting Up the Equilibrium for Weak Acids and Bases-Weak Base Example 2: Ammonia in Water What is the pH of a solution of 0.100 mol/L of ammonia in water? What are the equilibrium concentration of all species in equilibrium in this system? Kb = 1.8 *10-5 Step 1: Balanced Equation NH3 (aq) + H2O(l)⇄ NH4+(aq) + OH- (aq) Step 2: Equilibrium Expression Kb= NO H2O(l)

  37. x*x [0.100-x ] Setting Up the Equilibrium for Weak Acids and Bases-Weak Base Example 2: Ammonia in Water NH3 (aq) + H2O(l)⇄ NH4+(aq) + OH- (aq) Step 3: I.C.E. Table Step 4: Equilibrium Expression Kb=

  38. x2 [0.100 ] Setting Up the Equilibrium for Weak Acids and Bases-Weak Base Example 2: Ammonia in Water NH3 (aq) + H2O(l)⇄ NH4+(aq) + OH- (aq) Step 5: Can we approximate? Since Kb is so small, we can assume that x will also be very small and that 0.100-x ≈ 0.100 Step 6: Solve with approximation and then check to make sure the approximation is valid! 1.8 * 10-5=

  39. x2 [0.100 ] Setting Up the Equilibrium for Weak Acids and Bases-Weak Base Example 2: Ammonia in Water NH3 (aq) + H2O(l)⇄ NH4+(aq) + OH- (aq) X = 1.3 * 10-3 mol/L 0.100 – 0.0013 ≈ 0.100 The approximation is valid! [NH3] = 9.9 * 10-2 mol/L [NH4+] = 1.3 *10-3 mol/L [OH-] = 1.3 * 10-3 mol/L 1.8 * 10-5=

  40. Setting Up the Equilibrium for Weak Acids and Bases-Weak Base Example 2: Ammonia in Water NH3 (aq) + H2O(l)⇄ NH4+(aq) + OH- (aq) [NH3] = 9.9 * 10-2 mol/L [NH4+] = 1.3 *10-3 mol/L [OH-] = 1.3 * 10-3 mol/L pOH = -log[OH-] = -log[1.3 *10-3 ] = 2.88 pH + pOH = 14 14- pOH = pH pH = 11.12

  41. Equilibrium for Weak Acids and BasesPractice, Practice, Practice! Weak Acid Problems What is the pH of the resulting solution when 0.500 moles of acetic acid are dissolved in water and diluted to 1.00 L?Unbalanced reaction: CH3COOH(aq)  H+(aq) + CH3COO-(aq)     Ka = 1.8x10-5 at 25oC. What is the pH of the resulting solution when 100.0 mL of 0.500 M of hypochlorous acid is mixed with 100.0 mL water?Unbalanced reaction: HOCl(aq)  H+(aq) + OCl-(aq)     Ka = 3.0x10-8 at 25oC. What is the pH of the resulting solution when 0.500 moles of trichloroacetic acid are dissolved in water and diluted to 1.00 L?Unbalanced reaction: CCl3COOH(aq)  H+(aq) + CCl3COO-(aq)     Ka = 1.3x10-1 at 25oC. Weak Base Problems What is the pH of the resulting solution when 250.0 mL of a 0.300 M solution of ammonia added to 750.0 mL of water?Unbalanced reaction: NH3(aq) + H2O  NH4+(aq) + OH-(aq)     Kb = 1.8x10-5 at 25oC. Calculate Kb for methylamine if a solution prepared by dissolving 0.100 moles of methylamine in 1.00 L of water has a measured pH of 11.80.Unbalanced reaction: CH3NH2(aq) + H2O  CH3NH3+(aq) + OH-(aq)

  42. Relationship Between Ka and Kb • The bigger the Ka of an acid, the smaller the Kb of its conjugate base • The bigger the Kb of a base, the smaller the Ka of its conjugate acid

  43. [A-][H3O+] [HA][OH-] [HA] [A-] How Can We Relate Ka and Kb For Conjugate Acids and Bases? • Acid Reaction HA (aq) + H2O (l)  H3O+ (aq) + A- (aq) • Conjugate Base Reaction A- (aq) + H2O (l)  HA (aq) + OH- (aq) Ka= Kb=

  44. [A-][H3O+] [HA][OH-] [HA] [A-] How Can We Relate Ka and Kb For Conjugate Acids and Bases? • Recall: When we add the 2 reactions together, we multiply their K’s • HA (aq) + H2O (l)  H3O+ (aq) + A- (aq) Ka • A- (aq) + H2O (l)  HA (aq) + OH- (aq) Kb • 2H2O (l)  H3O+(aq) + OH-(aq) Ka*Kb Ka*Kb = x = [H3O+]x[OH-] = Kw

  45. How Can We Relate Ka and Kb For Conjugate Acids and Bases? • Therefore, we can say the following relationship between Ka and Kb exists for CONJUGATE PAIRS • Kw=Ka*Kb • Ka=Kw/Kb • Kb=Kw/Ka • So if only Ka is given, we can find the Kb of its conjugate base when needed • This formula shows that the stronger the weak acid/base, the weaker it’s conjugate base/acid is • Stronger weak acid = weaker conjugate base • Stronger weak base = weaker conjugate acid

  46. How Can We Relate Ka and Kb For Conjugate Acids and Bases?Example • Example 1: The Kb for the rocket fuel hydrazine, N2H4(g) is 1.7 x 10-6. What is the Ka of its conjugate acid N2H5+? • 5.9 x 10-9

  47. Mark received [Ion in Solution] Total Marks Possible [Total Weak Acid] Percent Ionization for Weak AcidsAnother Way of Determining/Reporting [H3O+] • Percent Ionization is • The amount of acid that has dissociated or ionized in solution • We can calculate the percent of weak acid that has ionized in solution in the same manner we calculate the percent we receive on a test… • Percent on test = x 100% • Percent Ionization = x 100%

  48. [Ion in Solution] [H3O+(aq)] [OH-(aq)] [Total Weak Acid/Weak Base in Solution] [HA(aq)] [B(aq)] Percent Ionization Percent Ionization = x 100% Percent Ionization Acid = x 100% Percent Ionization Base= x 100%

  49. Percent Ionization for Weak AcidsExamples • Example 1: The pH of a 0.10mol/L methanoic acid (HCOOH) solution is 2.38. Calculate the percent ionization of methanoic acid • 4.2% • Example 2: Calculate the acid ionization constant, Ka, of acetic acid (CH3COOH) if a 0.1000mol/L solution at equilibrium has a percent ionization of 1.3% • 1.7 x 10-5

  50. Salt Solutions • Salts are formed when an acid reacts with a base • The result is a salt and water HCl + NaOH NaCl + H2O acid base salt water • Salts may either form neutral, acidic or basic solutions depending on where the ions came from • This means that when added to a solution, salts may or may not affect the pH of that solution

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