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SCH4U Unit #2: EQUILIBRIUM

SCH4U Unit #2: EQUILIBRIUM . Ms. Cornacchione Mon Mar 17 th 2014. AGENDA. Recap, Yield, and Reaction Quotient (Q) EQM HOMEWORK QUIZ MARKING Unit 2 Acid Base EQM. Kyler and Roger to write Test 2. Equilibrium Law & Equilibrium Constant.

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SCH4U Unit #2: EQUILIBRIUM

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  1. SCH4UUnit #2: EQUILIBRIUM Ms. Cornacchione Mon Mar 17th 2014

  2. AGENDA • Recap, Yield, and Reaction Quotient (Q) • EQM HOMEWORK QUIZ MARKING • Unit 2 Acid Base EQM Kyler and Roger to write Test 2

  3. Equilibrium Law & Equilibrium Constant The value of KC is a constant for a given reaction at a certain Temperature, regardless of initial concentration EXOTHERMIC – as T , KC  ENDOTHERMIC - as T, KC 

  4. Percentage Yield (Gr 11 Recap) • Percentage Yield = Actual Yield (mol) x 100% Theoretical Yield (mol)

  5. Percentage Yield & EQM • % Yield = Actual Yield (mol) x 100% Theoretical Yield (mol) Example:H2(g) + I2(g) ⇌ 2 HI (g) Starting with 1 mol of H2 and I2in a 1.0 L reactor, Theoretical Yield = 2 mol of HI But with KC = 51 (at 448°C), [H2]EQM = 0.22 mol/L [I2]EQM = 0.22 mol/L Actual Yield [HI]EQM = 1.56 mol/L Percentage Yield = 1.56 mol HI 2 mol HI = 78 %

  6. Reaction Quotient, Q • The Reaction Quotient (Q)tells us whether a system is at equilibrium or not Q [ ] • Reaction Quotient (Q)is calculated exactly the same way as KC. IT IS ALSO UNITLESS!!  [ ] CURRENT CONCENTRATIONS!! (not necessarily at equilibrium) A, B, C, D are reactants and products in (g) or (aq) state ONLY a, b, c, d are the coefficients in the balanced chemical equation

  7. Reaction Quotient, Q H2(g) + I2(g) ⇌ 2 HI (g) Is the chemical system at equilibrium yet?

  8. Reaction Quotient, Q Products NOW NOW EQM EQM Reactants NOW NOW EQM EQM Q

  9. Reaction Quotient EXAMPLE 2 CIF3(g) ⇌ Cl2(g) + 3 F2(g) KC = 3.72 x 10-2 1 mole of ClF3(g) is placed in a closed vessel. After 40 minutes, concentrations were found to be: [CIF3] = 0.20 M, [Cl2] = 0.08 M, [F2] = 0.10 M Is this chemical system at equilibrium yet? If not, predict the direction in which the equilibrium will shift and explain why A: NOT AT EQM (Q<K), therefore must shift to the RIGHT (PRODUCTS)

  10. Unit #2: Acid-Base Equilibrium (Ch 8)TOPICS • Brønsted Lowry Acids & Bases (8.1) • Strong and Weak Acids & Bases • Acid Calculations • Base Calculations • Acid-Base Properties of Salts • Acid-Base Titration • Buffer Systems

  11. Brønsted-Lowry Acid An acid is a hydrogen ion, H+ (proton) donor A base is a hydrogen ion, H+ (proton) acceptor Which one is the H+ donor? Which one is the H+ acceptor?

  12. Brønsted-Lowry Base An acid is a hydrogen ion, H+ (proton) donor A base is a hydrogen ion, H+ (proton) acceptor Which one is the H+ donor? Which one is the H+ acceptor?

  13. BL Conjugate Acid-Base Pairs Conjugate base– the substance that forms when an acid loses an H+ (acid becomes CB) Conjugate acid– the substance that forms when a base accepts an H+ (base becomes CA)

  14. Brønsted-Lowry vs. Arrhenius Acid What about conjugate acid-base pairs? Base

  15. Non-Aqueous Reactions Acid What about conjugate acid-base pairs? Base

  16. Amphiprotic (Amphoteric) Notice that water can be an acid or a base: acid base

  17. Amphiprotic (Amphoteric) Amphiprotic (amphoteric) – being able to donate or accept a hydrogen ion (proton) depending on what its reacting with, thus can be a BL acid or a BL base 2. Explain why the hydrogen carbonate ion is amphiprotic, but the fluoride ion is not. 1. Which is the amphiprotic substance given these two reactions?

  18. Acid-Base Dynamic Equilibrium The reaction of a weak acid, HA, with water (acting as a base) forms a dynamic equilibrium involving a conjugate base, A- and a conjugate acid, H3O+ (the hydronium ion): The Acid Ionization Constant, Ka (also called the acid dissociation constant)

  19. The Acid Ionization Constant, Ka Tylenol (acetaminophen) Ka = 1.2 x 10-10 pH 5.3 at 0.25 mol/L Aspirin (acetylsalicylic acid, ASA) Ka = 3.27 x 10-4 pH 2.0 at 0.25 mol/L Knowing that acid-base reactions are dynamic equilibria, what does the magnitude of Ka tell you?

  20. Practice Makes Perfect!! • Page 494 Q#1-9 • Review Section 8.1 Sample Problems

  21. Unit #2: Acid/Base EquilibriumTOPICS • Bronsted Lowry Acids & Bases (8.1) • Strong and Weak Acids & Bases (8.2) • Acid Calculations • Base Calculations • Acid-Base Properties of Salts • Acid-Base Titration • Buffer Systems

  22. Strong and Weak Acids Strong acid – ionizes almost 100% in water, producing hydrogen ions (almost all HA molecules have broken apart to produce ions) Weak acid – only partly ionizes in water, producing hydrogen ions (most of HA is in equilibrium)

  23. Strong and Weak Acids

  24. Acids and Their Conjugate Bases • The stronger the acid, the weaker its conjugate base • The weaker the acid, the stronger its conjugate base Example: HCN (Ka = 6.2 x 10-10) is a relatively weak acid, and CN- is a relatively strong base HCN(aq) ⇌ H+(aq) + CN-(aq)

  25. Strong and Weak Bases Strong base – dissociates completed in water, producing hydroxide ions (eg. NaOH, KOH, Ca(OH)2) Weak base – undergoes an equilibrium reaction with water to produce hydroxide ions (eg. NH3, methylamine, pyridine) Note that even though BL bases don’t have to contain the hydroxide ion, they increase the hydroxide ion concentration because of their reaction with water

  26. The Base Ionization Constant, Kb A base, B, reacts with water (acting as an acid) to produce OH-(aq) ions (conjugate base) and a conjugate acid, BH+ The Base Ionization Constant, Kb (also called the base dissociation constant)

  27. The Base Ionization Constant, Kb (small) Knowing that acid-base reactions are dynamic equilibria, what does the magnitude of Kb tell you?

  28. Autoionization of Water H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq) Autoionization of water– transfer of H+ from one H2O to another • One water molecule acts as a BL acid (releases H+) and the other acts as a BL base (accepting H+) The Ion-Product Constant for Water, Kw – the equilibrium constant for the autoionization of water

  29. Relationship Between Kaand Kb For a weak acid and its conjugate base or for a weak base, and its conjugate acid

  30. Relationship Between Kaand Kb - EXAMPLE

  31. pH and pOH • Since the product of [H+] and [OH-] must always be Kw (1.0 x 10-14), there are three possible situations: 1 x 10 -7 [OH-(aq)] = 10 -pOH [H+(aq)] = 10 -pH pOH + pH = 14

  32. Acid-Base Indicator Acid-Base Indicator – a substance that changes colour within a specific pH range

  33. EXAMPLES – pOH from pH

  34. EXAMPLES – [H+] from pH

  35. EXAMPLES – pH of Strong Acid

  36. EXAMPLES – acidic or basic?

  37. Practice Makes Perfect!! • Page 509 Q#1-10 • Review Section 8.2 Sample Problems

  38. Unit #2: Acid/Base EquilibriumTOPICS • Bronsted Lowry Acids & Bases (8.1) • Strong and Weak Acids & Bases (8.2) • Acid Calculations (8.4) • Base Calculations • Acid-Base Properties of Salts • Acid-Base Titration • Buffer Systems

  39. Acid Calculations • Strong acids (ionize completely) • Can assume [H+] = [HA(aq)] to calculate pH • Weak acids (ionize partially) • Must consider EQM (Initial, Change, and EQM Concs) and Ka, to calculate pH • Percent Ionization – percentage of solute that ionizes when it dissolves in a solvent

  40. EXAMPLES – Strong Acid

  41. EXAMPLES – % Ionization from pH

  42. EXAMPLES – CalcKa from % Ionization (TIP: Use an ICE Table)

  43. EXAMPLES – Calc pH from Ka (TIP: Use an ICE Table)

  44. EXAMPLES – CalcKafrom pH

  45. Polyprotic Acids Monoprotic acids – possesses only one ionizable (acidic) hydrogen atom (eg. HCl(aq)) Polyprotic acids – possesses more than one ionizable (acidic) hydrogen atom (eg. H2SO4(aq)) H2A ⇌ H+ + HA- Ka1= [H+][HA-] [H2A] HA-⇌ H+ + A- Ka2 = [H+ ][A-] [HA- ]

  46. EXAMPLES – pH for a Polyprotic Acid

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