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Chemistry 30 – Unit 4 Part 2 Acid-Base Equilibrium Systems

To accompany Inquiry into Chemistry. PowerPoint Presentation prepared by Robert Schultz robert.schultz@ei.educ.ab.ca. Chemistry 30 – Unit 4 Part 2 Acid-Base Equilibrium Systems. Chapter 17, Section 17.1. • Acid rain: pH < 5 “normal rain” is still acidic due to dissolved CO 2 :

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Chemistry 30 – Unit 4 Part 2 Acid-Base Equilibrium Systems

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  1. To accompany Inquiry into Chemistry PowerPoint Presentation prepared by Robert Schultz robert.schultz@ei.educ.ab.ca Chemistry 30 – Unit 4 Part 2Acid-Base Equilibrium Systems

  2. Chapter 17, Section 17.1 • Acid rain: pH < 5 “normal rain” is still acidic due to dissolved CO2: CO2(g) + H2O(l)  H2CO3(aq) H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq) • Acid Deposition: acid rain, acid snow, acid hail, acid fog, acid dust • Causes: Mainly SO2(g) & SO3(g) “sox” and NO(g) and NO2(g) “nox”

  3. Chapter 17, Section 17.1 • “sox” from combustion of coal and natural gas containing sulfur and from smelting metal ores to produce pure metals See equations page 680: “sox” ultimately becomes H2SO3(aq) and H2SO4(aq) “scrubber” to remove SO2(g) from smokestacks • “nox” from automobile exhaust see equations, page 681

  4. Chapter 17, Section 17.1 • 1st step: N2(g) + O2(g) + energy  2 NO(g) Kc very small – barely occurs at all at room temperature At the temperature in a car engine Kc much larger Lowering temperature would cause engines to run very inefficiently, car engines produce NO(g) which ultimately gets converted to HNO2(aq) and HNO3(aq)

  5. Chapter 17, Section 17.1 • Acid deposition serious problem for lakes Some organisms can’t tolerate lower pH’s • Acidic lakes also leach minerals from their lakebeds (happens more in granitic lakebeds in Canadian Shield than in alkaline soils of Alberta) • Toxic metals can accumulate in the bodies of aquatic organisms

  6. Chapter 17, Section 17.1 • Acid shock – spring run-off from acid snow Treatment – liming – calcium carbonate (limestone) Raises pH; reduces solubility of toxic metal ions; adds calcium ions (plant nutrient)

  7. Chapter 17, Section 17.2 • The Brønsted-Lowry Theory acid: proton (H+) donor base: proton (H+) acceptor • allows explanation of non-aqueous acid-base reactions • acids and bases defined by their behaviour in a chemical reaction, not by their empirical properties

  8. Chapter 17, Section 17.2 conjugate acid/base pair: two chemical entities differing by 1 proton (H+) All Brøntsted-Lowry acid-base reactions contain 2 conjugate acid/base pairs In above reaction conjugate acid/base pairs: HCl(aq)/Cl‾(aq) and H3O+(aq)/H2O(l) conjugate pair conjugate pair acid base conjugate conjugate acid base

  9. Chapter 17, Section 17.2 • second example – bottom of page 685 do Practice Problems 2 and 4, page 687

  10. Chapter 17, Section 17.2 • Equations can be written for the reaction of water with some substances that will produce H3O+(aq) and OH‾(aq), each with a familiar balancing entity e.g. Called amphiprotic substances familiar: HCO3‾(aq) + H2O(l) H3O+(aq) + CO32-(aq) HCO3‾(aq) + H2O(l) OH‾(aq) + H2CO3(aq)

  11. Chapter 17, Section 17.2 • Examples of amphiprotic substances: • H2O(l) • anions containing an H that can ionize as H+(aq): • HSO4‾(aq) • HOOCCOO‾(aq) • C3H5O(COOH)2COO‾(aq) • C3H5OCOOH(COO)22‾(aq) • HSO3‾(aq) • H2PO4‾(aq) • HPO42‾(aq) H2PO4‾(aq)

  12. Chapter 17, Section 17.2 • Polyprotic acid: acid that can donate 2 or more protons (note that amphiprotic ions are between polyprotic acids and polyprotic bases) e.g. H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3‾(aq) polyprotic acid amphiprotic base HCO3‾(aq) + H2O(l)  H3O+(aq) + CO32‾(aq) amphiprotic acid polyprotic base Polyprotic acid: H2CO3(aq) can donate 2 protons Polyprotic base: CO32‾(aq) can accept 2 protons Amphiprotic entity: HCO3‾(aq) can donate or accept a proton (depending on what it’s reacting with)

  13. Chapter 17, Section 17.2 • Do Practice Problems 5-7, page 688

  14. Chapter 17, Section 17.2 • Predicting equilibrium position for acid-base reactions: 2 methods; see example HF(aq) + CH3COO‾(aq)  CH3COOH(aq) + F‾(aq) A B A B Look at position of acid and base on left side of equation If acid base If base acid In example HF(aq) is above CH3COO‾(aq) equilibrium favours products favours products favours reactants Using your acid-base chart, compare the acid and base on the left to the acid and base on the right Stronger pair will transfer a proton better and equilibrium will favour the opposite side In the example HF(aq) is stronger acid; CH3COO‾(aq) is stronger base – equilibrium favours products

  15. Chapter 17, Section 17.2 • Example: Practice Problem 8a, page 690 H3PO4(aq) is stronger acid; NH3(aq) is stronger base, therefore equilibrium favours reactants • It is unnecessary to compare both acids and bases – the stronger acid will always be paired with the stronger base – why? • the stronger an acid, the weaker its conjugate base NH4+(aq) + H2PO4‾(aq)  NH3(aq) + H3PO4(aq)

  16. Chapter 17, Section 17.2 • Do WS 67

  17. Chapter 17, Section 17.3 • When dealing with acid ionizations (or reactions with water) we use Ka: For the acid HA(aq) HA(aq) + H2O(l)  H3O+(aq) + A‾(aq) or HA(aq)  H+(aq) + A‾(aq) Kais a special case of Kc

  18. Chapter 17, Section 17.3 • Strong acids have large Ka; weak acids have very small Ka •Ka’s are present in your acid-base chart on pages 8-9 of the Data Booklet • % ionizations are valid for acids only over a very narrow concentration range; Ka’s better but ………. You may be required to calculate % ionization for the Diploma Exam

  19. Chapter 17, Section 17.3 • Sample Problem, page 692-3 initial [CH3COOH] = 0.10 mol/L pH at equilibrium = 2.96 Find Ka and % ionization Review: -0.0011 +0.0011 +0.0011 0.099 0.0011 • equilibrium [H3O+(aq)] = 10-2.96 = 0.0011 mol/L CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO‾(aq) I 0.10 0 0C E 0.0011

  20. Chapter 17, Section 17.3 • Try Practice Problems 14 and 17, page 693 14. equilibrium [H3O+] = HC4H3N2O3(s) + H2O(l)  H3O+(aq) + C4H3N2O3‾(aq) I 0.10 0 0C -0.0032 +0.0032 +0.0032E0.097 0.0032 0.0032

  21. Chapter 17, Section 17.3 • 17. Equilibrium [H3O+] = C6H4NH2COOH(aq) + H2O(l)  H3O+(aq) + C6H4NH2COO‾(aq) I 0.034 0 0C -6.5 x 10-4 +6.5 x 10-4 +6.5 x 10-4E 0.034 6.5 x 10-46.5 x 10-4

  22. Chapter 17, Section 17.3 • Ion product constant for water, Kw 2 H2O(l) H3O+(aq) + OH‾(aq) or H2O(l) H+(aq) + OH‾(aq) • True for all aqueous solutions • For pure water, [H3O+] = [OH-] = 1.0 x 10-7 mol/L

  23. Chapter 17, Section 17.3 • Recall from Chem 20: pH = -log [H3O+]* pOH = -log [OH‾] pH + pOH = 14.00 (a log form of Kw) [H3O+]* = [OH‾] = * H3O+ or H+ • pH Review Questions – WS 63 Recall special deal with significant digits and pH

  24. Chapter 17, Section 17.3 • Another special of Kc:Kb • B(aq) + H2O(l)  HB+(aq) + OH‾(aq) example: NH3(aq) NH3(aq) + H2O(l) NH4+(aq) + OH‾(aq)

  25. Chapter 17, Section 17.3 • Practice Problems 19 c, 20c, 19 d, 20d, 22, page 698 22. Since pH = 10.26, codeine is a base C18H21NO3(s) + H2O(l)  OH‾(aq) + HC18H22NO3+(aq) I 0.020 0 0C-1.8 x 10-4+1.8 x 10-4 +1.8 x 10-4E0.020 1.8 x 10-41.8 x 10-4

  26. Chapter 17, Section 17.3 • Finding pH of weak acids or bases Weak acid: Example: Practice Problem 24, page 699 CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO‾(aq) I 0.83 0 0C-x +x +xE 0.83 – x x x Ka = 1.8 x 10-5(Data Booklet) can approximate since 0.83 > 1000 times 1.8 x 10-5 this will always be true in your questions

  27. Chapter 17, Section 17.3 • Shortcut*: pH of a weak acid WA = weak acid • 2 notes about shortcut: • Never use to find Kaor [WA] – this is an approximation; no need to solve a quadratic to get Kaor [WA]; don’t approximate • You cannot eliminate all ICE boxes; only those solving for H3O+ using Ka WS 65

  28. Chapter 17, Section 17.3 • Weak Base: Example: Practice Problem 28, page 701 NH3(aq) + H2O(l) NH4+(aq) + OH‾(aq) I 0.105 0 0C-x +x +xE0.105 - xx x (approximation)

  29. Chapter 17, Section 17.4 • Shortcut*: pH of a weak base WB = weak base Same cautions as for weak acid shortcut! Try Practice Problem 29, page 701

  30. Chapter 17, Section 17.3 Practice Problem 29, page 701 N2H4(l) + H2O(l)  N2H5+(aq) + OH-(aq) I 5.9x10-2 0 0 C -x +x +x E 5.9x10-2 – x x x okay

  31. Chapter 17, Section 17.3 • In previous example, you were given Kb for ammoniaThis wasn’t really necessary • For any conjugate acid-base pair in your Data Booklet, Kax Kb= Kw Data BookletKa for NH4+ = 5.6 x 10-10 • Kb for NH3 • Discuss question 37, page 703, do Review 4a, 4c, page 704

  32. Chapter 17, Section 17.3 • 4a 4c Conjugate base: C6H7O6‾(aq) Conjugate base: HCOO‾(aq)

  33. Chapter 17, Section 17.4 • Acid-base indicators: conjugate acid-base pairs ( HIn(aq)/In‾(aq) ) where acid and its conjugate base are different colours, for example: bromothymol blue HBb yellow/ Bb‾ blue • Chart on page 10 of Data Booklet • Indicators chosen for titrations to change colour over appropriate pH range formulas organic and complex (see p. 705)so abbreviations used

  34. Chapter 17, Section 17.3 • Note: because of Ka’s on indicator chart and acid/base chart of Data Booklet you could be expected to write an equation and predict colour of an indicator in a solution Example: what would be expected if bromocresol green in its base form, Bg‾(aq), was added to a benzoic acid solution?

  35. Chapter 17, Section 17.3 • As before, make your list: Bg‾(aq), C6H5COOH(aq), H2O(l) SB SA A, B C6H5COOH(aq) + Bg‾(aq)  HBg(aq) + C6H5COO‾(aq) Kaof HBg(aq) (1.3 x 10-5) is slightly lower than that of C6H5COOH(aq) (6.3 x 10-5), therefore equilibrium will favour products and colour of HBg/Bg- will be green or yellow

  36. Chapter 17, Section 17.4 • Titration curves: review of Chem 20 acid with base base with acid strong acid /strong base* *Chem 20 VNaOHor VHCl (mL)

  37. Chapter 17, Section 17.4 • polyprotic acid titrated with strong base weakacid Note: only quantitative reactions have detectable endpoints strongacid weak acid and strong acid titrated with strong base

  38. Chapter 17, Section 17.4 • titration of weak acid versus strong: SA with SB* pH Explains why pH > 7 for equivalence point of WA with SB volume of base favours products e.g. CH3COOH(aq) + OH-(aq)  H2O(l) + CH3COO-(aq) WA with SB WA with WB >7 7

  39. Chapter 17, Section 17.4 • titration of weak base versus strong: pH 7 <7 WB with WA SB with SA* WB with SA volume of acid

  40. Chapter 17, Section 17.4 • polyprotic acid with strong base • polyprotic base with strong acid pH Figures 17.14 and 17.15, pages 709-10

  41. Chapter 17, Section 17.4 • Buffer regions In buffer regions there is a mixture of a weak acid and its conjugate base Addition of small amounts of acid or base has little effect on the pH Figures 17.12 and 17.13, pages 707-8

  42. Chapter 17, Section 17.4 • x CH3COOH(aq) only 50/50 mixture of CH3COOH(aq) and CH3COO‾(aq) (page 712) Example: CH3COOH(aq)/CH3COO‾(aq) buffer

  43. Chapter 17, Section 17.4 • example continued: Buffer equilibrium: Add acid: [H3O+] , equilibrium shifts left to consume added H3O+(aq) and convert some CH3COOˉ(aq) into CH3COOH(aq) Add base: since, CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COOˉ(aq) essentially 100% OHˉ(aq) + H3O+(aq) H2O(l), the buffer equilibrium will shift right as H3O+(aq) is consumed

  44. Chapter 17, Section 17.4 • Buffer demo: pH 7 versus water as concentrated HCl(aq ) is added Also show buffer capacity as more HCl(aq) is added

  45. Chapter 17, Section 17.4 • Buffers in your body: read pages 714-717 Human blood buffered to pH = 7.4 Most important buffer in human blood: HCO3ˉ(aq)/CO32-(aq) Without buffering, the amount of acid in a glass of orange juice could kill you

  46. Chapter 17, Section 17.4 • In internal cell fluid, most important buffer is H2PO4ˉ(aq)/HPO42-(aq) acidosis – decreased blood pH alkalosis – increased blood pH respiratory metabolic

  47. Chapter 17, Section 17.4 •

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