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Acid-Base Equilibria

Acid-Base Equilibria. The reaction of weak acids with water, OR the reaction of weak bases with water, always results in an equilibrium!! The equilibrium constant for the reaction of a weak acid with water is K a. [H 3 O+] [F - ]. K a =. [HF]. Acid-Base Equilibria.

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Acid-Base Equilibria

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  1. Acid-Base Equilibria The reaction of weak acids with water, OR the reaction of weak bases with water, always results in an equilibrium!! • The equilibrium constant for the reaction of a weak acid with water is Ka

  2. [H3O+] [F-] Ka = [HF] Acid-Base Equilibria H3O+(aq) + F-(aq) eg. HF(aq) + H2O(l) º Keq = ?

  3. [H3O+] [conjugate base] Ka = [weak acid] Acid-Base Equilibria • For any weak acid • Why is H2O(l) omitted from the Ka expression?

  4. Acid-Base Equilibria • the equilibrium constant for the reaction of a weak base with water is Kb HS-(aq) + H2O(l) º Kb = H2S(aq) + OH-(aq)

  5. [OH-] [conjugate acid] Kb = [weak base] Acid-Base Equilibria • For any weak base

  6. [OH-] [HS-] Kb = [S2-] eg. Write the expression for Kb for S2-(aq) ANSWER: S2-(aq) + H2O(l)º Worksheet #5 HS-(aq) + OH-(aq)

  7. 5.a) Use Ka to find [H3O+] for 0.100 mol/L HF(aq) HF(aq) + H2O(l)º H3O+(aq) + F-(aq) Ka = 6.6 x 10-4

  8. 1st try - Ignore x x 2 = (0.100)(6.6 x 10-4) x 2 = 6.6 x 10-5 x = 8.1 x 10-3 mol/L

  9. 2nd try– Include x x 2 = (0.0919)(6.6 x 10-4) x 2 = 6.0654 x 10-5 x = 7.8 x 10-3 mol/L

  10. 3rd try– Include new x x 2 = (0.0922)(6.6 x 10-4) x 2 = 6.0852 x 10-5 x = 7.8 x 10-3 mol/L [H3O+] = 7.8 x 10-3 mol/L

  11. 5.b) find [H3O+] for 0.250 mol/L CH3COOH(aq CH3COOH(aq) + H2O(l)º H3O+(aq) + CH3COO-(aq) Ka = 1.8 x 10-5

  12. 1st try - Ignore x x 2 = (0.250)(1.8 x 10-5) x 2 = 4.5 x 10-6 x = 2.1 x 10-3 mol/L

  13. 2nd try– Include x x 2 = (0.2479)(1.8 x 10-5) x 2 = 4.462 x 10-6 x = 2.1 x 10-3 mol/L [H3O+] = 2.1 x 10-3 mol/L

  14. pH of a weak acid Step #1: Write a balanced equation Step #2: ICE table OR assign variables Step #3: Write the Ka expression Step #4: Check (can we ignore dissociation) Step #5: Substitute into Ka expression

  15. pH of a weak acid eg. Find pH of 0.100 mol/L HF(aq). Step #1: Write a balanced equation HF(aq) + H2O(l)º H3O+(aq) + F-(aq)

  16. Step #2: Equilibrium Concentrations Let x = [H3O+] at equilibrium [F-] = x [HF] = 0.100 - x

  17. [H3O+] [F-] Ka = [HF] Step #3: Ka expression

  18. If > 500 [weak acid] Ka Step #4: Check (can we ignore dissociation) dissociation (- x) may be IGNORED We have to use the – x = 151 (0.100) 6.6 x 10-4 Acid dissociation CANNOT be IGNORED in this question.

  19. Step #5: Substitute into Ka expression Quadratic Formula!! x2 = 6.6 x 10-5 - 6.6 x 10-4 x x2 + 6.6 x 10-4 x - 6.6 x 10-5 = 0 a = 1 b = 6.6 x 10-4 c = -6.6 x 10-5

  20. Ignore negative roots

  21. Try these: a) Find the [H3O+] in 0.250 mol/L HCN(aq) Check: 4.0 x 108 x = 1.24 x 10-5 [H3O+] = 1.24 x 10-5 b) Calculate the pH of 0.0300 mol/L HCOOH(aq) Check: 167 x = 2.24 x 10-3 pH = 2.651

  22. Practice Formic acid, HCOOH, is present in the sting of certain ants. What is the [H3O+] of a 0.025 mol/L solution of formic acid? (0.00203 mol/L) Calculate the pH of a sample of vinegar that contains 0.83 mol/L acetic acid. ( [H3O+] = 3.87 x 10-3 pH = 2.413 ) What is the percent dissociation of the vinegar in 2.? % diss = 0.466 %

  23. Practice A solution of hydrofluoric acid has a molar concentration of 0.0100 mol/L. What is the pH of this solution? ( [H3O+] = 0.00226 pH = 2.646 ) The word “butter” comes from the Greek butyros. Butanoic acid gives rancid butter its distinctive odour. Calculate the [H3O+] of a 1.0 × 10−2 mol/L solution of butanoic acid. (Ka = 1.51 × 10−5 ) (3.89 x 10-4 mol/L)

  24. pH of a weak base • same method as acids • calculate Kb • ignore dissociation if

  25. pH of a weak base Calculate the pH of 0.0100mol/L Na2CO3(aq)

  26. pH of a weak base Calculate the pH of 0.500 mol/L NaNO2(aq)

  27. Calculating Ka from [weak acid] and pH See p. 591 #6 & 8 eg. The pH of a 0.072 mol/L solution of benzoic acid, C6H5COOH, is 2.68. Calculate the numerical value of the Ka for this acid. • Equation • Find [H3O+] from pH • Subtract from [weak acid] • Substitute to find Ka

  28. (0.00209)(0.00209) Ka = (0.06991) C6H5COOH(aq) + H2O(l)º H3O+(aq) + C6H5COO-(aq) [H3O+] = 10-2.68 = 0.00209 mol/L [C6H5COO-] = 0.00209 mol/L [C6H5COOH] = 0.072 – 0.00209 = 0.06991 mol/L Find Ka = 6.2 x 10-5

  29. Calculating Ka from [weak acid] and pH [H3O+] = 10-2.68 = 0.00209 mol/L See p. 591 #’s 5 & 6 = 2.9 % eg. The pH of a 0.072 mol/L solution of benzoic acid, C6H5COOH, is 2.68. Calculate the % dissociation for this acid.

  30. Calculate the acid dissociation constant, Ka , and the percent dissociation for each acid: • a) 0.250 mol/L chlorous acid, HClO2(aq); pH = 1.31 • 0.012 19.5% • b) 0.150 mol/L cyanic acid, HCNO(aq); pH = 2.15 • 0.00035 4.7% • c) 0.100 mol/L arsenic acid, H3AsO4(aq); pH = 1.70 • 0.0050 20% • 0.500 mol/L iodic acid, HIO3(aq); pH = 0.670 • 0.160 42.8%

  31. More Practice: • Weak Acids: pp. 591, 592 #’s 6 -8 • Weak Bases: p. 595 #’s 11 - 16 (Kb’s on p. 592) Worksheet #8 1.a) 1.4 x 10-10 b) 0.0014 % 2.a) 2.5 x 10-9 b) 0.0080 % 3.a) 1.6 x 10-9 b) 0.0080 % 4.a) 2.7 x 10-9 b) 0.042 %

  32. Acid-Base Stoichiometry Solution Stoichiometry (Review) • Write a balanced equation • Calculate moles given ( OR n = CV) • Mole ratios • Calculate required quantity OR OR m = nM

  33. eg. 25.0 mL of 0.100 mol/L H2SO4(aq) was used to neutralize 36.5 mL of NaOH(aq). Calculate the molar concentration of the NaOH solution. H2SO4(aq) + NaOH(aq) 2 → H2O(l) + Na2SO4(aq) 2 nH2SO4 = nNaOH = CNaOH =

  34. Acid-Base Stoichiometry pp. 600, 601 – Sample Problems p. 602 #’s 17 - 20 Worksheet #9

  35. Dilution • Given 3 of the four variables • Only one solution • CiVi = CfVf Stoichiometry • Given 3 of the four variables • Two different solutions • 4 step method

  36. Excess Acid or Base To calculate the pH of a solution produced by mixing an acid with a base: • write the B-L equation (NIE) • calculate the moles of H3O+ and OH- • subtract to determine the moles of excess H3O+ or OH- • divide by total volume to get concentration • calculate pH

  37. eg. 20.0 mL of 0.0100 M Ca(OH)2(aq) is mixed with 10.0 mL of 0.00500 M HCl(aq). Determine the pH of the resulting solution. ANSWER: Species present: Ca2+ OH- H3O+ Cl- H2O SB SA

  38. NIE: OH- + H3O+ → 2 H2O 0.00500 mol/L 0.0100 L 0.0200 mol/L 0.0200 L n = CV 4.00 x 10-4 mol OH- 5.0 x 10-5 mol H3O+ 3.5 x 10-4 mol excess OH-

  39. = 0.01167 mol/L [OH-] = 0.01167 mol/L pOH = 1.933 pH = 12.067 Worksheet #10

  40. Indicators • An indicator is a weak acid that changes color with changes in pH • To choose an indicator for a titration, the pH of the endpoint must be within the pH range over which the indicator changes color

  41. HIn(aq) + H2O(l)º H3O+(aq) + In-(aq) Colour #1 Colour #2 • HIn is the acid form of the indicator. • Adding H3O+ causes colour 1 (LCP) • Adding OH- removes the H3O+ & causes colour #2 LCP

  42. methyl orange HMo(aq) + H2O(l)º H3O+(aq) + Mo-(aq) red yellow bromothymol blue HBb(aq) + H2O(l)º H3O+(aq) + Bb-(aq) yellow blue

  43. Acid-Base Titration (p. 603 → ) • A titration is a lab technique used to determine an unknown solution concentration. • A standard solution is added to a known volume of solution until the endpoint of the titration is reached.

  44. Acid-Base Titration • The endpoint occurs when there is a sharp change in colour • The equivalence point occurs when the moles of hydronium equals the moles of hydroxide • The colour change is caused by the indicator added to the titration flask.

  45. Acid-Base Titration • An indicator is a chemical that changes colour over a given pH range (See indicator table) • A buret is used to deliver the standard solution

  46. Acid-Base Titration standard solution - solution of known concentration primary standard - a standard solution which can be made by direct weighing of a stable chemical. Titration Lab – pp. 606, 607

  47. Multi-Step Titrations (p. 609 - 611) • Polyprotic acids donate their protons one at a time when reacted with a base. eg. Write the equations for the steps that occur when H3PO4(aq) is titrated with NaOH(aq) H3PO4(aq) + OH-(aq) H2PO4-(aq) + OH-(aq) HPO42-(aq) + OH-(aq)

  48. Multi-Step Titrations H3PO4(aq) + OH-(aq)→ H2PO4-(aq) + H2O(l) H2PO4-(aq) +OH-(aq)→ HPO42-(aq) + H2O(l) HPO42-(aq) +OH-(aq)ºPO43-(aq) + H2O(l) H3PO4(aq) + 3 OH-(aq)ºPO43-(aq) + 3 H2O(l)

  49. Multi-Step Titrations Write the balanced net ionic equations, and the overall equation, for the titration of Na2S(aq) with HCl(aq). p. 611 #’s 21.b), 22, & 23 LAST TOPIC!! Titration Curves

  50. Acids and Bases • Properties / Operational Definitions • Acid-Base Theories and Limitations • Arrhenius – H-X and X-OH • Modified – react with water → hydronium • BLT – proton donor/acceptor (CA and CB) • Writing Net Ionic Equations (BLT) • Strong vs. Weak • pH & pOH calculations • Equilibria (Kw, Ka,Kb) • Titrations/Indicators/Titration Curves • Dilutions and Excess Reagent questions

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