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Chapter 16. Acid –Base Equilibria

Chapter 16. Acid –Base Equilibria

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Chapter 16. Acid –Base Equilibria

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  1. Chapter 16. Acid –Base Equilibria • . • .

  2. Equilibria in Solutions of Weak Acids • The dissociation of a weak acid is an equilibrium situation with an equilibrium constant, called the acid dissociation constant, Ka based on the equation • HA (aq) + H2O (l)H3O+ (aq) + A- (aq)

  3. Equilibria in Solutions of Weak Acids • The acid dissociation constant, Ka is always based on the reaction of one mole of the weak acid with water. • If you see the symbol Ka, it always refers to a balanced equation of the form • HA (aq) + H2O (l)H3O+ (aq) + A- (aq)

  4. Problem • The pH of 0.10 mol/L HOClis 4.23. Calculate Kafor hypochlorous acid. • HOCl (aq) + H2O (l)H3O+ (aq) + ClO- (aq)

  5. Calculating Equilibrium Concentrations in Solutions of Weak Acids • We can calculate equilibrium concentrations of reactants and products in weak aciddissociationreactions with known values for Ka. • To do this, we will often use the ICE table technique we saw in the last chapter on equilibrium.

  6. Calculating Equilibrium Concentrations in Solutions of Weak Acids • We need to figure out what is an acid and what is a base in our system. • For example, if we start with 0.10 mol/L HCN, then HCN is an acid, and water is a base. • HCN (aq) + H2O (l)H3O+ (aq) + CN- (aq) • Ka = 4.9 x 10-10

  7. Like our previous equilibrium problems, we then create a table of the initial concentrations of all chemicals, the change in their concentration, and their equilibrium concentrationsin terms of known and unknown values.

  8. We canALWAYS solve this equation using thequadratic formula and get the right answer, but itmight be possibleto do it more simply.

  9. Every time we do an weak acidequilibrium problem, divide the initial concentration of the acid by Ka. • For this example • 0.10 / 4.9 x 10-10 = 2 x 108

  10. 0.10 / 4.9 x 10-10 = 2 x 108 • Since this value isgreater than 100, we can assumethat theinitial concentration of the acid and the equilibrium concentration of the acidare the same. • This assumption will lead to answers with less than 5% error since this pre-check is greater than 100.

  11. The assumption we will make is that • x << [HCN]i so [HCN]eqm[HCN]I • 4.9 x 10-10 = x2 / 0.10 • x2 = (4.9 x 10-10)(0.10) • x = 4.9 x 10-11 • x = 7.0 x 10-6 mol/L

  12. Based on the assumption we’ve made, at equilibrium • x = [H3O+]eqm=[CN-]eqm • =7.0 x 10-6 mol/L • (-ve value isn’t physically possible) • [HCN]eqm = 0.10 mol/L.

  13. . • Any time we make an assumption, • we MUST check it. • We assumed x << [HCN]i • To check the assumption, we divide x by [HCN]i and express it as a percentage

  14. As long as the assumption check is less than 5%, then • the assumption is valid! • If the assumption was not valid, we would have to go back and use the quadratic formula!

  15. Remember! • H2O (l)+H2O (l) H3O+ (aq) + OH- (aq) • is alwaystaking place inwaterwhether or not we have added an acid or base. • This reaction also contributes • H3O+ (aq) and OH- (aq) • to our systemat equilibrium

  16. Remember! • Since at 25 C • Kw=[H3O+][OH-]=1.0 x 10-14 • it turns out that if ouracid-base equilibrium we’re interested ingives a pH value between about6.8 and 7.2 then the auto-dissociation of water contributes a significant amount of [H3O+] and [OH-]to our system and the real pH would not be what we calculated in the problem.

  17. Problem • Acetic acid CH3COOH (or HAc) is the solute that gives vinegar its characteristic odour and sour taste. Calculatethe pHand the concentration of all species present in: • a) 1.00 mol/L CH3COOH • b) 0.00100 mol/L CH3COOH

  18. Problem a) Let’s check the initial acid concentration / Ka ratio. 1.00 / 1.8 x 10-555000 is larger than 100.

  19. Problem a) We can probably assume that x << [HAc]i so [HAc]eqm[HAc]i 1.8 x 10-5 = x2 / 1.00 x2 = (1.8 x 10-5)(1.00) x = 1.8 x 10-5 x = 4.2 x 10-3 mol/L (but must be + value sincex =[H3O+])

  20. Problem a) • So at equilibrium, • [H3O+] = [CH3COO-] = 4.2 x 10-3 mol/L • [CH3COOH] = 1.00 mol/L. • The assumption was valid and so • pH = - log [H3O+] • pH = - log 4.2 x 10-3 • pH = 2.38

  21. 0.00100 0.00100 Problem b) Let’s check the initial acid concentration / Ka ratio. 0.00100 / 1.8 x 10-556 is smaller than 100.

  22. 0.00100 0.00100 Problem b) We can probably CAN NOT assume that x << [HAc]i so [HAc]eqm[HAc]i

  23. Problem b)

  24. Problem b) Since [H3O+] = x we must use the positive value, so [H3O+] = [CH3COO-]= 1.3 x 10-4 mol/L [CH3COOH] = 0.00100 mol/L – 1.3 x 10-4 mol/L = 0.00087 mol/L.

  25. Problem b) Let’s confirm that x << [HAc]i IS NOT TRUE pH = - log [H3O+] pH = - log 4.2 x 10-4 pH = 3.38

  26. Problem • A vitamin C tablet containing 250 mg of ascorbic acid (C6H8O6; Ka = 8.0 x 10-5 is dissolved in a 250 mL glass of water to give a solution where [C6H8O6] = 5.68 x 10-3 mol/L. • What is the pH of the solution?

  27. Problem • Check the initial acid concentration / Ka ratio. • 5.68 x 10-3/ 8.0 x 10-5 71 • which is not larger than 100 so

  28. Problem

  29. Problem • Since [H3O+] = x the answer must be the positive value • [H3O+] =[C6H7O6-]= 6.3 x 10-4 mol/L • [C6H8O6] = (5.68 x 10-3 - 6.3 x 10-4)mol/L • = 5.05 x 10-3 mol/L • pH = - log [H3O+] • pH = - log 6.3 x 10-4 • pH = 3.20

  30. Degree of ionization • The pH of a solution of a weak acid like acetic acid will depend on the initial concentration of the weak acid andKa. Therefore, we can define a second measure of the strength of a weak acid by looking of the • degree (or percent) ionization of theacid. • %ionization = [HA]ionized / [HA]initial x 100%

  31. Percent ionization • In part a) of an earlier problem an acetic acid solution with initial concentration of 1.00 mol/Lat equilibrium had • [H3O+]eqm=[HA]ionized=4.2 x 10-3 mol/L • %ionization= [HA]ionized / [HA]initial x 100% • %ionization = 4.2 x 10-3 mol/L / 1.00 mol/L x 100% • %ionized= 0.42%

  32. Percent ionization • In part b) of an earlier problem an acetic acid solution with initial concentration of 0.00100 mol/Lat equilibrium had • [H3O+] = [HA]ionized = 1.3 x 10-4 mol/L • %ionization= [HA]ionized / [HA]initial x 100% • %ionization = 1.3 x 10-4 mol/L / 0.00100 mol/L x 100% • %ionization= 13%

  33. Figure

  34. Equilibria in Solutions of Weak Bases • The dissociation of a weak base is an equilibrium situation with an equilibrium constant, called the base dissociation constant, Kb based on the equation • B (aq) + H2O (l)BH+ (aq) + OH- (aq)

  35. Equilibria in Solutions of Weak Bases • The base dissociation constant, Kb is always based on the reaction of one mole of the weak base with water. • If you see the symbol Kb, it always refers to a balanced equation of the form • B (aq) + H2O (l)BH+ (aq) + OH- (aq)

  36. Equilibria in Solutions of Weak Bases • Our approach to solving equilibria problems involving bases is exactly the same as for acids. • 1. Set up the ICE table • 2. Establish the equilibrium constant expression • 3. Make a simplifying assumption when possible • 4. Solve for x, and then for eq’m amounts

  37. Problem • Strychnine (C21H22N2O2),a deadly poison used for killing rodents,is aweak base having Kb = 1.8 x 10-6. Calculate the pH if • [C21H22N2O2]initial = 4.8 x 10-4 mol/L

  38. Problem Check the initial base concentration / Kb ratio 4.8 x 10-4 / 1.8 x 10-6267 which is greater than 100 We are probably good to make a simplifying assumption that x << [C21H22N2O2]i

  39. The assumption we will make is that • x << [C21H22N2O2]i so • [C21H22N2O2]eqm[C21H22N2O2]I • 1.8 x 10-6 = x2 / 4.8 x 10-4 • x2 = (1.8 x 10-6)(4.8 x 10-4) • x = 8.64 x 10-10 • x = 2.94 x 10-5 mol/L

  40. Problem • Since x = [OH-], the answer must be the positive value, • x =[C21H23N2O2+]= [OH-] =2.9 x 10-5 mol/L • [C21H22N2O2] = 4.8 x 10-4 mol/L – 2.9 x 10-5 mol/L • = 4.5 x 10-4 mol/L. • We should check the assumption!

  41. Problem • In this case, the error is more than 5%. • I will leave it to you to go back and use the quadratic formula. • Compare the two answers

  42. Problem • To continue towards the answer of the problem AS IF the assumption WERE VALID • pOH = - log [OH-] • pOH = - log 2.9 x 10-5 • pOH = 4.54 • pH + pOH = 14.00 • pH = 14.00 - pOH • pH = 14.00 - (4.54) • pH = 9.46

  43. Relation Between Ka and Kb • The strength of an acid in water is expressed throughKa, while the strength of a base can be expressed through Kb • Since Brønsted-Lowryacid-base reactions involve conjugateacid-basepairsthereshould be a connectionbetween the • Ka value and the Kb value of a • conjugateacid-basepair.

  44. Relation Between Ka and Kb • HA (aq) + H2O (l)H3O+ (aq) + A- (aq) • A- (aq) + H2O (l)OH- (aq) + HA (aq)

  45. Since these reactions take place in the same beaker at the same time let’s • add them together

  46. The sum of the reactions is the dissociation of water reaction, which has the ion-product constant for water • Kw=[H3O+][OH-]=1.0 x 10-14 at 25 °C • Closer inspection shows us that

  47. As the strength of an acid increases(larger Ka) the strength of the conjugate base must decrease(smaller Kb) because their product must always be thedissociation constant for water Kw.

  48. Strong acidsalways havevery weak conjugate bases. Strong basesalways havevery weak conjugate acids. • Since Ka x Kb = Kw • then Ka = Kw / Kb • and Kb = Kw / Ka

  49. Problem • a) – Piperidine (C5H11N) is an amine found in black pepper. Find Kb for piperidine in Appendix C, and then calculate Ka for the C5H11NH+ cation. • Kb = 1.3 x 10-3 • b) Find Ka for HOCl in Appendix C, and then calculate Kbfor OCl-. • Ka = 3.5 x 10-8

  50. Acid-Base Properties of Salts • When acids and bases react with each other, • they form ionic compounds called salts. • Salts, when dissolved in water, can lead to acidic, basic, or neutral solutions, depending on the relative strengths of the acid and base we derive them from. • Strong acid + Strong baseNeutral salt solution • Strong acid + Weak baseAcidic salt solution • Weak acid + Strong baseBasic salt solution