The Danger of Antifreeze: Aqueous Ionic Equilibrium Lecture Presentation

1. Lecture Presentation Chapter 16Aqueous Ionic Equilibrium Sherril Soman Grand Valley State University

2. The Danger of Antifreeze • Each year, thousands of pets and wildlife species die from consuming antifreeze. • Most brands of antifreeze contain ethylene glycol. • Sweet taste • Initial effect drunkenness • Metabolized in the liver to glycolic acid • HOCH2COOH

3. Why Is Glycolic Acid Toxic? • In high enough concentration in the bloodstream, glycolic acid overwhelms the buffering ability of the HCO3− in the blood, causing the blood pH to drop. • Low blood pH compromises its ability to carry O2. • Acidosis HbH+(aq) + O2(g)  HbO2(aq) + H+(aq) • One treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol.

4. Buffer Solutions • Resist changes in pH when an acid or base is added. • Act by neutralizing acid or base that is added to the buffered solution. • Contain either • Significant amounts of a weak acid and its conjugate base • Significant amounts of a weak base and its conjugate acid • Blood has a mixture of H2CO3 and HCO3−

5. Making an Acidic Buffer Solution It must contain significant amounts of both a weak acid and its conjugate base.

6. An Acidic Buffer Solution • If a strong base is added, it is neutralized by the weak acid (HC2H3O2) in the buffer. NaOH(aq) + HC2H3O2(aq) H2O(l) + NaC2H3O2(aq) • If the amount of NaOH added is less than the amount of acetic acid present, the pH change is small.

7. An Acidic Buffer Solution • If a strong acid is added, it is neutralized by the conjugate base (NaC2H3O2) in the buffer. HCl(aq) + NaC2H3O2  HC2H3O2(aq) + NaCl(aq) • If the amount of HCl is less than the amount of NaC2H3O2 present, the pH change is small.

8. How Acid Buffers Work: Addition of Base HA(aq) + H2O(l) A−(aq) + H3O+(aq) • Buffers work by applying Le Châtelier’s principle to weak acid equilibrium. • Buffer solutions contain significant amounts of the weak acid molecules, HA . • These molecules react with added base to neutralize it. HA(aq) + OH−(aq) → A−(aq) + H2O(l) • You can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium.

9. How Buffers Work H2O new A− A− HA A− HA  H3O+ + Added HO−

10. How Acid Buffers Work: Addition of AcidHA(aq) + H2O(l) A−(aq) + H3O+(aq) • The buffer solution also contains significant amounts of the conjugate base anion, A−. • These ions combine with added acid to make more HA. H+(aq) + A−(aq) → HA(aq) • After the equilibrium shifts, the concentration of H3O+ is kept constant.

11. How Buffers Work H2O new HA HA HA A− A−  H3O+ + Added H3O+

12. Common Ion Effect HA(aq) + H2O(l) A−(aq) + H3O+(aq) • Adding a salt containing the anion NaA, which is the conjugate base of the acid (the common ion), shifts the position of equilibrium to the left. • This causes the pH to be higher than the pH of the acid solution, lowering the H3O+ ion concentration.

13. Common Ion Effect

14. Henderson–Hasselbalch Equation • An equation derived from the Ka expression that allows us to calculate the pH of a buffer solution. • The equation calculates the pH of a buffer from the pKa and initial concentrations of the weak acid and salt of the conjugate base, as long as the “x is small” approximation is valid.

15. Do I Use the Full Equilibrium Analysis or the Henderson–Hasselbalch Equation? • The Henderson–Hasselbalch equation is generally good enough when the “x is small” approximation is applicable. • Generally, the “x is small” approximation will work when both of the following are true: • The initial concentrations of acid and salt are not very dilute. • The Ka is fairly small. • For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000 times larger than the value of Ka.

16. How Much Does the pH of a Buffer Change When an Acid or Base Is Added? • Calculating the new pH after adding acid or base requires breaking the problem into two parts: • A stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other. • Added acid reacts with the A− to make more HA • Added base reacts with the HA to make more A− • An equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]

17. Action of a Buffer Figure 16.3 pg 762

18. Basic BuffersB:(aq) + H2O(l)  H:B+(aq) + OH−(aq) • Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl− H2O(l) + NH3(aq) NH4+(aq) + OH−(aq)

19. Henderson–Hasselbalch Equation for Basic Buffers • The Henderson–Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product. • The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product. B: + H2O  H:B+ + OH− • To apply the Henderson–Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction. H:B+ + H2O  B: + H3O+ • This does not affect the concentrations, just the way we are looking at the reaction.

20. Relationship between pKa and pKb • Just as there is a relationship between the Ka of a weak acid and Kb of its conjugate base, there is also a relationship between the pKa of a weak acid and the pKb of its conjugate base. Ka Kb = Kw = 1.0 x 10−14 −log(Ka Kb) = −log(Kw) = 14 −log(Ka) + −log(Kb) = 14 pKa + pKb = 14

21. Henderson–Hasselbalch Equation for Basic Buffers • The Henderson–Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product. • The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product. B: + H2O  H:B+ + OH− • We can rewrite the Henderson–Hasselbalch equation for the chemical equation of the basic buffer in terms of pOH.

22. Buffering Effectiveness • A good buffer should be able to neutralize moderate amounts of added acid or base. • However, there is a limit to how much can be added before the pH changes significantly. • The buffering capacityis the amount of acid or base a buffer can neutralize. • The buffering rangeis the pH range the buffer can be effective. • The effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base.

23. Buffering Capacity A concentrated buffer can neutralize more added acid or base than a dilute buffer. A concentrated buffer can neutralize more added acid or base than a dilute buffer.

24. Effectiveness of Buffers • A buffer will be most effective when the [base]:[acid] = 1. • Equal concentrations of acid and base • A buffer will be effective when 0.1 < [base]:[acid] < 10. • A buffer will be most effective when the [acid] and the [base] are large.

25. Buffering Range • We have said that a buffer will be effective when 0.1 < [base]:[acid] < 10. • Substituting into the Henderson–Hasselbalch equation we can calculate the maximum and minimum pH at which the buffer will be effective. Lowest pH Highest pH Therefore, the effective pH range of a buffer is pKa± 1. When choosing an acid to make a buffer, choose one whose pKa is closest to the pH of the buffer.

26. Buffering Capacity • Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in pH. • The buffering capacity increases with increasing absolute concentration of the buffer components.

27. Buffering Capacity • As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves. • Buffers that need to work mainly with added acid generally have [base] > [acid]. • Buffers that need to work mainly with added base generally have [acid] > [base].

28. Titration • In an acid–base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete. • When the reaction is complete we have reached the endpoint of the titration. • An indicator may be added to determine the endpoint. • An indicator is a chemical that changes color when the pH changes. • When the moles of H3O+ = moles of OH−, the titration has reached its equivalence point.

29. Titration

30. Titration Curve • It is a plot of pH versus the amount of added titrant. • The inflection point of the curve is the equivalence point of the titration. • Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH. • The pH of the equivalence point depends on the pH of the salt solution. • Equivalence point of neutral salt, pH = 7 • Equivalence point of acidic salt, pH < 7 • Equivalence point of basic salt, pH > 7 • Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH.

31. Titration Curve: Strong Base Added to Strong Acid

32. Titration of a Strong Base with a Strong Acid • If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown.

33. Titration of a Weak Acid with a Strong Base • Titrating a weak acid with a strong base results in differences in the titration curve at the equivalence point and excess acid region. • The initial pH is determined using the Ka of the weak acid. • The pH in the excess acid region is determined as you would determine the pH of a buffer. • The pH at the equivalence point is determined using the Kb of the conjugate base of the weak acid. • The pH after equivalence is dominated by the excess strong base. • The basicity from the conjugate base anion is negligible.

34. Titrating Weak Acid with a Strong Base • The initial pH is that of the weak acid solution. • Calculate like a weak acid equilibrium problem • For example, 15.5 and 15.6 • Before the equivalence point, the solution becomes a buffer. • Calculate mol HAinit and mol A−init using reaction stoichiometry • Calculate pH with Henderson–Hasselbalch using mol HAinit and mol A−init • Half-neutralization pH = pKa

35. Titrating Weak Acid with a Strong Base • At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established. • Mol A− = original mole HA • Calculate the volume of added base as you did in Example 4.8. • [A−]init = mol A−/total liters • Calculate like a weak base equilibrium problem • For example, 15.14 • Beyond equivalence point, the OH is in excess. • [OH−] = mol MOH xs/total liters • [H3O+][OH−] = 1 × 10−14

36. Titration Curve of a Weak Base with a Strong Acid

37. Titration of a Polyprotic Acid • If Ka1 >> Ka2, there will be two equivalence points in the titration. • The closer the Ka’s are to each other, the less distinguishable the equivalence points are. Titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH

38. Monitoring pH During a Titration • The general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H3O+]. • Using a probe that specifically measures just H3O+ • The endpoint of the titration is reached at the equivalence point in the titration—at the inflection point of the titration curve. • If you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with anindicator.

39. Monitoring pH During a Titration Figure 16.10 pg 780

40. Indicators • Many dyes change color depending on the pH of the solution. • These dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solution. HInd(aq) + H2O(l) Ind(aq) + H3O+(aq) • The color of the solution depends on the relative concentrations of Ind:Hind. • When Ind:HInd ≈ 1, the color will be a mix of the colors of Ind and HInd . • When Ind:HInd > 10, the color will be a mix of the colors of Ind. • When Ind:HInd < 0.1, the color will be a mix of the colors of Hind.

41. Phenolphthalein

42. Methyl Red

43. Monitoring a Titration with an Indicator • For most titrations, the titration curve shows a very large change in pH for very small additions of titrant near the equivalence point. • An indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH. • pKa of HInd ≈ pH at equivalence point

44. Acid–Base Indicators Table 16.1 pg 783

45. Solubility Equilibria • All ionic compounds dissolve in water to some degree. • However, many compounds have such low solubility in water that we classify them as insoluble. • We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water.

46. Solubility Product • The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp. • For an ionic solid MnXm, the dissociation reaction is MnXm(s)  nMm+(aq) + mXn−(aq). • The solubility product would be Ksp = [Mm+]n[Xn−]m. • For example, the dissociation reaction for PbCl2 is PbCl2(s)  Pb2+(aq) + 2 Cl−(aq). • And its equilibrium constant is Ksp = [Pb2+][Cl−]2.

48. Molar Solubility • Solubilityis the amount of solute that will dissolve in a given amount of solution at a particular temperature. • The molar solubilityis the number of moles of solute that will dissolve in a liter of solution. • The molarity of the dissolved solute in a saturated solution For the general reaction MnXm(s)  nMm+(aq) + mXn−(aq)

49. Ksp and Relative Solubility • Molar solubility is related to Ksp. • But you cannot always compare solubilities of compounds by comparing their Ksps. • To compare Ksps, the compounds must have the same dissociation stoichiometry.

50. The Effect of Common Ion on Solubility • Addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt. • For example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2. PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) Addition of Cl− shifts the equilibrium to the left.