1. HOMEWORK – DUE THURSDAY 12/10/15 HOMEWORK – DUE THURSDAY 12/10/15 HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) HW-BW 18 CH 18 #’s 9-18 (all), 21, 23-30 (all), 41-52 (all) NO EXTRA CREDIT FOR CHAPTER 18 HW NO EXTRA CREDIT FOR CHAPTER 18 HW Lab Lab Wednesday/Thursday – Locker check-out and questions about the final Wednesday/Thursday – Locker check-out and questions about the final

2. Sig., Figs., and Logs When you take the log of a number written in scientific notation, the digits before the decimal point come from the exponent, and the digits after the decimal point come from the decimal part of the number. When you take the log of a number written in scientific notation, the digits before the decimal point come from the exponent, and the digits after the decimal point come from the decimal part of the number. Because the part of the scientific notation number that determines the significant figures is the decimal part, the sig. figs. are the digits after the decimal point in the log. Because the part of the scientific notation number that determines the significant figures is the decimal part, the sig. figs. are the digits after the decimal point in the log.

3. What Does the pH Number Imply? The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution. 1 pH unit corresponds to a factor of 10 difference in acidity. Normal range of pH is 0 to 14. pH 0 is [H 3 O + ] = 1 M; pH 14 is [OH – ] = 1 M. pH can be negative (very acidic) or larger than 14 (very alkaline).

4. pOH Another way of expressing the acidity/basicity of a solution is pOH. pOH = −log[OH  ], [OH  ] = 10 −pOH pOH water = −log[10 −7 ] = 7 Need to know the [OH  ] concentration to find pOH pOH 7 is acidic; pOH = 7 is neutral. pH + pOH = 14.0

5. Relationship between pH and pOH pH + pOH = 14.00 a t 25 °C. pH + pOH = 14.00 a t 25 °C. You can use pOH to find the pH of a solution. You can use pOH to find the pH of a solution.

6. pKpKpKpK What does “p” mean? What does “p” mean? p___ = -log___ (pH = - log H +, pOH = -logOH - ) p___ = -log___ (pH = - log H +, pOH = -logOH - ) A way of expressing the strength of an acid or base is pK. A way of expressing the strength of an acid or base is pK. pK a = −log(K a ), K a = 10 −pKa pK a = −log(K a ), K a = 10 −pKa pK b = −log(K b ), K b = 10 −pKb pK b = −log(K b ), K b = 10 −pKb The stronger the acid, the smaller the pK a. The stronger the acid, the smaller the pK a. Larger K a = smaller pK a Larger K a = smaller pK a The stronger the base, the smaller the pK b. The stronger the base, the smaller the pK b. Larger K b = smaller pK b Larger K b = smaller pK b

7. [H 3 O + ] and [OH − ] in a Strong Acid or Strong Base Solution There are two sources of H 3 O + in an aqueous solution of a strong acid—the acid and the water. There are two sources of H 3 O + in an aqueous solution of a strong acid—the acid and the water. There are two sources of OH − in an aqueous solution of a strong acid—the base and the water. There are two sources of OH − in an aqueous solution of a strong acid—the base and the water. For a strong acid or base, the contribution of the water to the total [H 3 O + ] or [OH − ] is negligible. For a strong acid or base, the contribution of the water to the total [H 3 O + ] or [OH − ] is negligible. The [H 3 O + ] acid shifts the K w equilibrium so far that [H 3 O + ] water is too small to be significant. The [H 3 O + ] acid shifts the K w equilibrium so far that [H 3 O + ] water is too small to be significant. Except in very dilute solutions, generally < 1 × 10 −4 M Except in very dilute solutions, generally < 1 × 10 −4 M

8. Finding pH of a Strong Acid or Strong Base Solution For a monoprotic strong acid [H + ] = [HA]. For a monoprotic strong acid [H + ] = [HA]. For 0.10 M HCl, [H + ] = 0.10 M and pH = 1.00 For 0.10 M HCl, [H + ] = 0.10 M and pH = 1.00 For 0.025 M HNO 3, [H + ] = 0.025 M and pH = 1.60 For 0.025 M HNO 3, [H + ] = 0.025 M and pH = 1.60 For a strong ionic base, [OH − ] = (number OH − ) × [Base]. For a strong ionic base, [OH − ] = (number OH − ) × [Base]. For 0.15 M Ca(OH) 2, [OH − ] = 0.30 M and pH = 13.48 For 0.15 M Ca(OH) 2, [OH − ] = 0.30 M and pH = 13.48 For 0.15 M KOH, [OH − ] = 0.15 M and pH = 13.18 For 0.15 M KOH, [OH − ] = 0.15 M and pH = 13.18

9. pH of an Acid/Base Mixture 17.20 mL of a 0.09000 M barium hydroxide solution is added to 25.00 mL of 0.1234 M hydrochloric acid. 1)What are the pH and pOH of the original barium hydroxide? 2)What is the pH of the original hydrochloric acid solution? 3)What is the pH of the solution after the two solutions are mixed? pOH Ba(OH) 2(aq) = – log(0.18000) = 0.74473 pH Ba(OH) 2(aq) = 14 – 0.74473 = 13.25527 pH HCl (aq) = – log(0.1234) = 0.9087

10. pH of an Acid/Base Mixture 17.20 mL of a 0.09000 M barium hydroxide solution is added to 25.00 mL of 0.1234 M hydrochloric acid. 1)What are the pH and pOH of the original barium hydroxide? 2)What is the pH of the original hydrochloric acid solution? 3)What is the pH of the solution after the two solutions are mixed? pOH = – log(2.607x10 –4 ) = 3.5839 pH = 14 – 3.583920 = 10.4161

11. Percent Ionization Another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water; this is called the percent ionization. Another way to measure the strength of an acid is to determine the percentage of acid molecules that ionize when dissolved in water; this is called the percent ionization. The higher the percent ionization, the stronger the acid. The higher the percent ionization, the stronger the acid. Because [ionized acid] equil = [H 3 O + ] equil

12. Strong Bases The stronger the base, the more willing it is to accept H +. For ionic bases, practically all units are dissociated into OH – or accept H +. Strong electrolyte Multi-OH strong bases completely dissociated

13. Weak Bases In weak bases, only a small fraction of molecules accept H +. Weak electrolyte Most of the weak base molecules do not take H from water. Much less than 1% ionization in water [HO – ] << [weak base] Finding the pH of a weak base solution is similar to finding the pH of a weak acid.

14. Table 15.8 page 721 Common Weak Bases

15. Relationship between K a of an Acid and K b of Its Conjugate Base Many reference books only give tables of K a values because K b values can be found from them. Many reference books only give tables of K a values because K b values can be found from them. When you add equations, you multiply the K’s.

16. What is the pH of 0.100 M NH 3(aq) ? (K a = 5.68x10 -10 ) NH 3(aq) ICEICE 0.100 M NH 4 + (aq) OH - (aq) + 0 M – x+ x 0.100 – xxx HOH (l) +

17. Structure of Amines

18. Acid–Base Properties of Ions and Salts The conjugate base of a weak acid is basic. The conjugate base of a weak acid is basic. NaHCO 3 solutions are basic. NaHCO 3 solutions are basic. HCO 3 − is the conjugate base of the weak acid H 2 CO 3. HCO 3 − is the conjugate base of the weak acid H 2 CO 3. The conjugate base of a strong acid is neutral. The conjugate base of a strong acid is neutral. The conjugate acid of a weak base is acidic. The conjugate acid of a weak base is acidic. NH 4 Cl solutions are acidic. NH 4 Cl solutions are acidic. NH 4 + is the conjugate acid of the weak base NH 3. NH 4 + is the conjugate acid of the weak base NH 3. The conjugate acid of a strong base is neutral. The conjugate acid of a strong base is neutral. HCO 3 - (aq) + HOH (l) H 2 CO 3 (aq) + OH - (aq) Cl - (aq) + HOH (l) HCl (aq) + OH - (aq) Na + (aq) + HOH (l) NaOH (aq) + H + (aq) NH 4 + (aq) + HOH (l) NH 3 (aq) + H 3 O + (aq)

19. The stronger the acid, the weaker the conjugate base. The stronger the acid, the weaker the conjugate base. The stronger the base, the weaker the conjugate acid. The stronger the base, the weaker the conjugate acid. Acid–Base Properties of Ions and Salts acidKaKa conjugate base HIO 3 1.7x10 -1 IO 3 - HF3.5x10 -4 F-F- HN 3 2.5x10 -5 N3-N3- HClO2.9x10 -8 ClO - HBrO2.8x10 -9 BrO - HIO2.3x10 -11 IO - stronger acidweaker base

20. Titration In an acid–base titration, a solution of known concentration from a burette (titrant) is slowly added to a solution of unknown concentration (analyte) until the reaction is complete. In an acid–base titration, a solution of known concentration from a burette (titrant) is slowly added to a solution of unknown concentration (analyte) until the reaction is complete. When the reaction is complete we have reached the endpoint of the titration. When the reaction is complete we have reached the endpoint of the titration. An indicator may be added to determine the endpoint. An indicator may be added to determine the endpoint. When the moles of H + = moles of OH −, the titration has reached its equivalence point. When the moles of H + = moles of OH −, the titration has reached its equivalence point.

21. Titration

22. Titration Curve A plot of pH versus the amount of added titrant. A plot of pH versus the amount of added titrant. Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH. Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH.

23. Titration Curve A plot of pH versus the amount of added titrant. A plot of pH versus the amount of added titrant. The inflection point of the curve is the equivalence point of the titration. The inflection point of the curve is the equivalence point of the titration.

24. Titration Curve A plot of pH versus the amount of added titrant. A plot of pH versus the amount of added titrant. The inflection point of the curve is the equivalence point of the titration. The inflection point of the curve is the equivalence point of the titration. The pH of the equivalence point depends on the pH of the salt solution. The pH of the equivalence point depends on the pH of the salt solution. Equivalence point of neutral salt, pH = 7 Equivalence point of neutral salt, pH = 7 Equivalence point of acidic salt, pH < 7 Equivalence point of acidic salt, pH < 7 Equivalence point of basic salt, pH > 7 Equivalence point of basic salt, pH > 7

25. Titration Curve A plot of pH versus the amount of added titrant. A plot of pH versus the amount of added titrant. Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH. Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH.

26. Titration Curve: Strong Base Added to Strong Acid

27. 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. 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.

28. 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. Titrating a weak acid with a strong base results in differences in the titration curve at the equivalence point and excess acid region.

29. 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. 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 K a of the weak acid. The initial pH is determined using the K a of the weak acid. The pH in the excess acid region is determined as you would determine the pH of a buffer. 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 K b of the conjugate base of the weak acid. The pH at the equivalence point is determined using the K b of the conjugate base of the weak acid. The pH after equivalence is dominated by the excess strong base. The pH after equivalence is dominated by the excess strong base. The basicity from the conjugate base anion is negligible. The basicity from the conjugate base anion is negligible.

30. Titrating Weak Acid with a Strong Base The initial pH is that of the weak acid solution. The initial pH is that of the weak acid solution. Calculate like a weak acid equilibrium problem (ICE) Calculate like a weak acid equilibrium problem (ICE) Before the equivalence point, the solution becomes a buffer. Before the equivalence point, the solution becomes a buffer. Calculate mol HA init and mol A − init using reaction stoichiometry (ICF) Calculate mol HA init and mol A − init using reaction stoichiometry (ICF) Calculate pH with Henderson–Hasselbalch using mol HA init and mol A − init Calculate pH with Henderson–Hasselbalch using mol HA init and mol A − init Half-neutralization pH = pK a Half-neutralization pH = pK a

31. 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. 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 Mol A − = original mole HA Calculate the volume of added base as you did in Example 4.8. Calculate the volume of added base as you did in Example 4.8. [A − ] init = mol A − /total liters [A − ] init = mol A − /total liters Calculate like a weak base equilibrium problem Calculate like a weak base equilibrium problem For example, 15.14 For example, 15.14 Beyond equivalence point, the OH is in excess. Beyond equivalence point, the OH is in excess. [OH − ] = mol MOH xs/total liters [OH − ] = mol MOH xs/total liters [H 3 O + ][OH − ] = 1 × 10 −14 [H 3 O + ][OH − ] = 1 × 10 −14

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

33. Titration of a Polyprotic Acid If K a1 >> K a2, there will be two equivalence points in the titration. If K a1 >> K a2, there will be two equivalence points in the titration. Titration of 25.0 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH

34. Titration of a Polyprotic Acid If K a1 >> K a2, there will be two equivalence points in the titration. If K a1 >> K a2, there will be two equivalence points in the titration.

35. Titration of a Polyprotic Acid If K a1 >> K a2, there will be two equivalence points in the titration. If K a1 >> K a2, there will be two equivalence points in the titration. The closer the K a ’s are to each other, the less distinguishable the equivalence points are. The closer the K a ’s are to each other, the less distinguishable the equivalence points are.

36. 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 [H 3 O + ]. The general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H 3 O + ]. Using a probe that specifically measures just H 3 O + Using a probe that specifically measures just H 3 O + The endpoint of the titration is reached at the equivalence point in the titration—at the inflection point of the titration curve. 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 an indicator. If you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator.

37. Monitoring pH During a Titration Figure 16.10 pg 780

38. Indicators Many dyes change color depending on the pH of the solution. Many dyes change color depending on the pH of the solution. These dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution. These dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution. HInd (aq) + H 2 O (l)  Ind  (aq) + H 3 O + (aq) The color of the solution depends on the relative concentrations of Ind  :Hind. 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 ≈ 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 > 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. When Ind  :HInd < 0.1, the color will be a mix of the colors of Hind.

39. Phenolphthalein

40. Methyl Red

41. 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. 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. 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. pK a of HInd ≈ pH at equivalence point pK a of HInd ≈ pH at equivalence point

42. Acid–Base Indicators Table 16.1 pg 783

43. A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate: a.the volume required to reach the equivalence point

44. A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate: b.the pH after adding 5.00 mL of KOH (K a = 4.57x10 -4 ) ICFICF 0.004000.00100 – 0.00100 0.003000 + 0.00100 HNO 2(aq) OH – (aq) +NO 2 - (aq) 0 0.00100 +HOH

45. A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate: c. the pH at one-half the equivalence point (pK a = 3.34) ICFICF 0.004000.00200 – 0.00200 0.002000 + 0.00200 HNO 2(aq) OH – (aq) +NO 2 - (aq) 0 0.00200 +HOH

46. A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate: d.the pH at the equivalence point (K a = 4.57x10 -4 ) ICFICF 0.00400 – 0.00400 00 + 0.00400 HNO 2(aq) OH – (aq) +NO 2 - (aq) 0 0.00400 +HOH

47. A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate: d.the pH at the equivalence point (K a = 4.57x10 -4 ) ICEICE 00 + x xx – x HNO 2(aq) OH – (aq) +NO 2 - (aq) 0.0667 M 0.0667– x + HOH

48. A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate: e.The pH after adding 25.0 mL of KOH ICFICF 0.004000.00500 – 0.00400 00.00100 + 0.00400 HNO 2(aq) OH – (aq) +NO 2 - (aq) 0 0.00400 +HOH