1. Daniel L. Reger Scott R. Goode David W. Ball www.cengage.com/chemistry/reger Chapter 16 Reactions Between Acids and Bases

2. Titration of Strong Acids and Bases Titration: a method used to determine the concentration of a substance known as the analyte by adding another substance, the titrant, which reacts in a known manner with the analyte. analyte + titrant → products

3. Laboratory Titrations (a) A known volume of acid is measured into a flask. (b) Standard base is added from a buret. (c) The endpoint is indicated by a color change. (d) The volume of base is recorded.

4. Titration Curve: a graph of pH of a solution as titrant is added. For a titration of a strong acid with a strong base, the pH will start at a very low value and stay low as long as strong acid is still present. Titration: Strong Acid and Base

5. The pH will rise sharply to 7 at the equivalence point, where the acid and base are present in stoichiometrically equivalent amounts. After excess strong base has been added, the pH levels off at a high value. Titration: Strong Acid and Base

6. Titration Curve for a Strong Acid with a Strong Base

7. Calculate the equivalence point in the titration of 20.00 mL of 0.1252 M HCl with 0.1008 M NaOH. HCl + NaOH  H 2 O + NaCl The Equivalence Point in a Titration

8. Calculate the equivalence point in the titration of 40.00 mL of 0.2387 M HNO 3 with 0.3255 M NaOH. Test Your Skill

9. Millimole: one thousandth of a mole. If molarity is expressed in moles/liter (M) and volume in milliliters (mL), n will be in millimoles (mmol).. Liters cancel but the milli- multiplier remains. Units of Millimoles

10. Calculating a Titration Curve Calculate the pH in the titration of 20.0 mL of 0.125 M HCl with 0.250 M NaOH after 0, 2.00, 10.00, and 20.00 mL base are added.

11. Calculate the pH in the titration of 20.0 mL of 0.125 M HCl with 0.250 M NaOH after 5.00 mL and 12.00 mL base are added. Test Your Skill

12. Calculating a Titration Curve

13. Titration curve of 50.00 mL 0.500 M KOH with 1.00 M HCl. Strong Base + Strong Acid Curve

14. 10 mL of two different 0.100 M acids titrated with 0.100 M NaOH. Stoichiometry and Titration Curves

15. Estimating the pH of Mixtures Fill in first 3 three lines of sRf table. Look at the final solution (f-line). If a strong acid is present, the solution will be strongly acidic. If a strong base is present the solution will be strongly basic. If only water is present, the solution will be neutral. SolutionEstimate of pH Strongly acidic1 Neutral7 Strongly basic13

16. Buffers Buffer: a solution that resists changes in pH. A buffer is a mixture of a weak acid or base and its conjugate partner. HA + OH - → H 2 O + A - Weak acid reacts with any added OH -. A - + H 3 O → HA + H 2 O Weak base reacts with any added H 3 O +.

17. The pH of a Buffer System For the chemical reaction HA + H 2 O → H 3 O + + A -

18. The pH of a Buffer System Calculate the pH of a solution of 0.50 M HCN and 0.20 M NaCN, K a = 4.9 x 10 -10.

19. Calculate the pH of a solution of 0.40 M NH 3 and 0.10 M NH 4 Cl. For NH 3 K b = 1.8 x 10 -5. The pH of a Buffer System

20. Calculate the pH of a buffer that is 0.25 M HCN and 0.15 M NaCN, K a = 4.9 x 10 -10. Test Your Skill

21. Calculate the amount of sodium acetate that must be added to 250 mL of 0.16 M acetic acid in order to prepare a pH 4.68 buffer.K a = 1.8 x 10 -5 The Composition of a pH Buffer

22. Test Your Skill How many moles of NaCN should be added to 100 mL of 0.25 M HCN to prepare a buffer with pH = 9.40? K a = 4.9 x 10 -10.

23. Determining the Response of a Buffer to Added Acid or Base Calculate the initial and final pH when 10 mL of 0.100 M HCl is added to (a) 100 mL of water, and (b) 100 mL of a buffer which is 1.50 M CH 3 COOH and 1.20 M CH 3 COONa.

24. Test Your Skill Calculate the final pH when 10 mL of 0.100 M NaOH is added to 100 mL of a buffer which is 1.50 M CH 3 COOH and 1.25 M CH 3 COONa.

25. Before any base added. (a) Part way to equivalence point. (b) Equivalence point. (c) Beyond equivalence point. Qualitative Aspects: Titration: Weak Acid + Strong Base

26. Titration: Weak Acid + Strong Base HA + OH -  A - + H 2 O (a) Before any base is added the solution is a weak acid has a low pH. Estimated pH = 3  pH 2-4 is typical –Depends on the concentration of the acid. –Depends on the value of K a.

27. Titration: Weak Acid + Strong Base HA + OH -  A - + H 2 O (b) After some base is added, but before the equivalence point is reached. The solution is a mixture of the weak acid HA and its conjugate base A - ; therefore, the solution is a buffer. The estimated pH is equal to pK a.

28. Titration: Weak Acid + Strong Base HA + OH -  A - + H 2 O (c) At the equivalence point the solution is salt of A -, all the HA having been consumed by the stoichiometric amount of OH -. A - is the weak conjugate base of HA. The estimated pH is 10.

29. Titration: Weak Acid + Strong Base HA + OH -  A - + H 2 O (d) After excess strong base is added OH - is in excess. The estimated pH is 13.

30. pH Estimates SolutionEstimate of pH Strongly acidic1 Weakly acidic3 Neutral7 Weakly basic11 Strongly basic13 Buffer (acidic to basic) pK a 4-10 typical

31. Titration: Weak Acid + Strong Base

32. Titration Curves for Acids of Different Strengths

33. Calculate the pH in the titration of 20.00 mL of 0.500 M formic acid (HCOOH K a =1.8 x 10 -4 ) with 0.500 M NaOH after 0, 10.00, 20.00, and 30.00 mL of base have been added. The titration reaction is HCOOH + OH -  HCOO - + H 2 O Calculating the Titration Curve for a Weak Acid

34. Titration of 25.00 mL of 0.500 M Formic Acid with 0.500 M NaOH

35. Test Your Skill Calculate the pH in the titration of 12.00 mL of 0.100 M HOCl with 0.200 M NaOH after 0, 3.00, 6.00, and 9.00 mL of base have been added.

36. Titration of 20.00 mL of 0.500 M Methylamine with 0.500 M HCl

37. pH Indicators Indicator: a substance that changes color at the endpoint of a titration. pH indicators are weak acids or bases whose conjugate species are a different color.

38. HIn + H 2 O  H 3 O + + In - pK In = -log(K In ) pH Indicators

39. When pH is lower than pK In, the indicator will be in the acid form. When pH is greater than pK In, the indicator will be in the base form. An indicator should be chosen which changes at or just beyond the equivalence point. pH Indicators

40. * Thymol blue is polyprotic and has three color forms. NameAcid ColorBase ColorpH Rangep K In Thymol blue*RedYellow1.2–2.81.6 Methyl orangeRedYellow3.1–4.43.5 Methyl redRedYellow4.2–6.35.0 Bromthymol blueYellowBlue6.2–7.67.3 PhenolphthaleinClearPink8.3–10.08.7 Thymol blue*YellowBlue8.0–9.69.2 Properties of Indicators

41. Titration Curves for Strong and Weak Acids

42. Polyprotic acids provide more than one proton when they ionize. Polyprotic acids ionize in a stepwise manner. H 2 A + H 2 O ⇌ H 3 O + + HA - Step 1 HA - + H 2 O ⇌ H 3 O + + A 2- Step 2 Polyprotic Acids

43. There is a separate acid ionization constant for each step H 2 A + H 2 O ⇌ H 3 O + + HA - Step 1 HA - + H 2 O ⇌ H 3 O + + A 2- Step 2 Polyprotic Acids

44. HA - is the conjugate base of H 2 A, so it is a weaker acid than H 2 A. K a1 is always larger than K a2. For triprotic acids (such as H 3 PO 4 ), K a2 is always larger than K a3. Polyprotic Acids

45. When successive K a values differ by a factor of 1000 or more, each step can be assumed to be essentially unaffected by the occurrence of the subsequent step. Calculating Concentrations of Species in Polyprotic Acid Solutions

46. Calculate the concentrations of all species in 0.250 M malonic acid, K a = 1.6 x 10 -2. Consider the first ionization and solve by usual approach. H 2 C 3 H 2 O 4 + H 2 O ⇌ HC 3 H 2 O 4 - + H 3 O + Concentrations of Species in Polyprotic Acid Solutions

47. The second step is needed only to calculate the concentration of C 3 H 2 O 4 2- because the concentration of H 3 O + is determined by the first step. You can ignore the effect of the second step on the pH because the K a1 is so much larger than K a2. Concentrations of Species in Polyprotic Acid Solutions

48. Test Your Skill Calculate the pH of a 0.040 M solution of ascorbic acid. (K a1 = 8.0 x 10 -5, K a2 = 1.6 x 10 -12 )

49. Amphoteric Species Amphoteric: having both acidic and basic properties. Conjugate bases of weak polyprotic acids are amphoteric. The hydrogen oxalate ion, HC 2 O 4 -, is a weak acid (K a2 = 1.6 ×10 -4 ). HC 2 O 4 - + H 2 O ⇌ C 2 O 4 2- + H 3 O +

50. Amphoteric Species Weak Acid The hydrogen oxalate ion, HC 2 O 4 -, is a weak acid. HC 2 O 4 - + H 2 O ⇌ C 2 O 4 2- + H 3 O + K a2 = 1.6 x 10 -4 Weak Base The hydrogen oxalate ion, HC 2 O 4 -, can also act as a weak base. HC 2 O 4 - + H 2 O → H 2 C 2 O 4 + OH - K b = K w /K a1 = 1.0 x 10 -14 / 5.6 x 10 -2 K b = 1.9 x 10 -13 Since K a > K b, the ion will act as a weak acid in water. When comparing K a to K b note that K a is K a2 and K b is K w /K a1.

51. Test Your Skill K a for the hydrogen malonate ion, HC 3 H 2 O 4 -, is 2.1 x 10 -6. Is a solution of sodium hydrogen malonate acidic or basic?

52. Factors That Influence Solubility The pH affects the solubility of salts of weak acids. Complex ion formation affects the solubility of salts of transition metal cations.

53. Salts of Anions of Weak Acids The solubility of salts of anions of weak acids is enhanced by lowering the pH. Cd(CN) 2 (s) ⇌ Cd 2+ (aq) + 2CN - (aq) K sp = 1.0 x 10 -8 Adding acid reduces [CN - ] in solution, by the reaction H 3 O + (aq) + CN - (aq) ⇌ HCN (aq) + H 2 O (l)

54. Salts of Transition Metal Cations Transition metal cations form complexes with Lewis bases such as H 2 O, NH 3, or OH -. Formation of a complex reduces the concentration of metal ion and increases the solubility of the salt.

55. Solubility of Amphoteric Species Amphoteric species, such as Be(OH) 2, Al(OH) 3, Sn(OH) 2, Pb(OH) 2, Cr(OH) 3, Ni(OH) 2, Cu(OH) 2, Zn(OH) 2, and Cd(OH) 2 react with acid or base to form the soluble metal ion or complex ions M(OH) x + xH +  M x+ + xH 2 Ox = 2,3 M(OH) x + yOH -  M(OH) x y+ x = 2,3, y = 1,2