1. Chapter FifteenPrentice-Hall ©2002Slide 1 of 31 920203 http:\\asadipour.kmu.ac.ir........57 slides 1

2. Equilibrium In Solutions Of Weak Acids And Weak Bases weak acid:HA + H 2 O  H 3 O + + A - [H 3 O + ][A - ] K a = [HA] weak base:B + H 2 O  HB + + OH - [HB + ][OH - ] K b = [B] 920203 http:\\asadipour.kmu.ac.ir........57 slides 2

3. Some Acid-Base Equilibrium Calculations cHA≈[HA] [H 3 O + ][A - ] [H 3 O + ][A - ] K a = --------------------= ---------------- cHA –[H 3 O + ] cHA cHA > [HA] Analytical C> Equilibrium C - the calculations can be simplified. - When M acid /K a or M base /K b > 100, - When K a or K b <1×10 -4 (In usual Conc.) 920203 http:\\asadipour.kmu.ac.ir........57 slides 3

4. 920203 http:\\asadipour.kmu.ac.ir........57 slides 4

5. An Example 1. Determine the concentrations of H 3 O +, CH 3 COOH and CH 3 COO -, and the pH of 1.00 M CH 3 COOH solution. K a = 1.8 x 10 -5. 2. What is the pH of a solution that is 0.200 M in methylamine, CH 3 NH 2 ? K b = 4.2 x 10 -4. 920203 http:\\asadipour.kmu.ac.ir........57 slides 5

6. Are Salts Neutral, Acidic or Basic? Salts are ionic compounds formed in the reaction between an acid and a base. 1. NaCl Na + is from NaOH, a strong base Cl - is from HCl, a strong acid H 2 O NaCl (s) → Na + (aq) + Cl - (aq) Na + and Cl - ions do not react with water. The solution is neutral. 920203 http:\\asadipour.kmu.ac.ir........57 slides 6

7. Are Salts Neutral, Acidic or Basic ? 2.KCN K + is from KOH, a strong base CN - is from HCN, a weak acid H 2 O KCN (s) → K + (aq) + CN - (aq) K + ions do not react with water, but CN - ions do. CN - + H 2 O  HCN + OH - hydrolysis The OH - ions are produced, so the solution is basic. 920203 http:\\asadipour.kmu.ac.ir........57 slides 7

8. Are Salts Neutral, Acidic or Basic? 3.NH 4 Cl NH 4 + is from NH 3, a weak base Cl - is from HCl, a strong acid H 2 O NH 4 Cl (s) → NH 4 + (aq) + Cl - (aq) Cl - ions do not react with water, but NH 4 + ions do. NH 4 + + H 2 O  H 3 O + + NH 3 hydrolysis The H 3 O + ions are produced, so the solution is acdic. 920203 http:\\asadipour.kmu.ac.ir........57 slides 8

9. Are Salts Neutral, Acidic or Basic? 3.NH 4 CN NH 4 + is from NH 3, a weak base CN - is from HCN, a weak acid H 2 O NH 4 CN (s) → NH 4 + (aq) + CN - (aq) NH 4 + + H 2 O  H 3 O + + NH 3 K a hydrolysis CN - + H 2 O  HCN + OH - K b hydrolysis (K a >K b,Acidic)’’’(K a < K b,Basic)‘’’ (K a = K b,Nutral) 920203 http:\\asadipour.kmu.ac.ir........57 slides 9

10. Ions As Acids And Bases Certain and anion ions can cause an aqueous solution to become acidic or basic due to hydrolysis. Salts of strong acids and strong bases form neutral solutions. Salts of weak acids and strong bases form basic solutions. Salts of strong acids and weak bases form acidic solutions. Salts of weak acids and weak bases form solutions that are acidic in some cases, neutral or basic in others. 920203 http:\\asadipour.kmu.ac.ir........57 slides 10

11. Strong Acids And Strong Bases Strong acids: HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4 Strong bases: Group IA and IIA hydroxides 920203 http:\\asadipour.kmu.ac.ir........57 slides 11

12. An Example Indicate whether the solutions (a) Na 2 S and (b) KClO 4 are acidic, basic or neutral. 920203 http:\\asadipour.kmu.ac.ir........57 slides 12

13. The pH Of a Salt Solution What is the pH of 0.1M NaCN solution? What is the pH of 0.1M NH 4 Cl solution? What is the pH of 0.1M NH 4 CN solution? K a of HCN=1.0×10 -9. K b for NH3=1.0×10 -5 K a x K b = K w so, K b = K w /K a 920203 http:\\asadipour.kmu.ac.ir........57 slides 13

14. Chapter FifteenPrentice-Hall ©2002Slide 14 of 31 920203 http:\\asadipour.kmu.ac.ir........57 slides 14

15. Common Ion Effect Illustrated ((1.00 M CH 3 COOH)) ((1.00 M CH 3 COOH + 1.00 M CH 3 COONa)) yellow: pH < 3.0 blue-violet: pH > 4.6 920203 http:\\asadipour.kmu.ac.ir........57 slides 15 CH 3 COOH  CH 3 COO - + H +

16. The Common Ion Effect Calculate the pH of 0.10 M CH 3 COOH solution. K a of CH 3 COOH=1.0×10 -5 Calculate the pH of 0.10 M CH 3 COONa solution. Calculate the pH of 0.10 M CH 3 COOH/ 0.10 M CH 3 COONa solution. 920203 http:\\asadipour.kmu.ac.ir........57 slides 16

17. Depicting Buffer Action 920203 http:\\asadipour.kmu.ac.ir........57 slides 17

18. Buffer Solutions A buffer solution is a solution that changes pH only slightly when small amounts of a strong acid or a strong base are added. A buffer contains CH 3 COOH  CH 3 COO - Acidic buffer NH 3  NH 4 + Alkalin buffer 920203 http:\\asadipour.kmu.ac.ir........57 slides 18

19. How A Buffer Solution Works The acid component of the buffer can neutralize small added amounts of OH -, and the basic component can neutralize small added amounts of H 3 O +. CH 3 COOH  CH 3 COO - + H + 920203 http:\\asadipour.kmu.ac.ir........57 slides 19

20. Chapter FifteenPrentice-Hall ©2002Slide 20 of 31 Ionization constant of an acid Taking log of the equation on both sides, 920203 http:\\asadipour.kmu.ac.ir........57 slides 20

21. Chapter FifteenPrentice-Hall ©2002Slide 21 of 31 Ionization constant of an acid Multiplying both sides of the equation by -1 920203 http:\\asadipour.kmu.ac.ir........57 slides 21 Henderson-Hasselbach equation

22. Henderson-Hasselbalch Equation For Buff Solutions [conjugate base] pH = pK a + log [conjugate acid] If [conjugate acid] = [conjugate base], pH = pK a Requirement: -[B] / [A] between 0.10 and 10 920203 http:\\asadipour.kmu.ac.ir........57 slides 22

23. Buffer Capacity There is a limit to the capacity of a buffer solution to neutralize added acid or base, and this limit is reached before all of one of the buffer components has been consumed. In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize. As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal. [B]=[A] [Buffer]=[Acid]+[Base] 920203 http:\\asadipour.kmu.ac.ir........57 slides 23

24. Buffer Capacity [Buffer]=[Acid]+[Base] [Acid] ↑ & [Base] ↑ Capacity ↑ In equimolar buffersis is important Capacity ↑ 920203 http:\\asadipour.kmu.ac.ir........57 slides 24

25. Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH 3 (pK b =5)and 1.0 M NH 4 Cl. (a)What is the pH of this buffer? 920203 http:\\asadipour.kmu.ac.ir........57 slides 25

26. Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH 3 (pK b =5)and 1.0 M NH 4 Cl. (a)What is the pH of this buffer? (b)If 5 mmol NaOH is added to 0.500 L of this solution, what will be the pH? 920203 http:\\asadipour.kmu.ac.ir........57 slides 26

27. Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH 3 (pK b =5)and 1.0 M NH 4 Cl. (a)What is the pH of this buffer? (b)If 5 mmol NaOH is added to 0.500 L of this solution, what will be the pH? (c)If 5 mmol HCl is added to 0.500 L of this solution, what will be the pH? 920203 http:\\asadipour.kmu.ac.ir........57 slides 27

28. Calculations in Buffer Solutions 1) A buffer solution is 0. 1 M NH 3 (pK b =5)and 1.0 M NH 4 Cl. (a)What is the pH of this buffer? (b)If 5 mmol NaOH is added to 0.500 L of this solution, what will be the pH? (c)If 5 mmol HCl is added to 0.500 L of this solution, what will be the pH? (d)If 5 mmol NH 4 Cl is added to 0.500 L of this solution, what will be the pH? 920203 http:\\asadipour.kmu.ac.ir........57 slides 28

29. Chapter FifteenPrentice-Hall ©2002Slide 29 of 31 Calculations in Buffer Solutions 2) What concentration of acetate ion in 500 ml of 0.500 M CH 3 COOH (pK a =5) produces a buffer solution with pH = 4.00? 920203 http:\\asadipour.kmu.ac.ir........57 slides 29

30. Calculations in Buffer Solutions 2) What concentration of acetate ion in500 ml of 0.500 M CH 3 COOH (pK a =5) produces a buffer solution with pH = 4.00? 920203 http:\\asadipour.kmu.ac.ir........57 slides 30 How many mg?

31. Acid-Base Indicators An acid-base indicator is a weak acid having one color and the conjugate base of the acid having a different color. One of the “colors” may be colorless. HIn + H 2 O  H 3 O + + In - Acid-base indicators are often used for applications in which a precise pH reading isn’t necessary. A common indicator used in chemistry laboratories is Phenolphetalein. 920203 http:\\asadipour.kmu.ac.ir........57 slides 31

32. Chapter FifteenPrentice-Hall ©2002Slide 32 of 31 920203 http:\\asadipour.kmu.ac.ir........57 slides 32

33. Chapter FifteenPrentice-Hall ©2002Slide 33 of 31 920203 http:\\asadipour.kmu.ac.ir........57 slides 33

34. Neutralization Reactions Neutralization is the reaction of an acid and a base. Titration is a common technique for conducting a neutralization. At the equivalence point in a titration, the acid and base have been brought together in exact stoichiometric proportions. The point in the titration at which the indicator changes color is called the end point. The indicator endpoint and the equivalence point for a neutralization reaction can be best matched by plotting a titration curve, a graph of pH versus volume of titrant. In a typical titration, 50 mL or less of titrant that is 1 M or less is used. 920203 http:\\asadipour.kmu.ac.ir........57 slides 34

35. Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the some points and draw the curve. 4 essential points. 1)initial point 2)equivalence point 3)before the equivalence point 4)beyond the equivalence point Ml تیترانت pH محیط 0 15 19 19.5 19.9 20 20.1 20.5 21 25 40 920203 http:\\asadipour.kmu.ac.ir........57 slides 35

36. Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O 4 questions. 1)What are the present compounds? 2)Which of them is effective on pH? 3)How much are the concentrations? 4)What is the relationship between their Conc. And pH? 920203 http:\\asadipour.kmu.ac.ir........57 slides 36

37. Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. (a)initial pH. (Before the addition of any NaOH). Answer Q1. There are:HCl & H2O Answer Q2. HCl Answer Q3. [HCl] Answer Q4. pH=-log[H + ] 920203 http:\\asadipour.kmu.ac.ir........57 slides 37

38. Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. b)equivalence point. Answer Q1. There are:NaCl & H 2 O Answer Q2. H 2 O Answer Q3. Answer Q4. pH=7 920203 http:\\asadipour.kmu.ac.ir........57 slides 38

39. Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. c)before the equivalence point. Answer Q1. There are:HCl,NaCl & H 2 O Answer Q2. HCl Answer Q3. Answer Q4. [H + ]=N pH=-log[H + ] 920203 http:\\asadipour.kmu.ac.ir........57 slides 39

40. Drawing titration Curve For Strong Acid - Strong Base HCl + NaOH → NaCl +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH. d)after the equivalence point. Answer Q1. There are:NaOH,NaCl & H 2 O Answer Q2. NaOH Answer Q3. Answer Q4. [OH - ]=N pOH=-log[OH - ] pH=14-pOH 920203 http:\\asadipour.kmu.ac.ir........57 slides 40

41. Titration Curve For Strong Acid - Strong Base pH is low at the beginning. pH changes slowly until just before equivalence point. pH changes sharply around equivalence point. pH = 7.0 at equivalence point. Further beyond equivalence point, pH changes slowly. Any indicator whose color changes in pH range of 4 – 10 can be used in titration. 920203 http:\\asadipour.kmu.ac.ir........57 slides 41

42. Chapter FifteenPrentice-Hall ©2002Slide 42 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the some points and draw the curve. K a =1×10 -5 5 essential points. 1)initial point 2)equivalence point 3)beyond the initial point 4)before the equivalence point 5)beyond the equivalence point 920203 http:\\asadipour.kmu.ac.ir........57 slides 42

43. Chapter FifteenPrentice-Hall ©2002Slide 43 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O 4 questions. 1)What are the present compounds? 2)Which of them is effective on pH? 3)How much are the concentrations? 4)What is the relationship between their Conc. And pH? 920203 http:\\asadipour.kmu.ac.ir........57 slides 43

44. Chapter FifteenPrentice-Hall ©2002Slide 44 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. (a)initial pH. (Before the addition of any NaOH). Answer Q1. There are: CH 3 COOH & H 2 O Answer Q2. CH 3 OOH Answer Q3. CH 3 OOH Answer Q4. pH=-log[H + ] 920203 http:\\asadipour.kmu.ac.ir........57 slides 44

45. Chapter FifteenPrentice-Hall ©2002Slide 45 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. b)equivalence point. Answer Q1. There are: CH3COO -, Na + & H 2 O Answer Q2. CH3COO - Answer Q3. Answer Q4. pOH=-log[OH - ] K a ×K b =K w 920203 http:\\asadipour.kmu.ac.ir........57 slides 45

46. Chapter FifteenPrentice-Hall ©2002Slide 46 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. c)beyond the initial point. Answer Q1. There are: CH 3 COOH, CH3COO -,Na + & H 2 O Answer Q2. CH 3 COOH, CH3COO - Answer Q3. Answer Q4. 920203 http:\\asadipour.kmu.ac.ir........57 slides 46

47. Chapter FifteenPrentice-Hall ©2002Slide 47 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. d)before the equivalence point. Answer Q1. There are: CH 3 COOH, CH3COO -,Na + & H 2 O Answer Q2. CH 3 COOH, CH3COO - Answer Q3. Answer Q4. 920203 http:\\asadipour.kmu.ac.ir........57 slides 47

48. Chapter FifteenPrentice-Hall ©2002Slide 48 of 31 Drawing titration Curve For weak acid- Strong Base CH 3 COOH + NaOH → CH3COO - + Na + +H 2 O Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH. e)after the equivalence point. Answer Q1. There are:NaOH, CH3COO -, Na + & H 2 O Answer Q2. NaOH Answer Q3. Answer Q4. [OH - ]=N pOH=-log[OH - ] pH=14-pOH 920203 http:\\asadipour.kmu.ac.ir........57 slides 48

49. Titration Curve For Weak Acid - Strong Base The initial pH is higher because weak acid is partially ionized. At the half-neutralization point, pH = pKa. pH >7 at equivalence point because the anion of the weak acid hydrolyzes. The steep portion of titration curve around equivalence point has a smaller pH range. The choice of indicators for the titration is more limited. 920203 http:\\asadipour.kmu.ac.ir........57 slides 49

50. Chapter FifteenPrentice-Hall ©2002Slide 50 of 31 920203 http:\\asadipour.kmu.ac.ir........57 slides 50

51. Chapter FifteenPrentice-Hall ©2002Slide 51 of 31 Application of K a The K a of nicotinic acid, HNic, is 1.4e-5. A solution containing 0.22 M HNic. What is its pH? What is the degree of ionization? Solution: HNic  H + + Nic – 0.22-x x x x 2 K a = ———— = 1.4e-5 0.22 – x(use approximation, small indeed) x =  (0.22*1.4e-5) = 0.0018pH = – log (0.0018) = 2.76 Degree of ionization = 0.0018 / 0.22 = 0.0079 = 0.79% 920203 http:\\asadipour.kmu.ac.ir........57 slides 51

52. Chapter FifteenPrentice-Hall ©2002Slide 52 of 31 Determine K a and percent ionization Nicotinic acid, HNic, is a monoprotic acid. A solution containing 0.012 M HNic, has a pH of 3.39. What is its K a ? What is the percent of ionization? Solution: HNic  H + + Nic – 0.012-x x x x = [H + ] = 10 –3.39 = 4.1e-4 [HNic] = 0.012 – 0.00041 = 0.012 (4.1e-4) 2 K a = ————— = 1.4e-5 0.012 Degree of ionization = 0.00041 / 0.012 = 0.034 = 3.4% 920203 http:\\asadipour.kmu.ac.ir........57 slides 52

53. Chapter FifteenPrentice-Hall ©2002Slide 53 of 31 Using the quadratic formula The K a of nicotinic acid, HNic, is 1.4e-5. A solution containing 0.00100 M HNic. What is its pH? What is the degree of ionization? Solution: HNic  H + + Nic – 0.001-x x x x 2 K a = —————— = 1.4e-5x 2 + 1.4e-5 x – 1.4e-8 = 0 0.00100 – x –1.4e–5 +  (1.4e–5) 2 + 4*1.4e-8 x = —————————————————— = 0.000111 M 2 pH = – log (0.000111) = 3.95 Degree of ionization = 0.000111/ 0.001 = 0.111 = 11.1% 920203 http:\\asadipour.kmu.ac.ir........57 slides 53

54. Chapter FifteenPrentice-Hall ©2002Slide 54 of 31 Degree of or percent ionization The degree or percent of ionization of a weak acid always decreases as its concentration increases, as shown from the table given earlier. Concentration of acid % ionization 920203 http:\\asadipour.kmu.ac.ir........57 slides 54 Deg.’f ioniz’n 0.220 0.8% 0.012 3.4 % 0.001 11.1 %

55. Chapter FifteenPrentice-Hall ©2002Slide 55 of 31 Polyprotic acids Polyprotic acids such as sulfuric and carbonic acids have more than one hydrogen to donate. H 2 SO 4 → H + + HSO 4 – K a1 very large completely ionized HSO 4 –  H + + SO 4 2– K a2 = 0.012 H 2 CO 3  H + + HCO 3 – K a1 = 4.3e-7 HCO 3 –  H + + CO 3 2– K a2 = 4.8e-11 Ascorbic acid (vitamin C) is a diprotic acid, abundant in citrus fruit. Others: H 2 S, H 2 SO 3, H 3 PO 4, H 2 C 2 O 4 (oxalic acid) … 920203 http:\\asadipour.kmu.ac.ir........57 slides 55

56. Chapter FifteenPrentice-Hall ©2002Slide 56 of 31 Species concentrations of diprotic acids Evaluate concentrations of species in a 0.10 M H 2 SO 4 solution. Solution: H 2 SO 4 → H + + HSO 4 – completely ionized (0.1–0.1) 0.10 0.10 HSO 4 –  H + + SO 4 2– K a2 = 0.012 0.10–y 0.10+y yAssume y = [SO 4 2– ] (0.10+y) y ————— = 0.012 (0.10-y) [SO 4 2– ] = y = 0.01M [H+] = 0.10 + 0.01 = 0.11 M; [HSO 4 – ] = 0.10-0.01 = 0.09 M 920203 http:\\asadipour.kmu.ac.ir........57 slides 56

57. Chapter FifteenPrentice-Hall ©2002Slide 57 of 31 Species concentrations of weak diprotic acids Evaluate concentrations of species in a 0.10 M H 2 S solution. Solution: H 2 S = H + + HS – K a1 = 1.02e-7 (0.10–x) x+y x-yAssume x = [HS – ] HS – = H + + S 2– K a2 = 1.0e-13 x–y x+y yAssume y = [S 2– ] (x+y) (x-y) (x+y) y ————— = 1.02e-7 ———— = 1.0e-13 (0.10-x)(x-y) [H 2 S] = 0.10 – x = 0.10 M [HS – ] = [H + ] = x  y = 1.0e–4 M; [S 2– ] = y = 1.0e-13 M 0.1>> x >> y: x+ y = x-y = x x =  0.1*1.02e-7 = 1.00e-4 y = 1e-13 920203 http:\\asadipour.kmu.ac.ir........57 slides 57