1. 119 Ionic Equilibria 離子平衡 : Part II Buffers 緩衝液 and Titration Curves 滴定曲線 花青素

2. 2 Chapter Goals 1.The Common Ion Effect and Buffer Solutions ( 共同離子效應及緩衝溶液 ) 2.Buffering Action ( 緩衝作用 ) 3.Preparation of Buffer Solutions ( 緩衝溶液的製備 ) 4.Acid-Base Indicators ( 酸鹼指示劑 ) Titration Curves ( 滴定曲線 ) 5.Strong Acid/Strong Base Titration Curves 6.Weak Acid/Strong Base Titration Curves 7.Weak Acid/Weak Base Titration Curves 8.Summary of Acid-Base Calculations

3. 3 The Common Ion Effect and Buffer Solutions Common ion effect 共同離子效應 – When a solution of a weak electrolyte is altered by adding one of its ions from another source, the ionization of the weak electrolyte is suspressed.( 當弱電解質溶液中加 入具有相同離子的強電解質時，弱電解質的電離平衡 會移動，使弱電解質的電離度下降，這種現象叫共同 離子效應 ) If a solution is made in which the same ion is produced by two different compounds the common ion effect is exhibited. Buffer solutions are solutions that resist changes in pH when acids or bases are added to them. – Buffering is due to the common ion effect.

4. 4 The Common Ion Effect and Buffer Solutions Buffer solution 緩衝溶液 – Resists changes in pH when strong acids or strong bases are added.( 當加入強酸或強鹼時,pH 的變化不大 ) – contains a conjugate acid-base pair in reasonable concentrations. It can react with added base or acid ( 具共軛酸鹼對 ) – Buffer solution contain ( 緩衝溶液含有 :) A weak acid and a soluble ionic salt of the weak acid ( 弱酸和具此弱酸根的可溶性鹽類 ) – CH 3 COOH plus NaCH 3 COO A weak base and a soluble ionic salt of the weak base ( 弱鹼和具此弱鹼根的可溶性鹽類 ) – NH 3 plus NH 4 Cl

5. 5 The Common Ion Effect and Buffer Solutions 1.Solutions made of weak acids plus a soluble ionic salt of the weak acid – Solution that contain a weak acid plus a salt of weak acid are always less acidic than solutions that contain the same concentration of weak acid alone – One example of this type of buffer system is: The weak acid - acetic acid CH 3 COOH The soluble ionic salt - sodium acetate NaCH 3 COO CH 3 COOH  H + + CH 3 COO - 部分解離 Na + CH 3 COO - Na + + CH 3 COO - 完全解離  100% [CH 3 COO - ] 增加  反應向左  減少 [H + ]  pH 增加

6. 6 The Common Ion Effect and Buffer Solutions Example 19-1: Calculate the concentration of H + and the pH of a solution that is 0.15 M in acetic acid and 0.15 M in sodium acetate. – This is another equilibrium problem with a starting concentration for both the acid and anion. CH 3 COOH  H + + CH 3 COO - Na + CH 3 COO - Na + + CH 3 COO -  100% (0.15-x) Mx M 0.15 M Ka=Ka= [H + ] [CH 3 COO - ] [CH 3 COOH] =1.8x10 -5 = (x)(0.15+x) (0.15-x) (0.15+x)  0.15 and (0.15-x)  0.15 = 1.8x10 -5 (x)(0.15) (0.15) x =1.8x10 -5 = [H + ] pH=4.74

7. 7 The Common Ion Effect and Buffer Solutions Compare the acidity of a pure acetic acid solution and the buffer described in Example 19-1. Solution[H + ]pH 0.15 M CH 3 COOH 1.6 x 10 -3 2.80 0.15 M CH 3 COOH & 0.15 M NaCH 3 COO buffer 1.8 x 10 -5 4.74 [H + ] is 89 times greater in pure acetic acid than in buffer solution.

8. 8 The Common Ion Effect and Buffer Solutions The general expression for the ionization of a weak monoprotic acid is: The generalized ionization constant expression for a weak acid is: HA  H + + A - Ka=Ka= [H + ] [A - ] [HA]

9. 9 The Common Ion Effect and Buffer Solutions If we solve the expression for [H + ], this relationship results: By making the assumption that the concentrations of the weak acid and the salt are reasonable, the expression reduces to: [H + ] = K a x [A - ] [HA] acid salt [H + ] = K a x [salt] [acid] The relationship developed in the previous slide is valid for buffers containing a weak mono protic acid and a soluble, ionic salt. If the salt’s cation is not univalent the relationship changes to: [H + ] = K a x n[salt] [acid] Where n=charge on cation

10. 10 The Common Ion Effect and Buffer Solutions Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation The Henderson-Hasselbach equation is one method to calculate the pH of a buffer given the concentrations of the salt and acid. log [H + ] =log K a + log [salt] [acid] Multiply by -1 -log [H + ] =-log K a + log [acid] [salt] pH= pK a + log [acid] [salt]

11. Example 19-1-1: Use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution in Example 19-1 The Common Ion Effect and Buffer Solutions pH= pK a + log [acid] [salt] pH= 4.74 + log 0.15M Example 19-1: Calculate the concentration of H + and the pH of a solution that is 0.15 M in acetic acid and 0.15 M in sodium acetate. pKa =-logKa =-log(1.8x10 -5 ) = 4.74 pH= 4.74 + log1 = 4.74 + 0 = 4.74 [H + ] = K a x [salt] [acid] [H + ] = 1.8x10 -5 x (0.15) [H + ] = 1.8x10 -5 pH=4.74

12. 12 Weak Bases plus Salts of Weak Bases 2.Buffers that contain a weak base plus the salt of a weak base One example of this buffer system is ammonia plus ammonium nitrate. NH 3 + H 2 O  NH 4 + + OH - NH 4 NO 3 NH 4 + + NO 3 -  100% Kb=Kb= [NH 4 + ] [OH - ] [NH 3 ] =1.8x10 -5

13. 13 Weak Bases plus Salts of Weak Bases Example 19-2: Calculate the concentration of OH - and the pH of the solution that is 0.15 M in aqueous ammonia, NH 3, and 0.30 M in ammonium nitrate, NH 4 NO 3. NH 3 + H 2 O  NH 4 + + OH - NH 4 NO 3 NH 4 + + NO 3 -  100% Kb=Kb= [NH 4 + ] [OH - ] [NH 3 ] (0.15-x) Mx M 0.30 M =1.8x10 -5 = (x)(0.30+x) (0.15-x) (0.30+x)  0.30 and (0.15-x)  0.15 = 1.8x10 -5 (x)(0.30) (0.15) x =9.0x10 -6 = [OH - ] pOH=5.05, pH=8.95

14. 14 Weak Bases plus Salts of Weak Bases A comparison of the aqueous ammonia concentration to that of the buffer described above shows the buffering effect. Solution[OH - ]pH 0.15 M NH 3 1.6 x 10 -3 M 11.20 0.15 M NH 3 & 0.15 M NH 4 NO 3 buffer 9.0 x 10 -6 M 8.95 The [OH - ] in aqueous ammonia is 180 times greater than in the buffer.

15. 15 Weak Bases plus Salts of Weak Bases We can derive a general relationship for buffer solutions that contain a weak base plus a salt of a weak base similar to the acid buffer relationship. – The general ionization equation for weak bases is: B + H 2 O  BH + + OH - Where B represents a weak base The general form of the ionization expression is: Solve for the [OH - ] Kb=Kb= [BH + ] [OH - ] [B] [OH - ] = K b x [BH + ] [B] base salt

16. 16 Weak Bases plus Salts of Weak Bases For salts that have univalent ions: For salts that have divalent or trivalent ions: [OH - ] = K b x [salt] [base] [OH - ] = K b x n[salt] [base] Where n= charge on anion

17. 17 Weak Bases plus Salts of Weak Bases Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation log [OH - ] =log K b + log [salt] [base] Multiply by -1 -log [OH - ] =-log K b + log [base] [salt] pOH= pK b + log [base] [salt]

18. 18 Example 19-3: If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the HCl. 1. Calculate the pH of the original buffer solution. [OH - ] = K b x [NH 4 Cl] [NH 3 ] [OH - ] = 1.8x10 -5 x 0.20M 0.10M [OH - ] = 9.0x10 -6 pOH=5.05 pH=8.95 2. Next, calculate the concentration of all species after the addition of the gaseous HCl. – The HCl will react with some of the ammonia and change the concentrations of the species. – This is another limiting reactant problem. HCl + NH 3 → NH 4 Cl Initial0.02 mol0.1 mol0.2 mol change-0.02 mol +0.02 mol After rxn0 mol0.08 mol0.22 mol M NH 3 = 1.0L 0.08mol =0.08M M NH 4 Cl ==0.22M [OH - ] = K b x [NH 4 Cl] [NH 3 ] = 1.8x10 -5 x 0.22M 0.08M = 6.5x10 -6 pOH=5.19 pH=8.81  pH=pH new -pH original =8.81-8.95=-0.14 1.0L 0.22mol ?mol NH 3 =0.1M x 1L=0.1mol?mol NH 4 Cl=0.2M x 1L=0.2mol

19. 19 Buffering Action Example 19-4: If 0.020 mole of NaOH is added to 1.00 liter of solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the solid NaOH. pH of the original buffer solution is 8.95, from above. 1. First, calculate the concentration of all species after the addition of NaoH. – NaOH will react with some of the ammonium chloride. – The limiting reactant is the NaOH. NH 4 Cl + NaOH → NH 3 + H 2 O + NaCl Initial0.2 mol0.02 mol0.1 mol change-0.02 mol +0.02 mol After rxn0.18 mol0 mol0.12 mol M NH 3 = =0.12M M NH 4 Cl = 1.0L 0.18mol =0.18M [OH - ] = K b x [NH 4 Cl] [NH 3 ] = 1.8x10 -5 x 0.18M 0.12M = 1.2x10 -5 pOH=4.92 pH=9.08  pH=pH new -pH original =9.08-8.95=0.13 1.0L 0.12mol

20. 20 Buffering Action This table is a summary of examples 19-3 and 19-4. Notice that the pH changes only slightly in each case. Original Solution Original pH Acid or base added New pH  pH 1.00 L of solution containing 0.100 M NH 3 and 0.200 M NH 4 Cl 8.95 0.020 mol NaOH 9.08+0.13 0.020 mol HCl 8.81-0.14

21. 21 Preparation of Buffer Solutions Example 19-5: Calculate the concentration of H + and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions. Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction. ?mmol CH 3 COOH =200ml x 0.15M = 30.0 m mol ?mmol NaOH =100ml x 0.1M = 10.0 m mol NaOH + CH 3 COOH → NaCH 3 COO + H 2 O Initial10.0 mmol30.0 mmol change-10.0 mmol +10.0 mmol After rxn0.0 mmol20.0 mmol10.0 mmol After the two solutions are mixed, the total volume of the solution is 300 mL (100 mL of NaOH + 200 mL of acetic acid). – The concentrations of the acid and base are: M CH 3 COOH = =0.0667M M NaCH 3 COO = =0.0333M 300mL 20 mmol 300mL 10 mmol Ka=Ka= [H + ] [CH 3 COO - ] [CH 3 COOH] =1.8x10 -5 = [H + ] [0.0333] [0.0667] [H + ]=3.6x10 -5 M pH =4.44

22. 22 For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH. Example 19-6: Calculate the number of moles of solid ammonium chloride, NH 4 Cl, that must be used to prepare 1.00 L of a buffer solution that is 0.10 M in aqueous ammonia, and that has a pH of 9.15. Because pH = 9.15Then pOH = 14.00 - 9.15 = 4.85 [OH - ] = 10 -4.85 = 1.4x10 -5 M NH 3 + H 2 O  NH 4 + + OH - NH 4 Cl → NH 4 + + Cl - (0.1-1.4x10 -5 ) M(1.4x10 -5 ) M x M Kb=Kb= [NH 4 + ] [OH - ] [NH 3 ] = 1.8x10 -5 Kb=Kb= (1.4x10 -5 +x)(1.4x10 -5 ) (0.1-1.4x10 -5 ) = (x)(1.4x10 -5 ) (0.1) = 1.8x10 -5 x= 0.13 M = [NH 4 Cl] original 0.13Mx1L =0.13 mol

23. 23 Acid-Base Indicators The point in a titration at which chemically equivalent amounts of acid and base have reacted is called the equivalence point 當量點. The point in a titration at which a chemical indicator changes color is called the end point 終點.

24. Three common indicators in solutions that cover the pH range 3 to 11. a. Methyl red 甲基紅 Red at pH4 and below Yellow at pH 7 and above Red → orange → yellow b. Bromthymol blue 溴瑞香草酚藍 yellow at pH6 and below blue at pH 8 and above yellow → green → blue c. Phenolphthalein 酚酞 (most common use) Colorless below pH8 Bright pink above pH10

25. 25 Acid-Base Indicators Many acid-base indicators are weak organic acid, HIn, where “In” represents various complex organic gups A symbolic representation of the indicator’s color change at the end point is: HIn  H + + In - color1 color2 The equilibrium constant expression for an indicator would be expressed as: Ka=Ka= [H + ] [In - ] [HIn] HIn represents nonionized acid molecules In - represents the anion (conjugate base) of HIn [In - ] [HIn] = Ka [H + ]

26. 26 Acid-Base Indicators Color change ranges of some acid-base indicators Indicator Color in acidic rangepH range Color in basic range Methyl violet 甲基紫 Yellow0 - 2Purple Methyl orangePink3.1 – 4.4Yellow Litmus 石蕊 Red4.7 – 8.2Blue PhenolphthaleinColorless8.3 – 10.0Red

28. 28 Titration Curves Strong Acid/Strong Base Titration Curves – These graphs are a plot of pH vs. volume of acid or base added in a titration. – As an example, consider the titration of 100.0 mL of 0.100 M perchloric acid with 0.100 M potassium hydroxide. In this case, we plot pH of the mixture vs. mL of KOH added. Note that the reaction is a 1:1 mole ratio. HClO 4 + KOH → KClO 4 + H 2 O

29. 29 Strong Acid/Strong Base Titration Curves Before any KOH is added the pH of the HClO 4 solution is. – Remember perchloric acid is a strong acid that ionizes essentially 100%. HClO 4 H + + ClO 4 -  100% 0.10M [H + ] =0.10M pH =-log(0.10)=1.0

30. 30 Strong Acid/Strong Base Titration Curves After a total of 20.0 mL 0.100 M KOH has been added the pH of the reaction mixture is ___? HClO 4 + KOH → KClO 4 + H 2 O Start10.0 mmol2.0 mmol change-2.0 mmol +2.0 mmol After rxn8.0 mmol0.0 mmol2.0 mmol ? mmol KOH = 20ml x (0.1M) =2.0 mmol ? mmol HClO 4 = 100ml x (0.1M) =10.0 mmol M HClO 4 = =0.067M 120mL 8.0 mmol [H + ]=0.067M pH = 1.17

31. 31 Strong Acid/Strong Base Titration Curves After a total of 50.0 mL of 0.100 M KOH has been added the pH of the reaction mixture is ___? HClO 4 + KOH → KClO 4 + H 2 O Start10.0 mmol5.0 mmol change-5.0 mmol +5.0 mmol After rxn5.0 mmol0.0 mmol5.0 mmol M HClO 4 = =0.033M 150mL 5.0 mmol [H + ]=0.033M pH = 1.48 ? mmol KOH = 50ml x (0.1M) =5.0 mmol

32. 32 Strong Acid/Strong Base Titration Curves After a total of 90.0 mL of 0.100 M KOH has been added the pH of the reaction mixture is ____? HClO 4 + KOH → KClO 4 + H 2 O Start10.0 mmol9.0 mmol change-9.0 mmol +9.0 mmol After rxn1.0 mmol0.0 mmol9.0 mmol M HClO 4 = =0.0053M 190mL 1.0 mmol [H + ]=0.0053M pH = 2.28 ? mmol KOH = 90ml x (0.1M) =9.0 mmol

33. 33 Strong Acid/Strong Base Titration Curves After a total of 100.0 mL of 0.100 M KOH has been added the pH of the reaction mixture is ___? HClO 4 + KOH → KClO 4 + H 2 O Start10.0 mmol change-10.0 mmol +10.0 mmol After rxn0.0 mmol 10.0 mmol No acid or base  Neutral  pH=7.0 ? mmol KOH = 100ml x (0.1M) =10.0 mmol

34. 34 Strong acid-strong base

35. 35 Strong Acid/Strong Base Titration Curves We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

36. 36 Weak Acid/Strong Base Titration Curves As an example, consider the titration of 100.0 mL of 0.100 M acetic acid, CH 3 COOH, (a weak acid) with 0.100 M KOH (a strong base). – The acid and base react in a 1:1 mole ratio. CH 3 COOH + KOH → K + CH 3 COO - + H 2 O 1mol 1mmol Before the equivalence point is reached, both CH 3 COOH and KCH 3 COO are present in solution forming a buffer. – The KOH reacts with CH 3 COOH to form KCH 3 COO. – A weak acid plus the salt of a weak acid form a buffer. Hypothesize how the buffer production will effect the titration curve.

37. 37 Weak Acid/Strong Base Titration Curves Before the equivalence point is reached, both CH 3 COOH and KCH 3 COO are present in solution forming a buffer. – The KOH reacts with CH 3 COOH to form KCH 3 COO. A weak acid plus the salt of a weak acid form a buffer. Hypothesize how the buffer production will effect the titration curve.

38. 38 1. Determine the pH of the acetic acid solution before the titration is begun. CH 3 COOH  CH 3 COO - +H + (0.1-x) Mx M Ka=Ka= [H + ] [CH 3 COO - ] [CH 3 COOH] = 1.8x10 -5 = (x)(x) (0.1-x) x 2 = 1.8x10 -6 x= 1.3x10 -3 =[H + ] pH= 2.89 After a total of 20.0 mL of KOH solution has been added, the pH is: KOH + CH 3 COOH → K + CH 3 COO - + H 2 O Start2.0 mmol10.0 mmol change-2.0 mmol +2.0 mmol After rxn0.0 mmol8.0 mmol2.0 mmol M CH 3 COOH = =0.067M 120mL 8.0 mmol ? mmol KOH = 20ml x (0.1M) =2.0 mmol ? mmol CH 3 COOH = 100ml x (0.1M) =10.0 mmol M CH 3 COO - = =0.017M 120mL 2.0 mmol = (1.8x10 -5 ) x [H + ] = K a x [CH 3 COOH] [CH 3 COO - ] 0.067 0.017 = 7.1x10 -5 pH= 4.15

39. 39 Weak Acid/Strong Base Titration Curves At the equivalence point, the solution is 0.500 M in KCH 3 COO, the salt of a strong base and a weak acid which hydrolyzes to give a basic solution. – This is a solvolysis process as discussed in Chapter 18. – Both processes make the solution basic. The solution cannot have a pH=7.00 at equivalence point. Let us calculate the pH at the equivalence point.

40. 40 Weak Acid/Strong Base Titration Curves Set up the equilibrium reaction: KOH + CH 3 COOH → K + CH 3 COO - + H 2 O Start10.0 mmol change-10.0 mmol +10.0 mmol After rxn0.0 mmol 10.0 mmol M KH 3 COOH = =0.05M 200mL 10.0 mmol  0.05M CH 3 COO - CH 3 COO - +H 2 O  CH 3 COOH +OH - (0.05-x) Mx M Kb=Kb= [OH - ] [CH 3 COOH] [CH 3 COO - ] = (x)(x) (0.05-x) =5.6x10 -10 x 2 =2.8x10 -11 x=5.27x10 -6 = [OH - ] pOH=5.28 pH=8.72 the equivalence point

41. 41 Weak Acid/Strong Base Titration Curves After the equivalence point is reached, the pH is determined by the excess KOH just as in the strong acid/strong base example. KOH + CH 3 COOH → K + CH 3 COO - + H 2 O Start11.0 mmol10.0 mmol change-10.0 mmol +10.0 mmol After rxn1.0 mmol0.0 mmol10.0 mmol M KOH = =4.8x10 -3 M 210mL 1.0 mmol [OH - ]=4.8x10 -3 M pOH=2.32 pH=11.68

42. 42 Weak acid-strong base

44. 44 Strong Acid/Weak Base Titration Curves Titration curves for Strong Acid/Weak Base Titration Curves look similar to Strong Base/Weak Acid Titration Curves but they are inverted.

45. 45 Weak Acid/Weak Base Titration Curves Weak Acid/Weak Base Titration curves have very short vertical sections. The solution is buffered both before and after the equivalence point. Visual indicators cannot be used.

46. Table 19-7a, p. 763

47. Table 19-7b, p. 763