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Ionic Equilibria 離子平衡 : Part II Buffers 緩衝液 and Titration Curves 滴定曲線

19. Ionic Equilibria 離子平衡 : Part II Buffers 緩衝液 and Titration Curves 滴定曲線. 花青素. Chapter Goals. The Common Ion Effect and Buffer Solutions ( 共同離子效應及緩衝溶液 ) Buffering Action ( 緩衝作用 ) Preparation of Buffer Solutions ( 緩衝溶液的製備 ) Acid-Base Indicators ( 酸鹼指示劑 ) Titration Curves ( 滴定曲線 )

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Ionic Equilibria 離子平衡 : Part II Buffers 緩衝液 and Titration Curves 滴定曲線

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  1. 19 Ionic Equilibria離子平衡: Part II Buffers 緩衝液and Titration Curves滴定曲線 花青素

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

  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. 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) • CH3COOH plus NaCH3COO • 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) • NH3 plus NH4Cl

  5. The Common Ion Effect and Buffer Solutions • 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 CH3COOH • The soluble ionic salt - sodium acetate NaCH3COO CH3COOH  H++CH3COO-部分解離 100% Na+CH3COO- Na+ + CH3COO-完全解離 [CH3COO-] 增加  反應向左  減少[H+]  pH增加

  6. (x)(0.15) (x)(0.15+x) [H+] [CH3COO-] Ka= = =1.8x10-5 = 1.8x10-5 (0.15-x) (0.15) [CH3COOH] 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. CH3COOH  H+ + CH3COO- (0.15-x) M x M x M 0.15 M 0.15 M 0.15 M 100% Na+CH3COO- Na+ + CH3COO- (0.15+x)  0.15 and (0.15-x)  0.15 x =1.8x10-5= [H+] pH=4.74

  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. • [H+] is 89 times greater in pure acetic acid than in buffer solution.

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

  9. 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: [acid] [acid] [HA] [H+] = Ka x [H+] = Ka x [H+] = Ka x n[salt] [A-] [salt] The Common Ion Effect and Buffer Solutions acid salt • The relationship developed in the previous slide is valid for buffers containing a weak monoprotic acid and a soluble, ionic salt. • If the salt’s cation is not univalent the relationship changes to: Where n=charge on cation

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

  11. pH= pKa + log [salt] [acid] [acid] [salt] [H+] = Ka x (0.15) 0.15M (0.15) 0.15M The Common Ion Effect and Buffer Solutions Example 19-1-1: Use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution in Example 19-1 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 [H+] = 1.8x10-5 x pH= 4.74 + log [H+] = 1.8x10-5 pH=4.74 pH= 4.74 + log1 = 4.74 + 0 = 4.74

  12. [NH4+] [OH-] Kb= [NH3] Weak Bases plus Salts of Weak Bases • Buffers that contain a weak base plus the salt of a weak base • One example of this buffer system is ammonia plus ammonium nitrate. NH3 + H2O  NH4+ + OH- 100% NH4NO3NH4+ + NO3- =1.8x10-5

  13. [NH4+] [OH-] (x)(0.30+x) (x)(0.30) Kb= = = 1.8x10-5 (0.15) (0.15-x) [NH3] 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, NH3, and 0.30 M in ammonium nitrate, NH4NO3. NH3 + H2O  NH4+ + OH- (0.15-x) M x M x M 0.30 M 0.30 M 0.30 M =1.8x10-5 100% NH4NO3NH4+ + NO3- (0.30+x)  0.30 and (0.15-x)  0.15 x =9.0x10-6= [OH-] pOH=5.05, pH=8.95

  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. The [OH-] in aqueous ammonia is 180 times greater than in the buffer.

  15. [BH+] [OH-] Kb= [B] [B] base [OH-] = Kb x [BH+] salt 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 + H2O  BH+ + OH- Where B represents a weak base • The general form of the ionization expression is: • Solve for the [OH-]

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

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

  18. 0.08mol 0.22mol 1.0L 1.0L [NH3] [NH3] [OH-] = Kb x [OH-] = Kb x [NH4Cl] [NH4Cl] Initial 0.02 mol 0.1 mol 0.2 mol change -0.02 mol -0.02 mol +0.02 mol 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. 0.10M [OH-] = 1.8x10-5 x 0.20M [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. ?mol NH3=0.1M x 1L=0.1mol ?mol NH4Cl=0.2M x 1L=0.2mol HCl + NH3→ NH4Cl =0.08M MNH3= MNH4Cl= =0.22M After rxn 0 mol 0.08 mol 0.22 mol 0.08M = 1.8x10-5 x = 6.5x10-6 0.22M pH=pHnew-pHoriginal =8.81-8.95=-0.14 pOH=5.19 pH=8.81

  19. [NH3] 0.12mol [OH-] = Kb x [NH4Cl] 1.0L Initial 0.2 mol 0.02 mol 0.1 mol change -0.02 mol -0.02 mol +0.02 mol 0.18mol 1.0L Buffering Action Example 19-4: If 0.020 mole of NaOHis 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 is8.95, 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. NH4Cl + NaOH → NH3 + H2O + NaCl MNH3= =0.12M MNH4Cl= After rxn 0.18 mol 0 mol 0.12 mol =0.18M 0.12M = 1.8x10-5 x = 1.2x10-5 0.18M pH=pHnew-pHoriginal =9.08-8.95=0.13 pOH=4.92 pH=9.08

  20. This table is a summary of examples 19-3 and 19-4. Notice that the pH changes only slightly in each case. Buffering Action

  21. Initial 10.0 mmol 30.0 mmol change -10.0 mmol -10.0 mmol +10.0 mmol 10 mmol 20 mmol [H+] [CH3COO-] [H+] [0.0333] 300mL 300mL Ka= [CH3COOH] [0.0667] 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 NaOH =100ml x 0.1M = 10.0 m mol ?mmol CH3COOH =200ml x 0.15M = 30.0 m mol NaOH + CH3COOH → NaCH3COO + H2O After rxn 0.0 mmol 20.0 mmol 10.0 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: MNaCH3COO= MCH3COOH= =0.0667M =0.0333M =1.8x10-5= [H+]=3.6x10-5M pH =4.44

  22. [NH4+] [OH-] Kb= [NH3] (1.4x10-5+x)(1.4x10-5) (x)(1.4x10-5) 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, NH4Cl, 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 = 9.15 • Then pOH = 14.00 - 9.15 • = 4.85 • [OH-] = 10-4.85 • = 1.4x10-5M NH3 + H2O  NH4+ + OH- (0.1-1.4x10-5) M (1.4x10-5) M (1.4x10-5) M NH4Cl → NH4+ + Cl- x M x M x M = 1.8x10-5 Kb= = = 1.8x10-5 (0.1-1.4x10-5) (0.1) x= 0.13 M = [NH4Cl]original 0.13Mx1L =0.13 mol

  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. a. Methyl red 甲基紅 • Red at pH4 and below • Yellow at pH 7 and above • Red → orange → yellow Three common indicators in solutions that cover the pH range 3 to 11. • 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. 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: [In-] Ka [H+] [In-] Ka= [HIn] [H+] [HIn] HIn  H+ + In- HIn represents nonionized acid molecules In- represents the anion (conjugate base) of HIn color2 color1 • The equilibrium constant expression for an indicator would be expressed as: = 25

  26. Acid-Base Indicators Color change ranges of some acid-base indicators

  27. 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. HClO4 + KOH → KClO4 + H2O

  28. 100% HClO4 H+ + ClO4- Strong Acid/Strong Base Titration Curves • Before any KOH is added the pH of the HClO4 solution is . • Remember perchloric acid is a strong acid that ionizes essentially 100%. 0.10M 0.10M 0.10M [H+] =0.10M pH =-log(0.10)=1.0

  29. Start 10.0 mmol 2.0 mmol change -2.0 mmol -2.0 mmol +2.0 mmol 8.0 mmol 120mL 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 ___? ? mmol HClO4 = 100ml x (0.1M) =10.0 mmol ? mmol KOH = 20ml x (0.1M) =2.0 mmol HClO4 + KOH → KClO4 + H2O After rxn 8.0 mmol 0.0 mmol 2.0 mmol MHClO4= =0.067M [H+]=0.067M pH = 1.17

  30. Start 10.0 mmol 5.0 mmol change -5.0 mmol -5.0 mmol +5.0 mmol 5.0 mmol 150mL 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 ___? ? mmol KOH = 50ml x (0.1M) =5.0 mmol HClO4 + KOH → KClO4 + H2O After rxn 5.0 mmol 0.0 mmol 5.0 mmol MHClO4= =0.033M [H+]=0.033M pH = 1.48

  31. Start 10.0 mmol 9.0 mmol change -9.0 mmol -9.0 mmol +9.0 mmol 1.0 mmol 190mL 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 ____? ? mmol KOH = 90ml x (0.1M) =9.0 mmol HClO4 + KOH → KClO4 + H2O After rxn 1.0 mmol 0.0 mmol 9.0 mmol MHClO4= =0.0053M [H+]=0.0053M pH = 2.28

  32. Start 10.0 mmol 10.0 mmol change -10.0 mmol -10.0 mmol +10.0 mmol 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 ___? ? mmol KOH = 100ml x (0.1M) =10.0 mmol HClO4 + KOH → KClO4 + H2O After rxn 0.0 mmol 0.0 mmol 10.0 mmol • No acid or base • Neutral • pH=7.0

  33. Strong acid-strong base

  34. 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.

  35. Weak Acid/Strong Base Titration Curves • As an example, consider the titration of 100.0 mL of 0.100 M acetic acid, CH3COOH, (a weak acid) with 0.100 M KOH (a strong base). • The acid and base react in a 1:1 mole ratio. 1mol 1mol 1mol CH3COOH + KOH → K+CH3COO- + H2O 1mmol 1mmol 1mmol • Before the equivalence point is reached, both CH3COOH and KCH3COO are present in solution forming a buffer. • The KOH reacts with CH3COOH to form KCH3COO. • A weak acid plus the salt of a weak acid form a buffer. • Hypothesize how the buffer production will effect the titration curve.

  36. Weak Acid/Strong Base Titration Curves • Before the equivalence point is reached, both CH3COOH and KCH3COO are present in solution forming a buffer. • The KOH reacts with CH3COOH to form KCH3COO. • A weak acid plus the salt of a weak acid form a buffer. • Hypothesize how the buffer production will effect the titration curve.

  37. (x)(x) = (0.1-x) Start 2.0 mmol 10.0 mmol change -2.0 mmol -2.0 mmol +2.0 mmol 8.0 mmol 2.0 mmol [H+] [CH3COO-] 120mL 120mL Ka= [CH3COOH] [CH3COOH] [H+] = Ka x [CH3COO-] CH3COOH  CH3COO- +H+ (0.1-x) M x M x M 1. Determine the pH of the acetic acid solution before the titration is begun. = 1.8x10-5 x= 1.3x10-3=[H+] x2= 1.8x10-6 pH= 2.89 • After a total of 20.0 mL of KOH solution has been added, the pH is: ? mmol CH3COOH = 100ml x (0.1M) =10.0 mmol ? mmol KOH = 20ml x (0.1M) =2.0 mmol KOH + CH3COOH→ K+CH3COO- + H2O After rxn 0.0 mmol 8.0 mmol 2.0 mmol MCH3COO-= =0.017M MCH3COOH= =0.067M 0.067 = (1.8x10-5) x = 7.1x10-5 0.017 pH= 4.15

  38. Weak Acid/Strong Base Titration Curves • At the equivalence point, the solution is 0.500 M in KCH3COO, 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.

  39. (x)(x) [OH-] [CH3COOH] = Kb= [CH3COO-] Start 10.0 mmol 10.0 mmol change -10.0 mmol -10.0 mmol +10.0 mmol 10.0 mmol 200mL Weak Acid/Strong Base Titration Curves • Set up the equilibrium reaction: KOH + CH3COOH→ K+CH3COO- + H2O After rxn 0.0 mmol 0.0 mmol 10.0 mmol MKH3COOH= =0.05M 0.05M CH3COO- CH3COO- +H2O  CH3COOH+OH- (0.05-x) M x M x M =5.6x10-10 (0.05-x) x2=2.8x10-11 x=5.27x10-6= [OH-] pOH=5.28 pH=8.72 the equivalence point

  40. Start 11.0 mmol 10.0 mmol change -10.0 mmol -10.0 mmol +10.0 mmol 1.0 mmol 210mL 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 + CH3COOH→ K+CH3COO- + H2O After rxn 1.0 mmol 0.0 mmol 10.0 mmol MKOH= =4.8x10-3 M [OH-]=4.8x10-3 M pOH=2.32 pH=11.68

  41. Weak acid-strong base

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

  43. Weak Acid/Weak BaseTitration 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.

  44. Table 19-7a, p. 763

  45. Table 19-7b, p. 763

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