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Ionic Equilibria I 離子平衡 : Acids and Bases

18. 維他命 C Ascorbic acid. Ionic Equilibria I 離子平衡 : Acids and Bases. 檸檬酸 Citric acid. Chapter Goals. A Review of Strong Electrolytes ( 強電解質 ) The Autoionization of Water ( 水的自身離子化 ) The pH and pOH Scales (pH 值和 pOH 值 )

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Ionic Equilibria I 離子平衡 : Acids and Bases

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  1. 18 維他命C Ascorbic acid Ionic EquilibriaI 離子平衡: Acids and Bases 檸檬酸 Citric acid

  2. Chapter Goals • A Review of Strong Electrolytes (強電解質) • The Autoionization of Water(水的自身離子化) • The pH and pOH Scales(pH值和pOH值) • Ionization Constants for Weak Monoprotic Acids and Bases(弱單質子酸及鹼的解離常數) • Polyprotic Acids(多質子酸) • Solvolysis(溶劑分解) • Salts of Strong Bases and Strong Acids(強鹼及強酸形成的鹽) • Salts of Strong Bases and Weak Acids(強鹼及弱酸形成的鹽) • Salts of Weak Bases and Strong Acids(弱鹼及強酸形成的鹽) • Salts of Weak Bases and Weak Acids (弱鹼及弱酸形成的鹽) • Salts That Contain Small, Highly Charged Cations

  3. A Review of Strong Electrolytes • This chapter details the equilibria of weak acids and bases. • We must distinguish weak acids and bases from strong electrolytes. • Weak acids and bases ionize or dissociate partially, much less than 100%. • In this chapter we will see that it is often less than 10%! • Strong electrolytes ionize or dissociatecompletely.(強電解質完全解離) • Strong electrolytes approach 100% dissociation in aqueous solutions.(強電解質在水溶液中幾近100%解離)

  4. Aqueous Solutions: An Introduction Acid: a substance that produces hydrogen ions, H+, in aqueous solutions Base: a substance that produces hydroxide ions, OH-, in aqueous solutions Salt: a compound that contains a cation other than H+, and an anion other than hydroxide ion, OH- ,or oxide ion, O2- • The reason nonelectrolytes do not conduct electricity is because they do not form ions in solution. • ions conduct electricity in solution 4

  5. Aqueous Solutions: An Introduction 檸檬酸 • Classification of solutes • strong electrolytes強電解質- conduct electricityextremely well in dilute aqueous solutions • Examples of strong electrolytes • HCl, HNO3, etc. • strong soluble acids • NaOH, KOH, etc. • strong soluble bases • NaCl, KBr, etc. • soluble ionic salts • ionize in water essentially 100% 5

  6. Aqueous Solutions: An Introduction • Classification of solutes • weak electrolytes - conduct electricity poorly in dilute aqueous solutions • CH3COOH, (COOH)2 • weak acids • NH3, Fe(OH)3 • weak bases • some soluble covalent salts • ionize in water much less than 100% 6

  7. Aqueous Solutions: An Introduction HCl(g) H+(aq) + Cl-(aq) HNO3 + H2O H3O+(aq) + NO-3(aq) 100% 100% H2O HNO3 H+(aq) + NO-3(aq) Strong and Weak Acids • Acids are substances that generate H+ in aqueous solutions. • Strong acids ionize 100% in water. 7

  8. Aqueous Solutions: An Introduction strong acids--ionize almost 100% 氫氯酸 氫溴酸 氫碘酸 硝酸 過氯酸 氯酸 硫酸 8

  9. Aqueous Solutions: An Introduction Weak acids--Typically ionize 10% or less! 氫氟酸 醋酸 氰化氫 亞硝酸 碳酸 亞硫酸 磷酸 草酸 9

  10. Aqueous Solutions: An Introduction NaOHNa+(aq) + OH-(aq) H2O H2O Ba(OH)2Ba+2(aq) + 2OH-(aq) Strong Bases, Insoluble Bases, and Weak Bases • Strong Bases • Characteristic of common inorganic bases is that they produce OH- ions in solution. • Similarly to strong acids, strong bases ionize 100% in water. 10

  11. Aqueous Solutions: An Introduction 氫氧化鋰 氫氧化鈉 氫氧化鈣 氫氧化鉀 氫氧化鍶 氫氧化銣 氫氧化銫 氫氧化鋇 Notice that they are all hydroxides of IA and IIA metals 11

  12. Aqueous Solutions: An Introduction NH3(g) + H2O(l) NH4+(aq) + OH-(aq) Strong Bases, InsolubleBases, and Weak Bases • Insoluble bases • Ionic compounds that are insoluble in water, consequently, not very basic. • Cu(OH)2, Zn(OH)2, Fe(OH)2氫氧化亞鐵, Fe(OH)3 • Weak bases • are covalent compounds that ionize slightly in water. • Ammonia is most common weak base-- NH3 12

  13. A Review of Strong Electrolytes Most Water Soluble Salts The solubility guidelines from Chapter 4 will help you remember these salts. 100% 100% H2O NaCl(s)Na+(aq) + Cl-(aq) H2O Ca(NO3)2Ca+2(aq) + 2NO3-(aq)

  14. HNO3 + H2O H3O+(aq) + NO-3(aq) A Review of Strong Electrolytes • The calculation of ion concentrations in solutions of strong electrolytes is easy. Example 18-1: Calculate the concentrations of ions in 0.050 M nitric acid, HNO3. 100% 0.050M 0.050M 0.050M

  15. A Review of Strong Electrolytes Example 18-1 Calculation of Concentrations of Ions Calculation the molar concentration of Ba2+ and OH- ions in 0.03M barium hydroxide. initial 0.03M change -0.03M +0.03M +2x(0.03)M final 0.0M 0.03M +0.06M [Ba2+]=0.03M [OH-]=0.06M H2O Ba(OH)2Ba2+(aq) + 2OH-(aq) Exercise 4 and 6

  16. A Review of Strong Electrolytes Example 18-2: Calculate the concentrations of ions in 0.020 M strontium hydroxide, Sr(OH)2, solution. initial 0.02M change -0.02M +0.02M +2x(0.02)M final 0.0M 0.02M +0.04M [Sr2+]=0.02M [OH-]=0.04M H2O Sr(OH)2Sr2+(aq) + 2OH-(aq)

  17. H2O(l)+ H2O(l) H3O+(aq) + OH-(aq) The Autoionization of Water水的自身離子化[反應] • Pure water ionizes very slightly. • The concentration of the ionized water is less than one-millionth molar at room temperature. • We can write the autoionization of water as a dissociation reaction similar to those previously done in this chapter. • Because the activity of pure water is 1, the equilibrium constant for this reaction is: Kc=[H3O+][OH-]

  18. The Autoionization of Water • Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7 M at 25oC. • Note that this is at 25oC, not every temperature! • We can determine the value of Kc from this information. Kc=[H3O+][OH-] = (1.0x10-7)x(1.0x10-7) = 1.0x10-14

  19. The Autoionization of Water • This particular equilibrium constant is called the ion-product for water and given the symbol Kw. • Kw is one of the recurring expressions for the remainder of this chapter and Chapters 19 and 20. Kw=[H3O+][OH-] = 1.0x10-14

  20. HNO3 + H2O H3O+(aq) + NO3- (aq) 2 H2O(l) H3O+(aq) + OH-(aq) Example 18-2 Calculation of Ions Concentrations Calculation the concentrations of H3O+ and OH- ions in 0.05M HNO3 solution. Strong acid initial 0.05M change -0.05M +0.05M +0.05M At equil 0.0M 0.05M +0.05M [H3O+]=[NO3-]=0.05M 0.05M initial +xM change -2xM +xM +xM At equil (0.05+x)M Kw=[H3O+][OH-] 1.0x10-14 =(0.05+x)(x) (very small number) 1.0x10-14 =(0.05)(x) x=2.0x10-13 M=[OH-] Exercise 14

  21. HCl + H2O H3O++ Cl- 2 H2O(l) H3O+(aq) + OH-(aq) Example 18-3: Calculate the concentrations of H3O+ and OH- in 0.050 M HCl. initial 0.05M change -0.05M +0.05M +0.05M At equil 0.0M 0.05M +0.05M [H3O+]=[Cl-]=0.05M 0.05M initial +xM change -2xM +xM +xM At equil (0.05+x)M Kw=[H3O+][OH-] 1.0x10-14 =(0.05+x)(x) (very small number) 1.0x10-14 =(0.05)(x) x=2.0x10-13 M=[OH-]

  22. The pH and pOH scales • A convenient way to express the acidity and basicity of a solution is the pH and pOH scales. • The pH of an aqueous solution is defined as: pH=-log[H3O+] or [H3O+]=10-pH pOH=-log[OH-] or [OH-]=10-pOH

  23. The pH and pOH scales • If either the [H3O+] or [OH-] is known, the pH and pOH can be calculated. Example 18-4: Calculate the pH of a solution in which the [H3O+] =0.030 M. pH=-log[H3O+] pH=-log(3.0x10-2) pH=1.52

  24. The pH and pOH scales Example 18-5: The pH of a solution is 4.597. What is the concentration of H3O+? pH=-log[H3O+] 4.597=-log[H3O+] [H3O+]=10-4.597 [H3O+]=2.53x10-5M

  25. Example 18-3 Calculation of pH Calculate the pH of a solution in which the H3O+ concentration is 0.050mol/L. pH=-log[H3O+] pH=-log(5.0x10-2) pH=1.3 Exercise 22 Example 18-4 Calculation of H3O+ concentration from pH The pH of a solution is 3.301. What is the concentration of H3O+ in this solution? pH=-log[H3O+] 3.301=-log[H3O+] [H3O+]=10-3.301 [H3O+]=5.0x10-4M Exercise 24

  26. A convenient relationship between pH and pOH may be derived for all dilute aqueous solutions at 250C. Taking the logarithm of both sides of this equation gives: The pH and pOH scales [H3O+][OH-]= 1.0x10-14 log[H3O+]+log[OH-]= -14.0 • Multiplying both sides of this equation by -1 gives: -log[H3O+]+(-log[OH-])= 14.0 • Which can be rearranged to this form: pH + pOH = 14.0

  27. The pH and pOH scales • Remember these two expressions!! • They are key to the next three chapters! [H3O+][OH-]= 1.0x10-14 pH + pOH = 14.0

  28. The usual range for the pH scale is 0 to 14. And for pOH the scale is also 0 to 14 but inverted from pH. pH = 0 has a pOH = 14 and pH = 14 has a pOH = 0. The pH and pOH scales [H3O+]=1.0 M to [H3O+]= 1.0x10-14M pH=0 to pH=14.0 [OH-]= 1.0x10-14 up to [OH-]=1.0M pOH=14.0 to pOH=0

  29. HNO3 + H2O H3O+(aq) + NO3- (aq) The pH and pOH scales Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a 0.020 M HNO3 solution. • Is HNO3 a weak or strong acid? • What is the [H3O+] ? Strong acid initial 0.02M change -0.02M +0.02M +0.02M At equil 0.0M 0.02M +0.02M [H3O+]= 2x10-2 M pH=-log(2x10-2 M) pH=1.70 Kw=[H3O+][OH-]=1.0x10-14 [OH-]=1.0x10-14/[H3O+] =1.0x10-14/2.0x10-2 =5.0x10-13 M pOH=-log[OH-] =-log(5.0x10-13) =12.30

  30. The pH and pOH scales • To help develop familiarity with the pH and pOH scale we can look at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x 10-14M.

  31. Ca(OH)2 Ca2++ 2OH-(aq) Example 18-6 Calculations Involving pH and pOH Calculate the [H3O+], pH, [OH-], and pOH for a 0.015 M Ca(OH)2 solution. initial 0.015M change -0.015M +0.015M +2x(0.015)M At equil 0.0M 0.015M +0.03M [OH-]= 3x10-2 M pOH=-log(3x10-2 M) pOH=1.52 pH=14-pOH =14-1.52 =12.48 Kw=[H3O+][OH-]=1.0x10-14 [H3O+]=1.0x10-14/[OH-] =1.0x10-14/3.0x10-2 =3.3x10-13 M Exercise 26 and 37

  32. 胃酸 炭酸飲料 檸檬 醋 蕃茄 啤酒 尿液 牛奶 唾液 血漿 蛋白 鎂乳(瀉藥) 氨水

  33. pH meter Universal indicator Universal pH paper

  34. Strong acids ionize completely in dilute aqueous, whereas weak acids ionize only slightly Let’s look at the dissolution of acetic acid, a weak acid, in water as an example. The equation for the ionization of acetic acid is: The equilibrium constant for this ionization is expressed as: Ionization Constants for Weak Monoprotic Acids單質子酸 and Bases CH3COOH + H2O  H3O+ + CH3COO- [H3O+] [CH3COO-] Kc= [CH3COOH] [H2O]

  35. Ionization Constants for Weak Monoprotic Acids and Bases • The water concentration in dilute aqueous solutions is very high. • 1 L of water contains 55.5 moles of water. • Thus in dilute aqueous solutions: [H2O]  55.5M

  36. [H3O+] [CH3COO-] Kc= [CH3COOH] [H2O] [H3O+] [CH3COO-] [H3O+] [CH3COO-] [CH3COOH] [CH3COOH] Ionization Constants for Weak Monoprotic Acids and Bases • The water concentration is many orders of magnitude greater than the ion concentrations. • Thus the water concentration is essentially that of pure water. • Recall that the activity of pure water is 1. Kc [H2O]= K

  37. [H3O+] [CH3COO-] [CH3COOH] Ionization Constants for Weak Monoprotic Acids and Bases • We can define a new equilibrium constant for weak acid equilibria that uses the previous definition. • This equilibrium constant is called the acid ionization constant 酸解離常數. • The symbol for the ionization constant is Ka. =1.8x10-5 Ka= for acetic acid

  38. =1.8x10-5 Ka= [H+] [CH3COO-] [CH3COOH] Ionization Constants for Weak Monoprotic Acids and Bases • In simplified form the dissociation equation and acid ionization expression are written as: CH3COOH  H+ + CH3COO-

  39. Ionization Constants for Weak Monoprotic Acids單質子酸 and Bases Carbonic acid H2CO3 Citric acid C3H5O(COOH)3

  40. From the above table we see that the order of increasing acid strength for these weak acids is: The order of increasing base strength of the anions (conjugate bases) of these acids is: Ionization Constants for Weak Monoprotic Acids and Bases HF > HNO2 > CH3COOH > HClO > HCN F- < NO2- < CH3COO- < ClO- < CN-

  41. [H+] [CN-] [HCN] Ionization Constants for Weak Monoprotic Acids and Bases Example 18-8: Write the equation for the ionization of the weak acid HCN and the expression for its ionization constant. HCN  H+ + CN- =4.0x10-10 Ka=

  42. [H+] [Y-] [H+] [Y-] [HY] [HY] HY  H+ + Y- Example 18-9: In a 0.12 M solution of a weak monoprotic acid, HY, the acid is 5.0% ionized. Calculate the ionization constant for the weak acid. Ka= • Since the weak acid is 5.0% ionized, it is also 95% unionized. • Calculate the concentration of all species in solution. [H+]=[Y-]=0.05 x (0.12M) =0.006M =6x10-3M [HY]=0.95 x (0.12M) =0.11M (6.0x10-3)(6.0x10-3) Ka= = (0.11) Ka=3.3x10-4

  43. [H+] [A-] [HA] (1.1x10-3)(1.1x10-3) 0.1 Example 18-10: The pH of a 0.10 M solution of a weak monoprotic acid, HA, is found to be 2.97. What is the value for its ionization constant? pH = 2.97 so [H+]= 10-pH [H+]= 10-2.79 [H+]=1.1x 10-3 • Use the [H3O+] and the ionization reaction to determine concentrations of all species. HA  H+ + A- At equil (0.1-1.1x10-3)M 1.1x10-3M 1.1x10-3M  0.1M • Calculate the ionization constant from this information. = Ka= Ka=1.2x10-5

  44. Example 18-7 Calculations of Ka and pKa from equilibrium Concentrations Nicotinic acid is a weak monoprotic oganic acid that we can represent as HA. A dilute solution of nicotinic acid was found to contain the following concentrations at equilibrium at 25oC. What are the Ka and pKa values? [HA]=0.049M; [H3O+]=[A-]=8.4x10-4M. [H3O+] [A-] [HA] (8.4x10-4)(8.4x10-4) 0.049 HA + H2O  H3O+ + A- = Ka= Ka=1.4x10-5 pKa=-log(1.4x10-5) =4.85 Exercise 30

  45. Example 18-8 Calculations of Ka from Percent Ionization In 0.01M solution, acetic acid is 4.2% ionized. Calculate its ionization constant. [H3O+] [CH3COO-] [CH3COOH] CH3COOH  H+ + CH3COO- • Since the weak acid is 4.2% ionized, it is also 95.8% unionized. • Calculate the concentration of all species in solution. [H+]=[CH3COO-]=0.042 x (0.01M) =4.2x10-4M [CH3COOH]=0.958 x (0.01M) =9.58x10-3M Ka= (4.2x10-4)(4.2x10-4) = (9.58x10-3) = 1.8x10-5 Exercise 38

  46. Example 18-9 Calculations of Ka from pH The pH of a 0.115M solution of chloroacetic acid, ClCH2COOH, is measure to be 1.92. Calculate Ka from this weak momoprotic acid. [H3O+] [A-] Ka= [HA] pH=-log[H3O+] [H3O+]=10-pH=10-1.92=0.012M HA + H2O  H3O+ + A- initial 0.115M change -0.012M +0.012M +0.012M At equil 0.103M 0.012M 0.012M (0.012)(0.012) = (0.103) = 1.4x10-3 Exercise 40

  47. [H3O+] [CH3COO-] Ka= =1.8x10-5 [CH3COOH] CH3COOH  H+ + CH3COO- initial 0.15M change +xM +xM -xM Example 18-11: Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution. At equil (0.15-x)M xM xM (x)(x) x is small enough to ignore compare to 0.15M Ka= =1.8x10-5 (0.15-x) x2(0.15)x(1.8x10-5) x22.7x10-6 x=1.6x10-3 M=[H3O+]=[CH3COO-] [CH3COOH]=0.15-1.6x10-3  0.15M

  48. [CH3COOH]ionized [CH3COOH]original Ionization Constants for Weak Monoprotic Acids and Bases • Let us now calculate the percent ionization for the 0.15 M acetic acid. From Example 18-11, we know the concentration of CH3COOH that ionizes in this solution. The percent ionization of acetic acid is % ionization= x100% (1.6x10-3M) x100% % ionization= 0.15M =1.1%

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