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Chapter 17: Acid–Base Equilibria in Aqueous Solutions

Chapter 17: Acid–Base Equilibria in Aqueous Solutions. Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop. Ionization of Water and pH Concept. Lots of weak acids and bases How can we quantify their relative strengths? Need reference Choose H 2 O

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Chapter 17: Acid–Base Equilibria in Aqueous Solutions

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  1. Chapter 17: Acid–Base Equilibriain Aqueous Solutions Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop

  2. Ionization of Water and pH Concept • Lots of weak acids and bases • How can we quantify their relative strengths? • Need reference • Choose H2O • Water under right voltage • Slight conductivity • Where does conductivity come from?

  3. Where Does Slight Conductivity of H2O Come From? • Trace ionization  self-ionization of water • H2O + H2O H3O+(aq) + OH–(aq) acidbaseacidbase • Equilibrium law is: • But [H2O]pure = = 55.6 M [H2O] = constant even for dilute solutions

  4. Where Does Slight Conductivity of H2O Come From? H2O + H2O H3O+(aq) + OH–(aq) • Since [H2O] = constant, equilibrium law is • Kw = is called the ion product of water • Often omit second H2O molecule and write • H2O H+(aq) + OH–(aq)

  5. Numeric Value of Kw = 1.0 × 10–14 H2O H+(aq) + OH–(aq) • For pure H2O at 25 °C • [H+] = [OH–] = 1.0 × 10–7M • Kw= (1.0 × 10–7)(1.0 × 10–7) = 1.0 × 10–14 • See Table 17.1 for Kwat various temperatures • H2O auto-ionization occurs in all solutions • When other ions present • [H+] is usually NOT equal to [OH–] • But Kw= [H+][OH–] = 1.0 × 10–14

  6. In aqueous solution, Product of [H3O+] and [OH–] equals Kw [H3O+] and [OH–] may not actually equal each other Solutions are classified on the relative concentrations of [H3O+] and [OH–] Solution Classification Acid and Basic Solutions

  7. Learning Check Ex. 1 In a sample of blood at 25 °C, [H+] = 4.6  10–8M. Find [OH–] and determine if the solution is acidic, basic or neutral. • So 2.2 × 10–7M > 4.6 × 10–8M • [OH–] > [H3O+] so the solution is basic

  8. The pH Concept • Arrhenius (of kinetics fame) • Sought an easy way to write the very small numbers associated with [H+] and [OH–] • Developed the “p” notation where p stands for the –log mathematical operation • Result is a simple number • pH is defined as: • Define pOH as: • Define pKw as: • Take anti-log to obtain [H+], [OH–] or Kw

  9. General Properties of Logarithms Using Logarithms • Start with • Taking –log of both sides of eqn. gives • So at 25 °C:

  10. Redefine Acidic, Basic and Neutral Solutions in Terms of pH • As pHincreases, [H+] decreases; pOH decreases, and [OH–] increases • As pH decreases, [H+] increases; pOHincreases, and [OH–] decreases

  11. Your Turn! Kw increases with increasing temperature. At 50 °C, Kw = 5.476 ×10–14. What is the pH of a neutral solution at 50 °C ? A. 7.00 B. 6.63 C. 7.37 D. 15.3

  12. Learning Check What are [H+] and [OH–] of pH = 3.00 solution? • [H+] = 10–3.00 = 1.0  10–3M • [OH–] = = 1.0  10–11M What are [H+] and [OH–] of pH = 4.00 solution? • pH = 4.00 [H+] = 1.0  10–4M • [OH–] = = 1.0  10–10M • Or pH 4.00 solution has 10 times less H+ than pH 3.00 solution

  13. How do we Measure pH? • pH meter • Most accurate • Calibrate with solutions of known pH before use • Electrode sensitive to [H+] • Accurate to  0.01 pH unit • Acid-base indicator • Dyes, change color depending on [H+] in solution • Used in pH paper and titrations • Give pH to  1 pH unit • Litmus paper • RedpH  4.7 acidic • Blue pH  4.7 basic • Strictly acidic vs. basic

  14. Indicators Help Us Estimate pH

  15. Sample pH Calculations Calculate pH and pOH of blood in Ex.1. We found [H+] = 4.6 × 10–8M [OH–] = 2.2 × 10–7 M pH = –log(4.6 × 10–8) = 7.34 pOH = –log(2.2 × 10–7) = 6.66 14.00 = pKw Or pOH = 14.00 – pH = 14.00 – 7.34 = 6.66

  16. Sample pH Calculations (cont’d) What is the pH of NaOH solution at 25 °C in which the OH– concentration is 0.0026 M? [OH–] = 0.0026 M pOH = –log(0.0026) = 2.59 pH = 14.00 – pOH = 14.00 – 2.59 = 11.41

  17. Your Turn! A sample of fresh pressed apple juice has an [H+] = 1.7 × 10–4M. Calculate the pH of this apple juice. • –3.76 • 7.60 • 10.24 • –3.00 • 3.76 pH = –log [H+] = –log(1.7 × 10–4) = 3.76

  18. Learning Check What is the [H3O+] and pH of a solution that has [OH–] = 3.2 × 10–3M? [H3O+][OH–] = 1 × 10–14 [H3O+] = = 3.1 × 10–12M pH = –log [H3O+] = –log(3.1 × 10–12)= 11.50

  19. What is the [OH–] and pH of a solution that has [H3O+] = 2.3 × 10–5 M? Your Turn!

  20. What is the pOH and the [H3O+] of a solution that has a pH of 2.33? Learning Check pOH = 11.67 [H3O+]= 4.7×10–3

  21. What is the pH and the [H3O+] of a solution that has a pOH of 1.89? Your Turn!

  22. Your Turn! A sample of fresh pressed apple juice has a pH of 3.76. Calculate [H+]. • 7.6 × 10–3M • 3.76 M • 10.24 M • 5.9 × 10–9M • 1.7 × 10–4M = 10–3.76 = 1.7 × 10–4M

  23. Strong Acids: pH of Dilute Solutions • Assume 100% dissociated in solution • Good ~ if dilute • Makes calculating [H+] and [OH] easier • 1 mole H+ for every 1 mole HX • So [H+] = CHX for strong acids • Thus, if 0.040 M HClO4 • [H+] = 0.040 M • And pH = –log (0.040) = 1.40

  24. pH of Dilute Solutions of Strong Bases • 1 mole OH– for every 1 mole MOH • [OH–] = CMOH for strong bases • 2 mole OH– for 1 mole M(OH)2 • [OH–] = for strong bases

  25. Learning Check Calculate the pH of 0.00011 M Ca(OH)2. Ca(OH)2(s) + H2O  Ca2+(aq) + 2OH–(aq) • [OH–] = 2 × C(Ca(OH)2) • = 2 × 0.00011 M = 0.00022 M • pOH = – log (0.00022) = 3.66 • pH = 14.00 – pOH • = 14.00 – 3.66 = 10.34 • What is this in the [H+] of the solution? • [H+] = 10–12.34 = 4.6 × 10–11M

  26. Calculations of Strong Acids and Bases • The auto-ionization of H2O will always add to [H+] and [OH–] of an acid or base. Does this have an effect on the last answer? • The previous problem had 0.00022 M [OH–] from the Ca(OH)2 but the [H+] must have come from water. If it came from water an equal amount of [OH–] comes from water and the total [OH–] is • [OH–]total = [OH–]from Ca(OH)2 + [OH–]from H2O • [OH–]total= 0.00022 M + 4.6 × 10–11 M = 0.00022 M (when properly rounded)

  27. Learning Check Show that the contribution of water self-ionization to the total [H+] is negligible for a 0.020 M HCl solution. [H+]total = [H+]HCl + [H+] from H2O = 0.020 M + ? • [OH–] in this solution comes only from auto-ionization of H2O and Kw = [H+][OH–] • So can use [OH–] to determine [H+] from H2O

  28. [H+] from Self Ionization (cont’d) • So [H+] from H2O must also be 5.0  10–13 M • [H+]total = 0.020 M + (5.0  10–13M) = 0.020 M (when properly rounded) • So we see that [H+]from H2O will be negligible except in very dilute solutions of acids and bases

  29. Learning Check What is the pH of 0.10 M HCl? HCl is a strong acid and dissociates completely, we can set up an ICE table HCl(aq) + H2O  H+(aq) + Cl–(aq) I 0.10 N/A 0 0 C –0.10 –0.10 0.10 0.10 E 0 N/A 0.10 0.10 pH = –log(0.10) =1.00

  30. Learning Check What is the pH of 0.5 M Ca(OH)2? A strong base and dissociates completely Ca(OH)2(aq) Ca2+(aq) + 2 OH–(aq) I 0.5 0 0 C -0.5 +0.5 +0.5  2 E 0 0.5 1.0 pOH = –log(1.0) = 0.00 pH = 14.00 – pOH = 14.00 – 0.00 = 14.00

  31. Weak Acids and Bases • Incompletely ionized • Molecules and ions exist in equilibrium • HA = any weak acid; B = any weak base HA(aq) + H2O A–(aq) + H3O+(aq) B(aq) + H2OBH+(aq) + OH–(aq) CH3COOH(aq) + H2O CH3COO–(aq) + H3O+(aq) HSO3–(aq) + H2O SO32–(aq) + H3O+(aq) NH4+(aq) + H2O NH3 (aq) + H3O+ (aq)

  32. Weak Acid/Base Equilibria Or generally HA(aq) + H2O A–(aq) + H3O+(aq) • But [H2O] = constant (55.6 M) so rewrite as • Where Ka = acid ionization constant Acid + Water Conjugate Base + Hydronium Ion

  33. Weak Acid/Base Equilibria • Often simplify as • HA (aq) A–(aq) + H+(aq)

  34. Table 17.2 Weak Monoprotic Acids at 25 °C

  35. Learning Check What is the pKa of HF if Ka = 3.5 × 10–4? HF(aq) + H2O F–(aq) + H3O+(aq) or HF(aq) F–(aq) + H+(aq) = 3.5 × 10–4 pKa = –log Ka = –log(3.5 × 10–4) = 3.46

  36. Reaction of a Weak Base with Water CH3COO–(aq) + H2O CH3COOH(aq) + OH–(aq) NH3(aq) + H2O NH4+(aq) + OH–(aq) • Or generally B(aq) + H2O BH+(aq) + OH–(aq) But [H2O] = constant so rewrite as Where Kb = base ionization constant

  37. Learning Check What is the pKb of C5H5N if Kb = 1.7 × 10–9? C5H5N(aq) + H2O C5H5NH+(aq) + OH–(aq) =1.7 × 10–9 pKb= –log Kb = –log(1.7 × 10–9) = 8.76

  38. Table 17.3 Weak Bases at 25 °C

  39. Conjugate Acid-Base Pairs and Values of Ka and Kb 1. Consider ionization reaction of generic acid and water HA(aq) + H2O A–(aq) + H3O+(aq) 2. Consider reaction of a salt containing anion of this acid (its conjugate base) with water A–(aq) + H2O HA(aq) + OH–(aq)

  40. Now adding reactions we get HA(aq) + H2O A–(aq) + H3O+(aq) A–(aq) + H2O HA(aq) + OH–(aq) 2H2O H3O+(aq) + OH–(aq) For any conjugate acid base pair: (at 25 °C)

  41. Using Properties of Logarithms • Then taking –logof both sides of equation gives: • Earlier we learned the inverse relationship of conjugate acid-base strengths, now we have numbers to illustrate this. • The stronger the conjugate acid, the weaker the conjugate base. So (at 25 °C)

  42. Equilibrium Calculations • Need to develop strategy for dealing with weak acid/base equilibrium calculations Two general types of calculations: • Calculating Ka or Kb from initial concentrations of acid or base and measured pH in solution • Calculating equilibrium concentrations given Ka or Kb and initial concentrations

  43. 1. Calculating Ka and Kb from Initial Concentrations and Equilibrium Data • Need to evaluate mass action expression = reaction quotient Q when the reaction is at equilibrium • Since Q= Kc when at equilibrium • Can either be given initial concentrations and one equilibrium concentration • Usually pH or pOH • OR can be given initial concentrations and % ionization

  44. Ex. 1 Calculating Ka • Niotinic acid (niacin) is a monoprotic acid with the formula HC6H4NO2. A solution that is 0.012 M in nicotinic acid has a pH of 3.39 at 25 °C. What are the acid-ionization constant, Ka, and pKa for this acid at 25 °C? What is the degree of ionization of nicotinic acid in this solution? HC6H4NO2 + H2O C6H4NO2– + H3O+

  45. Ex. 1 Calculating Ka(cont.) • What is value of x? We know the pH so we can calculate [H+] which is equal to x. • x = antilog(–pH) = 10–pH = 10–3.39 = 4.1 × 10–4= [H+]

  46. Ex. 1 Calculating Ka (cont.) x = 0.00041 is inserted into the ICE table. • Notice if Cinitial>> Ka, then equilibrium concentration of acid is the initial concentration since x (in this case 0.00041) must be zero when properly rounded to calculate [HC6H4NO2]

  47. Ex. 1 Calculating Ka (cont) • Now ready to calculate Ka. • Next calculate:

  48. 2. Calculating Equilibrium Concentrations from Ka (or Kb) and Initial Concentrations Almost any problem where you are given Ka or Kb falls into one of three categories: • Only solute is a weak acid • Only solute is a weak base • Two solutes, one a weak acid, the other its conjugate base (this is a buffer problem solved in section 17.7)

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