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Section 16 Acid-Base Equilibria

Section 16 Acid-Base Equilibria. Solutions of a Weak Acid or Base. The simplest acid-base equilibria are those in which a single acid or base solute reacts with water. we will first look at solutions of pure weak acids or bases.

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Section 16 Acid-Base Equilibria

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  1. Section 16 Acid-Base Equilibria Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  2. Solutions of a Weak Acid or Base • The simplest acid-base equilibria are those in which a single acid or base solute reacts with water. • we will first look at solutions of pure weak acids or bases. • Next, we will also consider solutions of salts, which can have acidic or basic properties as a result of the reactions of their ions with water. Salts can be neutral, acidic, or basic. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  3. Acid-Ionization Equilibria • Acid ionization (or acid dissociation) is the reaction of an acid with water to produce hydronium ion (hydrogen ion) and the conjugate base anion (hydrolysis). • Because acetic acid is a weak electrolyte, it ionizes to a small extent in water, hence double arrow; not 100% like strong acid which wrote single arrow. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  4. Acid-Ionization Equilibria • For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant, Ka). • Consider the generic monoprotic acid, HA. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  5. Acid-Ionization Equilibria • The corresponding equilibrium expression is: • Since the concentration of water remains relatively constant (liquid = 1), we rearrange the equation to get: Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  6. Acid-Ionization Equilibria • Thus, Ka , the acid-ionization constant, equals the constant [H2O]Kc. • acid + water is Ka Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  7. Experimental Determination of Ka • The degree of ionization of a weak electrolyte is the fraction of molecules that react with water to give ions. • Electrical conductivity or some other colligative property can be measured to determine the degree of ionization. • With weak acids, the pH can be used to determine the equilibrium composition of ions in the solution because it gives the concentration of hydronium ions at equil. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  8. A Problem To Consider • Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C. • It is important to realize that the solution was made 0.012 M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than 0.012 M, just a small amount less. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  9. A Problem To Consider • Let x be the moles per liter of product formed. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  10. A Problem To Consider • We can obtain the value of x from the given pH. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  11. Note first, however, that A Problem To Consider • Substitute this value of x in our equilibrium expression. • the concentration of unionized acid remains virtually unchanged. Important point to remember for later. • Substitute this value of x in our equilibrium expression. HW 28 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  12. A Problem To Consider • To obtain the degree of ionization: • weak acids typically less than 5%. The lower the %ionized, less H+ formed, the lower Ka, weaker the acid. Ka is measure of strength of acid or Kb if base. Strength based on ionization not pH directly. Lower pH more acidic, does not mean stronger acid Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  13. Ka is a measure of strength of acid. It indicates the amount ionized to form hydronium ions. The more hydronium ions formed, naturally the more acidic and lower pH. However, strength of acid is measure of amount ionized, not conc of acid. Conc can vary which ultimately will vary the pH of solution. To determine the strength of an acid you must compare Ka values not pH of solution. pH is dependent on Ka as well as conc. of solution. Must compare apples and apples. 10M HC2H3O2 Ka = 1.8 x 10-5 pH = 1.87 0.1M HF Ka = 6.9x10-4 pH = 2.08 Which stronger acid? Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  14. Calculations With Ka • Once you know the value of Ka, you can calculate the equilibrium concentrations of species HA, A-, and H3O+ for solutions of different molarities. • The general method for doing this is same way we did Kc or Kp except we can simplify the math because of percent ionize is so small compared to initial conc of acid; remember 0.012-x=0.012. This wasn't the case for Kc or Kp problems so can't neglect in those type problems Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  15. Calculations With Ka • Note that in our previous example, the degree of dissociation was so small that “x” was negligible compared to the concentration of nicotinic acid. • It is the small value of the degree of ionization that allowed us to ignore the subtracted x in the denominator of our equilibrium expression. • The degree of ionization of a weak acid depends on both the Ka and the concentration of the acid solution Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  16. Calculations With Ka • How do you know when you can use this simplifying assumption? • It can be shown that if the acid concentration, Ca, divided by the Ka exceeds 100, that is, • then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5% otherwise must do quadratic formula. Another way look at it factor of two - three difference in power of 10 is needed between Ka and conc of acid. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  17. ex. Calculate the pH in a 1.00 M solution of nitrous acid, HNO2, for which Ka = 6.0 x 10-4. HNO2 + H2O <--> H3O+ + NO2- Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  18. A Problem To Consider • Really two equil: Kw as well but neglect most of the time as long as [H+] is greater than 10-6 M from the acid. • Really [H+]tot = [H+]acid +[H+]H2O (water so small neglect) Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  19. ex. Calculate the pH of 10.00 mL of 1.00 M solution of nitrous acid , HNO2, diluted to 100.0 mL with water for which Ka = 6.0 x 10-4. HNO2 + H2O <--> H3O+ + NO2- Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  20. A Problem To Consider • What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetyl salicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C. • The molar mass of HC9H7O4 is 180.2 g. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  21. A Problem To Consider • What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C. • These data are summarized below. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  22. A Problem To Consider • Note that • which is less than 100 or tell factor 10-3 vs 10-4 not factor 2-3; therefore, won't work, so we must solve the equilibrium equation exactly . Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  23. If we substitute the equilibrium concentrations and the Ka into the equilibrium constant expression, we get Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  24. You can solve this equation exactly by using the quadratic formula. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  25. A Problem To Consider • Taking the upper sign, we get and neglect water OK • Now we can calculate the pH. HW 29 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  26. Calculate the pH of a solution that has a concentration of 1.0 x 10-8 M NaOH. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  27. Polyprotic Acids • Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. • Sulfuric acid, for example, can lose two protons in aqueous solution. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  28. Polyprotic Acids • For a weak diprotic acid like carbonic acid, H2CO3, two simultaneous equilibria must be considered. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  29. For the loss of the first proton Polyprotic Acids • Each equilibrium has an associated acid-ionization constant. • For the loss of the second proton Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  30. Polyprotic Acids • In general, the second ionization constant, Ka2, for a polyprotic acid is smaller than the first ionization constant, Ka1. Harder to pull proton off charged species. Total H+ is sum of all equil H+ but usually neglect all but the principal rxn. • In the case of a triprotic acid, such as H3PO4, the third ionization constant, Ka3, is smaller than the second one, Ka2. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  31. Polyprotic Acids • When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions. • However, reasonable assumptions can be made that simplify these calculations as we show in the next example. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  32. A Problem To Consider • Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12. • For diprotic acids, Ka2 is so much smaller than Ka1 that the smaller amount of hydronium ion produced in the second reaction can be neglected. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  33. A Problem To Consider • The pH can be determined by simply solving the equilibrium problem posed by the first ionization. • the first ionization is: Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  34. A Problem To Consider • Assuming that x is much smaller than 0.10 (1265), you get Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  35. A Problem To Consider • Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O62- ? The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12. • The ascorbate ion, C6H6O62-, which we will call y, is produced only in the second ionization of H2C6H6O6. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  36. Assume the starting concentrations for HC6H6O6- and H3O+ to be those from the first equilibrium. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  37. Substituting into the equilibrium expression • Assuming y is much smaller than 0.0028 (1.75 x 109 > 100), the equation simplifies to • Hence, • typically, the concentration of the anion ion equals Ka2 if starting with reactants of first equil. HW 30 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  38. Ammonia, for example, ionizes in water as follows. • The corresponding equilibrium constant is: Base-Ionization Equilibria • Equilibria involving weak bases are treated similarly to those for weak acids. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  39. Ammonia, for example, ionizes in water as follows. Base-Ionization Equilibria • Equilibria involving weak bases are treated similarly to those for weak acids. • The concentration of water is nearly constant. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  40. Base-Ionization Equilibria • Equilibria involving weak bases are treated similarly to those for weak acids. Higher Kb stronger base, ionize more OH-, more basic. • In general, a weak base B with the base ionization • has a base ionization constant equal to HW 31 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  41. A Problem To Consider • What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. • As before, we will follow the three steps in solving an equilibrium. • Write the equation and make a table of concentrations. • Set up the equilibrium constant expression. • Solve for x = [OH-]. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  42. Pyridine ionizes by picking up a proton from water (as ammonia does). A Problem To Consider • What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  43. A Problem To Consider • Note that • which is much greater than 100, so we may use the simplifying assumption that (0.20-x)  (0.20). Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  44. A Problem To Consider • The equilibrium expression is Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  45. A Problem To Consider • Using our simplifying assumption that the x in the denominator is negligible, we get Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  46. A Problem To Consider • Solving for pOH • Since pH + pOH = 14.00 HW 32 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  47. Consider a solution of sodium cyanide, NaCN. Acid-Base Properties of a Salt Solution • One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. • A 0.1 M solution has a pH of 11.1 and is therefore fairly basic. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  48. Acid-Base Properties of a Salt Solution • Sodium ion, Na+, is unreactive with water, but the cyanide ion, CN-, reacts to produce HCN and OH- making it a basic solution • From the Brønsted-Lowry point of view, the CN- ion acts as a base, because it accepts a proton from H2O. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  49. Acid-Base Properties of a Salt Solution • You can also see that OH- ion is a product, so you would expect the solution to have a basic pH. This explains why NaCN solutions are basic. • The reaction of the CN- ion with water is referred to as the hydrolysis of CN-. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

  50. The CN- ion hydrolyzes to give the conjugate acid and hydroxide. Acid-Base Properties of a Salt Solution • The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

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