1 / 57

Chapter 14 Acid/Base Equilibrium

AP Chemistry Unit 10. Chapter 14 Acid/Base Equilibrium. 1. Some Definitions. Arrhenius An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions.

nanda
Télécharger la présentation

Chapter 14 Acid/Base Equilibrium

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. AP Chemistry Unit 10 Chapter 14Acid/Base Equilibrium 1

  2. Some Definitions • Arrhenius • An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. • A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions.

  3. Some Definitions • Brønsted-Lowry • An acid is a proton donor. • A base is a proton acceptor.

  4. A Brønsted-Lowry acid… …must have a removable (acidic) proton. A Brønsted-Lowry base… …must have a pair of nonbonding electrons.

  5. If it can be either… …it is amphiprotic. HCO3- HSO4- H2O

  6. What Happens When an Acid Dissolves in Water? • Water acts as a Brønsted-Lowry base and abstracts a proton (H+) from the acid. • As a result, the conjugate base of the acid and a hydronium ion are formed.

  7. Conjugate Acids and Bases • The term conjugate comes from the Latin word “conjugare,” meaning “to join together.” • Reactions between acids and bases always yield their conjugate bases and acids.

  8. Example 15.1 Identify the Brønsted–Lowry acids and bases and their conjugates in: (a) H2S + NH3 NH4+ + HS– (b) OH– + H2PO4– H2O + HPO42–

  9. Acid and Base Strength • Strong acids are completely dissociated in water. • Their conjugate bases are quite weak. • Weak acids only dissociate partially in water. • Their conjugate bases are weak bases.

  10. Acid and Base Strength • Substances with negligible acidity do not dissociate in water. • Their conjugate bases are exceedingly strong.

  11. Acid and Base Strength • In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H2O (l) H3O+(aq) + Cl-(aq) • H2O is a much stronger base than Cl-, so the equilibrium lies so far to the right that K is not measured (K>>1).

  12. CH3CO2H (aq) + H2O (l) H3O+(aq) + CH3CO2-(aq) Acid and Base Strength • In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. • Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1).

  13. Factors Affecting Acid Strength • The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound. • So acidity increases from left to right across a row and from top to bottom down a group.

  14. Factors Affecting Acid Strength In oxyacids, in which an -OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.

  15. Factors Affecting Acid Strength For a series of oxyacids, acidity increases with the number of oxygens.

  16. Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

  17. Example 15.2 Select the stronger acid in each pair: (a) nitrous acid, HNO2, and nitric acid, HNO3 (b) Cl3CCOOH and BrCH2COOH

  18. Strengths of Amines as Bases • Aromatic amines are much weaker bases than aliphatic amines. • This is due in part to the fact that the p electrons in the benzene ring of an aromatic molecule are delocalized and can involve the nitrogen atom’s lone-pair electrons in the resonance hybrid. • As a result, the lone-pair electrons are much less likely to accept a proton. • Electron-withdrawing groups on the ring further diminish the basicity of aromatic amines relative to aniline.

  19. Example 15.3 Select the weaker base in each pair:

  20. Lewis Acids • Lewis acids are defined as electron-pair acceptors. • Atoms with an empty valence orbital can be Lewis acids.

  21. Lewis Bases • Lewis bases are defined as electron-pair donors. • Anything that could be a Brønsted-Lowry base is a Lewis base. • Lewis bases can interact with things other than protons, however.

  22. H2O (l) + H2O (l) H3O+(aq) + OH-(aq) Autoionization of Water • As we have seen, water is amphoteric. • In pure water, a few molecules act as bases and a few act as acids. • This is referred to as autoionization.

  23. Ion-Product Constant • The equilibrium expression for this process is Kc = [H3O+] [OH-] • This special equilibrium constant is referred to as the ion-product constant for water, Kw. • At 25C, Kw = 1.0  10-14

  24. pH pH is defined as the negative base-10 logarithm of the concentration of hydronium ion. pH = -log [H3O+]

  25. pH • In pure water, Kw = [H3O+] [OH-] = 1.0  10-14 • Since in pure water [H3O+] = [OH-], [H3O+] = 1.0  10-14 = 1.0  10-7

  26. pH • Therefore, in pure water, pH = -log (1.0  10-7) = 7.00 • An acid has a higher [H3O+] than pure water, so its pH is <7. • A base has a lower [H3O+] than pure water, so its pH is >7.

  27. pH These are the pH values for several common substances.

  28. Other “p” Scales • The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions). • Some similar examples are • pOH: -log [OH-] • pKw: -log Kw

  29. Watch This! Because [H3O+] [OH-] = Kw = 1.0  10-14, we know that -log [H3O+] + -log [OH-] = -log Kw = 14.00 or, in other words, pH + pOH = pKw = 14.00

  30. Example 15.4 By the method suggested in Figure 15.5, a student determines the pH of milk of magnesia, a suspension of solid magnesium hydroxide in its saturated aqueous solution, and obtains a value of 10.52. What is the molarity of Mg(OH)2 in its saturated aqueous solution? The suspended, undissolved Mg(OH)2(s) does not affect the measurement. Example 15.5 A Conceptual Example Is the solution 1.0 x 10–8 M HCl acidic, basic, or neutral?

  31. How Do We Measure pH? • For less accurate measurements, one can use • Litmus paper • “Red” paper turns blue above ~pH = 8 • “Blue” paper turns red below ~pH = 5 • Or an indicator.

  32. How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

  33. Strong Acids • You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. • These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. • For the monoprotic strong acids, [H3O+] = [acid].

  34. Strong Bases • Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+). • Again, these substances dissociate completely in aqueous solution.

  35. Kc = [H3O+] [A-] [HA] HA (aq) + H2O (l) A-(aq) + H3O+(aq) Dissociation Constants • For a generalized acid dissociation, the equilibrium expression would be • This equilibrium constant is called the acid-dissociation constant, Ka.

  36. Dissociation Constants The greater the value of Ka, the stronger is the acid.

  37. Weak Bases Bases react with water to produce hydroxide ion.

  38. [HB] [OH-] [B-] Kb = Weak Bases The equilibrium constant expression for this reaction is where Kb is the base-dissociation constant.

  39. Weak Bases Kb can be used to find [OH-] and, through it, pH.

  40. Kaand Kb Kaand Kb are related in this way: Ka Kb = Kw Therefore, if you know one of them, you can calculate the other.

  41. Equilibrium in Solutions of Weak Acids and Weak Bases These calculations are similar to the equilibrium calculations performed in Unit 10. • An equation is written for the reversible reaction. • Data are organized, often in an ICE format. • Changes that occur in establishing equilibrium are assessed. • Simplifying assumptions are examined (the “5% rule”). • Equilibrium concentrations, equilibrium constant, etc. are calculated.

  42. Example 15.6 Ordinary vinegar is approximately 1 M CH3COOH and as shown in Figure 15.6, it has a pH of about 2.4. Calculate the expected pH of 1.00 M CH3COOH(aq), and show that the calculated and measured pH values are in good agreement.

  43. Example 15.7 What is the pH of 0.00200 M ClCH2COOH(aq)?

  44. Example 15.8 What is the pH of 0.500 M NH3(aq)? Example 15.9 The pH of a 0.164 M aqueous solution of dimethylamine is 11.98. What are the values of Kb and pKb? The ionization equation is (CH3)2NH + H2O (CH3)2NH2+ + OH–Kb = ? Dimethylamine Dimethylammonium ion Example 15.10 A Conceptual Example Without doing detailed calculations, indicate which solution has the greater [H3O+], 0.030 M HCl or 0.050 M CH3COOH.

  45. [H3O+]eq [HA]initial Calculating Percent Ionization • Percent Ionization =  100

  46. Polyprotic Acids • A monoprotic acid has one ionizable H atom per molecule. • A polyprotic acid has more than one ionizable H atom per molecule. • Sulfuric acid, H2SO4 Diprotic • Carbonic acid, H2CO3 Diprotic • Phosphoric acid, H3PO4 Triprotic • The protons of a polyprotic acid dissociate in steps, each step having a value of Ka. • Values of Ka decrease successively for a given polyprotic acid. Ka1 > Ka2 > Ka3 , etc.

  47. Polyprotic Acids If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

  48. Example 15.11 Calculate the following concentrations in an aqueous solution that is 5.0 M H3PO4: (a) [H3O+] (b) [H2PO4–] (c) [HPO42–] (d) [PO43–] Example 15.12 What is the approximate pH of 0.71 M H2SO4?

  49. Ions as Acids and Bases • HCl is a strong acid, therefore Cl– is so weakly basic in water that a solution of chloride ions (such as NaCl) is virtually neutral. • Acetic acid, CH3COOH, is a weak acid, so acetate ion, CH3COO–, is significantly basic in water. • A solution of sodium acetate (which dissociates completely into sodium and acetate ions in water) is therefore slightly basic: CH3COO– + H2O  CH3COOH + OH–

More Related