Download
chapter 14 acids and bases n.
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 14 Acids and Bases PowerPoint Presentation
Download Presentation
Chapter 14 Acids and Bases

Chapter 14 Acids and Bases

113 Vues Download Presentation
Télécharger la présentation

Chapter 14 Acids and Bases

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chapter 14Acids and Bases

  2. Some Definitions • Arrhenius • Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions. • Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions.

  3. Some Definitions • Brønsted–Lowry • Acid: Proton donor • Base: 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. Acid Dissociation (Ionization) Reactions • Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids: • a. Hydrochloric acid • b. Acetic acid • c. Ammonium ion • d. Anilinium ion (C6H5NH3) • e. Hydrated aluminum (III) ion [Al(H2O)6]3+

  8. Conjugate Acids and Bases: • From the Latin word conjugare, meaning “to join together.” • Reactions between acids and bases always yield their conjugate bases and acids.

  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 K is not measured (K>>1).

  12. C2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Acid and Base Strength Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1).

  13. Relative Base Strength Using the following Ka values, arrange the following species according to their strength as bases:

  14. 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.

  15. 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

  16. Calculating [H+] & [OH-] • Calculate [H+] & [OH-] as required for each of the following solutions at 250C, & state whether the solution is neutral, acidic, or basic. • a. = 1.0 x 10-5 M OH- • b. a. = 1.0 x 10-7 M OH- • c. 10.0 M H+

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

  18. pH • In pure water, Kw = [H3O+] [OH−] = 1.0  10−14 • Because in pure water [H3O+] = [OH−], [H3O+] = (1.0  10−14)1/2 = 1.0  10−7

  19. 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.

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

  21. Other “p” Scales • The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions). • Some similar examples are • pOH −log [OH−] • pKw−log Kw

  22. 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

  23. Calculating pH, pOH pH = -log10(H3O+) pOH = -log10(OH-) Relationship between pH and pOH pH + pOH = 14 Finding [H3O+], [OH-] from pH, pOH [H3O+] = 10-pH [OH-] = 10-pOH

  24. Calculating pH & pOH • Calculate pH & pOH for each of the following solutions at 250C. • a. = 1.0 x 10-3 M OH- • b. a. = 1.0 M H+

  25. Calculating pH • The pH a sample of human blood was measured to be 7.41 at 250C. Calculate pOH, [H+], & [OH-] for the sample.

  26. 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 • An indicator

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

  28. 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].

  29. pH of Strong Acids • Calculate pH of 0.10 M HNO3. • Calculate pH of 1.0 x 10-10 M HCl.

  30. 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.

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

  32. [H3O+] [COO−] [HCOOH] Ka = Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • We know that

  33. Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • To calculate Ka, we need the equilibrium concentrations of all three things. • We can find [H3O+], which is the same as [HCOO−], from the pH.

  34. Calculating Ka from the pH pH = −log [H3O+] 2.38 = −log [H3O+] −2.38 = log [H3O+] 10−2.38 = 10log [H3O+] = [H3O+] 4.2  10−3 = [H3O+] = [HCOO−]

  35. Calculating Ka from pH Now we can set up a table…

  36. [4.2  10−3] [4.2  10−3] [0.10] Ka = Calculating Ka from pH = 1.8  10−4

  37. Solving Weak Acid Equilibrium Problems • 1. List the major species in the solution. • 2. Choose the species that can produce H+ and write balanced equations for the reactions producing H+ . • 3. Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H+ . • 4. Write the equilibrium expression for the dominant equilibrium • 5. ICE the problem • 6. Substitute the equilibrium [ ] into the equilibrium expression • 7. Solve for x the “easy” way; that is, by assuming [HA]0-x  [HA]0 • 8. Use the 5% rule to verify whether the approximation is valid • 9. Calculate [H+] and pH

  38. The pH of Weak Acids • The hypochlorite ion (OCl-) is a strong oxidizing agent often found in household bleaches & disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is a much stronger base than Cl-for example) & forms the weakly acidic hypochlorous acid (HOCl, Ka = 3.5 x 10-8 ). Calculate pH of 0.100 M aqueous solution of hypochlorous acid.

  39. The pH of Weak Acids continued • HOCl, Ka = 3.5 x 10-8 Calculate pH of 0.100 M aqueous solution of hypochlorous acid.

  40. The pH of Weak Acid Mixtures • Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 x 10-10) and 5.00 M HNO2 (Ka = 4.0 x 10-4) . Also calculate the concentration of cyanide ion in this solution at equilibrium.

  41. Percent Dissociation • In general, the more dilute the weak acid solution, the greater the percent dissociation of the weak acid.

  42. Calculate Percent Dissociation • Calculate the percent dissociation of acetic acid (Ka = 1.8x 10-5) in each of the following solutions. • a. 1.00 M HC2H3O2 • b. 0.100 M HC2H3O2

  43. Calculation Ka from Percent Dissociation • Lactic acid (HC3H5O3) is a waste product that accumulates in muscle tissue during exertion, leading to pain & feeling of fatigue. In a 0.100 M aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of Ka for this acid.

  44. 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.

  45. The pH of strong bases • Calculate the pH of a 5.0x 10-2 M NaOH solution.

  46. Weak Bases • Many types of proton acceptors (bases) do not contain hydroxide ions. When dissolved in water, they increase the concentration of hydroxide ions because of their reaction with water. • Ex. • Bases such as ammonia typically have at least one unshared pair of electrons that is capable of forming a bond with a proton.

  47. Weak Bases • We will solve weak base problems in the same manner we solved weak acid problems (look back over the steps you were given • We will use Kb instead of Ka and will find [OH-] instead of [H+] • Remember the process for “switching” from pOH to pH &/or from [OH-] [H+]

  48. pH of a Weak Base (I) • Calculate the pH for a 15.0 M NH3 solution (Kb = 1.8x 10-5).

  49. pH of a Weak Base (II) • Calculate the pH for a 1.0 M methylamine solution (Kb = 4.38x 10-4).

  50. Polyprotic Acids • Some acids furnish more than one acidic proton such as H2SO4, H3PO4. • Ex. H2CO3