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Chapter 14 Acids and Bases

Chapter 14 Acids and Bases. 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. Some Definitions. Brønsted – Lowry

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Chapter 14 Acids and Bases

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  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

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