Chapter 15 Lecture Outline


15.3 Definitions of Acids and Bases

The Arrhenius Definition

Bronsted-Lowry Acid-Base Reactions

  1. General Equation
  2. AcidBaseConjugate
    Base of HA
    Congugate
    acid of B
    HA+B-->A-+HB+

  3. Strong Acid and Strong Base
  4. AcidBaseConjugate
    Base of HA
    Congugate
    acid of B
    HCl(aq)+NaOH(aq)-->Cl-+Na1++H2O (aq)

  5. Weak Acid and Weak Base
  6. AcidBaseConjugate
    Base of HA
    Congugate
    acid of B
    CH3COOH(aq) + H2O(l) <--> CH3COO-(aq) + H3O+(aq)

  7. Autoionization of water:
  8. AcidBaseConjugate
    Base of HA
    Congugate
    acid of B
    H2O(l) + H2O(l) <--> OH-(aq) + H3O+(aq)

  9. Polyprotic Acid
  10. AcidBaseConjugate
    Base of HA
    Congugate
    acid of B
    H2SO4 + H2O(l) --> HSO41- + H3O+(aq)
    HSO41- + H2O(l) <--> SO42- + H3O+(aq)


15.4 Acid Strength and the Acid Ionization Constant (Ka)

  1. Binary acids X-H bond strength:
  2. Oxyacids, more O, stronger acid (Electron withdrawing by X in X-OH)

    AcidFormulaAcid Dissociation Constant
    hypochlorous acidHOClKa = 10-8
    chlorous acidHOClOKa = 10-2
    chloric acidHOClO2Ka = 101
    perchloric acidHOClO3Ka = 108

  3. Metal hydroxides, X is electroposative
    1. NaOH strong base
    2. Be(OH)2 amphoteric

  4. Carboxylic acids R-COOH

15.5 Autoionization of Water and pH

AcidBaseConjugate
Base of HA
Congugate
acid of B
H2O(l) + H2O(l) --> OH-(aq) + H3O+(aq)

  1. The equilibrium concentrations can be described with a constant K for the autoionization of water:

  2. This is usually rearranged since [H2O] is essentially constant. and given as Kw:

  3. At 25 C, Kw = 1.0 * 10-14
  4. Calculate [OH-] and [H3O+]
    1. 0.1 M HCl
    2. 0.1 M NaOH

  5. pH and pOH
    1. pH = - log[H3O+]
    2. pOH = - log[OH-]
    3. pH + pOH = 14 (Show algebra)


15.6 Finding the [H3O+] and pH of Strong and Weak Acid Solutions

  1. Generic acid reaction
    1. HA + H2O <--> A- + H3O+

    2. Look up Ka in table

  2. Acetic acid:
    1. For the reaction CH3COOH + H2O <--> CH3COO- + H3O1+

    2. Since [H2O] is constant, this is typically expressed as Ka
    3. Ka = 1.8 x 10-5

  3. Calculating pH (Given Solution concentration)
  4. Solution[H3O+]pH [OH-]pOH
    10 M HCl101-110-1515
    0.1 M HCl10-1110-1313
    10-8 M HCl10-8810-66
    10 M NaOH 10-1515101 -1
    0.1 M NaOH10-13 13 10-11
    10-8 M NaOH10-6610-88

    1. Note: pH + pOH = 14
    2. Note: 10-8 M HCl and NaOH make no sense, why?
    3. Significant figures and logs

  5. Calculate [H3O+] and [OH-] given pH or pOH
    1. Vinegar pH=2.9
    2. ammonia pH = 11.5

Polyprotic Acids

Example Problems: Solving a Weak Acid Equlibrium System

  1. For a nitrous acid (Ka = 4.5 x 10-4) solution find the pH and pOH for initial nitrous acid concentrations of
    1. 1.0 M
    2. 0.001 M
    3. 10-8 M.
    4. Solutions

  2. Acetic Acid Equlibrium
    1. Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of acetic acid is added to 250.0 mL of water.
    2. Solution

  3. Nitrous Acid Equlibrium
    1. Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of nitrous acid is added to 250.0 mL of water.
    2. Solution

  4. Nitric Acid Equlibrium
    1. Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of nitric acid is added to 250.0 mL of water.
    2. Solution


15.7 Base Solutions

  1. Ka =

  2. Kb =

  3. Kw =

  4. Kw = Ka * Kb

15.8 The Acid-Base Properties of Ions and Salts

  1. Chemical System
    1. Reaction #1 CH3COOH + H2O <-> CH3COO- + H3O+
    2. Reaction #2
      1. step 1: NaCH3COO --> Na1+ + CH3COO1-
      2. step 2: CH3COO1- + H2O <-> CH3COOH + OH-

    3. TOTAL rxn: 2 H2O <-> H3O+ + OH-
  2. Equlibrium for reaction #1 is Ka for CH3COOH
    1. For a 0.100 M CH3COOH at equilibrium, calculate Ka if at equlibrium [H3O] = 1.318*10-3 M:
    2. CH3COOH H2O <--> CH3COO- H3O+
      Initial (M) 0.100     0 0
      Change (M) -1.318*10-3     +1.318*10-3 +1.318*10-3
      Final (M) 0.0987     1.318*10-3 1.318*10-3

  3. Flow Chart

  4. Examples. What happens when...
    1. sodium acetate dissolves in water.
    2. potassium nitrate dissolves in water.
    3. ammonium chloride dissolves in water.
    4. sodium fluoride dissovles in water.
    5. potassium hydroxide dissolves in water.

Example Problems

  1. Sodium Acetate Equlibrium
    1. Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of sodium acetate is added to 250.0 mL of water.
    2. Solution

  2. Ammonium Chloride Equlibrium
    1. Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of ammonium chloride is added to 250.0 mL of water.
    2. Solution

  3. Sodium Nitrite Equlibrium
    1. Calculate the H3O+, OH-, acid concentration, base concentration, pH and pOH when 1.00 g of sodium nitrite is added to 250.0 mL of water.
    2. Solution


15.9 Acid Strength and Molecular Structure


15.10 Lewis Acids and Bases


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