Chapter 17 Lecture Outline

Review of Bronsted-Lowry Acid-Base Reactions

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

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

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

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

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

Strengths of acids and bases

  1. See Table 17.3, page 702

  2. Binary acids X-H bond strength:

  3. 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

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

  5. Carboxylic acids R-COOH

What Will React

  1. Strongest acid will donate protons
  2. Strongest base will accept protons
  3. A Strong Acid will react completely
  4. A Strong Base will react completely
  5. Weak Acids and Bases are at equlibrium

Acid Base Equilibrium Expression

  1. Water
    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

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

    2. Which is typically expressed as Ka

    3. Ka = 1.8 x 10-5

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

    2. Look up Ka in table (Appendix F, page A-20)

Hydrolysis Reactions:

  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.

Water and pH Scale

  1. Autoionization of water and KW

    1. Equilibrium: H2O + H2O <--> H3O+ + OH-

    2. Kw = [H3O+][OH-] = 1.0*10-14

    3. Calculate [OH-] and [H3O+]
      1. 0.1 M HCl
      2. 0.1 M NaOH

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

  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

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

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.
  2. Solutions

Acid-Base Equlibrium Problems

  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

Polyprotic Acids

Introduction to Buffers

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