CHAPTER 14

Chapter 14: Acids & Bases

14.1: Nature of Acids & Bases

Arrhenius Model
  • Definition:
    • Acids: Substances that produce H+ ions in aqueous solutions.
    • Bases: Substances that produce OH- ions in aqueous solutions.
  • Limitations:
    • Only applicable to aqueous solutions and hydroxides as bases.
Brønsted-Lowry Model
  • Definition:
    • Acid: Proton (H+) donor.
    • Base: Proton acceptor.
  • Characteristics:
    • Not limited to aqueous solutions; applicable in gases as well.
Lewis Model
  • Definition:
    • Acids: Electron pair acceptor.
    • Bases: Electron pair donor.

Arrhenius Acids and Bases

Arrhenius Acid
  • Reactions:
    • Increases the concentration of H+ ions in aqueous solution:
      HA{(aq)} ightarrow H^{+}{(aq)} + A^{-}_{(aq)}
Arrhenius Base
  • Reactions:
    • Increases the concentration of OH- ions in aqueous solution:
      BOH{(aq)} ightarrow B^{+}{(aq)} + OH^{-}_{(aq)}

14.2: Brønsted-Lowry Theory Examples

Proton Transfer
  • Example of a Brønsted-Lowry Acid-Base Reaction:
    • Acid: HCl (Hydrochloric acid)
    • Base: NH3 (Ammonia)
    • Conjugates:
    • Conjugate Acid: NH4+ (Ammonium)
    • Conjugate Base: Cl- (Chloride)
Other Examples
  • H2SO4 + H2O → H3O+ + HSO4-
  • HNO3 + H2O → H3O+ + NO3-

Lewis Acid-Base Reactions

Mechanism
  • Reaction Form:
    • A + B
      ightarrow A-B
    • Curved Arrow Representation:
    • Lewis Acid reacts with Lewis Base to form an adduct.
  • Characteristics:
    • A Lewis acid typically has an empty atomic orbital.
    • A Lewis base has a lone pair of electrons available for bonding.

14.11: Lewis Acid – Base Model

Electron Pair Interaction
  • A Lewis Acid has an empty orbital that can accept an electron pair.
Summary of Definitions
ModelDefinition of AcidDefinition of Base
ArrheniusH+ producerOH- producer
Brønsted-LowryH+ donorH+ acceptor
LewisElectron pair acceptorElectron pair donor

14.3: pH Scale

Definitions
  • pH Calculation:
    • extpH=−extlog[H+]ext{pH} = - ext{log}[H^{+}]
    • extpOH=−extlog[OH−]ext{pOH} = - ext{log}[OH^{-}]
    • extpK=−extlogKext{pK} = - ext{log}K
  • Change in pH Unit Importance:
    • A change of 1 unit in pH corresponds to a 10-fold change in H+ concentration.
Example Calculation
  • 1.45×10−4 M concentration:
    • [H+]=1.45imes10−4[H^{+}] = 1.45 imes 10^{-4}
    • extpH=3.83,ext{pH} = 3.83,
    • extpOH=14.00−3.83=10.17.ext{pOH} = 14.00 - 3.83 = 10.17.

14.4: Calculating the pH of Strong Acid Solutions

Strong Acid Characteristics
  • Strong acids dissociate completely in aqueous solutions.
  • Example:
    • $HNO3
      ightarrow H+ + NO3-$
Example Calculation for Strong Acid
  • 0.10 M HNO3 calculation:
    • extpH=−extlog(0.10)=1.00ext{pH} = - ext{log}(0.10) = 1.00
  • 1.0×10−10 M solution:
    • Major species is H2O, so extpH=7ext{pH} = 7.

14.5: Calculating pH of Weak Acids

Weak Acid Characteristics
  • Weak acids do not dissociate completely.
  • Calculation Strategy:
    • Use the expression Ka=[H+][A−][HA]K_a = \frac{[H^{+}][A^{-}]}{[HA]}
  • Example: Calculate pH for 1.00 M HF
  • Given $K_a = 7.2 imes 10^{-4}$, use simplifications for weak acids to find equilibrium concentrations.
Systematic Dissociation Reaction for HF
  • HF ightleftharpoons H^{+} + F^{-}
    • Initial:
    • [HF] = 1.00 M, [H+] = 0, [F-] = 0
    • Change:
    • -x, +x, +x
    • Equilibrium:
    • [HF] =