Acid-Base Equilibria and Salt Hydrolysis

Determination of Kb from Equilibrium Concentrations

  • Caffeine (C<em>8H</em>10N<em>4O</em>2C<em>8H</em>{10}N<em>4O</em>2) is a weak base.
  • To find KbK_b for caffeine, use equilibrium concentrations:
    • [C<em>8H</em>10N<em>4O</em>2{C<em>8H</em>{10}N<em>4O</em>2}] = 0.050 M
    • [C<em>8H</em>10N<em>4O</em>2H+{C<em>8H</em>{10}N<em>4O</em>2H^+}] = 5.0×10−35.0 × 10^{−3} M
    • [OH−{OH^-}] = 2.5×10−32.5 × 10^{−3} M

Calculating Equilibrium Concentrations in a Weak Base Solution

  • Find [OH−{OH^-}], pOH, and pH of a 0.25-M solution of trimethylamine ((CH<em>3)</em>3N(CH<em>3)</em>3N), a weak base.
  • Kb=6.3×10−5K_b = 6.3 × 10^{−5}

Hydrolysis of Salts

  • Salt hydrolysis: Ions from salt dissociation react with water to produce OH−OH^- or H3O+H_3O^+.
  • Basic salts (conjugates of weak acids):
    • F−(aq)+H2O(l)⇌HF(aq)+OH−(aq)F^−(aq) + H_2O(l) \rightleftharpoons HF(aq) + OH^−(aq)
  • Acidic salts (conjugates of weak bases):
    • NH<em>4+(aq)+H</em>2O(l)⇌NH<em>3(aq)+H</em>3O+(aq)NH<em>4^+(aq) + H</em>2O(l) \rightleftharpoons NH<em>3(aq) + H</em>3O^+(aq)

Acid and Base Ionization

  • Strong acid (e.g., HCl) ionizes completely:
    • HCl(aq)→H+(aq)+Cl−(aq)HCl(aq) \rightarrow H^+(aq) + Cl^−(aq)
    • Cl−Cl^− has no affinity for H+H^+, making it a weak conjugate base.
  • Weak acid (e.g., HF) ionizes partially:
    • HF(aq)⇌H+(aq)+F−(aq)HF(aq) \rightleftharpoons H^+(aq) + F^−(aq)
    • F−F^− has strong affinity for H+H^+, making it a strong conjugate base.

Types of Ions

  • Neutral Ions:
    • Conjugate bases from 6 strong acids
    • Cations from strong bases (hydroxides)
  • Basic Ions:
    • Conjugate bases from weak acids
  • Acidic Ions:
    • Conjugate acids from weak bases
    • Highly charged metal ions

Aluminum Ion Reaction with Water

  • Hydrated aluminum ion becomes a weak acid:
    • [Al(H<em>2O)</em>6]3++H<em>2O⇌[Al(H</em>2O)<em>5OH]2++H</em>3O+[Al(H<em>2O)</em>6]^{3+} + H<em>2O \rightleftharpoons [Al(H</em>2O)<em>5OH]^{2+} + H</em>3O^+

Classifying Salt Solutions

  • Neutral Salts:
    • Cation from strong base, anion from strong acid (e.g., NaCl, Ca(NO<em>3)</em>2Ca(NO<em>3)</em>2, KBr)
  • Basic Salts:
    • Cation from strong base, anion from weak acid (e.g., NaF, Ca(C<em>2H</em>3O<em>2)</em>2Ca(C<em>2H</em>3O<em>2)</em>2, KNO2KNO_2)
  • Acidic Salts:
    • Cation from weak base, anion from strong acid (e.g., NH4ClNH_4Cl)
    • Highly charged metal ion, anion from strong acid (e.g., Al(NO<em>3)</em>3Al(NO<em>3)</em>3)

pH of Salt Solutions

  • Depends on the relative strengths of the weak acid and base if both cation and anion hydrolyze.
  • Qualitative predictions using K<em>bK<em>b (anion) and K</em>aK</em>a (cation):
    • K<em>b>K</em>aK<em>b > K</em>a: Basic solution
    • K<em>b<K</em>aK<em>b < K</em>a: Acidic solution
    • K<em>b≈K</em>aK<em>b ≈ K</em>a: Neutral solution

Relationship Between Ka and Kb

  • For a conjugate acid-base pair:
    • K<em>a×K</em>b=KwK<em>a × K</em>b = K_w

Calculating pH of an Acidic Salt Solution

  • Example: Anilinium chloride ([C<em>6H</em>5NH3]Cl[C<em>6H</em>5NH_3]Cl)
    • K<em>a=K</em>wKb=1.0×10−144.3×10−10=2.3×10−5K<em>a = \frac{K</em>w}{K_b} = \frac{1.0 × 10^{-14}}{4.3 × 10^{-10}} = 2.3 × 10^{-5}
    • Equilibrium: 2.3×10−5=(x)(x)0.233−x≈x20.2332.3 × 10^{−5} = \frac{(x)(x)}{0.233 − x} ≈ \frac{x^2}{0.233}
    • x = 0.0023 M
    • pH = −log[H3O+{H_3O^+}] = −log(0.0023) = 2.64

Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base

  • Determine acetic acid concentration in a solution with [CH<em>3CO</em>2−{CH<em>3CO</em>2}^−] = 0.050 M and [OH−{OH}^−] = 2.5×10−62.5 × 10^{−6} M at equilibrium.

Determining Acidic or Basic Nature of Salts

  • Examples:
    • KBr: Neutral
    • NaHCO3NaHCO_3: Basic
    • Na<em>2HPO</em>4Na<em>2HPO</em>4: Basic
    • NH4FNH_4F: Acidic

pH of NaCHO2 Solution

  • Find the pH of 0.100 M NaCHO<em>2(aq)NaCHO<em>2(aq) solution (K</em>aK</em>a of CHO2HCHO_2H = 1.8×10−41.8 × 10^{−4}).

Hydrolysis of [Al(H2O)6]3+

  • Calculate the pH of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion [Al(H<em>2O)</em>6]3+[Al(H<em>2O)</em>6]^{3+} in solution.

Polyprotic Acids

  • Monoprotic: One ionizable hydrogen atom.
  • Polyprotic: More than one ionizable hydrogen atom (diprotic, triprotic).
  • Ionize in steps; KaK_a decreases with each step.

Ionization of a Diprotic Acid

  • Carbonated water example with initial [H<em>2CO</em>3{H<em>2CO</em>3}] = 0.033 M:
    • H<em>2CO</em>3(aq)+H<em>2O(l)⇌H</em>3O+(aq)+HCO<em>3−(aq)H<em>2CO</em>3(aq) + H<em>2O(l) \rightleftharpoons H</em>3O^+(aq) + HCO<em>3^−(aq), K</em>a1=4.3×10−7K</em>{a1} = 4.3 × 10^{−7}
    • HCO<em>3−(aq)+H</em>2O(l)⇌H<em>3O+(aq)+CO</em>32−(aq)HCO<em>3^−(aq) + H</em>2O(l) \rightleftharpoons H<em>3O^+(aq) + CO</em>3^{2−}(aq), Ka2=4.7×10−11K_{a2} = 4.7 × 10^{−11}
    • [H<em>3O+{H<em>3O^+}] = [HCO</em>3−{HCO</em>3}^−] = 1.2×10−41.2 × 10^{−4} M
    • [CO32−{CO_3}^{2−}] = 4.7×10−114.7 × 10^{−11} M