CFR 16: Acid/Base Balance in the Body

Acid-Base Balance in the Body

Learning Objectives

• Define pKa and pKb: Understand their significance when comparing the strengths of acids and bases.
• Equation: Recall that $pKa + pKb = 14$.
• Types of Acids:
  • Monoprotic: Can donate one proton (H+).
  • Diprotic: Can donate two protons.
  • Triprotic: Can donate three protons.
• Salts and Hydrolysis:
  • Understand how salts undergo hydrolysis and predict pH of salt solutions derived from:
  • Strong acid/strong base.
  • Weak acid/weak base.
  • Strong acid/weak base.
  • Strong base/weak acid.
• Calculate pH of salt solutions:
  • Salt from strong acid/weak base or strong base/weak acid.

Properties of Acids and Bases

• Strong Acids: Completely ionize in solution.
  • Example: ext{HCl(aq)}
    ightarrow ext{Cl}^-(aq) + ext{H}_3 ext{O}^+(aq)
• Strong Bases: Completely ionize in solution.
  • Example: ext{NaOH(aq)}
    ightarrow ext{Na}^+(aq) + ext{OH}^-(aq)
• Weak Acids: Partially ionize.
  • Example: ext{CH}3 ext{COOH(aq)} + ext{H}2 ext{O}(l)
    ightleftharpoons ext{CH}3 ext{COO}^-(aq) + ext{H}3 ext{O}^+(aq)
• Weak Bases: Partially ionize.
  • Example: ext{NH}3(aq) + ext{H}2 ext{O}(l)
    ightleftharpoons ext{NH}_4^+(aq) + ext{OH}^-(aq)

Key Constants

• Ion Product of Water: $K<em>w = [ ext{H}</em>3 ext{O}^+][ ext{OH}^-] = 1 imes 10^{-14} ext{ at } 25^ ext{°C}$
• For Weak Acids:
  • $K<em>a = \frac{[ ext{H}</em>3 ext{O}^+][ ext{A}^-]}{[ ext{HA}]}$
• For Weak Bases:
  • $K_b = \frac{[ ext{HB}^+][ ext{OH}^-]}{[ ext{B}]}$
• Relationship between acid and base strengths: $K<em>a imes K</em>b = 10^{-14}$

pKa and pKb

• Calculating pKa: $pKa = - ext{log}(K_a)$
• Example: For Acetic Acid: K_a = 8.1 imes 10^{-5} ightarrow pKa = 4.74
  • Smaller pKa indicates a stronger acid.
• Calculating pKb: $pKb = - ext{log}(K_b)$
  • Example: For Phenol (C6H5OH): K_b = 1.3 imes 10^{-10}
    ightarrow pKb = 9.90
  • Smaller pKb indicates a stronger base.

Acid and Base Strength Comparison

• Weak Acid Comparison:
  • HCN: $K_a = 4.0 imes 10^{-10}, pKa = 9.40$
  • CH3COOH: $K_a = 1.8 imes 10^{-5}, pKa = 4.74$
  • Smaller pKa (CH3COOH) denotes stronger acid.
• Weak Base Comparison:
  • C6H5NH2: $K_b = 4.2 imes 10^{-10}, pKb = 9.38$
  • CH3NH2: $K_b = 4.4 imes 10^{-4}, pKb = 3.36$
  • Smaller pKb (CH3NH2) denotes stronger base.

Hydrolysis of Salts

• Salts in Water: Completely dissociate into ions:
  • Example: ext{NaCl}
    ightarrow ext{Na}^+ + ext{Cl}^-
• When acids and bases react:
  • Hydrolysis occurs when salts produce ions from weak acids/bases.
• Types of Hydrolysis:
  • Neutral: Produced from strong acid and strong base (e.g., NaCl) - solution pH = 7.
  • Basic: Produced from weak acid and strong base (e.g., $ext{CH}_3 ext{COONa}$) - solution pH > 7.
  • Acidic: Produced from strong acid and weak base (e.g., $ext{NH}_4 ext{Cl}$) - solution pH < 7.

Calculating pH of Salt Solutions

1. Example 1: Calculate pH of 0.1 M $CH_3COONa$:

   • ext{CH}3 ext{COONa} ightarrow ext{CH}3 ext{COO}^- + ext{Na}^+
   • Hydrolyzes: ext{CH}3 ext{COO}^- + ext{H}2 ext{O}
     ightleftharpoons ext{CH}_3 ext{COOH} + ext{OH}^-
   • Calculate Kb from Ka: Use $K<em>b = \frac{K</em>w}{K_a}$.
   • Follow equilibrium steps to find hydroxide concentration and then pH.

2. Example 2: Calculate pH of 0.1 M $NH_4Cl$:

   • NH4Cl ightarrow NH4^+ + Cl^-
   • Hydrolyzes: NH4^+ + H2O
     ightarrow NH3 + H3O^+
   • Set up equations similarly to find pH.

Summary: pH of Salt Solutions

• Neutral: Derived from strong acid/strong base (pH ≈ 7)
• Acidic: Derived from strong acid/weak base (pH < 7)
• Basic: Derived from weak acid/strong base (pH > 7)
• Use $K<em>a$ and $K</em>b$ relationships to compute and predict pH.