Acid/Base Balance in the Body Notes

Lecture Learning Objectives

• Define $pKa$ and $pKb$ and understand their significance when comparing the strengths of acids and bases.
• Recall $pKa + pKb = 14$.
• Differentiate between monoprotic, diprotic, and triprotic acids.
• Understand how salts undergo hydrolysis and be able to predict the pH of salt solutions where the salt may be derived from a strong acid/strong base, weak acid/weak base, strong acid/weak base, and strong base/weak acid.
• Calculate the pH of salt solutions where the salt is derived from a strong acid/weak base or a strong base/weak acid.

Properties of Acids & Bases

• STRONG ACIDS ionize fully. Example: $HCl(aq) <br />\nrightarrow Cl^-(aq) + H_3O^+(aq)$. The single-headed arrow indicates complete ionization.
• STRONG BASES ionize fully. Example: $NaOH(aq) <br />\nrightarrow Na^+(aq) + OH^-(aq)$. The single-headed arrow indicates complete ionization.

Properties of Weak Acids & Weak Bases

• WEAK BASES ionize partially, indicated by an equilibrium arrow. Example: NH3(aq) + H2O(l)
  rightleftharpoons NH_4^+(aq) + OH^-(aq).
• WEAK ACIDS ionize partially, indicated by an equilibrium arrow. Example: CH3COOH(aq) + H2O(l)
  rightleftharpoons CH3COO^-(aq) + H3O^+(aq).

Key Equations and Relationships

• Kw = [H3O^+][OH^-] = 1 x 10^{-14} @ 25°C, where 2H2O
  rightleftharpoons H_3O^+ + OH^-.
• Kw = Ka
  x Kb.
• For a weak acid: $Ka = [H<em>3O^+][A^-]/[HA]$, where HA + H2O
  rightleftharpoons H_3O^+ + A^-.
• For its conjugate base: $Kb = [HA][OH^-]/[A^-]$, where A^- + H_2O
  rightleftharpoons HA + OH^-.

pH and pOH

• $pH + pOH = 14$.
• $pH = -log[H_3O^+]$ and $pOH = -log[OH^-]$.

pKa Scale

• $pKa = -log(Ka)$.
• Example: If [H3O^+] = 1.0 x 10^{-5}, then pH = -log{10}(1.0
  x 10^{-5}) = 5.
• For $CH_3COOH$, Ka = 8.1
  x 10^{-4} and $pKa = 4.74$.

pKb Scale

• $pKb = -log_{10}(Kb)$.
• Example: For $C<em>6H</em>5OH$, Kb = 1.3
  x 10^{-10} and $pKb = 9.90$.

Relationship between Ka and Kb

• Ka
  x Kb = 10^{-14}.
• $pKa + pKb = 14$.

Comparing Acid and Base Strengths

• The smaller the $pKa$, the stronger the acid.
• The smaller the $pKb$, the stronger the base.
• Example comparing two weak acids:
  • HCN: Ka = 4.0
    x 10^{-10}, $pKa = 9.40$.
  • $CH_3COOH$: Ka = 1.8
    x 10^{-5}, $pKa = 4.74$.
• Example comparing two weak bases:
  • $C<em>6H</em>5NH_2$: Kb = 4.2
    x 10^{-10}, $pKb = 9.38$.
  • $CH<em>3NH</em>2$: Kb = 4.4
    x 10^{-4}, $pKb = 3.36$.

Summary Table for Weak Acids and Weak Bases

• Acids: Weaker acids have small $Ka$ and large $pKa$, while stronger acids have large $Ka$ and small $pKa$.
  • Example: HCN (Ka = 4.0
    x 10^{-10}, $pKa = 9.40$) vs. $CH_3COOH$ (Ka = 1.8
    x 10^{-5}, $pKa = 4.74$).
• Bases: Weaker bases have small $Kb$ and large $pKb$, while stronger bases have large $Kb$ and small $pKb$.
  • Example: $C<em>6H</em>5NH<em>2$ (Kb = 4.2 x 10^{-10}, $pKb = 9.38$) vs. $CH</em>3NH_2$ (Kb = 4.4
    x 10^{-4}, $pKb = 3.36$).

Types of Acids

• MONOPROTIC: 1 available proton.
• POLYPROTIC: many protons or $H^+$.
  • DIPROTIC: 2 available protons.
  • TRIPROTIC: 3 available protons.

Examples of Polyprotic Acids

• DIPROTIC Example 1: $H<em>2SO</em>4$
  1. $H<em>2SO</em>4 + H<em>2O \nrightarrow HSO</em>4^- + H_3O^+$ (Ka)
  2. HSO4^- + H2O
     rightleftharpoons SO4^{2-} + H3O^+ (Ka = 1.2
     x 10^{-2})
• DIPROTIC Example 2: $H<em>2CO</em>3$
  1. H2CO3 + H2O rightleftharpoons HCO3^- + H_3O^+ (Ka1 = 4.3
     x 10^{-7})
  2. HCO3^- + H2O
     rightleftharpoons CO3^{2-} + H3O^+ (Ka2 = 4.7
     x 10^{-11})
• TRIPROTIC Example: $H<em>3PO</em>4$
  1. H3PO4 + H2O rightleftharpoons H2PO4^- + H3O^+ (Ka1 = 7.5
     x 10^{-3}, $pKa1 = 2.12$)
  2. H2PO4^- + H2O rightleftharpoons HPO4^{2-} + H_3O^+ (Ka2 = 6.2
     x 10^{-8}, $pKa2 = 7.21$)
  3. HPO4^{2-} + H2O
     rightleftharpoons PO4^{3-} + H3O^+ (Ka3 = 4.8
     x 10^{-13}, $pKa3 = 12.32$)
  • In general, Ka1 >>> Ka2 > Ka3.

Hydrolysis of Salts

• Salts are produced by the reaction of an acid with a base and contain parts of each. Example: $NaCl <br />\nrightarrow Na^+ + Cl^-$.
• When a salt dissolves in water, it completely dissociates into its acid part and base part.
• Hydrolysis is the reaction of a compound with the $H^+$ ion or the $OH^-$ ion derived from water.

Hydrolysis of Salts - Strong vs. Weak

• Ions derived from strong acids and strong bases do not hydrolyze. Example: $HCl <br />\nrightarrow H^+ + Cl^-$ and $NaOH <br />\nrightarrow Na^+ + OH^-$.
• Ions derived from weak acids or weak bases WILL hydrolyze.
• If a salt dissociates and produces an ion derived from a weak acid, that ion will hydrolyze (react with water) to form the corresponding acid.
  • Example: CH3COOH rightleftharpoons CH3COO^- + H_3O^+
  • H2O + CH3COO^-
    rightleftharpoons CH_3COOH + OH^-.
• If a salt dissociates and produces an ion that is derived from a weak base, then that ion will hydrolyse (react with water) to form the corresponding base:
  • Example: NH3 rightleftharpoons NH4^+ + OH^-
  • H2O + NH4^+
    rightleftharpoons NH3 + H3O^+.

Hydrolysis of Salts - Examples

• Salt from a strong acid and a strong base: Example: $NaCl \nrightarrow Na^+ + Cl^-$.
  • The resulting solution is neutral because neither $Na^+$ nor $Cl^-$ hydrolyzes in water.
• Salt from a weak acid and a strong base: Example: $CH<em>3COONa \nrightarrow Na^+ + CH</em>3COO^-$.
  • The resulting solution is basic.
  • The weaker the acid, the greater the extent of hydrolysis.
  • CH3COO^- + H2O
    rightleftharpoons CH_3COOH + OH^-.
• Salt from a strong acid and a weak base: Example: $NH<em>4Cl \nrightarrow NH</em>4^+ + Cl^-$.
  • The resulting solution is acidic.
  • The weaker the base, the greater the extent of hydrolysis.
  • NH4^+ + H2O
    rightleftharpoons NH3 + H3O^+.
• Salt from a weak acid and a weak base: Example: $CH<em>3COONH</em>4 \nrightarrow CH<em>3COO^- + NH</em>4^+$.
  • NH4^+ + H2O
    rightleftharpoons NH3 + H3O^+ (Ka).
  • CH3COO^- + H2O
    rightleftharpoons CH_3COOH + OH^- (Kb).
  • If $Ka = Kb$, the solution is neutral.
  • Both hydrolyze to the same extent, and Kb = 5.6
    x 10^{-10}.

Summary: pH of Salt Solutions

Calculating pH of Salt Solutions - Example 1

• Calculate the pH of 0.1 M CH_3COONa$($pKa is 4.75).
• CH3COONa rightarrow CH3COO^- + Na^+.
• CH3COO^- + H2O
  rightleftharpoons CH_3COOH + OH^-.
• Initial concentrations: 0.1 M, 0, 0.
• Equilibrium concentrations: 0.1 - x, x, x.
• Kb = [CH3COOH][OH^-]/[CH3COO^-] = x^2/0.1.
• Kb = Kw/Ka = (1
  x 10^{-14})/(1.8
  x 10^{-5}) = 5.56
  x 10^{-10}.
• x = 7.46
  x 10^{-6}
• [OH^-] = 7.46
  x 10^{-6}
• pOH = -log_{10}(7.46
  x 10^{-6}) = 5.13.
• pH = 14 - 5.13 = 8.87.

Calculating pH of Salt Solutions - Example 2

• Calculate the pH of 0.1 M NH4Cl$($Ka(NH4^+) = 5.6
  x 10^{-10}).
• NH4Cl rightarrow NH4^+ + Cl^-.
• NH4^+ + H2O
  rightleftharpoons NH3 + H3O^+.
• Initial concentrations: 0.1 M, 0, 0.
• Equilibrium concentrations: 0.1 - x, x, x.
• Ka = [NH3][H3O^+]/[NH_4^+] = x^2/0.1 = 5.6
  x 10^{-10}.
• x = 7.5
  x 10^{-6} M = [H_3O^+].
• pH = -log_{10}(7.5
  x 10^{-6}) = 5.12.
• % Dissociation: (7.5
  x 10^{-6}/0.1)
  x 100 = 0.0075%.

Summary: pH of Salt Solutions