Chemical Analysis Notes

Monoprotic Acid-Base Equilibria

  • Overview: Chapter 9 focuses on the analysis of monoprotic acid-base equilibria in quantitative chemical analysis.

Strong Acids and Bases (Section 9-1)

  • Dissociation and Equilibrium:

    • Large dissociation constants (KaK_a) signify strong acids.

    • General dissociation reaction:
      HX+H<em>2O⇌H</em>3O++X−HX + H<em>2O \rightleftharpoons H</em>3O^+ + X^-

    • Example acids with KaK_a values:

      • HCl: Ka=103.9K_a = 10^{3.9}

      • HBr: Ka=105.8K_a = 10^{5.8}

      • HI: Ka=1010.4K_a = 10^{10.4}

      • HNO3: Ka=101.4K_a = 10^{1.4}

    • For Sulfuric Acid (H2SO4), only the first dissociation is complete:
      H<em>2SO</em>4→H++HSO4−H<em>2SO</em>4 \rightarrow H^+ + HSO_4^-

    • Second dissociation: Ka2=1.0imes10−2K_{a2} = 1.0 imes 10^{-2}

  • Common Strong Bases:

    • Hydroxides of alkali metals (e.g., LiOH, NaOH, KOH, RbOH, CsOH)

    • Quaternary ammonium hydroxides (e.g., tetrabutylammonium hydroxide)

pH Calculations

  • pH of Strong Acid:

    • Calculation for 0.10 M HBr:

    1. Reaction completion assumption:

    • [H3O+]=0.10M[H_3O^+] = 0.10 M

    1. pH Calculation:
      extpH=−extlog[H3O+]=−extlog(0.10)=1.00ext{pH} = - ext{log}[H_3O^+] = - ext{log}(0.10) = 1.00

  • pH of Strong Base:

    • Calculation for 0.10 M KOH:
      [OH−]=0.10M[OH^-] = 0.10 M

    • pH can be found from [H+][H^+] using the relationship with K<em>wK<em>w: K</em>w=[H+][OH−]=1.0imes10−14K</em>w = [H^+][OH^-] = 1.0 imes 10^{-14}

Example Calculation with Activity Coefficient

  • pH of 0.10 M HBr with activity coefficient (β\beta):

    • Ionic strength (c):

    • Calculate pH using:
      extpH=−extlog(β[H+])ext{pH} = - ext{log}(\beta [H^+])

    • Use:
      β=0.83\beta = 0.83

Dilute Base Calculations

  • pH of 1 × 10^{-8} M KOH and water contribution:

    • Systematic treatment is required as the concentration of OH−OH^- from water ionization can influence pH calculations.

Weak Acids and Bases (Section 9-2)

  • Dissociation Constants:

    • Acid dissociation constant (KaK_a) for weak acids reflects partial dissociation in water.

    • Base hydrolysis constant (KbK_b) reflects weak base reactions.

  • Understanding pK Values:

    • Negative logarithm of equilibrium constants:
      pK=−extlog(K)pK = - ext{log}(K)

    • In general, higher KK results in lower pKpK values, correlating strength.

Weak Acid Equilibria (Section 9-3)

  • Typical Weak Acid Problem:

    • Reaction illustrative approach with equilibrium constants and charge/mass balance equations:

    1. HA⇌H++A−HA \rightleftharpoons H^+ + A^-

    2. Charge Balance: [H+]=[A−]+[OH−][H^+] = [A^-] + [OH^-]

    3. Mass Balance: Total concentration in weak acid form, FF.

    4. Incorporate equilibrium properties:
      Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]}

  • Fraction of Dissociation:

    • (α = Fraction of acid dissociated).

    • Concentration's relation leads to simplifications in calculations at different dilution states.

Buffers Section (Section 9-5)

  • Buffer Properties:

    • A buffered solution minimizes pH changes upon acid/base addition or dilution, relying on equilibria between weak acids and their conjugate bases.

    • Ideal mix: acid + conjugate base in near-equal concentrations.

  • Henderson-Hasselbalch Equation:

    • Used for calculating pH of buffer solutions,

    • For Acid:
      extpH=pKa+extlog([A−][HA])ext{pH} = pK_a + ext{log}\left(\frac{[A^-]}{[HA]}\right)

    • Changes in pH based on the ratio of acid/base concentrations.

  • Example - Buffer Calculation:

    • If given a pH and pK, calculate the concentration ratio of conjugate species using the Henderson-Hasselbalch equation.