Lecture Notes: Polyprotic Acids

Introduction to Polyprotic Acids

• Polyprotic acids: Acids that can donate more than one proton.
• Example: Sulfuric acid, where the first ionization is strong and the second is weak.

Ionization Steps

• Polyprotic acids ionize in successive steps.
• Each step has its own ionization constant, Ka.
• Ka values decrease in successive steps because it's harder to remove additional protons.
  • Example: Sulfurous acid
    • First step: $H_2SO_3 → H^+ + HSO_3^-$ with $Ka_1 = 1.6 \times 10^{-2}$
    • Second step: $HSO_3^- → H^+ + SO_3^{2-}$ with $Ka_2 = 6.4 \times 10^{-8}$

Ka Values in Polyprotic Acids

• Sulfuric Acid
  • Ka1 is very strong (100% ionized in water), no value provided.
  • $Ka_2 = 1.2 \times 10^{-2}$
• Phosphoric Acid (H3PO4) has three Ka values.

Calculating pH of Polyprotic Acids

• Approximation Conditions:
  1. $Ka_1$ must be much greater than $Ka_2$ (about 300 times greater).
  2. Concentration must be greater than 0.05 M.
• H+ from step one usually dominates, so first ionization is mainly considered.
• Le Chatelier's Principle: Additional H+ from ionization suppresses subsequent ionizations.

Example Calculation: H2SO4

• Given: 0.010 M $H_2SO_4$
• First ionization: Strong, ionizes completely.
  • Initial $[H^+]$ = 0.010 M
• Second ionization: Weak
  • Use a RICE table to calculate ion concentrations.
  • Solve quadratic equation for x.
  • $[H^+] = 0.010 + 0.0045 = 0.0145$
  • pH Calculation: $pH = -\log(0.0145) = 1.84$

Additional Considerations

• Dilute Solutions (Concentration closer to 1 x 10^-6 or 1 x 10^-7):
  • Must account for $K_W$ from water autoionization.
  • Not required for this lecture.
• Calculating every ionization step is more precise but often unnecessary unless concentrations are low.

Conclusion

• Understanding the ionization of polyprotic acids is crucial in calculating pH accurately.
• In practice, focus on the primary ionization unless conditions demand otherwise.