Apr 18, 2025
pH and pOH Calculations:
-log[H₃O⁺]-log[OH⁻]pH + pOH = 14Weak Acids and Bases:
[Products]/[Reactants]ICE Tables:
Relation Between Ka and Kb:
Ka × Kb = Kw (1 x 10⁻¹⁴) at 25°C.[H₃O⁺].-log[H₃O⁺].14 - pOH.Example: Ammonium Chloride
[H₃O⁺] using Ka.Example: Sodium Fluoride
[OH⁻] using Kb.Formula: (x / [HA]) × 100%
x is concentration of ionized acid/base.Example Problem: Hydrofluoric Acid (HF)
[H₃O⁺].pH = -log[H₃O⁺]pOH = -log[OH⁻](x / [Initial Acid/Base]) × 100%