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Drill: Calculate the pH of: [H + ] = 3.0 x 10 -5 M [HI] = 2.0 x 10 -4 M. Acid-Base Equilibria. Ionization of Water. H 2 O (l) H + (aq) + OH - (aq) 2 H 2 O (l) H 3 O + (aq) + OH - (aq). Water Ionization Constant. K w = [H + ][OH - ] or K w = [H 3 O + ][OH - ]
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Drill: Calculate the pH of: [H+] = 3.0 x 10-5 M [HI] = 2.0 x 10-4 M
Ionization of Water H2O(l) H+(aq)+ OH-(aq) 2 H2O(l) H3O+(aq) + OH-(aq)
Water Ionization Constant Kw = [H+][OH-] or Kw = [H3O+][OH-] Kw = 1.00 x 10-14 Memorize this
Kw Calculations In pure water [H+] = [OH-] Kw =[H+][OH-] = 1.0 x 10-14 Let [H+] = x = [OH-]
Kw Calculations Kw =[H+][OH-] = x2 Kw = x2 =1.0 x 10-14 Thus x = 1.0 x 10-7 M [H+] = x = 1.0 x 10-7 M
pH of Pure Water [H+] = 1.0 x 10-7 M pH = -log[H+] pH = -log[1.0 x 10-7] pH = 7 -log 1 Thus pH = 7.00
Weak Acid Ionization HA(aq) H+(aq) + A-(aq) HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Acid Dissociation Constant • HA(aq)H+(aq) + A-(aq) • [H+][A-] • [HA] Ka =
Weak Base Ionization NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Base Dissociation Constant • NH3(aq) + H2O(l) • NH4+(aq) + OH-(aq) • [NH4+][OH-] • [NH3] Kb =
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