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Common Ion Effect

Common Ion Effect. consider 0.10M HC 2 H 3 O 2 What happens to pH if we add NaC 2 H 3 O 2 ? add product in acid dissociation rxn equilibrium shifts [H 3 O + ] pH. Reaction?. to left. decreases. increases. Common Ion Effect. consider 0.10M NaC 2 H 3 O 2

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Common Ion Effect

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  1. Common Ion Effect consider 0.10M HC2H3O2 What happens to pH if we add NaC2H3O2? add product in acid dissociation rxn equilibrium shifts [H3O+] pH Reaction? to left decreases increases

  2. Common Ion Effect consider 0.10M NaC2H3O2 What happens to pH if we add HC2H3O2? add product in base rxn equilibrium shift [OH-] [H+] pH Reaction? to left decreases increases decreases

  3. Common Ion Effect

  4. Common Ion Effect What is the pH of a 0.30M HCOOH solution? What is the pH of a solution containing 0.30M HCOOH and 0.52M KCOOH?

  5. HCOOH + H2O  H3O+ + COOH- 0.30 0 0 -x +x +x 0.30-x x x 1.8 x 10-4 = x2 / 0.30 x = 7.3x10-3M= [H+] pH = 2.13 What is the pH of a 0.30M HCOOH solution?

  6. Common Ion Effect What is the pH of a solution containing 0.30M HCOOH and 0.52M KCOOH? HCOOH + H2O  H3O+ + COOH- 0.30 0 0.52 -x +x +x 0.30-x x 0.52+x 1.8 x 10-4 = 0.52x / 0.30 x = 1.0 x 10-4 = [H+] pH = 3.98

  7. Henderson Hasselbach Equation If you have a solution containing significant amounts of an acid - base pair: HA + H2O  H3O+ + A- Ka = [H+][A-] [HA] [HA] • -log [H+] = -log Ka – log • [A-] [A-] • pH = pKa + log [HA] [HA] • [H+] = Ka x [A-] • pH = pKa + log ( B / A ) The “x is small” approximation is implicit in the H-H equation.

  8. Henderson Hasselbach Equation What is the pH of a solution containing 0.30M HCOOH and 0.52M KCOOH? pH = pKa + log ( [base] / [acid] ) = -log 1.8 x 10-4 + log (0.52 / 0.3) = 3.98

  9. Henderson Hasselbach Equation What is the pH of a solution of 0.12M lactic acid (HC3H5O3) and 0.10M sodium lactate? The Ka for lactic acid is 1.4 x 10-4. pH = -log 1.4 x 10-4 + log (0.10 / 0.12) = 3.77 To test the H-H equation, solve this problem using an ICE table.

  10. Henderson Hasselbach Equation Can also use initial number of moles, rather than initial molarities: pH = pKa + log (nA- / nHA) Using number of moles (instead of molarity) is only allowable when solving using H-H Equation. You CANNOT use moles in a Ka or Kb expression!!!

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