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Acids & Bases

Acids & Bases. Calculating Ka from % Dissociation Polyprotic Acids Properties of Salts. Example. Lactic acid (HC 3 H 5 O 3 ) is a waste product that accumulates in the muscle during exertion, leading to pain & a feeling of fatigue. A 0.100 M solution is 3.7% dissociated. Calculate K a.

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Acids & Bases

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  1. Acids & Bases Calculating Ka from % Dissociation Polyprotic Acids Properties of Salts

  2. Example • Lactic acid (HC3H5O3) is a waste product that accumulates in the muscle during exertion, leading to pain & a feeling of fatigue. A 0.100 M solution is 3.7% dissociated. Calculate Ka.

  3. Example HC3H5O3  H+ + C3H5O3-

  4. Example HC3H5O3  H+ + C3H5O3- 0.100 x 3.7% = 0.0037

  5. Example • Ka = [H+][C3H5O3-] / [HC3H5O3] • Ka = [0.0037][0.0037] / [0.9963] • Ka = 1.37 x 10-5

  6. Polyprotic Acids • More that one ionizable H+ • H+ come off one at a time • Each one has a specific Ka value • Ka1>>>>Ka2>>Ka3 • When calculating the pH of a polyprotic acid, you only use the 1st Ka because all of the others are so small

  7. Complete Ionization • Write the steps for the complete ionization of H2Se • H2Se • H2Se + H20  H30+ + HSe- Ka1 = # • HSe- + H20  H30+ + Se-2 Ka2 = #

  8. Complete Ionization • Write the steps for the complete ionization of H3AsO • H3AsO + H2O  H2AsO - + H3O + Ka1 = # • H2AsO - + H2O  HAsO -2 + H3O + Ka2 = # • HAsO -2 + H2O  AsO -3 + H3O + Ka3 = #

  9. Example • Calculate the pH & the [ ] of all species of a 3.0 M solution of H3PO4. • Ka1 = 7.5 x 10-3, Ka2 = 6.2 x 10-8, Ka3 = 4.8 x 10 -13

  10. Example H3PO4 H+ + H2PO4- Ka1 = 7.5 x 10-3

  11. Example H3PO4 H+ + H2PO4- Ka1 = 7.5 x 10-3 7.5 x 10-3 = x2 / (3-x) 7.5 x 10-3 (3-x) = x2 x = 0.146

  12. Example H2PO4 - H+ + HPO4-2 Ka1 = 6.2 x 10-8

  13. Example H2PO4 - H+ + HPO4-2 Ka2 = 6.2 x 10-8 6.2 x 10-8 =(0.146+x)x / (0.146-x) x = 6.2 x 10-8

  14. Example HPO4 -2 H+ + PO4-3 Ka3 = 4.8 x 10-13

  15. Example HPO4-2 H+ + PO4-3 Ka3 = 4.8 x 10-13 4.8x 10-8 =(0.146+x)x/(6.2x10-8-x) x = 2.0x10-19

  16. Example • pH = 0.836 • [H3PO4] = 2.85M • [H2PO4-] = 0.146M • [HPO4-2] = 6.2x10-8 • [PO4-3] = 2.0x10-19M • [H+] = 0.146M

  17. Acid Base Properties of Salts • Salts of weak acids produce basic solutions • NaC2H3O2 Na+ + C2H3O2- • NaOH HC2H3O2 • SB WA • C2H3O2- + H2O  HC2H3O2 + OH- • Resulting - basic solution

  18. Acid Base Properties of Salts • Salts of weak bases produce acidic solutions • NH4Cl  NH4+ + Cl- • NH3 HCl • WB SA • NH4+ + H2O  NH3 + H3O+ • Resulting – acidic solution

  19. Acid Base Properties of Salts • Salts of strong acids & Bases produce neutral solutions • NaCl  Na+ + Cl- • NaOH HCl • SB SA • Resulting solution is neutral

  20. Calculations with salts • KaxKb = Kw • Can find Ka or Kb with this equation • On AP formula sheet

  21. Example • What is the pH if a 0.140M solution of NaC2H3O2 (KaHC2H3O2 = 1.8x10-5) • NaC2H3O2 Na+ + C2H3O2- • C2H3O2- + H2O  HC2H3O2 + OH-

  22. Example • What is the pH if a 0.140M solution of NaC2H3O2 (KaHC2H3O2 = 1.8x10-5) • NaC2H3O2 Na+ + C2H3O2- • C2H3O2- + H2O  HC2H3O2 + OH-

  23. Example • C2H3O2- + H2O  HC2H3O2 + OH- • Ka HC2H3O2 = 1.8x10-5 but we are starting with a base C2H3O2– • Ka x Kb = Kw • (1.8x10-5)(Kb) = 1 x 10-14 • Kb = 5.56x10-10

  24. Example • Kb = [HC2H3O2][OH-] / [C2H3O2-] • 5.56x10-10 = [x2]/ [0.14-x] • X=8.82x10-6 = [OH-] • pOH = 5.05 • pH = 8.95

  25. Another example • What is the pH of a 0.140M solution of diethylaminochloride (C2H5)2NH3Cl. • Kb (C2H5)2NH3+ = 1.3x10-3 • (C2H5)2NH3Cl  (C2H5)2NH3+ + Cl- • (C2H5)2NH3+ + H2O  (C2H5)2NH2 + H3O +

  26. Another example • What is the pH of a 0.140M solution of diethylaminochloride (C2H5)2NH3Cl. • Kb (C2H5)2NH2 = 1.3x10-3 • (C2H5)2NH3Cl  (C2H5)2NH3+ + Cl- • (C2H5)2NH3+ + H2O  (C2H5)2NH2 + H3O +

  27. Another Example (C2H5)2NH3+ + H2O  (C2H5)2NH2 + H3O+ • Kb (C2H5)2NH2 = 1.3x10-3 but we are starting with an acid (C2H5)2NH3+ • Ka x Kb = Kw • (1.3x10-3)(Kb) = 1 x 10-14 • Kb = 7.7x10-12

  28. Example • Kb = [(C2H5)2NH2][H3O+] / [(C2H5)2NH3+] • 7.7x10-12 = [x2]/ [0.14-x] • X=1.4x10-6 = [H3O+] • pH = 5.98

  29. Qualitative Predictions of pH • Ka>Kb  acidic • Kb>Ka  basic • Ka = Kb  neutral

  30. pH Predictions • Predict if an aqueous solution of the following are acidic, basic, or neutral • NH4C2H3O2 • Kb NH4+ = 1.8x10-5 • Ka HC2H3O2 = 1.8x10-5 • NH4C2H3O2  NH4+ + C2H3O2- • CA (Ka) CB (Kb) • 5.56x10-10 = 5.56x10-10 • neutral

  31. pH Predictions • Predict if an aqueous solution of the following are acidic, basic, or neutral • NH4CN • Kb NH4+ = 1.8x10-5 • Ka HCN= 6.2x10-10 • NH4CN NH4+ + CN- • CA (Ka) CB (Kb) • 5.56x10-10 < 1.6x10-5 • Basic

  32. Effect of Structure of Acid Base Properties • HF weakest because the H – X bond the strongest, so it won’t let the H pop off easily • HI strongest because the H – X bond the weakest, so the H pop off easily

  33. Effect of Structure of Acid Base Properties • HClO4 • The O’s are pulling the e- (they are e- hogs) leaving the H to easily pop off • HClO • The O is not pulling as much as HClO4, so it is harder for the H to pop off

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